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CAREER INSTITUTEKOTA (RAJASTHAN)

T M

FORM NUMBER

(ACADEMIC SESSION 2014-2015)

PAPER CODE

ALLEN AIPMT TEST DATE : 02 - 04 - 2015

0 1 C M 4 1 4 1 0 7

TARGET : PRE-MEDICAL 2015

Corporate Office

CAREER INSTITUTE

“SANKALP”, CP-6, Indra Vihar, Kota (Rajasthan)-324005

+91-744-2436001 [email protected]

CLASSROOM CONTACT PROGRAMME

1. A seat marked with Reg. No. will be allotted to each student. The student should ensure that he/she occupies the correct seat only.If any student is found to have occupied the seat of another student, both the students shall be removed from the examination andshall have to accept any other penalty imposed upon them.

2. Duration of Test is 3 Hours and Questions Paper Contains 180 Questions. The Max. Marks are 720.

31807203. Student can not use log tables and calculators or any other material in the examination hall.

4. Student must abide by the instructions issued during the examination, by the invigilators or the centre incharge.

5. Before attempting the question paper ensure that it contains all the pages and that no question is missing.

6. Each correct answer carries 4 marks, while 1 mark will be deducted for every wrong answer. Guessing of answer is harmful.

17. A candidate has to write his / her answers in the OMR sheet by darkening the appropriate bubble with the help of Blue / Black Ball

Point Pen only as the correct answer(s) of the question attempted.

OMR

8. Use of Pencil is strictly prohibited.

INSTRUCTIONS ()

www.allen.ac.in

ENTHUSIAST, LEADER & ACHIEVER COURSE

Test Type : MAJOR Test Pattern : AIPMT

Hindi

Do not open this Test Booklet until you are asked to do so /

TEST SYLLABUS : SYLLABUS - 3

Note : In case of any Correction in the test paper, please mail to [email protected] within 2 days along with Your FormNo. & Complete Test Details.

Correction Form No. Test Details [email protected] mail

1 4

(Phase - ENTHUSIAST, MLI, MLJ, MLSP, MAZA, MAZB & MAZC)

PHYSICS : Gravitation

Electrostatics and Capacitors

Current electricity

Magnetic effect of current and Magnetism

CHEMISTRY : Organic Chemistry : Some Basic Principles and Techniques

Hydrocarbons

Haloalkanes and Haloarens

Alcohols, Phenols and Ethers

Aldehydes, Ketones and Carboxylic Acids

Organic Compounds Containing Nitrogen(Amines)

BIOLOGY : Reproduction :

(i) Reproduction in Organisms

(ii) Sexual Reproduction in Flowering Plants

(iii) Human Reproduction

(iv) Reproductive Health

Genetics and Evolution :

(i) Principles of inheritance and Variation

(ii) Evolution

Biology in Human Welfare :

(i) Microbes in Human Welfare

Biotechnology :

(i) Biotechnology : Principles and Processes

(ii) Biotechnology and its Applications

ALLEN AIPMT TEST DATE : 02 - 04 - 2015

SYLLABUS – 3

ENTHUSIAST, LEADER & ACHIEVER COURSE(PHASE : ENTHUSIAST, MLI, MLJ, MLSP, MAZA, MAZB, MAZC)

Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015

H-1/45Kota/01CM414107

HAVE CONTROL HAVE PATIENCE HAVE CONFIDENCE 100% SUCCESS

BEWARE OF NEGATIVE MARKING

1. If r represents the radius of the orbit of a satellite

of mass m moving around a planet of mass M,

the velocity of the satellite is given by -

(1) 2 M

v gr

(2) 2 GMm

vr

(3) GM

vr

(4) 2 GM

vr

2. In the circuit shown, potential difference between

X and Y will be –

120V

X Y

(1) Zero (2) 20 V (3) 60 V (4) 120 V

3. The emf of a cell is balanced at 52 cm length of

the potentiometer wire. If 5 ohm resistance is

inserted from the resistance box connected with

the cell, then balancing length obtained is 40 cm.

The internal resistance of the cell will be :-

(1) 1.5 ohm (2) 2.0 ohm

(3) 1.8 ohm (4) 3.0 ohm

4. Two wires of same diameter having length x1 and

x2 and specific resistance r1 and r

2 respectively are

connected in series. Effective resistivity of

combination will be –

(1) 1 1 2 2

1 2

x x

x x

(2) 1 1 2 2

1 2

x x

x x

(3)1 1 2 2

1 2

x x

x x

(4) 1 1 2 2

1 2

x x

x x

5. Two identical wires A and B have the same length

L and carry the same current I. Wire A is bent

into a circle of radius R and wire B is bent to form

a square of side a. If B1 and B

2 are the values of

magnetic induction at the centre of the circle and

the centre of the square respectively, then the ratio

of B1/B2 is :

(1) (2/8) (2) 2 8 2

(3) (2/16) (4) 2 16 2

1. m r

M

(1) 2 M

v gr

(2) 2 GMm

vr

(3) GM

vr

(4) 2 GM

vr

2. X Y –

120V

X Y

(1) Zero (2) 20 V (3) 60 V (4) 120 V

3. 52 cm 5 ohm

40 cm

:-

(1) 1.5 ohm (2) 2.0 ohm

(3) 1.8 ohm (4) 3.0 ohm

4. x1 x2 1

2

–

(1) 1 1 2 2

1 2

x x

x x

(2) 1 1 2 2

1 2

x x

x x

(3)1 1 2 2

1 2

x x

x x

(4) 1 1 2 2

1 2

x x

x x

5. A B L

I A R

B a

B1 B2

B1/B

2

(1) (2/8) (2) 2 8 2

(3) (2/16) (4) 2 16 2

Kota/01CM414107H-2/45

Major Test For Target : Pre-Medical 2015/02-04-2015

6. The ratio of the radius of the earth to that of the

moon is 10. The ratio of acceleration due to

gravity on the earth and on the moon is 6. The

ratio of the escape velocity from the earth’s

surface to that from the moon is -

(1) 10 (2) 6

(3) Nearly 8 (4) 1.66

7. A potentiometer is used for the comparison of

e.m.f. of two cells E1 and E2 . For cell E1 the no

deflection point is obtained at 20 cm and for E2

the no deflection point is obtained at 30 cm. The

ratio of their e.m.f.'s will be –

(1) 2/3 (2) 1/2 (3) 1 (4) 2

8. The electric field intensity due to a charged

conducting plate at a distance 10 cm from it is

E. The electric field intensity at a distance 5 cm

from the plate will be

(1) 2 E (2) E (3) E/2 (4) E/4

9. The resistance of the voltmeter used in the circuit

here is 4k . Then the percentage error in the

measurement of the voltmeter is about –

(1) 33 % (2) 67 %

(3) 50 % (4) None of the above

10. A magnet of magnetic moment M is cut into two

equal parts. The two parts are placed

perpendicular to each other so that their north

poles touch each other. The resultant magnetic

moment is :

(1) 2 M (2) M

2

(3) 3M (4) M

3

11. If a body describes a circular motion under

inverse square field, the time taken to complete

one revolution T is related to the radius of the

circular orbit as -

(1) T r (2) T r2 (3) T2 r3 (4) T r4

6. 10

6

(1) 10 (2) 6

(3) Nearly 8 (4) 1.66

7. E1 E2 E1 20 E2 30–(1) 2/3 (2) 1/2 (3) 1 (4) 2

8. 10

E 5

(1) 2 E (2) E (3) E/2 (4) E/4

9.

4k

–

(1) 33 % (2) 67 %

(3) 50 % (4) 10. M

(1) 2 M (2) M

2

(3) 3M (4) M

3

11.

T r

(1) T r (2) T r2 (3) T2 r3 (4) T r4


H-3/45Kota/01CM414107

12. What is the maximum power output that can be

obtained from a cell of emf E and internal

resistance r –

(1)22E

r(2)

2E

r

(3) 2E

4r(4)

2E

2r

13. Graphs are drawn between current 'i' taken from

a cell and the terminal voltage V of the cell. Which

one is correct :–

(1) (2)

(3) (4)

14. Three charges of 0.1C each are placed on the

corners of an equilateral triangle of side 1.0m. If

the energy is supplied to this system at the rate

of 0.1 KW, how much time would be required to

move all the charges at infinite separation :-

(1) 1.5 × 107 sec (2) 1.8 × 105 sec

(3) 2.7 × 106 sec (4) 1.9 × 1010 sec

15. If the magnetic dipole moment of an atom of

diamagnetic material, paramagnetic material and

ferromagnetic material are denoted by d,

p and

f respectively, then :

(1) d = 0 and p 0 (2) d 0 and

p = 0

(3) p = 0 and f 0 (4) d 0 and f 0

16. ve and vp denotes the escape velocity from the

earth and another planet having twice the radius

and the same mean density as the earth. Then -

(1) ve = vp (2) ve = vp / 2

(3) ve = 2vp (4) ve = vp / 4

12. r E –

(1)22E

r(2)

2E

r

(3) 2E

4r(4)

2E

2r

13. 'i' V

(1) (2)

(3) (d)

14. 1.0

0.1 0.1

(1) 1.5 × 107 sec (2) 1.8 × 105 sec

(3) 2.7 × 106 sec (4) 1.9 × 1010 sec

15.

d,

p

f

(1) d = 0 p 0 (2) d 0 p = 0

(3) p = 0 f 0 (4) d 0 f 0

16. ve vp

(1) ve = vp (2) ve = vp / 2

(3) ve = 2vp (4) ve = vp / 4

Kota/01CM414107H-4/45


17. 2 4 12V 2 12V

–(1) 4 V (2) 8 V(3) 3.43 V (4) 6.86 V

18. Q S (1) (2)

(3) (4)

19. x, y, z V = 4x2 (1m, 0, 2m)

:-(1) 8, X–

(2) 8, X–

(3) 16, X–

(4) 16, Z–

20. I

a O

Z

Y

I

II O

a

X

(1) 0I

4a

(2)

20I 44 a

(3) 0 0I I

4a 2 a

(4) 0 2

I4

4 a

21. M R m

(1) 2GM

R(2)

GM2

R

(3) 2GMm

R(4)

GM

R

17. Two resistances of 2 ohm and 4 ohm are connected

in a series with a battery of 12V emf and negligible

internal resistance. On connecting a second battery

across the ends of 2 ohm resistance it is found that

the current drawn from the 12V battery does not

change. Then the emf of the second battery is –

(1) 4 V (2) 8 V

(3) 3.43 V (4) 6.86 V

18. In a balanced Wheatstone's network the resistance

in the arms Q and S are interchanged. As a result

(1) The galvanometer shows zero deflection

(2) The galvanometer and the cell must be

interchanged for balance

(3) The network is must balanced

(4) The network is not balanced

19. The electric potential V at any point x, y, z (all

in metres) in space is given by V = 4x2 volt. The

electric field at the point (1m, 0, 2m) in

volt/metre is ) :-

(1) 8 along negative X–axis

(2) 8 along positive X–axis

(3) 16 along negative X–axis

(4) 16 along positive Z–axis

20. A long wire bent as shown in the figure carries

current I. If the radius of the semi-circular portion

is a, the magnetic in-duction at the centre O is :Z

Y

I

II O

a

X

(1) 0I

4a

(2)

20I 44 a

(3) 0 0I I

4a 2 a

(4) 0 2

I4

4 a

21. The escape velocity of a sphere of mass m from

earth having mass M and radius R is given by-

(1) 2GM

R(2)

GM2

R

(3) 2GMm

R(4)

GM

R


H-5/45Kota/01CM414107

22. G = 20R1, R2, R3 3V, 15V, 150V

1mA

R1, R2, R3 –

0

R1

R2 R3

3V 15V 150V

(1) 3, 12, 135

(2) 2.98, 12, 135

(3) 2.98, 14.98, 149.98

(4)

23. 'a' 'b' A B A B :-

(1) a/b (2) b/a (3) a2/b2 (4) b2/a2

24. :-

(C)

(1) A (2) B(3) C (4) D

25. 5 × 106 m/s

B = 0.02 T (1) 105 V m-1 (2) 2.5 × 108 V m-1

(3) 1.25 × 1010 V m-1 (4) 2 × 103 V m-1

22. A voltmeter of variable ranges 3V, 15V, 150V is

to be designed by connecting resistances R1, R2,

R3 in series with a galvanometer of resistance

G = 20, as shwon in figure. The galvanometer

gives full scale deflection when a current of 1mA

pass through its coil. Then, the resistances R1 , R2

and R3 (in kilo ohms) should be, respectively –

0

R1

R2 R3

3V 15V 150V

(1) 3, 12, 135

(2) 2.98, 12, 135

(3) 2.98, 14.98, 149.98

(4) None of the above

23. Two spheres A and B of radius a and b

respectively are at the same potential. The ratio

of the surface charge density of A to B is :-

(1) a/b (2) b/a (3) a2/b2 (4) b2/a2

24. An uncharged sphere of metal is placed in betweentwo charged plates as shown. The lines of forcelook like :-

(C)

(1) A (2) B

(3) C (4) D

25. A beam of well collimated cathode rays travelling

with speed of 5 × 106 m/s enters a region of

mutually perpendicular electric and magnetic

fields and emerges undeviated from this region.

If B = 0.02 T, the magnitude of electric field is :

(1) 105 V m-1 (2) 2.5 × 108 V m-1

(3) 1.25 × 1010 V m-1 (4) 2 × 103 V m-1

Key Filling

Kota/01CM414107H-6/45


26. m h

R g

(1) 2gR

R h(2) gR

(3) gR

R h(4)

2gR

R h

27. K (Slab)

E1 E2E3

K

(1) E1 = E2 = E3 (2) E1 = E3 > E2

(3) E1 < E2 < E3 (4) E1 = E3 < E2

28. 16µF 1000V 8µF, 250V

(1) 8 (2) 32 (3) 16 (4) 64

29. n r I

h

(1) 2

2

3 h

2 r(2)

2

2

2 h

3 r(3)

2

2

3 r

2 h(4)

2

2

2 r

3 h

30. B (1) B (2) 2B (3) 4B (4) B/2

31.

( g = 9.8 m/sec2) -

(1) 4.9 m/sec2 (2) 8.9 m/sec2

(3) 19.6 m/sec2 (4) 29.4 m/sec2

26. An earth satellite of mass m revolves in a circular

orbit at a height h from the surface of the earth. R

is the radius of the earth and g is acceleration due

to gravity at the surface of the earth. The velocity

of the satellite in the orbit is given by -

(1) 2gR

R h(2) gR

(3) gR

R h(4)

2gR

R h

27. A dielectric slab of dielectric constant K is placed

between the plates then compare electric fields

E1, E2 and E3

E1 E2E3

K

(1) E1 = E2 = E3 (2) E1 = E3 > E2

(3) E1 < E2 < E3 (4) E1 = E3 < E2

28. How many capacitors each of 8µF and 250V are

required to form a composite capacitor of 16µF

and 1000V :-

(1) 8 (2) 32 (3) 16 (4) 64

29. The field normal to the plane of a wire of n turns

and radius r which carries a current I is measured

on the axis of the coil at a small distance h from

the centre of the coil. This is smaller than the field

at the centre by the fraction :

(1) 2

2

3 h

2 r(2)

2

2

2 h

3 r(3)

2

2

3 r

2 h(4)

2

2

2 r

3 h

30. A long solenoid carrying a current produces a

magnetic field B along its axis. If the current is

doubled and the number of turns per cm is halved,

then new value of the magnetic field is :

(1) B (2) 2B (3) 4B (4) B/2

31. If a planet consists of a satellite whose mass and

radius were both half that of the earth, the

acceleration due to gravity at its surface would

be (g on earth = 9.8 m/sec2) -

(1) 4.9 m/sec2 (2) 8.9 m/sec2

(3) 19.6 m/sec2 (4) 29.4 m/sec2

Use stop, look and go method in reading the question


H-7/45Kota/01CM414107

32. 30

:-

(1) 30V (2) 45V (3) 20V (4) Zero

33.

2 × 105 V/m

1 × 105 V/m –

(1)1/2 (2) 1 (3) 2 (4) 3

34.

I XOY

O r P :-

(1) 0I

r

(2) 02 I

r

Y

X

P

I

I

O45°

r

(3) 0I2 1

4 r

(4) 0 2I

2 14 r

35.

0.2T

(1) 0.2 (2) 0.4 (3) 0.6 (4) 0.8

36. (x,y,z) V = 2x + 3y – z –(1) x y (2) y z (3) z x (4)

37. Q c

(1) 6

0

Q10

6

(2)

0

Q

6

(3)0

Q

24 (4) 0

Q

8

32. If potential at centre of a charged solid

Non–conducting sphere is 30 volt, then potential

at its surface will be :-

(1) 30V (2) 45V (3) 20V (4) Zero

33. Two parallel plates have equal and opposite

charge. When the space between them is

evacuated, the electric field between the plates is

2 × 105 V/m. When the space is filled with

dielectric, the electric field becomes

1 × 105 V/m. The dielectric constant of the

dielectric material is –

(1)1/2 (2) 1 (3) 2 (4) 3

34. Current I flows through a long conducting wire

bent at right angles as shown in the figure. The

magnetic field at a point P on the right bisector

of the angle XOY at a distance r from O is :

(1) 0I

r

(2) 02 I

r

Y

X

P

I

I

O45°

r

(3) 0I2 1

4 r

(4) 0 2I

2 14 r

35. The electric current in a circular coil of two turns

produced a magnetic induction of 0.2T at its

centre. The coil is unwound and is rewound into

a circular coil of four turns. The magnetic

induction at the centre of the coil now is, in tesla,

(if same current flows in the coil) :

(1) 0.2 (2) 0.4 (3) 0.6 (4) 0.8

36. In a certain region the electric potential at a point

(x,y,z) is given by the potential function

V = 2x + 3y – z. Then the electric field in this

region –

(1) Increases with increase in x and y

(2) Increases with increase in y and z

(3) Increases with increase in z and x

(4) Remains constant

37. A charge Q c is placed at the centre of a cube,

the flux coming out from any surface will be :–

(1) 6

0

Q10

6

(2)

0

Q

6

(3)0

Q

24 (4) 0

Q

8

Kota/01CM414107H-8/45


38. (off) 3 –

(1) 3 : 1 (2) 5 : 1

(3) 3: 5 (4) 5 : 3

39.

30°

20 60°

15

(1) 3 3 : 8 (2) 16 : 9 3

(3) 4 : 9 (4) 2 3 : 9

40.

39°

30°

(1) 39° (2) 30°

(3) 39° (4) 39°

41. AB –

(1)5 (2) 2 (3) 3 (4) 4

38. Figure given below shows two identical parallelplate capacitors connected to a battery with switchS closed. The switch is now opened and the freespace between the plate of capacitors is filled witha dielectric of dielectric constant 3. What will bethe ratio of total electrostatic energy stored in bothcapacitors before and after the introduction of the

dielectric –

(1) 3 : 1 (2) 5 : 1

(3) 3: 5 (4) 5 : 3

39. A magnet is suspended in such a way that it

oscillates in the horizontal plane. It makes 20

oscillations per minute at place where dip angle

is 30° and 15 oscillations per minute at a place

where dip angle is 60°. Ratio of the total the earth

magnetic field at the two places is :

(1) 3 3 : 8 (2) 16 : 9 3

(3) 4 : 9 (4) 2 3 : 9

40. A dip circle is so set that the dip needle moves

freely in the magnetic meridian. In this position

the angle of dip is 39°. Now, the dip circle is

rotated so that the plane in which the needle

moves makes an angle of 30° with the magnetic

meridian. In this position, the needle will dip by

an angle :

(1) exactly 39° (2) 30°

(3) more than 39° (4) less than 39°

41. The effective resistance between the points A and

B in the figure is –

(1) 5 (2) 2 (3) 3 (4) 4


H-9/45Kota/01CM414107

42. Eight equals charged drops are combined to form

a big drop. If the potential on each drop is 10V

then potential of big drop will be:-

(1) 40V (2) 10V

(3) 30V (4) 20V

43. A constant voltage is applied between the two

ends of a uniform metallic wire. Some heat is

developed in it. If both length and radius of the

wire are halved then the heat developed in the

same duration will become –

(1) Half (2) Twice

(3) One-fourth (4) Same

44. A wire loop formed by joining two simi-circular

wires of radii R1 and R2 carries a current as shown

in the adjoining diagram. The magnetic induction

at the centre O is :

QR1

R2

SI

I

I

I

PR

(1) 0

1

I

4 R

(2) 0

2

I

4 R

(3) 0

1 2

1 1I

R R4

(4) 0

1 2

1 1I

R R4

45. The period of oscillation of a magnetic needle in a

magnetic field is 1.0 s. If the length of the needle is

halved by cutting it, then the time period will be :

(1) 1.0 sec (2) 0.5 sec

(3) 0.25 sec (4) 2.0 sec

42. 10V

(1) 40V (2) 10V

(3) 30V (4) 20V

43. –(1) (2) (3) (4)

44. R1 R2

O

QR1

R2

SI

I

I

I

PR

(1) 0

1

I

4 R

(2) 0

2

I

4 R

(3) 0

1 2

1 1I

R R4

(4) 0

1 2

1 1I

R R4

45. 1.0 s

(1) 1.0 sec (2) 0.5 sec

(3) 0.25 sec (4) 2.0 sec

Kota/01CM414107H-10/45


46. Which is most acidic hydrogen ?

O

H

H–NN

O

H

(1)

(2)

H(3)

(4)

(1) 1 (2) 2 (3) 3 (4) 4

47.

O

OH

CH3

OH

HIO4

?

(1) CH –C–CH –CH –CH –CH=O+CO3 2 2 2 2

O

(2) CH –C–CH –CH –CH –CH=O+HCOOH3 2 2 2

O

(3) CH –CH–CH –CH –CH –CH –OH+HCOOH3 2 2 2 2

OH

(4) CH –C–CH –CH –CH –COOH + CO3 2 2 2 2

O

48.OH

H

?, major ?

CH3

(1)

CH3

(2)

CH3

(3)

CH2

(4)

OHCH3

49. The reaction products of : C6H5OCH3 + HI

is :-

(1) C6H5OH + CH3I

(2) C6H5I + CH3OH

(3) C6H5CH3 + HOI

(4) C6H6+CH3OI

46.

O

H

H–NN

O

H

(1)

(2)

H(3)

(4)

(1) 1 (2) 2 (3) 3 (4) 4

47.

O

OH

CH3

OH

HIO4

?

(1) CH –C–CH –CH –CH –CH=O+CO3 2 2 2 2

O

(2) CH –C–CH –CH –CH –CH=O+HCOOH3 2 2 2

O

(3) CH –CH–CH –CH –CH –CH –OH+HCOOH3 2 2 2 2

OH

(4) CH –C–CH –CH –CH –COOH + CO3 2 2 2 2

O

48.

OHH

?,

CH3

(1)

CH3

(2)

CH3

(3)

CH2

(4)

OHCH3

49. C6H5OCH3 + HI

(1) C6H5OH + CH3I

(2) C6H5I + CH3OH

(3) C6H5CH3 + HOI

(4) C6H6+CH3OI

Take it Easy and Make it Easy


H-11/45Kota/01CM414107

50.

Br

Mgether2 (A); Product (A) is :-

(1) (2)

(3) (4)

OH

51. Which is false statement :-

(1) Carboxylic acid and phenol can be

differentiated by NaHCO3

(2) 2º Amine does not react with Hinsberg

reagent

(3) Ethyl alcohol and Acetone can be

differentiated by Brady's reagent

(4) 2-Alkyne does not give Tollen's reagent test

52. Decreasing reactivity order towards ethyl

alcohol?

(1) (CH3–CO)2O > CH3–COCl > CH3–CONH2

> CH3–COOR

(2) CH3–COCl > (CH3–CO)2O > CH3COOR >

CH3–CONH2

(3) CH3COCl > CH3COOR > CH3–CONH2 >

(CH3CO)2O


CH3CONH2

53. I2 A+yellow ppt

O

C–CH3

+NaOHO

A B. 'B' is :-

(1)

O

(2)

O

CH3

(3)

H

COCH3 (4) CH3–H

50.

Br

Mgether2 (A); (A)

(1) (2)

(3) (4)

OH

51. (1) NaHCO3

(2) 2º

(3)

(4) 2-52.

(1) (CH3–CO)2O > CH3–COCl > CH3–CONH2

> CH3–COOR


CH3–CONH2

(3) CH3COCl > CH3COOR > CH3–CONH2 >

(CH3CO)2O


CH3CONH2

53. I2 A+

O

C–CH3

+NaOHO

A B. 'B'

(1)

O

(2)

O

CH3

(3)

H

COCH3 (4) CH3–H

Kota/01CM414107H-12/45


54. What is the major product of the following

reaction ?

OH

CH3

FeBr3+Br2 ?

(1)

OH

CH3

Br

(2)

OH

BrCH2

(3)

OBr

CH3

(4)

OH

CH3Br

55. Which of the following is most stable ?

(1) (2)

(3) (4)

56. Which have maximum stable enol form :-

(1) CH –C–CH –C–CH3 2 3

O O

(2) CH –C–CH –C–O–CH3 2 3

O O

(3) CH –C–CH3 3

O

(4)O

57.

CHO

CHO

KOHA

H

B ?

(1)CH COOH2

CH –OH2

(2) O

O

(3) O

O

O

(4)

O

O

O

54.

OH

CH3

FeBr3+Br2 ??

(1)

OH

CH3

Br

(2)

OH

BrCH2

(3)

OBr

CH3

(4)

OH

CH3Br

55.

(1) (2)

(3) (4)

56. enol

(1) CH –C–CH –C–CH3 2 3

O O

(2) CH –C–CH –C–O–CH3 2 3

O O

(3) CH –C–CH3 3

O

(4)O

57.

CHO

CHO

KOHA

H

B ?

(1)CH COOH2

CH –OH2

(2) O

O

(3) O

O

O

(4)

O

O

O


H-13/45Kota/01CM414107

58. CH3–CN + CH3MgBr A H O2 B ?

(1) CH3–COCH3 (2) CH –C–CH3 3

CH3

OH

(3) CH3–CH=NH (4) CH –C–NH3 2

CH3

OH

59. Most reactive toward acid catalyst dehydration

is:-

(1)

OH

(2)

OH

(3)

OH

(4) OH


reaction ?

CH3

Br2

H O2 ?

(1)

CH3

Br

OH

(2)

CH3

OH

Br

(3)

CH3

Br (4)

CH3

Br

Br

61.O

HCNA

LiAH4 B ?

(1)

CH –NH2 2

(2)

CH3

(3)

OHCH –NH2 2 (4)

CH –OH2

NH2

58. CH3–CN + CH3MgBr A H O2 B ?

(1) CH3–COCH3 (2) CH –C–CH3 3

CH3

OH

(3) CH3–CH=NH (4) CH –C–NH3 2

CH3

OH

59.

(1)

OH

(2)

OH

(3)OH

(4)

OH

60.

CH3

Br2

H O2 ?

(1)

CH3

Br

OH

(2)

CH3

OH

Br

(3)

CH3

Br (4)

CH3

Br

Br

61.O

HCNA

LiAH4 B ?

(1)

CH –NH2 2

(2)

CH3

(3)

OHCH –NH2 2 (4)

CH –OH2

NH2

Kota/01CM414107H-14/45


62. Corrrect IUPAC name of

CH –CH–CH –C–Cl3 2

CH –CN2

O

is :-

(1) 3-cyanomethyl butanoyl chloride

(2) 5-cyano-3-methyl pentanoyl chloride

(3) 4-chloroformyl-2-methyl butanenitrile

(4) 4-cyano-3-methyl butanoyl chloride

63. Which is most basic :-

(1)N

(2) CONH2

(3)

NH

(4) NH–CH3

64. H O–H

CH3

Ph

CH I3 (B),SN2

Na(A) Product (B)

is:-

(1) H Ph

CH3

OCH3

(2) H O–CH3

CH3

Ph

(3) H O–CH3

Ph

CH3

(4) CH O3 H

Ph

CH3


reaction ?

HBrPeroxide

CH3

?

(1)

CH3

Br

(2)

CH Br2

(3)

CH3

Br

(4)

CH3

Br

62.

CH –CH–CH –C–Cl3 2

CH –CN2

O

IUPAC (1) 3-cyanomethyl butanoyl chloride

(2) 5-cyano-3-methyl pentanoyl chloride

(3) 4-chloroformyl-2-methyl butanenitrile

(4) 4-cyano-3-methyl butanoyl chloride

63.

(1)

N

(2) CONH2

(3)

NH

(4) NH–CH3

64. H O–H

CH3

Ph

CH I3 (B),SN2

Na(A) (B)

(1) H Ph

CH3

OCH3

(2) H O–CH3

CH3

Ph

(3) H O–CH3

Ph

CH3

(4) CH O3 H

Ph

CH3

65.

HBrPeroxide

CH3

?

(1)

CH3

Br

(2)

CH Br2

(3)

CH3

Br

(4)

CH3

Br


H-15/45Kota/01CM414107

66.

CHOCHO

+NH –OH2+NH –OH2

HH

AAP O2 5P O2 5

BBH O3

H O3

CCSOCl2 DD?

(1)

CH –Cl2

(2)

C–Cl

O

(3)

Cl

CHO

(4) Cl

CH3

67. Which does not show geometrical isomerism ?

(1)

CHCl

CH3 CH3

(2)

Cl

CH3

Cl

Br

(3) 1,1-diphenyl-2-butene

(4) 1,1-diphenyl propene

68. How many optically inactive compounds are

possible from CH –CH–CH–CH–CH3 3

OH Cl OH

:-

(1) 2 (2) 1 (3) 3 (4) 0

69. Which of the following compound give

Reimer-Teimann reaction ?

(1) (2)

NO2

(3)OMe

OH

(4)

NO2

NO2

70. Ph–CC–Me 20%D SO in D O/Hg2 4 2

++

(P) Principle

organic product is :-

(1) Ph–C–CD –Me2

O

(2) Ph–CD –C–Me2

O

(3) Ph–C–CH –CHD2 2

O

(4) Ph–C–C–CH D2

O H

D

66.

CHOCHO

+NH –OH2+NH –OH2

HH

AAP O2 5P O2 5

BBH O3

H O3

CCSOCl2 DD?

(1)

CH –Cl2

(2)

C–Cl

O

(3)

Cl

CHO

(4) Cl

CH3

67.

(1)

CHCl

CH3 CH3

(2)

Cl

CH3

Cl

Br

(3) 1,1-diphenyl-2-butene

(4) 1,1-diphenyl propene

68.

CH –CH–CH–CH–CH3 3

OH Cl OH

(1) 2 (2) 1 (3) 3 (4) 0

69.

(1) (2)

NO2

(3)OMe

OH

(4)

NO2

NO2

70. Ph–CC–Me 20%D SO in D O/Hg2 4 2

++

(P)

(1) Ph–C–CD –Me2

O

(2) Ph–CD –C–Me2

O

(3) Ph–C–CH –CHD2 2

O

(4) Ph–C–C–CH D2

O H

D

Kota/01CM414107H-16/45


71. Which are enantiomers :-

(1)

H Cl

H Br

COOH

COOCH3

and Br H

Cl H

COOCH3

COOH

(2) Br Cl and Br Cl

(3)

H CH3

HCH3

Br

Br

and

H Br

HCH3

CH3

Br

(4) Cl I

F

Br

and I F

Br

Cl

72. CH –CH –C–CH –CH3 2 2 3

O

and CH –C–CH–CH3 3

O CH3

are:-

(1) Metamers (2) Chain isomers

(3) Position isomers (4) Tautomers

73. CH3–CH2–CCH+H–Cl A H–IB ?

(1) CH –CH –CH–CH3 2 2

Cl I

(2) CH –C–CH –CH3 2 3

Cl

I

(3) CH –CH–CH –CH3 2 2

Cl I

(4) CH –CH–CH–CH3 3

Cl I

74.

OH

+CO21. NaOH2. H O/H2

(A), Product (A) is :-

(1)

OH

(2)

OH

CO Na2

(3)

OH

CO H2

(4)

OH

CHO

71.

(1)

H Cl

H Br

COOH

COOCH3

and Br H

Cl H

COOCH3

COOH

(2) Br Cl and Br Cl

(3)

H CH3

HCH3

Br

Br

and

H Br

HCH3

CH3

Br

(4) Cl I

F

Br

and I F

Br

Cl

72. CH –CH –C–CH –CH3 2 2 3

O

CH –C–CH–CH3 3

O CH3

(1) Metamers (2) Chain isomers

(3) Position isomers (4) Tautomers

73. CH3–CH2–CCH+H–Cl A H–IB ?

(1) CH –CH –CH–CH3 2 2

Cl I

(2) CH –C–CH –CH3 2 3

Cl

I

(3) CH –CH–CH –CH3 2 2

Cl I

(4) CH –CH–CH–CH3 3

Cl I

74.

OH

+CO21. NaOH2. H O/H2

(A), (A)

(1)

OH

(2)

OH

CO Na2

(3)

OH

CO H2

(4)

OH

CHO


H-17/45Kota/01CM414107

75. Most acidic hydrogen is present in :-

(1) (2)

(3) (4)

76. Incorrect statement for staggered and eclipsed

form of n butane, is ?

(1) Dihedral angle in staggered form is 180º

(2) Tortional strain in Eclipsed form is maximum

(3) Steric strain in staggered form is minimum

(4) Eclipsed and staggered both are optically

active

77.COOH

OH

+CH COCl3 Product ?

(1)COOCOCH3

OH

(2) COOH

OCOCH3

(3)

COOH

OH

COCH3

(4) COOH

COCH3

78. In the given reaction :-

Br

NO2

Br C H ONa2 5

C H OH2 5

(P), (P) is :-

(1)

OC H2 5

NO2

Br(2)

Br

NO2

OC H2 5

(3)

NO2

OC H2 5

OC H2 5

(4)

OC H2 5

Br

Br

75.

(1) (2)

(3) (4)

76. n

(1) 180º

(2)

(3)

(4)

77.COOH

OH

+CH COCl3 ?

(1)COOCOCH3

OH

(2) COOH

OCOCH3

(3)

COOH

OH

COCH3

(4) COOH

COCH3

78.

Br

NO2

Br C H ONa2 5

C H OH2 5

(P), (P) :-

(1)

OC H2 5

NO2

Br(2)

Br

NO2

OC H2 5

(3)

NO2

OC H2 5

OC H2 5

(4)

OC H2 5

Br

Br

Kota/01CM414107H-18/45


79. The best method for synthesis of given ether by

Williamson's ether synthesis is :-

O–CH=CH2

(1) O+H C=CH–Br2

(2) O+H C=CH–Br2

(3) Br+H C=CH–O2

(4) Br+H C=CH–O2

Br

80. 2CH3–CHO OH

?

(1) CH3–CH=CH–CHO

(2) CH2=CH–CH2–CHO

(3) CH –CH–CH–OH3

CH3

(4) CH =C–CHO2

CH3

81.CHCl3 A

Zn/B

Cl2 C

OH

AlCl3KOH ?

(1)

CHO

Cl

(2)

CH3

Cl

(3)

Cl

(4) Cl

OH

CH3

79.

O–CH=CH2

(1) O+H C=CH–Br2

(2) O+H C=CH–Br2

(3) Br+H C=CH–O2

(4) Br+H C=CH–O2

Br

80. 2CH3–CHO OH

?

(1) CH3–CH=CH–CHO

(2) CH2=CH–CH2–CHO

(3) CH –CH–CH–OH3

CH3

(4) CH =C–CHO2

CH3

81.CHCl3 A

Zn/B

Cl2 C

OH

AlCl3KOH ?

(1)

CHO

Cl

(2)

CH3

Cl

(3)

Cl

(4) Cl

OH

CH3


H-19/45Kota/01CM414107

82. Which is not correct matching :-

(1)

OCOCH3

AlCl3 Fries rearrangement

(2) CH –C–Cl3

O

H /Pd2

BaSO4

Rosenmund reduction

(3) AlCl3 Friedel craft Alkylation+CH COCl3

(4) CH –OH2 Pyridine

Darzen reaction+SOCl2

83. For the given reaction; R–C–X

R1

R2

HOHR–C–OH

R1

R2

Which substrate will give maximum racemisation?

(1) C H –C–Br6 5

CH3

C H2 5

(2) CH =CH–C–Br2

CH3

C H2 5

(3) OCH3CC H6 5

Br

CH3

(4) NO2CC H6 5

Br

84. PhMgBr + (x) 2H Ph–CH2–CH2–OH; (x) is:-

(1) CH3–CH=O (2) O

(3) EtOH (4) O

82.

(1)

OCOCH3

AlCl3 Fries rearrangement

(2) CH –C–Cl3

O

H /Pd2

BaSO4

Rosenmund reduction

(3) AlCl3 Friedel craft Alkylation+CH COCl3

(4) CH –OH2 Pyridine

Darzen reaction+SOCl2

83. R–C–X

R1

R2

HOHR–C–OH

R1

R2

(1) C H –C–Br6 5

CH3

C H2 5

(2) CH =CH–C–Br2

CH3

C H2 5

(3) OCH3CC H6 5

Br

CH3

(4) NO2CC H6 5

Br

84. PhMgBr + (x) 2H Ph–CH2–CH2–OH; (x)

(1) CH3–CH=O (2) O

(3) EtOH (4) O

Kota/01CM414107H-20/45


85. The IUPAC name of COOC H2 5

COCl

is :-

(1) 2-Chlorocarbonyl ethylbenzoate

(2) 2-Carboxyethyl benzoyl chloride

(3) Ethyl-2-(chlorocarbonyl) benzoate

(4) Ethyl-1-(chlorocarbonyl) benzoate

86. Which is most reactive towards diazotization:-

(1) NH2O N2 (2) NH2CCl3

(3) NH2CH3 (4) NH2CH –CH3 2

87. Which is least reactive towards hydrolysis.

(1) p-Nitro benzyl chloride

(2) p-Methyl benzyl chloride

(3) p-Chloro benzyl chloride

(4) p-methoxy phenyl chloride

88. Ph–CH–CH2

Br

Br

(X) NaNH2 Ph–C NaC

(X = No. of moles of NaNH )2

Value of X is :-

(1) 1 (2) 2 (3) 3 (4) 4


reaction ?

CH C CHCH +HBr3 3— —

CH3

CH3

heat

OH

(1) CH C CHCH3 3— —

CH3

CH3 Br

(2) CH C CH CH3 2 2— —

CH3

CH3

Br

(3) CH C CHCH3 3— —

Br

CH3

CH3

(4) CH C CH CH3 2 3— —

CH Br2

CH3

90.

O

O(A)

Cl

Cl

What is the optical rotation of

(A)?

(1) +50º (2) –50º (3) 0º (4) 100º

85.COOC H2 5

COCl

IUPAC

(1) 2-(2) 2-(3) -2-() (4) -1-()

86. :-

(1) NH2O N2 (2) NH2CCl3

(3) NH2CH3 (4) NH2CH –CH3 2

87. (1) p-Nitro benzyl chloride

(2) p-Methyl benzyl chloride

(3) p-Chloro benzyl chloride

(4) p-methoxy phenyl chloride

88. Ph–CH–CH2

Br

Br

(X) NaNH2 Ph–C NaC

(X = No. of moles of NaNH )2

X (1) 1 (2) 2 (3) 3 (4) 4

89.

CH C CHCH +HBr3 3— —

CH3

CH3

heat

OH

(1) CH C CHCH3 3— —

CH3

CH3 Br

(2) CH C CH CH3 2 2— —

CH3

CH3

Br

(3) CH C CHCH3 3— —

Br

CH3

CH3

(4) CH C CH CH3 2 3— —

CH Br2

CH3

90.

O

O(A)

Cl

Cl

(A) ?

(1) +50º (2) –50º

(3) 0º (4) 100º


H-21/45Kota/01CM414107

91. Match the following ?

(i) Microsporophyll (A) Carpel

(ii) Megasporangium (B) Ovule

(iii) Microsporangium (C) Stamen

(iv) Megasporophyll (D) Pollen sac

(1) (i) – (C), (ii) – (B), (iii) – (D), (iv) – (A)

(2) (i) – (C), (ii) – (B), (iii) – (A), (iv) – (D)

(3) (i) – (B), (ii) – (C), (iii) – (A), (iv) – (D)

(4) (i) – (B), (ii) – (C), (iii) – (D), (iv) – (A)

92. The figure below shows the structure of anther

with its wall layers labelled (A), (B), (C) and (D).

Select the part correctly matched with its

function:-

(a) Part A Epidermis Helps in protection

(b) Part D Tapetum Provide nutrition tomicrospore mothercells or developingmicrospores

(c) Part B Middle layer Secretion ofhormones andenzymes

(d) Part C Endothecium Help in dehiscenceof anther to releasepollen

(1) a, b, c (2) a, b

(3) a, c, d (4) Only d

93. Read the following statements :-

(a) Only birds and reptiles are oviparous

(b) The large amount of yolk provides the

nutrients for the developing embryo

(c) The shell protects the egg from dehydration

(d) Both O2 & CO

2 can diffuse through the shell.

How many of the above statements are correct

(1) 2 (2) 1 (3) 4 (4) 3

91. ?

(i) (A)

(ii) (B)

(iii) (C)

(iv) (D)

(1) (i) – (C), (ii) – (B), (iii) – (D), (iv) – (A)

(2) (i) – (C), (ii) – (B), (iii) – (A), (iv) – (D)

(3) (i) – (B), (ii) – (C), (iii) – (A), (iv) – (D)

(4) (i) – (B), (ii) – (C), (iii) – (D), (iv) – (A)

92. (A), (B), (C) (D) :-

(a) A

(b) D

(c) B

(d) C

(1) a, b, c (2) a, b

(3) a, c, d (4) d93. :-

(a) (b)

(c) (d) O2 CO2 (1) (2) (3) (4)

Kota/01CM414107H-22/45


94. Natural

selection

Peak shifts in one direction

This graph indicate which type of natural selection

(1) Directional (2) Stabilising

(3) Disruptive (4) Mutation

95. Select the correct statement ?

(1) Fitness is the end result of the ability to adapt

and get selected by nature.

(2) Natural selection and genetic variation are two

main key point of darwinism theory of evolution.

(3) Adaptive ability is always inherited

(4) Placental wolf and Tasmanian wolf are

example of homology.

96. 'Founder effect' is related to:-

(1) Gene recombination and Natural selection

(2) Genetic drift and origin of new species

(3) Isolation and Natural selection

(4) Hybridization and origin of new species

97. (A) A true breeding line is one that having

undergone continuous self pollination.

(B) Shows stable trait inheritance & expression for

several generations

(C) Mendel selected 14 true breeding pea plant

varieties as pairs which were dissimilar except

for one character with contrasting traits

(D) It increases hemozygosity for every character

that is under consideration.

Regarding to pure lines select out the correct

statements :-

(1) A, B, C (2) A, B, D

(3) A & C (4) B & C

94. Natural

selection

Peak shifts in one direction

(1) (2)

(3) (4)

95. ?

(1)

(2)

(3)

(4)

96. '' :-

(1)

(2)

(3)

(4)

97. (A)

(B) ()

(C) 14

(D)

:-

(1) A, B, C (2) A, B, D

(3) A & C (4) B & C


H-23/45Kota/01CM414107

98. In this given diagram which of the following is

not true for F2 generation :-

Round yellowRR YY

Wrinkled greenrr yy

Round yellowRrYy

Selfing

Gametes

P generation

F generation1

F generation2

(1) Types of gamets = 4

(2) Types of genotype = 9

(3) Types of phenotype = 4

(4) Types of zygote = 16

99. A boy is colour blind. His sister is normal and non

career.

(i) What are the chances that both his parents are

normal ?

(ii) What are the chances of mother being career?

(iii) What are the chances of his younger brother

being colour blind ?

(1) (i)-100%, (ii)-100%, (iii)-100%

(2) (i)-100%, (ii)-100%, (iii)-50%

(3) (i)-50%, (ii)-50%, (iii)-50%

(4) (i)-100%, (ii)-50%, (iii)-100%

100. Select out the incorrect pair regarding to

homology of PCR & natural Replication:-

Natural Replication PCR

(1) Helicase 94°C

(2) Primase 54°C

(3) DNA polymerase Taq polymerase

(4) Ligase 72°C

98. F2

:-

Round yellowRR YY

Wrinkled greenrr yy

Round yellowRrYy

Selfing

Gametes

P generation

F generation1

F generation2

(1) = 4

(2) = 9

(3) = 4

(4) = 16

99. (i)

(ii) ?(iii)

?(1) (i)-100%, (ii)-100%, (iii)-100%

(2) (i)-100%, (ii)-100%, (iii)-50%

(3) (i)-50%, (ii)-50%, (iii)-50%

(4) (i)-100%, (ii)-50%, (iii)-100%

100. PCR :-

PCR

(1) Helicase 94°C

(2) Primase 54°C

(3) DNA polymerase Taq polymerase

(4) Ligase 72°C

Kota/01CM414107H-24/45


101. Pollination by water is quite rare in flowering

plants & is limited to about 30 genera which are

mostly :-

(1) Monocotyledons

(2) Dicotyledons

(3) Polypetalae

(4) Gamopetalae

102. Find out the correct match from the following

table :-

Column - I Column - II

(i) Pollen exine Resistant to enzyme action

(ii) Parthenium Pollen allergy

(iii) Generative cell Bigger, Abundant foodreserve

(iv) Pollen Tablets Protection

(1) i, ii, iii (2) ii, iii, iv

(3) i, ii (4) iii, iv


(a) Pills are very effective with lesser side effects

(b) Saheli-contains a non steroidal preperation

(c) Pills inhibit ovulation and implantation

(d) Saheli is a 'once a week' pill

Choose option having correct statements

(1) a,b,c,d (2) a,b,c

(3) a,c,d (4) a,b,d

104. During evolution the animal which evolved in to

the first amphibian that lived on both land and

water were :-

(1) Sauropsids (2) Synapsids

(3) Lobefins (4) Therapsids

105. Study of history of life forms on earth is

called :-

(1) Exobiology

(2) Cosmology

(3) Evolutionary biology

(4) Ethology

106. Fountain head of evolution is :-

(1) Natural selection

(2) Mutation

(3) Genetic drift

(4) Reproductive isolation

101. 30 :-

(1) (2) (3) (4)

102. :-

- I - II

(i)

(ii)

(iii)

(iv)

(1) i, ii, iii (2) ii, iii, iv

(3) i, ii (4) iii, iv

103. :-(a)

(b) (c)

(d) '' (1) a,b,c,d (2) a,b,c

(3) a,c,d (4) a,b,d

104. :-

(1) (2) (3) (4)

105. :-(1) Exobiology

(2) Cosmology

(3) Evolutionary biology

(4) Ethology

106. (1) (2) (3) (4)


H-25/45Kota/01CM414107

107. Given set of diagrams are of sex determination.

Find out correct combination -

(a) (b)

(c)

(a) (b) (c)

(1)

(2)

(3)

(4)

108. Although Mendelian work was published in

1865 but for several reasons it remained

unrecognised till 1900. Which of the following

was not it's reason ?

(1) Poor communication

(2) Concept of factor was not accepted by others

(3) Mathematical approach of data analysis

(4) For factor he provided only physical proof but

no chemical evidence.

107. -

(a) (b)

(c)

(a) (b) (c)

(1)

(2)

(3)

(4)

108. 1865

1900

:-

(1)

(2)

(3)

(4)

Kota/01CM414107H-26/45


109. Genes A and b are present 20cM far from each

other on same chromosome. Then a dihybrid plant

will produce :

(1) 4 types of gametes in equal proportion

(2) More aB gametes as compared to AB gametes

(3) More aB gametes as compared to Ab gametes

(4) Ab gametes and ab gametes in equal propotion

110. Select the correct combination.

1

2

3

4

well 1 DNA + EcoR-I

well 2 DNA + Sal-I

well 3 DNA + Hind III

well 4 DNA + Bam HI

HighestRestriction sitefor associated

RestrictionEndonuclease

NoDNA

digestion

Lack ofrestriction sitefor associated

restrictionEndonuclease

maximumrestrictionsite are of

1 well (2) well (3) well (4) EcoR-I

2 well (2) well-3 well (3) Sal-I

3 well (2) well-3 well (3) Bam-HI


111. Non albuminous seeds is :-

(1) Seeds have no residual endosperm

(2) Seeds have retain a part of endosperm

(3) Seeds without fertilization

(4) Seeds without embryo

109. Gene A b 20cM

:

(1)

(2) AB aB

(3) Ab aB

(4) Ab ab

110. .

1

2

3

4

well 1 DNA + EcoR-I

well 2 DNA + Sal-I

well 3 DNA + Hind III

well 4 DNA + Bam HI

DNA

1 well (2) well (3) well (4) EcoR-I

2 well (2) well-3 well (3) Sal-I



111. :-

(1)

(2)

(3)

(4)


H-27/45Kota/01CM414107

112. Given below is the diagrammatic sketch of a

structure of maize seed. Identify the parts labelled

A, B, C and D and select the right option about

them :-

A

B

C

D

Part-A Part-B Part-C Part-D

(1) AleuroneLayer

Scutellum Coleorhiza Coleoptile

(2) Scutellum Aleuronelayer

Coleoptile Coleorhiza

(3) Scutellum Aleuronelayer

Coleorhiza Coleoptile

(4) Aleuronelayer

Scutellum Coleoptile Coleorhiza

113. Diaphragms are contraceptive devices used by the

females. Choose the correct option from the

statements given below :-

(i) They are introduced into the uterus

(ii) They are placed to cover the cervical rigion.

(iii) They act as physical barriers for sperm entry.

(iv) They act as spermicidal agent.

(1) (i) & (ii) (b) (i) & (iii)

(3) (ii) & (iii) (d) (iii) & (iv)

114. When did evolution of Australopithecus took

place ?

(1) Two million years ago

(2) Four million years ago

(3) One million years ago

(4) Five million years ago

115. The first living organism was :-

(1) Aerobic–heterotrophic

(2) Aerarobic–Autotrophic

(3) Anaerobic–heterotrophic

(4) Anaerobic–Autotrophic

112. A, B, C D :-

A

B

C

D

-A -B -C -D

(1)

(2)

(3)

(4)

113. :-(a) (b) (c)

(d) (1) (i) & (ii) (b) (i) & (iii)

(3) (ii) & (iii) (d) (iii) & (iv)

114. :-

(1) (2) (3) (4)

115. :-(1) (2) (3) (4)

Kota/01CM414107H-28/45


116. Which cannot be explain by Darwinism ?


(2) Struggle

(3) Survival of fittest

(4) Arrival of fittest

117. In independent asortment what actually assort

independently ?

(1) Characters

(2) Contrasting traits

(3) Allelic genes

(4) Non homologous chromosomes

118. Select out the incorrect statement :-

(1) If Mendel selected eight character for Pisum,

law of independent assortment might be

influenced.

(2) If 2n = 12 and Mendel selected 7 characters

for study law of independent assortment might

be influenced.

(3) If 2n = 14 and Mendel selected 7 pairs of

characters law of segregation generally not

influenced.

(4) If 2n = 14 and he selected 8 pairs then law

of independent assortment definitely get

influenced.

119. The figure below shows three types of crosses

(A, B, C) of sex linked inheritance. select the

option giving correct indentification :

A. c c cX X Married to X Y

B. A A c c S S cHb Hb X X Married to Hb Hb X Y

C. hXX Married to XY

(1) A - Carrier female married to normal male

(2) B - Colour blind female having normal

haemoglobin married to colour blind

man having abnormal haemoglobin.

(3) C - Haemophelic female married to normal

male.

(4) B - Female carrier for sickle cell anaemia and

colour blindness married to man suffering

from sickle cell anaemia.

116. (1) (2) (3) (4)

117. (1) (2) (3) (4)

118. :-

(1)

(2) 2n = 12 7

(3) 2n = 14 7

(4) 2n = 14 8

119. :

A. c c cX X Married to X Y

B. A A c c S S cHb Hb X X Married to Hb Hb X Y

C. hXX Married to XY

(1) A -

(2) B -

(3) C -

(4) B -


H-29/45Kota/01CM414107

120.

(D)Distinction of coding

and non codingsequences on chromosome

B

C

(A)

(C)

(B)

Chromosomes

In this given procedure, arrangement of base

sequences is immediately followed by :-

(1) Annotation (2) Alignment

(3) Assignment (4) Map formation

121. Dioecious plant prevents :-

(1) Autogamy

(2) Geitonogamy

(3) Both autogamy and geitonogamy

(4) Xenogamy

122. Which of the following statement is not correct :-

(1) Placenta acts as an endocrine tissue

(2) In the later phase of pregnancy, a hormone

called relaxin is also secreted by the ovary

(3) The Placenta is connected to the embryo

through an umbilical cord which helps in the

transport of substances to and from the

embryo

(4) hCG, hPL and relaxin are produced in women

all the time.

120.

(D)Distinction of coding

and non codingsequences on chromosome

B

C

(A)

(C)

(B)

Chromosomes

:-

(1) (2)

(3) (4)

121. :-

(1)

(2)

(3)

(4)

122. :-

(1)

(2)

(3)

(4) hcG, hPL

Kota/01CM414107H-30/45


123. Read the following sentences and select correct

option having all true statements:-

(a) During parturition, oxytocin induces intense

contractions in the uterus

(b) After parturition oxytocin stimulate milk

let-down.

(c) The laboour pain during child birth is due to

oxytocin

(d) Oxytocin is secreted from placenta and ovary

(1) b,c,d (2) a,b,c

(3) c & d (4) a & c

124. The brain capacity of Homo habilis was :-

(1) 1200 cc

(2) 1450 cc

(3) 650-800 cc

(4) 1200-1400 cc

125. Near about the carboniferous period all the present

day continents formed a single large land mass

called:-

(1) Alligators (2) Realms

(3) Pangaea (4) Country

126. According to theory of natural selection some

organism produce more offspring and other

produce fewer offspring. This is called

(1) Sexual selection

(2) Differential reproduction

(3) Genetic drift

(4) Arrival of fittest

127. Select the correct combination :-

DominanceIncompletedominance

Codominance

(1) Mendelianinheritance

Genotypicblending

Phenotypicblending


Phenotypicblending

Genotypicblending

(3) Non mendeliangenetics

Phenotypicblending

Phenotypicblending

(4) Mendeliangenetics

Phenotypicblending

Qualitativeinheritance

123. :-

(a)

(b)

(c)

(d)

(1) b,c,d (2) a,b,c

(3) c & d (4) a & c

124. :-(1) 1200 cc

(2) 1450 cc

(3) 650-800 cc

(4) 1200-1400 cc

125. :-

(1) (2)

(3) (4)

126.

(1)

(2)

(3)

(4)

127. :-

DominanceIncompletedominance

Codominance


Genotypicblending

Phenotypicblending


Phenotypicblending

Genotypicblending

(3) Non mendeliangenetics

Phenotypicblending

Phenotypicblending

(4) Mendeliangenetics

Phenotypicblending

Qualitativeinheritance


H-31/45Kota/01CM414107

128. In Lathyrus odoratus if double heterozygous

purple flowered plants are crossed then what

would by probability of double homozygous white

flowered plants ?

(1) 1/16 (2) 2/16

(3) 3/16 (4) 4/16

129. The possibility of a female becoming a haemophilic

is....A....rare because mother of such a female has

to be at least.....B.....and the father should be.....C.....

Choose the correct option for A, B and C.

(1) A-extremely, B-carrier, C-haemophilic

(2) A-extremely, B-carrier, C-carrier

(3) A-extremely, B-haemophilic, C-carrier

(4) A-extremely, B-haemophilic, C-haemophilic

130. Which of the following statements regarding

baculoviruses as bio-control agents is/are correct ?

I. Baculoviruses are pathogens that attack

insects and other arthropods.

II. Most of these biocontrol agents belong to the

genus Nucleopolyhedro virus.

III. They do harm plants, mammals, birds, fish

and other non-target insects.

IV. Baculoviruses are helpful in Integrated pest

Management (IPM) programme, in which

beneficial insects are conserved.

Choose the correct option.

(1) I, II and III

(2) I, II and IV

(3) II, III and IV

(4) All of these

131. In dicotyledonous embryo the zygote gives rise

to the proembryo and subsequently to the :-

(1) Heart shaped, globular, mature embryo

(2) Globular, heart shaped, mature embryo

(3) Heart shaped, mature embryo, globular

(4) Mature embryo, globular, heart shaped

128.

:-

(1) 1/16 (2) 2/16

(3) 3/16 (4) 4/16

129. ....A....

.....B..........C.....

A, B C

(1) A-, B-, C-

(2) A-, B-, C-

(3) A-, B-, C-

(4) A-, B-, C-

130. -

?

I.

II.

III.

IV.

(1) I, II and III

(2) I, II and IV

(3) II, III and IV

(4)

131.

:-

(1) , ,

(2) , ,

(3) , ,

(4) , ,

Kota/01CM414107H-32/45



(a) ICSI is another specialized procedure to form

embryo in the laboratory in which a sperm is

directly injected into the ovum

(b) Transfer of an ovum collected from a donor

into the fallopian tube of another female is

called GIFT.

(c) ZIFT, GIFT and IUT are invitro fertilization

technique

(d) In IUT method, embryo having less than 16

blastomeres is transferred into uterus.

Which of the above statements are not correct

(1) a,b,c (2) c,d (3) a,c,d (4) a,b

133. No labour pain occur in a woman after completing

her gestation period and the doctor injected a

hormone for starting labour pain, but due to

mistake, gestation period was increased. Which

hormone injection is responsible for the above

condition :-

(1) Oxytocin (2) Estrogen

(3) Progesterone (4) HCS

134. Among the following how many organisms are

marsupials present in Australia :-

Tiger cat, Wombat, lemur, Koala, Sugar glider

flying squirrel

(1) 3 (2) 4 (3) 2 (4) 1

135. Darwin's theory of evolution was based on :-

(1) Survival of fittest


(3) Mutation

(4) Descent with modifications

136. The character that proves that frogs have evolved

from fishes is :-

(1) The ability to swim in water

(2) The tadpole larva in frogs which resembles the

fishes in many character

(3) Similarity in the shape of the head

(4) The tadpole larva of frogs and fishes are

uricotelic

132. :-

(a) ICSI

(b)

GIFT

(c) ZIFT, GIFT IUT

(d) IUT 16

(1) a,b,c (2) c,d (3) a,c,d (4) a,b

133.

:-

(1) Oxytocin (2) Estrogen

(3) Progesterone (4) HCS

134.

:-

(1) 3 (2) 4 (3) 2 (4) 1

135. (1) (2) (3) (4)

136.

:-

(1)

(2)

(3)

(4)


H-33/45Kota/01CM414107

137. This given diagram is not a presentation of :-

Tall Dwarf

Tall Tall

Tall Tall Tall Dwarf

Parental

Selfing

F generation1

F generation2

(1) Monogenic inheritance

(2) Qualitative inheritance

(3) Mendelian inheritance

(4) Monogenetic extranuclear inheritance

138. Select out the odd one with respect to type of

phenotypic effect in pleiotropy :-

(1) Sickling of RBC

(2) Sickle cell crisis

(3) Bone marrow enlargement

(4) Replacement of glutamic acid with valine

139. Which of the following pedigree can be for

haemophilia ?

(1) (2)

(3) (4)

140. Which of the following enzyme is not involve inisolation of DNA :-

(1) Lysozyme (2) Cellulase(3) Chitinase (4) DNase

137. :-

Tall Dwarf

Tall Tall

Tall Tall Tall Dwarf

Parental

Selfing

F generation1

F generation2

(1) (2) (3) (4)

138. :-(1) RBC (2) (3) (4)

139.

(1) (2)

(3) (4)

140. DNA :-(1) (2) (3) (4)

Kota/01CM414107H-34/45


141. Radicle is enclosed in an undifferentiated sheath.

It is called :-

(1) Coleoptile (2) Coleorhiza

(3) Cotyledon (4) Plumule

142.

(B)

(A)

(C)(D)

(E)

See the diagram and Identify A,B,C,D,E :-

A B C D E

1. Infundibulum Ovary Fimbriae Cervix Vagina

2. Fallopiantube Infundibulum Cervix Vagina Isthmns

3. Isthmus Infundibulum Fimbriae Cervix Vagina

4. Cervix Ovary Vagina Fimbriae Isthmus

143. Consider the following four statements (a-d) and

select the option which includes all the correct one

only :-

(a) ZIFT is an invitro fertilization technique

(b) AI is an invivo fertilization technique

(c) "Durga"was the first test tube baby, born in

India

(d) ICSI is an invivo fertilization technique

(1) a, b, c (2) a, b, c, d

(3) a, c, d (4) a, b, d

144. Find out the correct match from the following

table.

Column-I Column-II Column-III

(i) AustralianMarupial

Sugarglider

Koala

(ii) Adaptiveradiation

Darwin'sfinches

AustralianMarsupial

(iii) Fertilehybrid

Tigon Lion

(1) (i) and (iii) (2) (ii) and (iii)

(3) (i) only (4) (i) and (ii)

141. :-

(1) (2)

(3) (4)

142.

(B)

(A)

(C)(D)

(E)

A,B,C,D,E :-

A B C D E

1. Infundibulum Ovary Fimbriae Cervix Vagina

2. Fallopiantube Infundibulum Cervix Vagina Isthmns

3. Isthmus Infundibulum Fimbriae Cervix Vagina

4. Cervix Ovary Vagina Fimbriae Isthmus

143. (a-d) :-

(a) ZIFT

(b) AI

(c)

(d) ICSI

(1) a, b, c (2) a, b, c, d

(3) a, c, d (4) a, b, d

144.

Column-I Column-II Column-III

(i)

(ii)

(iii)

(1) (i) (iii) (2) (ii) (iii)

(3) (i) (4) (i) (ii)


H-35/45Kota/01CM414107

145. Which of the following statements is correct with

respect to the hybrid to the hybrid Mule ?

(1) Mule is not a species

(2) Mule is a new species

(3) Horse and ass are two populations

(4) Mule represents a common ancestor of horse

and ass

146. Allelic gene located on :-

(1) Same locus of non homologous chromosomes

(2) Different locus of homologous chromosome

(3) Same locus of homologous chromosome

(4) None of these

147. Which of the following cytogenetical event is not

execuiting here :-

TallTT

Dwarftt

Gemetes Gemetes

GemetesGemetes

TallTt

TallTt

Selfing

F generation1

F generation2

Phenotypic ratio : tall : dwarf 3 : 1

Genotypic ratio :TT : Tt : tt 1 : 2 : 1

(1) Segregation of factors

(2) Pairing of factor

(3) Independent assortment of factors

(4) Descrete nature of factors

145.

(1)

(2)

(3)

(4)

146. :-

(1)

(2)

(3)

(4)

147. :-

TallTT

Dwarftt

Gemetes Gemetes

GemetesGemetes

TallTt

TallTt

Selfing

F generation1

F generation2

Phenotypic ratio : tall : dwarf 3 : 1

Genotypic ratio :TT : Tt : tt 1 : 2 : 1

(1)

(2)

(3)

(4)

Kota/01CM414107H-36/45


148. Phenotype Genotype Phenotype

A B

AaBb

AAbb

aaBB

P1

P2

P3

P1

A B

(1) Polygenic qualitative Polygenic quantitative

(2) Polygenic quantitative Monogenic quantitative

(3) Polygenic quantitative Polygenic qualitative

(4) Monogenic qualitative Polygenic qualitative

149. Figure below shows two conditions of Lac operon

in E. coli. Select the option giving correct

identification :-

A

B

(1) A - Repression of operon due to presence of lactose

(2) B - Induction of operon due to presence of lac

mRNA

(3) A - Repression of operon in absence of lactose

(4) B - Induction of operon due to presence of

-galactosidase

148. Phenotype Genotype Phenotype

A B

AaBb

AAbb

aaBB

P1

P2

P3

P1

A B

(1)

(2)

(3)

(4)

149. E. coli Lac operon :-

A

B

(1) A - Lactose

(2) B - Lac mRNA

(3) A - Lactose

(4) B - -galactosidase


H-37/45Kota/01CM414107

150. Consider the following statements about

therapeutic drugs :-

I. The recombinant DNA technology is used for

production of therapeutic drugs which are safe

and effective.

II. It avoid unwanted immunological responses,

commonly observed with similar products

isolated from non-human sources.

III. About thirty recombinant therapeutics have

been approved for human use in India.

Which of the statements given above are correct ?

(1) I and II (2) I and III

(3) II and III (4) I, II and III

151. In which process embryo is formed out side the

embryosac ?

(1) Diplospory

(2) Diploid apogamy

(3) Adventive embryony

(4) Apospory


(a) Spermatogenesis starts at the age of puberty

(b) Androgens stimulate the process of

spermatogenesis

(c) During spermiogenesis, sperm heads become

embedded in the sertoli cells

(d) The increased levels of GnRH acts at the

anterior pituitary gland and stimulates

secretion of two gonadotropins.

How many of the above statements are correct.

(1) Four (2) Three (3) Two (4) One

153. Testosterone is essential for :-

(a) Sperm formation

(b) ABP secretion

(c) Development of secondary sexual characters

(1) a,b (2) b,c (3) a,b,c (d) a,c

150. :-

I. DNA

II.

III. 30

-?

(1) I and II (2) I and III

(3) II and III (4) I, II and III

151.

?

(1)

(2)

(3)

(4)

152. :-

(a)

(b)

(c)

(d) GnRH

153. :-

(a)

(b) ABP

(c)

(1) a,b (2) b,c (3) a,b,c (d) a,c

Kota/01CM414107H-38/45


154. Which of the following statement is incorrect:-

(1) Any population has built in variation in

characteristics

(2) Natural selection is a mechanism of evolution

(3) The geological history of earth never correlates

with the biological history of earth

(4) According to Darwin, the fitness refers

ultimately and only to reproductive fitness

155. Role of mutation in evolution is :-

(1) Genetic drift

(2) Reproductive isolation

(3) Variation


156. About genotype Tt select out the incorrect

statement :-

(1) Allelic gene pair

(2) Can be present on non homologous chromosome

(3) They share common locus

(4) They share same characters

157. ABO blood group determining gene I leads to

addition of which chemical as determinant of

blood group :-

(1) Lipid (2) Protein

(3) Sugar (4) Phospholipid

158. Match the following :-

(A) Human Female (i) 2A + XY

(B) Male Drosophila (ii) 2A + XX

(C) Male Grass hopper (iii) 2A + ZW

(D) Female birds (iv) 2A + X

A B C D

(1) (ii) (i) (iii) (iv)

(2) (ii) (i) (iv) (iii)

(3) (i) (ii) (iv) (iii)

(4) (i) (ii) (iii) (iv)

159. Which of the following is a wrong statement with

reference to Lac operon ?

(1) Gene i is constitutive

(2) Galactose is the substrate for -galactosidase

(3) Polycistronic structural gene is regulated by

a common promoter and regulator gene.

(4) A very low level of expression of lac operon

is present all the time

154. :-(1)

(2) (3)

(4)

155. (1) (2) (3) (4)

156. Tt) :-(1) (2) (3) (4)

157. ABO -I :-

(1) (2) (3) (4)

158. :-(A) Human Female (i) 2A + XY

(B) Male Drosophila (ii) 2A + XX

(C) Male Grass hopper (iii) 2A + ZW

(D) Female birds (iv) 2A + X

A B C D

(1) (ii) (i) (iii) (iv)

(2) (ii) (i) (iv) (iii)

(3) (i) (ii) (iv) (iii)

(4) (i) (ii) (iii) (iv)

159.

(1) Gene i

(2) Galactose, -galactosidase

(3)

(4)


H-39/45Kota/01CM414107

160. Select the correct match :-

x1 x2 x3 x4

(1) Pvu I Sal I Eco R I Bam H I

(2) Pst I Bam H I Eco R I Sal I

(3) Pvu I Bam H I Eco R I Sal I

(4) Pvu I Hind III Eco R I Sal I

161. How many of the following are the parts of a

typical stamen ?

Style, Stigma, Filament, Anther, Ovule, Ovary.

(1) Three (2) One (3) Two (4) Four

160. :-

x1 x2 x3 x4

(1) Pvu I Sal I Eco R I Bam H I

(2) Pst I Bam H I Eco R I Sal I

(3) Pvu I Bam H I Eco R I Sal I

(4) Pvu I Hind III Eco R I Sal I

161. (1) (2) (3) (4)

Kota/01CM414107H-40/45


162. Given below is an incomplete flow chart showing

influence of hormone on gametogenesis in male,

observe the flow chart carefully and Identify

A,B,C,D :-

Pituitary

ICSH

Leydig cell

FSH

C Name of cell

factor stimulatesA

B

Stimulate

Name of Process

Name of hormone

D None of the process

A B C D

1. Testosterone Spermato-

genesisSertoli cell

Spermiog-

anesis

2. Androgen Spermiog-

enesis

Leydig

cell

Spermato-

genesis

3. FSH Oogenesis ICSH Spermatel-

iocis

4. LH Oogenesis Polar body Oogenesis

163. During genetic drift sometime the change in allele

frequency is so different in the new popultation

that they become a different species. This effect

is called :-

(1) Founder effect


(3) Hardy-Weinberg principle

(4) Bottlenect effect

164. Select the incorrect statement :-

(1) Lichens can be used as industrial pollution

indicators

(2) Evolution is a directed process in the sense of

determinism

(3) Evolution is a stochastic process based on

chance event in nature and chance mutation

in the organisms

(4) Similarities in proteins and genes performing

a given function among diverse organisms

give clues to common ancestory

Time Management is Life Management

162. A,B,C

D :-

Pituitary

ICSH

Leydig cell

FSH

C

factor stimulatesA

B

Stimulate

Name of ( )

Name of (

D

A B C D

1. Testosterone Spermato-

genesisSertoli cell

Spermiog-

anesis

2. Androgen Spermiog-

enesis

Leydig

cell

Spermato-

genesis

3. FSH Oogenesis ICSH Spermatel-

iocis

4. LH Oogenesis Polar body Oogenesis

163. :-

(1) (2) (3) - (4)

164. :-(1)

(2)

(3)

(4)


H-41/45Kota/01CM414107

165. A = Natural selection

B = variation & their inheritance

C = Survival of fittest

D = Struggle for existence

According to darwinism, correct sequence are :-

(1) B, C, A, D (2) D, B, C, A

(3) C, D, A, B (4) D, A, C, B

166. Dominance does not depends on :-

(1) Genotype

(2) Gene product-enzyme

(3) Production of phenotype from gene

(4) Type of organism

167. Select out the correct combination :-

Allele fromParent 1

Allele fromParent 2

Genotypeoffspring

Blood typesof offspring

IA IA A A

IA IB IA IB AB

IA i IA i B

IB IA C AB

IB IB IB IB B

IB i IB i D

i i i i O

A B C D

(1) IAIA AB IBIB B

(2) IAIB AB IAIA B

(3) IAIA A IAIB AB

(4) IAIA B IAIB A

168. Match the following :-

Column-I Column-II

(A) Microsatellite

(i) Chromosome structuredynamics and evolution

(B) Minisatellite (ii) Human history

(C) SNPs (iii) DNA finger printing

(D) Repeatitivesequences

(iv) Genetic and physicalmap

Options :-

A B C D

(1) (i) (ii) (iii) (iv)

(2) (iv) (iii) (ii) (i)

(3) (iv) (iii) (i) (ii)

(4) (iii) (iv) (ii) (i)

165. A = B = C = D = (1) B, C, A, D (2) D, B, C, A

(3) C, D, A, B (4) D, A, C, B

166. :-(1) (2) (3) (4)

167. :-

Allele fromParent 1

Allele fromParent 2

Genotypeoffspring

Blood typesof offspring

IA IA A A

IA IB IA IB AB

IA i IA i B

IB IA C AB

IB IB IB IB B

IB i IB i D

i i i i O

A B C D

(1) IAIA AB IBIB B

(2) IAIB AB IAIA B

(3) IAIA A IAIB AB

(4) IAIA B IAIB A

168. :-

Column-I Column-II

(A) Microsatellite

(i) Chromosome structuredynamics and evolution

(B) Minisatellite (ii) Human history

(C) SNPs (iii) DNA finger printing

(D) Repeatitivesequences

(iv) Genetic and physicalmap

:-A B C D

(1) (i) (ii) (iii) (iv)

(2) (iv) (iii) (ii) (i)

(3) (iv) (iii) (i) (ii)

(4) (iii) (iv) (ii) (i)

Kota/01CM414107H-42/45


169. Match the following-

(A) Arabidopsisthalliana

(i) Ist Animal to besequnced

(B) Caenorhabditiselegans

(ii) Suitable host forcloning of DNAfragments

(C) Yeast (iii) Ist plant to besequenced

(D) BAC (iv) Vector

(1) A-iv B-iii, C-ii D-i

(2) A-i B-ii C-iii D-iv

(3) A-iii B-i C-ii D-iv

(4) A-i B-iv C-iii D-ii

170. The aims and objectives of genetic engineering

approval committee are :

(i) To permit the use of genetically modified

organisms and their products for commercial

applications

(ii) To adopt the procedures for restriction,

production and application of GM organisms

(iii)Approval to conduct large scale field trails and

release of transgenic crops in the environment

(iv)GEAC make decisions about selection of

vectors for recombinant DNA technology

(v) To prevent exploitation of bioresources of under

developing contries by developed nations

Among given statements how many statements

are true

(1) Two (2) Three

(3) Four (4) Five

171. Consider the following statements (a-d) and select

the option which includes all the correct ones

only:-

(a) The microsporangia develop further and

become pollen sac. In anther these pollen sacs

extends transversely.

(b) The anther is a tetragonal structure.

(c) Typical angiosperm anther is bilobed with

each lobe having single theca.

(d) Cells of the tapetum possess dense cytoplasm

and generally have more than one nucleus.

Options :

(1) Statements a, b, d (2) Statements b, c, d

(3) Statements b, d (4) Statements a, b

169. -

(A)

(i)

(B)

(ii) DNA

(C) (iii)

(D) BAC (iv)

(1) A-iv B-iii, C-ii D-i

(2) A-i B-ii C-iii D-iv

(3) A-iii B-i C-ii D-iv

(4) A-i B-iv C-iii D-ii

170. :-

(i)

(ii) GM

(iii)

(iv)rDNA GEAC

(v)

(1) (2)

(3) (4) 171. (a-d)

:-(a)

(b) (c)

(d)

:(1) a, b, d (2) b, c, d

(3) b, d (4) a, b


H-43/45Kota/01CM414107

172. Which of the following pair is not correct :-

(1) Temporary contraceptive method-Nim-76

(2) Non medicated IUD's

(3) Male pill - Gossypol

(4) Female oral pill-DMPA

173. A process in which heritable variation enabling

better survival are enabled to reproduce and leave

greater number of progeny is called :-

(1) Genetic drift


(3) Founder effect

(4) Both (1) and (3)

174. Select the incorrect statements :-

(A) The essence of Darwinian theory of evolution

is natural selection

(B) Evolution is a directed process in the sense

of determinism

(C) The geological history of earth is not related

with the biological history of earth

(D) During evolution the rate of appearance of

new forms is linked to the life cycle

(1) A and B (2) B and C

(3) A and D (4) B and D

175. Two population of deer mouse are found one in

the middle of the forest and the other from a

nearest field habitat. The two populations show

no interbreeding naturally but can interbreed in

laboratory. This type of reproductive isolation is

called –

(1) Seasonal isolation

(2) Ecological isolation

(3) Physiological isolation

(4) Behavioural isolation

176. As these Mendelian factors represent the genetic

basis of inheritance, understanding of which of the

following become the focus of attention in biology

for next century :-

(1) Structure of genetic material

(2) Structural basis of genotype

(3) Phenotype conversion

(4) All the above

172. :-

(1) Nim-76

(2) IUD's

(3) - Gossypol

(4) -DMPA

173.

:-

(1)

(2)

(3)

(4) (1) (3) 174. :-

(A)

(B)

(C)

(D)

(1) A B (2) B C

(3) A D (4) B D

175. Deer mouse

–

(1)

(2)

(3)

(4)

176.

:-

(1)

(2)

(3)

(4)

Kota/01CM414107H-44/45


Your Target is to secure Good Rank in Pre-Medical 2015

Your moral duty

is to prove that is

177. Inheritance is the process by which characters are

passed from :-

(1) Cell to cell (2) Zygote to cells

(3) Parent to progony (4) Husband to wife

178. Since it is evident that control crosses that can be

performed in peaplants and some other organisms,

are not possible in human brings. So which of the

following provide and alternative :-

(1) Study of dizygotic twins

(2) Study of DNA sequences

(3) DNA finger printing

(4) Study of family history

179. In this give diagram accesibility of promotor

depends on interaction between:-

p i p z y a

p i p z y a

In absence of inducer

Repressor mRNA

Repressor

Transcription

Translationlac mRNA

-galactosidase permease transacetylase

Repressor mRNA

(Inactive repressor)

Inducer

C B

A

CAPE

D

(1) A & C (2) A & B

(3) E & B (4) E & C

180. Select the incorrect match :-

(1) Allternate heating and cooling – Competent

(2) Ori – Controlling copy number

(3) Primer in PCR – Extension

(4) Selectable Marker – Screening

177. :-(1) (2) (3) (4)

178. :-(1) (2) DNA (3) DNA (4)

179. :-

p i p z y a

p i p z y a

In absence of inducer

Repressor mRNA

Repressor

Transcription

Translationlac mRNA

-galactosidase permease transacetylase

Repressor mRNA

(Inactive repressor)

Inducer

C B

A

CAPE

D

(1) A & C (2) A & B

(3) E & B (4) E & C

180. :-(1) – (2) Ori – (3) PCR – (4) –


H-45/45Kota/01CM414107

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