Hindi PAPER CODE KOTA (RAJASTHAN) FORM NUMBER …Secure Site €¦ · 14. Three charges of 0.1C...
Transcript of Hindi PAPER CODE KOTA (RAJASTHAN) FORM NUMBER …Secure Site €¦ · 14. Three charges of 0.1C...
Your Target is to secure Good Rank in Pre-Medical 2015
Path to Success
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
(4) CH3–COCl > (CH3–CO)2O > CH3COOR >
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
(2) CH3–COCl > (CH3–CO)2O > CH3COOR >
CH3–CONH2
(3) CH3COCl > CH3COOR > CH3–CONH2 >
(CH3CO)2O
(4) CH3–COCl > (CH3–CO)2O > CH3COOR >
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
60. What is the major product of the following
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
65. What is the major product of the following
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
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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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
Major Test For Target : Pre-Medical 2015/02-04-2015
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
89. What is the major product of the following
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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
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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
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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
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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
103. Read the following statements :-
(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)
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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)
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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
4 well (4) 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
3 well (2) well-3 well (3) Bam-HI
4 well (4) well-3 well (3) Bam-HI
111. :-
(1)
(2)
(3)
(4)
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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)
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116. Which cannot be explain by Darwinism ?
(1) Natural selection
(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 -
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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
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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
(2) Mendelianinheritance
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
(1) Mendelianinheritance
Genotypicblending
Phenotypicblending
(2) Mendelianinheritance
Phenotypicblending
Genotypicblending
(3) Non mendeliangenetics
Phenotypicblending
Phenotypicblending
(4) Mendeliangenetics
Phenotypicblending
Qualitativeinheritance
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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) , ,
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132. Read the following statements :-
(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
(2) Natural selection
(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)
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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)
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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)
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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)
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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
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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
152. Read the following statements :-
(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
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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
(4) Natural selection
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)
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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)
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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
(2) Natural selection
(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)
Phase/ENTHUSIAST, MLI , MLJ, MLSP, MAZA, MAZB & MAZC/02-04-2015
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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)
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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
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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
(2) Natural selection
(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)
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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) –
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