Download BCECE Medical Entrance Exam

BCECE
Medical Entrance Exam
Solved Paper 2013
Physics
1. The dimensions of universal gravitational
constant are
(a) [M−2 L−3 T−2 ]
(c) [M−1 L3 T−2 ]
(b) [M−2 L2 T−1 ]
(d) [ML2 T−1 ]
2. If A × B = 3 A . B then the value of A + B
5. When capillary tubes of different radii ' r'
dipped in water, water rises to different
heights ' h' in them, then
(a) hr 2 = constant
h
(c) = constant
r
(b) hr = constant
h
(d) 2 = constant
r
is
(a) A+B
(b) (A 2 + B2 +
6. In the adjoining figure A, B, and C
represents three progressive waves. Which
of the following statement about the waves
is correct?
1/ 2
3 AB)
1/ 2
AB 

(c)  A 2 + B2 +


3
2
2
1/ 2
(d) (A + B + AB)
3. Two spheres of masses m and M are situated
in air and the gravitional force between
them is F. The space around the masses is
now filled with a liquid of specific gravity 3.
The new gravitational force will be
(a) F
(c) 3F
F
3
F
(d)
9
volume above the surface of water. If
placed in oil it floats with one third of its
volume above the surface of oil. The
density of oil is
3
4
2
(c)
3
A
C
(b)
4. A body floats in water with one fourth of its
(a)
B
4
9
9
(d)
8
(b)
(a) Wave C lags behind in phase by
π
from A and B
2
π
2
(b) Wave C leads in phase by π from A and B lags
behind by π
π
(c) Wave C leads in phase by from A and lags behind
2
π
by
2
(d) Wave C lags behind in phase by π from A and B
leads by π
leads by
2
| BCECE (Medical) l Solved Paper 2013
7. If 110 J of heat are added to a gaseous
system, whose internal energy is 40J, then
the amount of external work done is
(a) 40 J
(c) 110 J
(b) 70 J
(d) 150 J
(a) E
(b) B
(c) E × B
(d) B × E
13. The current in RCL circuit is maximum
where
8. A ray is incident at an angle of incidence i on
one surface of a prism of small angle A and
emerges normally from the opposite surface.
If the refractive index of the material of the
prism is µ, the angle of incidence i is nearly
equals to
(a) µ A/2
(c) µ A
direction of propagation of electromagnetic
waves is that of
(b) A/2 µ
(d) A/µ
(a) X L = 0
(b) X L = XC
(c) XC = 0
(d) X
2
L
+ X C2 = 1
14. What is the value of A + A in the boolean
algebra?
(a) 0
(b) 1
(c) A
(d) A
15. The truth table given below is for which gate?
9. Figure below shows four plates each of area
A
B
Y
S separated from one another by a distance d.
What is the capacitance between A and B ?
0
0
1
0
1
1
1
0
1
1
1
0
A
B
4 ε0 3
d
2 ε0 5
(c)
d
(a) XOR
(c) AND
3 ε0 5
d
ε 5
(d) 0
d
(a)
(b)
16. The magnetic flux through a circuit of
10. What is the value of current in the arm
containing 2π resistor in the adjoining
circuit?
resistance R changes by an amount ∆φ in a
time ∆t. Then the total quantity of electric
charge Q that passes any point in the circuit
during the time ∆t is represented by
∆φ
∆t
∆φ
(c) Q = R.
∆t
(a) Q =
10Ω
1.4A
25Ω
(a) 0.7 amp
(c) 1.5 amp
2Ω
1.4A
G
 20 

 m with constant tangential acceleration.
 π 
(b) 1.2 amp
(d) 1.0 amp
the electric field applied across it are related
as
vd ∝ E
vd ∝ E 2
vd ∝ E
1
vd ∝
E
If the velocity of the particle is 80 m/s at the
end of the second revolution after motion has
begun the tangential acceleration will be
(a) 40π m/s 2
(c) 160π m/s 2
(b) 40 m/s 2
(d) 640π m/s 2
18. The output of a OR gate is 1
12. If E and B be the electric and magnetic field
vectors
∆φ
R
1 ∆φ
(d) Q =
R ∆t
(b) Q =
17. A particle moves along a circle of radius
5Ω
11. For ohmic conductor the drift velocity vd and
(a)
(b)
(c)
(d)
(b) OR
(d) NAND
of
electromagnetic
waves,
the
(a)
(b)
(c)
(d)
if either input is zero
if both inputs are zero
only if both inputs are 1
if either or both inputs are 1
BCECE (Medical) l Solved Paper 2013 |
19. Two 220 V, 100 W bulbs are connected first
in series and then in parallel. Each time the
combination is connected to a 220 V AC
supply line, the power drawn by the
combination in each case respectively will be
(a) 50 W, 200 W
(c) 100 W, 50 W
(b) 50 W, 20 W
(d) 200 W, 150 W
20. Barrier potential of a p-n junction diode does
not depend on
(a) diode design
(c) temperature
momentum p making an angle θ with the
horizontal. The change in momentum of the
projectile on return to the ground will be
(b) 2 p sin θ
(d) 2 p
22. A weigthtless thread can bear tension up
to 3.7 kg wt. A stone of mass 500 g is tied to
it and revolved in a circular path of radius
4m in a vertical plane. If g = 10 m /s2 , then
the maximum angular velocity of the stone
will be
(a) 2 rad/s
(c) 16rad/s
(b) 4 rad/s
(d) 21 rad/s
23. Under a constant torque, the angular
momentum of a body changes from A to 4 A in
4 s. The torque is
(a) 3 A
(c)
3
A
4
coalesce under isothermal condition, then
the radius of new bubble is
(a) 4.3 cm
(b) 4.5 cm
(c) 5 cm
(d) 7 cm
26. Deuteron and α-particle are put 1Å apart in
air. Magnitude of intensity of electric field
due to deuteron of α-particle is
(a) zero
(c) 144
. × 1011 N/C
(b) 2.88 × 1011 N/C
(d) 576
. × 1011 N/C
20. The air column in pipe, which is closed at
(b) doping density
(d) farward bias
21. A projectile is projected with a linear
(a) 2 p tan θ
(c) 2 p cos θ
3
1
A
4
4
(d)
A
3
(b)
24. The mass of a planet is double and its radius
is half compared to that of earth. Assuming
g = 10m /s2 on earth, the acceleration due to
gravity at the planet will be
(a) 10 m/s 2
(b) 20 m/s 2
(c) 40 m/s 2
(d) None of these
25. A soap bubble in vacuum, has a radius of
3 cm and another soap bubble in vacuum
has a radius of 4 cm. If the two bubbles
one end will be in resonance with a vibrating
turning fork at a frequency of 260 Hz, if the
length of the air column is
(a) 31.73 cm
(c) 35.75 cm
(b) 62.5 cm
(d) 12.5 cm
28. A ball strikes against the floor and returns
with double the velocity. In which type of
collision is it possible?
(a) Inelastic
(c) Perfectly elastic
(b) Perfectly inclastic
(d) Not possible
29. A body is under the action of three force F1, F2,
and F3. In which case the body cannot under
go angular acceleration?
(a) F1 + F2 + F3 = 0
(b) F1, F2 and F3 are concurrent
(c) F1, and F2 act at the same point but, F3 acts at different
point
(d) F1 + F2 is parallel to F3, but the three forces are not
concurrent
30. Two beams of protons moving parallel in
same direction will
(a)
(b)
(c)
(d)
repel each other
exert no force
attract each other
deflect perpendicular to the plane of the beams
31. A photon and an electron possess same
de-Broglie wavelength given that C = speed
of light and v = space of electron, which of
the following relation is correct?
(here, E e = K. E of electron, E Ph = K. E of
photon, Pe = momentum of electron, Pph =
momentum of photon )
Pe
C
=
PPn 2 v
E
2c
(c) ph =
Ee
v
(a)
Ee
C
=
EPh 2 v
P
2c
(d) e =
PPh
v
(b)
4
| BCECE (Medical) l Solved Paper 2013
32. To get an output Y = 1 from circuit of
adjoining figure, the input must be
A
B
Y
C
A
(a) 0
(b) 1
(c) 1
(d) 1
B
1
0
0
1
magnet does not depend upon
(a) length of the magnet
(b) the pole strength of the magnet
(c) the horizontal component of magnetic field of
earth
(d) the length of the suspension
C
0
0
1
0
37. The magnetic permeability is defined as the
ratio of
(a)
(b)
(c)
(d)
33. Hubble’s law is expressed as
(here, v = speed of recession, r = distance of
galaxy, H = hubble constant)
(b) v = H2 r
(a) v = Hr
H
(c) v = 2
r
36. The time period of a freely suspended
(d) v = Hr 2
magnetic induction and magnetising field
intensity of magnetisation and magnetising field
intensity of magnetisation and magnetic field
None of the above
represents an area A = 05
. m2
situated in a uniform magnetic field
B = 20
. Wb /m 2 and making an angle of 60°
with respect to magnetic field. The value of
magnetic flux through the area will be
38. Figure
34. The displacement versus time graph for a
body moving in a straight line is shown in
figure. Which of the following regions
represents motor when no force is acting on
the body?
e
d
S
c
60°
(a) 0.5 Wb
3
(c) Wb
2
(b) 3 Wb
(d) 2.0 Wb
39. Of the given diodes, shown in the adjoining
diagrams, which one is reverse biased?
b
a
(a) ab
(c) cd
A
t
R
(a)
(b) bc
(d) be
–10 V
+5 V
R
35. Which of the following graphs shows the
variation of magnetic field B, with distance
from a long current carrying conductor?
(b)
–12 V
(a) B
(c)
(b) B
–5 V
r
r
(d)
(c) B
(d) B
r
+5 V
+10 V
40. Three particles each of mass m gram, are
r
situated at the vertices of an equilateral
triangle ABC of the side 1 cm. The moment
BCECE (Medical) l Solved Paper 2013 |
of inertia of the system (as shown in figure)
about a line AX perpendicular to AB and in
the plane of ABC in gram- cm 2 units will be
m
X
45. An electron moves with a velocity v in an
m
A
3 2
ml
2
(c) 2 ml 2
(a)
electric field E. If the angle between v and E
is neither 0 nor π, the path followed by the
electron is
l
m
l
(a) parabola
(c) straight line
C
3 2
ml
4
5
(d) ml 2
4
01
. π. If it is connected to a resistance of 3.9 π,
then the voltage across the cell will be (in
volts)
−4
divisions. A current of 4 × 10 A gives a
deflection of one division. To convert this
galvanometer into a voltmeter having a
range of 25 volt, it should be connected with
a resistance of
(a) 245π as shunt
(c) 2500π as shunt
(b) 2450π as series
(d) 2550π in series
42. A car is moving towards a high cliff. The car
driver sounds a horn of frequency f . The
reflected sound heard by the driver has the
frequency 2f . If v be the velocity of sound,
then the velocity of the car, in the same
velocity units will be
v
2
v
(d)
4
(b)
43. A plane glass slab is kept over various
coloured letters, the letter which appears
least raised is
(a) violet
(c) green
(b) circle
(d) ellipse
46. The internal resistance of a cell of emf 2 V is
(b)
41. A galvanometer of 50π resistance has 25
v
2
v
(c)
3
cos r
cos i
sin i
(d) v = C
sin r
(b) v = C
B
l
(a)
sin i
cos r
sin r
(c) v = C
sin i
(a) v = C
5
(b) blue
(d) red
44. If the angle of incidence is i and that of
refraction is r. Then the speed of light in the
medium to which the light is refracted from
air is
(a) 2 V
(c) 195
. V
(b) 0.5 V
(d) 2.5 V
47. An electric kettle takes 4 A current a t 220 V.
How much time will it take to boil 1 kg of
water from room temperature 20°C? (the
temperature of boiling water is 100°C)
(a) 12.6 min
(c) 6.3 min
(b) 12.8 min
(d) 6.4 min
48. The phenomenon of pair production is
(a)
(b)
(c)
(d)
ejection of an electron from a nucleus
ejection of an electron from a metal surface
ionisation of a neutral atom
the production of an electron and a positron from
γ-rays.
49. In an electron microscope if the potential is
increased from 20kV to 80 kV, the resolving
power R of the microscope will become
R
2
(c) 4R
(a)
(b) 2R
(d) 5R
50. Force acting upon a charged particle kept
between the plates of changed capacitor is F.
If one of the plates of the capacitor is
removed force acting on the same particle
will become
(a) 0
F
(c)
2
(b) F
(d) 2 F
6
| BCECE (Medical) l Solved Paper 2013
Chemistry
CH 3 OH
 
HBr
1. CH 3  C  CH CH 3 →
A (Predominant)

CH 3
Identify A.
(a)
(b)
(c)
(d)
(CH3 )2 C (Br) CH (CH3 )2
(CH3 )3 CCH (Br) CH3
(CH3 )3 COHBr CH3
None of the above
7. In HNC, which element has least value of
formal charge?
(a) H
(c) C
(b) N
(d) All have same value
8. Consider following unit values of energy
I. 1 L atm,
II. 1 erg,
III. 1 J
IV. kcal,
Increasing order of these values is
2. A β-hydroxy carbonyl compound is obtained
(a) I = II = III = IV
(c) II < III < I < IV
(b) I < II < III < IV
(d) IV < I < III < II
by the action of NaOH on
(a) C 6H5CHO
(c) CH3CHO
(b) HCHO
(d) (CH3 )3 C ⋅ CHO
3. Which of the following statements is correct?
(a)
(b)
(c)
(d)
+ I effect stabilises a carbanion
+ I effect stabilises a carbocation
− I effect stabilises a carbanion
− I effect stabilises a carbocation
∆
sodalime
Compound ‘C ’ is a hydrocarbon, occupying
approx 0.75 L volume per gram. Identify ‘A’
and ‘B ’.
Tartaric acid, propanoic acid
Succinic acid, succinic anhydride
Maleic anhydride, maleic acid
Methyl malonic acid, propanoic acid
5. What is effective nuclear charge and the
periphery of nitrogen atom when an extra
electron is added in the formation of an
anion?
(b) 2.45
(d) 5.95
6. A certain metal sulphide, MS2 , is used
extensively as a high temperature lubricant.
If MS2 has 40.06 % sulphur by weight,
atomic mass of M will be
(a) 100 amu
(c) 60 amu
(a) 2
(c) 5
(b) 4
(d) 6
copper corrodes to
∆
(a) 1.20
(c) 3.55
metal series Sc through Zn that have four
unpaired electrons in their + 2 state are
10. In an atmosphere with industrial smog,
4. C4H 6O4 A → C3 H 6 O2 B → C
(a)
(b)
(c)
(d)
9. The number of elements in the transition
(b) 96 amu
(d) 30 amu
I. Cu2 (OH)2 SO4
II. Cu2 (OH)2 CO3
III. CuSO4
IV. CuCO3
(a)
(b)
(c)
(d)
I and III
II and IV
I and IV
I and II
11. Mischmetal is
(a)
(b)
(c)
(d)
an alloy of lanthanide and nickel
an alloy of lanthanide and copper
an alloy of lanthanide, iron and carbon
an alloy of lanthanide, magnesium and nickel
12. What is the oxidation state of iron in
[Fe(H2O)5 NO]2 + ?
(a) 0
(c) + 2
(b) + 1
(d) + 3
13. Identify the final product of the reaction
140 ° C
NaBH 4
BCl 3 + NH4Cl → → ?
C 6 H 5 Cl
(a) B 3N3H6
(c) NaBCl 4
(b) B 2H6
(d) BN
BCECE (Medical) l Solved Paper 2013 |
7
14. 2.56 g of sulphur in 100 g of CS2 has
19. Osmotic pressure of insulin solution at
depression in freezing point of 0.010°C.
Atomicity of sulphur in CS2 is
(Given, K f = 0.1° molal − 1)
298 K is found to be 0.0072 atm. Hence,
height of water column due to this pressure
is [given d (Hg) = 13.6 g / mL]
(a) 2
(c) 6
(a) 0.76 cm
(c) 7.4 cm
(b) 4
(d) 8
15. For the zeroth order reaction, sets I and II
are given, hence x is
(b) 0.70 cm
(d) 76 cm
20. ∆G° and ∆H ° for a reaction at 300 K is
− 66.9 kJ mol − 1 and – 41.8 kJ mol − 1
respectively. ∆G° for the same reaction at
330 K is
(a) − 25.1 kJ mol −1
I.
(b) + 25.1 kJ mol −1
t = 2 min
t=0
(c) 18.7 kJ mol −1
(d) − 69.4 kJ mol −1
21. At 1000 K, from the data
II.
N2 ( g) + 3H2 ( g) → 2NH 3( g);
t = x min
t=0
(a) 2 min
(c) 6 min
(b) 4 min
(d) 8 min
16. Equilibrium constant for the reaction,
NH4OH + H +
NH4+ + H2O is 1.8 × 109 .
Hence,
equilibrium
constant
for
NH 3 (aq) + H2O
NH4+ + OH − is
f
f
−5
(a) 1.8 × 10
−9
(c) 1.8 × 10
5
∆H = − 123.77 kJ mol − 1
Substance
N2
H2
NH 3
P/ R
3.5
3.5
4
Calculate the heat of formation of NH 3 at
300 K.
(a) − 44.42 kJ mol − 1
(b) − 88.85 kJ mol − 1
(c) + 44.42 kJ mol − 1
(b) 1.8 × 10
−10
(d) 5.59 × 10
(d) + 88.85 kJ mol − 1
17. Calculate pH change when 0.01 mol
22. Match the electrode (in column I) with its
CH 3KCOO Na solution is added to 1 L of
001
. M CH 3COOH solution.
general name (in Column II) and choose the
correct option given below.
Ka (CH 3COOH) = 1.8 × 10−5 , pKa = 4.74
(a) 3.37
(c) 4.74
(b) 1.37
(d) 8.01
Column I
Column II
A.
Calomel
I.
B.
Glass
II.
Redox
C.
Hydrogen
III.
Membrane
are 4, 5 and 6 respectively. Aqueous solution
of 0.01 M has pH in the increasing order
D.
Quinhydrone
IV.
Gas
(a)
(b)
(c)
(d)
(a)
(b)
(c)
(d)
18. pK b of NH 3 is 4.74 and pK b of A − , B− and C −
NH4 A < NH4 B < NH4C
NH4C < NH4 B < NH4 A
NH4C < NH4 A < NH4 B
All have equal pH
A
I
I
III
II
B
II
III
II
IV
C
III
IV
IV
III
Reference
D
IV
II
I
I
8
| BCECE (Medical) l Solved Paper 2013
23. In a Daniell cell constructed in the
laboratory, the voltage observed was 0.9 V
instead of 1.1 V of the standard cell. A
possible explanation is
28. Phenol and NH 3 reacts in presence of ZnCl2
at 300°C to produce
(a) tertiary amine
(c) primary amine
(b) secondary amine
(d) All of these
(a) [Zn2 + ] > [Cu2 + ]
NaOH + I 2
29. A [C6H10O3 (keto ester) ] →
(b) [Zn2 + ] < [Cu2 + ]
(c) Zn electrode has twice the surface of Cu electrode
(d) mol ratio of Zn2 + : Cu2 + is 2 : 1
H
∆
ppt + B → C → CH 3COOH.
− CO 2
24. A quantity of electrical charge that brings
about the deposition of 4.5 g Al from Al 3 + at
the cathode will also produce the following
volume at (STP) of H2 ( g) from H + at the
cathode
(a) 44.8 L
(c) 11.2 L
yellow
∆
+
(b) 22.4 L
(d) 5.6 L
25. IUPAC name of the following compound is
Hence, A is
O
O


(a) CH3 C CH2 C OC 2H5
O
O


(b) CH3CH2 C CH2 C OCH3
(c) Both are correct
(d) None is correct
O
O


30. CH 3 C CH 3 can be converted into CH 3 C OH
by following methods
CH2CH3
O
I 2 / NaOH

I. CH 3 C CH 3 →
∆, H +
(a)
(b)
(c)
(d)
3-propyl cyclo [3, 6] octane
1-ethyl tricyclo [2, 3, 0] heptane
1-ethyl bicyclo [2, 2, 1] heptane
4-ethyl bicyclo [2, 2, 2] heptane
26. Identify Y in the following series of equation
Mg / ether
hv
→ H + Br2 → ‘X’ → ‘Y ’
D 2O
Br
→ D
(c) Br → D
(a)
→ D
(d) →
O
–

Cr2 O 27 / H +
II. CH 3 C CH 3 →
O
Ag(NH 3 ) +

2
III. CH 3 C CH 3 →
Which method are most effective?
(a) I, III
(c) I, II
(b)
Br
D
27. Arrange the following in the order of rate of
oxidation with periodic acid
31. Relative
stabilities of the
carbocation will be in order
II. CH 3CHOHCHOHCH 3
III. (CH 3)2 COHCOH (CH 3)2
(b) II > I > III
(d) I > III = II
⊕
CH2
CH3O
I
I. HOCH2CH2OH,
(a) I > II > III
(c) III > II > I
(b) II, III
(d) I, II, III
III
(a) IV < III < II < I
(c) II < IV < III < I
⊕
CH2
II
⊕
CH2
CH3
following
⊕
CH3CH2
IV
(b) IV < II < III < I
(d) I < II < III < IV
BCECE (Medical) l Solved Paper 2013 |
32. The correct statement in respect of protein
haemoglobin is that it
(a)
(b)
(c)
(d)
functions as a catalyst for biological reactions
maintains blood sugar level
acts as an oxygen carrier in the blood
forms antibodies and offers resistance to diseases
570 K
Main product of above reaction is
(b) CH3COCH3
(d) (CH3CO)2 O
34. Aniline is reacted with bromine water and
the resulting product is treated with an
aqueous solution of sodium nitrite in
presence of dilute hydrochloric acid. The
compound so formed is converted to a
tetrafluoroborate which is subsequently
heated. The final product is
(a)
(b)
(c)
(d)
38. The dipole moment of HBr is 2.60 × 10− 30 cm
and the interatomic spacing is 1.41Å. What
is the per cent ionic character of HBr?
(a) 50%
(b) 11.5%
(c) 4.01%
p-bromoaniline
p-bromofluorobenzene
1, 3, 5-tri bromobenzene
2, 4, 6-tribromofluorobenzene
39. An element is oxidised by fluorine and not
(a) Sodium
(c) Oxygen
(b) Aluminium
(d) Sulphur
40. The reduction of an oxide by aluminium is
called
(a)
(b)
(c)
(d)
Kroll’s process
van Arkel process
Ellingham process
Goldschmidt’s aluminothermite process
41. H2 gas is liberated at cathode and anode
both by electrolysis of the following solution
except in
(a) NaCl
(c) LiH
(b) NaH
(d) HCOONa
42. Identify ( A) and ( B) in the following
sequence of reactions.
35. With hard water, ordinary soap forms curdy
SnCl2 + 2NaOH → ( A) (white ppt)
precipitate of
NaOH
(a) (RCOO)2 Ca
(c) Both (a) and (b)
(b) (RCOO)2 Mg
(c) None of these
36. Which is matched incorrectly?
Structure
(a)
O
Compound
O
HO
CH (OH) CH2OH Ascorbic acid
→ ( B)
(excess)
(a)
(b)
(c)
(d)
Sn(OH)2 , Na 2SnO 3
Sn(OH)2 , Na 2SnO 2
Sn(OH)2 , Na 2 [Sn(OH)6 ]
Sn(OH)2 , no effect
43. Nature of nitride ion is
(a) acidic
(c) amphiprotic
(b) basic
(d) cannot predict
OH
44. Wave number of spectral line for a given
(b)
O
O (RCOO)2 Mg Coumarin
(c) Both (a) and (b)
(d) None of the above
37. Which base is normally found in RNA but
not in DNA?
(a)
(b)
(c)
(d)
(d) 1.19%
by chlorine. Identify the element.
MnO
33. CH 3COOH + HCOOH →
(a) CH3CHO
(c) HCHO
9
Uracil
Thymine
Guanine
Adenine
transition is x cm − 1 for He + , then its value
for Be 3 + (isoelectronic of He + ) for same
transition is
x
cm− 1
4
(c) 4 x cm− 1
(b) x cm− 1
(a)
(d) 16 x cm− 1
45. H2O2 oxidises MnO2 is MnO4− in basic
medium, H2O2 and MnO2 react in the molar
ratio of
(a) 1 : 1
(b) 2 : 1
(c) 2 : 3
(d) 3 : 2
10
| BCECE (Medical) l Solved Paper 2013
46. 25 mL of an aqueous solution of KCl was
found to require 20 mL of 1 M AgNO3
solution when titrated using a K2CrO4 as
indicator. Depression in freezing point of
KCl solution with 100% ionisation will be
[ K F = 2.0° mol − 1kg, molarity = molality]
(a) 3.2°
(c) 0.8°
(b) 1.6°
(d) 5.0°
0.3010
log p
(b) 0.6
(d) 0.2
50. From the following reactions at 298 K,
(b) BF3
(d) Al(CH3 )3
(A) CaC2 (s ) + 2H 2O (l) → Ca(OH) 2 (s )
48. On adding AgNO3 solution into KI solution,
a negatively charged colloidal
obtained when they are in
45°
A
(a) 0.8
(c) 0.4
47. Out of the following, select Lux-Flood Acid
(a) CO 2
(c) H+
log x
m
sol.
is
(a) 100 mL of 0.1 M AgNO 3 + 100 mL of 0.1 M KI
(b) 100 mL of 0.1 M AgNO 3 + 100 mL of 0.2 mKI
(c) 100 mL of 0.2 M AgNO 3 + 100 mL of 0.1 M KI
(d) 100 mL of 0.15 M AgNO 3 + 100 mL of 0.15 M KI
x
49. Graph between log   and log p is a
 m
straight line at angle 45° with intercept OA
x
as shown. Hence,   at a pressure of
 m
+ C2H 2 ( g ) DH ° (kJ mol − 1 ) − 127.9
1
(B) Ca (s ) + O 2 ( g ) → CaO(s ) − 635.1
2
(C) CaO (s ) + H 2O (I ) → Ca(OH) 2 (s ) – 65.2
(D) C ( s ) + O 2 (s ) → CO 2 (s ) − 393.5
5
(E) C2H 2 ( g ) + O 2 ( g ) → 2CO 2 ( g )
2
+ H 2O (e) − 1299.58
Calculate the heat of formation of CaC2 ( s) at
298 K.
(a) – 59.82 KJ mol –1
(c) – 190.22 KJ mol –1
(b) + 59.82 KJ mol –1
(d) +190.22 KJ mol –1
0.2 atm is
Biology
1. Pineapple (annanas) fruit develops from
3. The ‘blue baby’ syndrome results from
(a) a unilocular polycarpellary flower
(b) a multipistillate syncarpous flower
(c) a cluster of compactly borne flowers on a common
axis
(d) a multilocular monocarpellary flower
(a) excess of chloride
(b) methaemoglobin
(c) excess of dissolved oxygen
(d) excess of TDS (Total Dissolved Solids)
4. Which one of the following statements is
2. Parthenocarpic
tomato
fruits
can
be
produced by
(a) removing androecium of flowers before pollen grains
are released
(b) treating the plants with low concentrations of
gibberellic acid and auxins
(c) raising the plants from vernalized seeds
(d) treating the plants with phenylmercuric acetate
correct?
(a) Neurons regulate endocrine activity, but not vice versa
(b) Endocrine glands regulate neural activity and
nervous system regulates endocrine glands
(c) Neither hormones control neural activity nor the
neurons control endocrine activity
(d) Endocrine glands regulate neural activity, but not
vice versa
BCECE (Medical) l Solved Paper 2013 |
5. Farmers in a particular region were
concerned that pre-mature yellowing of
leaves of a pulse crop might cause decrease
in the yield. Which treatment could be most
beneficial to obtain maximum seed yield?
(a) Frequent irrigation of the crop
(b) Treatment of the paints with cytokinins along with a
small dose of nitrogenous fertiliser
(c) Removal of all yellow leaves and spraying the
remaining green leaves with 2, 4, 5-trichlorophenoxy
acetic acid
(d) Application of iron and magnesium to promote
synthesis of chlorophyll
6. Which one of the following amino acids was
not found to be synthesised in Miller’s
experiment?
(a) Glycine
(c) Glutamic acid
(b) Aspartic acid
(d) Alanine
7. Crop plants grown in monoculture are
(a) low in yield
(b) free from intraspecific competition
(c) characterised by poor root system,
(d) highly prone to pests
8. The formula for exponential population
growth is
(a) dt / dN = rN
(c) rN / dN = dt
(b) dN / rN = dt
(d) dN / dt = rN
9. In photosystem-I, the first electron acceptor
is
(a) ferredoxin
(c) plastocyanin
(b) cytochrome
(d) an iron-sulphur protein
10. Which one of the following is the most
suitable, medium for culture of Drosophila
melanogaster?
(a) Moist bread
(c) Ripe banana
11. Which
(b) Agar agar
(d) Cow dung
antibiotic inhibits interaction
between rRNA and mRNA during bacterial
protein synthesis?
(a) Erythromycin
(b) Neomycin
(c) Streptomycin
(d) Tetracycline
11
12. Photochemical smog pollution does not
contain
(a) ozone
(b) nitrogen dioxide
(c) carbon dioxide
(d) PAN (Peroxy Acyl Nitrate)
13. The thalloid body of a slime mould
(Myxomycetes) is known as
(a) protonema
(c) fruiting body
(b) Plasmodium
(d) mycelium
14. What type of placentation is seen in sweet
pea?
(a) Basal
(c) Free central
(b) Axile
(d) Marginal
15. Long filamentous threads protruding at the
end of a young cob of maize are
(a) anthers
(c) ovaries
(b) styles
(d) hairs
16. How many different kinds of gametes will be
produced by a plant having the genotype
AABbCC?
(a) Three
(c) Nine
(b) Four
(d) Two
17. Which of the following statements regarding
mitochondrial membrane is not correct?
(a) The outer membrane is permeable to all kinds of
molecules
(b) The enzymes of the electron transfer chain are
embedded in the outer membrane
(c) The inner membrane is highly convoluted forming a
series of infoldings
(d) The outer membrane resembles a sieve
18. An organic substance bound to an enzyme
and essential for its activity is called
(a) coenzyme
(c) apoenzyme
(b) holoenzyme
(d) isoenzyme
19. Bowman’s glands are found in
(a) olfactory epithelium
(b) external auditory ca.ial
(c) cortical nephrons only
(d) juxtamedullary nephrons
12
| BCECE (Medical) l Solved Paper 2013
20. The bacterium (Clostridium botulinum)
27. What is common about Trypanosoma,
that causes botulism is
Noctiluca, Monocystis and Giardia?
(a) a facultative anaerobe
(b) an obligate anaerobe
(c) a facultative aerobe
(d) an obligate aerobe
(a) These are all unicellular protists
(b) They have flagella
(c) They produce spores
(d) These are all parasites
21. Sulphur is an important nutrient for
optimum growth and productivity in
(a) pulse crops
(c) fibre crops
hot-spot of biodiversity in India ?
23. Curing of tea leaves is brought about by the
activity of
(b) mycorrhiza
(d) fungi
of
came with the development of electron
microscope. This is because
(a) the resolution power of the electron microscope is
much higher than that of the light microscope
(b) the resolving power of the electron microscope is
200 - 350 nm as compared to 0.1 - 0.2 nm for the light
microscope
(c) electron beam can pass through thick materials,
whereas light microscopy requires thin sections
(d) the electron microscope is more powerful than the
light microscope as it uses a beam of electrons
which has wavelength much longer than that of
photons
(a) Platyhelminthes and Arthropoda
(b) Echinodermata and Annelida
(c) Annelida and Arthropoda
(d) Mollusca and Chordata
polygenic inheritance?
(a) Flower colour in Mirabilis jalapa
(b) Production of male honey bee
(c) Pod shape in garden pea
(d) Skin colour in humans
neurotransmitter ?
(b) Epinephrine
(d) Cortisone
32.A steroid hormone which regulates glucose
25. A major break through in the studies of cells
is
(b) Indo-Gangetic plain
(d) Aravalli hills
30. Which one of the following is an example of
(a) Acetylcholine
(c) Nor epinephrine
(b) Siberian crane
(d) arctic tern
segmentation
characteristic of
(a) Western ghats
(c) Eastern ghats
31. Which one of the following not act as a
24. Annual migration does not occur in the case
26. Metameric
(b) Ancestry
(d) Ontogeny
29. Which of the following is considered a
(a) is partially parasitic on the gametophyte
(b) produces gametes that give rise to the gametophyte
(c) arises from a spore produced from the
gametophyte
(d) manufactures food for itself, as well as for the
gametophyte
(a) salmon
(c) salamander
known as
(a) Phylogeny
(c) Paleontology
(b) cereals
(d) oilseed crops
22. In a moss the sporophyte
(a) bacteria
(c) viruses
28. Evolutionary history of an organism is
the
metabolism is
(a) cortisol
(b) corticosterone
(c) 11-deoxycorticosterone
(d) cortisone
33. Which one of the following is not a second
messenger in hormone action?
(a) cGMP
(b) Calcium (c) Sodium (d) cAMP
34. In Mendel’s experiments with garden pea,
round seed shape (RR) was dominant over
wrinkled seeds (IT), yellow cotyledon (YY)
was dominant over green cotyledon (yy).
What are the expected phenotypes in the F2
generation of the cross RRYY x rryy?
(a)
(b)
(c)
(d)
Only round seeds with green cotyledons
Only wrinkled seeds with yellow cotyledons
Only wrinkled seeds with green cotyledons
Round seeds with yellow cotyledons and wrinkled
seeds with yellow cotyledons
BCECE (Medical) l Solved Paper 2013 |
35. One turn of the helix in a B-form DNA is
approximately
(a) 20 nm
(c) 3.4 nm
means that
(a) one strand turns anti-clockwise
(b) the phosphate groups of two DNA strands, at their
ends, share the same position
(c) the phosphate groups at the start of two DNA strands
are in opposite position (pole)
(d) one strand turns clockwise
37. Mast cells secrete
(b) myoglobin
(d) haemoglobin
38. If a colourblind woman marries a normal
visioned man, their sons will be
(a) all normal visioned
(b) one-half colourblind and one-half normal
(c) three-fourths colourbling and one-fourth normal
(d) all colourblind
(b) steroids
(d) glycoproteins
Pollution Control Board for the discharge of
industrial and municipal waste water into
natural surface water, is
(a) < 3.0 ppm
(b) < 10 ppm
(c) < 100 ppm
(d) < 30 ppm
tube members is supported by
(a) root pressure and transpiration pull
(b) P-proteins
(c) mass flow involving a carrier and ATP
(d) cytoplasmic streaming
45. An enzyme that can stimulate germination
of barley seeds is
(a) α-amylase
(c) protease
(b) lipase
(d) invertase
46. In a cereal grain the single cotyledon of
embryo is represented by
(b) scutellum
(d) coleoptile
our body cells is transported to the lungs
(a) dissolved in the blood
(b) as bicarbonates
(c) as carbonates
(d) attached to haemoglobin
48. In order to obtain virus-free plants through
tissue culture the best method is
(a) protoplast culture
(c) anther culture
41. Earthworms are
from the African population because
42. Withdrawal of which of the following
immediate
cause
(b) embryo rescue
(d) meristem culture
49. Sickle-cell anaemia has not been eliminated
(a) ureotelic when plenty of water is available
(b) uricotelic when plenty of water is available
(c) uricotelic under conditions of water scarcity
(d) ammonotelic when plenty of water is available
(a) Oestrogens
(b) FSH
(c) FSH-RH
(d) Progesterone
44. The translocation of organic solutes in sieve
47. The majority of carbon dioxide produced by
40. Limit of BOD prescribed by Central
hormones is the
menstruation?
(a) The residual air in lungs slightly decreases the
efficiency of respiration in mammals
(b) The presence of non-respiratory air sacs, increases
the efficiency of respiration in birds
(c) In insects, circulating body fluids serve to distribute
oxygen to tissues
(d) The principle of countercurrent flow facilitates
efficient respiration in gills of fishes
(a) coleorhiza
(c) prophyll
39. Antibodies in our body are complex
(a) lipoproteins
(c) prostaglandins
43. Which one of the following statements is
incorrect?
(b) 0.34 nm
(d) 2 nm
36. Antiparallel strands of a DNA molecule
(a) hippurin
(c) histamine
13
of
(a) it is controlled by recessive genes
(b) it is not a fatal disease
(c) it provides immunity against malaria
(d) it is controlled by dominant genes
50. Both sickle-cell anaemia and Huntington’s
chorea are
(a) bacteria-related diseases
(b) congenital disorders
(c) pollutant-induced disorders
(d) virus-related diseases
Answers
Physics
1.
11.
21.
31.
41.
(c)
(c)
(b)
(c)
(b)
(d)
(c)
(b)
(c)
(c)
3.
13.
23.
33.
43.
(a)
(b)
(c)
(a)
(d)
4.
14.
24.
34.
44.
(d)
(b)
(d)
(b)
(c)
5.
15.
25.
35.
45.
(b)
(d)
(b)
(c)
(a)
6.
16.
26.
36.
46.
(a)
(b)
(c)
(d)
(c)
7.
17.
27.
37.
47.
(b)
(b)
(a)
(a)
(c)
8.
18.
28.
38.
48.
(c)
(d)
(d)
(a)
(d)
9.
19.
29.
39.
49.
(c)
(a)
(b)
(b)
(b)
10.
20.
30.
40.
50.
(d)
(a)
(c)
(d)
(c)
2.
12.
22.
32.
42.
(c)
(b)
(b)
(c)
(b)
3.
13.
23.
33.
43.
(b)
(a)
(a)
(a)
(b)
4.
14.
24.
34.
44.
(d)
(d)
(d)
(d)
(c)
5.
15.
25.
35.
45.
(c)
(c)
(c)
(c)
(d)
6.
16.
26.
36.
46.
(b)
(a)
(b)
(d)
(a)
7.
17.
27.
37.
47.
(c)
(b)
(a)
(a)
(a)
8.
18.
28.
38.
48.
(c)
(b)
(v)
(b)
(b)
9.
19.
29.
39.
49.
(a)
(c)
(a)
(c)
(a)
10.
20.
30.
40.
50.
(d)
(d)
(c)
(d)
(a)
2.
12.
22.
32.
42.
(b)
(c)
(a)
(a)
(d)
3.
13.
23.
33.
43.
(b)
(b)
(a)
(c)
(a)
4.
14.
24.
34.
44.
(a)
(d)
(c)
(d)
(c)
5.
15.
25.
35.
45.
(d)
(b)
(a)
(c)
(a)
6.
16.
26.
36.
46.
(c)
(d)
(c)
(c)
(b)
7.
17.
27.
37.
47.
(d)
(b)
(a)
(c)
(d)
8.
18.
28.
38.
48.
(d)
(a)
(a)
(d)
(d)
9.
19.
29.
39.
49.
(d)
(a)
(a)
(d)
(b)
10.
20.
30.
40.
50.
(c)
(b)
(d)
(b)
(b)
2.
12.
22.
32.
42.
Chemistry
1.
11.
21.
31.
41.
(a)
(c)
(a)
(b)
(b)
Biology
1.
11.
21.
31.
41.
(c)
(d)
(a)
(d)
(d)
BCECE (Medical) l Solved Paper 2013 |
15
Hints & Solutions
Physics
1. F =
Gm, m2
r2
Fr 2
[MLT −2 ][L2 ]
G=
=
m1m2
[M2 ]
⇒
2.
8. For a small angle prism, δ = (u − 1) A ...(i)
= [M−1 L3 T −2 ]
Given, A × B = 3 A . B
AB sin θ = 3 AB cos θ
⇒
A+ δ =i + e
We have,
i = A+δ
or
6 = ( i − A)
Hence,
A+B =
A2 + B 2 + 2 AB cos θ
=
A2 + B 2 + 2 AB cos 60 °
=
1
A + B + 2 AB
2
2
2
= ( A2 + B 2 + AB)1/ 2
3. The gravitational force will remain the same as
it is independent of density of medium.
4. Let the volume of the body be V and density
be d. Then for the case of water
3
V × d × g = × v × d water × g
4
2
For oil V × d × g = × V × doil × g
3
doil
9
⇒
=
d water 8
(i − A) = (µ − i )A
h. r = constant
6. According to the given figure, A attains mean
position T/4 time earlier than C. So, C lags
π
behind A by . Also B attains the mean
2
T
position time earlier.
2
π
Hence, B leads A by .
2
7. We have, ∆Q = ∆U + ∆W
Here, ∆Q = 110 J , ∆U = 40 J
∴
∆W = 70 J
i = µA
⇒
9. Two plates of B are joined together and
therefore cannot form a capacitor. Thus, there
are only two capacitors in parallel – one
between the upper most plate and the next
one and second between the lower most plate
and the one above it.
ε 5 ε 5 2ε 5
Hence, C = C1 + C2 = 0 + 0 = 0
d
d
d
10. As the given condition is of balanced wheat
stone bridge.
I1 × 12 = (14
. − I1 ) × 30
∴
⇒
⇒
I1 × 42 = 14
. × 30
30
14
. ×
= 1 A.
42
11. We have I = ne Avd
2σ cos θ
We have, h =
rρg
⇒
...(ii)
From Eqs. (i) and (ii)
tan θ = 3 or θ = 60 °
⇒
5.
Since, the ray emerges normally, ∴ e = 0
By the relation,
and
I=
V
R
∴
V
= ne A vd
R
or
V = ne A vd . R

= neA vd ρ

∴
⇒
l

A
V
= E = neρlvd
l
E ∝ vd
12. Poynting vector which is in the direction of
travel of electromagnetic wave is along the
vector E × B.
16
| BCECE (Medical) l Solved Paper 2013
13. The current an a LCR circuit is given as
I=
E0 sin (ωt ± φ)
R 2 + ( XL − XC )2
Current is maximum if,
R 2 + ( XL − XC )2 = minimum
⇒
or
14. A + A should always be equal to one as
addition of quantity like 1 to 0 or 0 to 1 always
gives 1.
15. The given truth table is of NAND gate.
∆φ
∆t
∴ Current, I =
e
∆φ
=
R R∆φ
So, charge Q = I . ∆t =
17. We have, ν 2 = u 2 + 2 as
⇒
⇒
⇒
20 

(80 )2 = 0 + 2 (a)  4π × 

π
[as two revolutions are completed]
a = 40 m/s 2
18. The output of a OR gate is
γ = A+ B
P
V2
19. In series, power, P =
= = 50 W
2R
2
2
V
In parallel, power, P =
= 2P = 20 W
R
2
20. Barrier potential of a p -n junction diode does
not depend on diode design.
21. (Pf − Pi ) = Change in momentum
0.5v 2
+ 0.5 × 10
4
v = 16 m/ s
v 16
ω= =
= 4 rad/s
r
4
23. Change in angular momentum = Torque × time
4A− A 3
= A
4
4
GM
24. By the relation, g = 2 we have
R
2
M
gP
R
p
=
× e = 2 × (2 )2
ge Me Rp2
⇒ Torque =
∴
∆φ .
∆φ
∆t =
R∆φ
R
mv 2
+ mg
r
37
. × 10 =
∴
XL − XC = 0
XL = XC
16. We have, e =
22. Maximum tension =
g p = 8 × 10 = 80 m/s 2
25. Total volume is conserved.
4π 3 4π 3
4π 3
(3) +
(4) =
R
3
3
3
or
R ≅ 4.5 cm
q
26. (b) E = 9 × 10 9 ×
n
16
.
× 10 −19
= 9 × 10 9
= 144
. × 1011 M/C
(1 × 10 −10 )2
∴
27. For Q, closed pipe. n = (2 K + 1) C/ 4
for K = 0, it gives
L = 1 × (10 2 × 330 ) / 4 × 260
= 317
. cm
28. This is not possible as it is against the law of
conservation of energy.
29. Only if the three forces are concurrent, they
cannot produce a net clockwise
anticlockwise
momentum,
which
neccessary for rotation.
or
is
30. Moving beams of protons are like currents.
q
= P 2 + P 2 + 2 P .P cos (π − 2 θ )
= P 2 + 2(− cos θ ) = P 2 (2 sin 2 θ )
= 2P sin θ
Two parallel currents attaract each other.
h
h
31. We have , λ Ph =
and λ e =
PPh
Pe
Given,
∴ We get,
λ Ph = λ e
PPh = Pe
h
= mv
λ Ph
BCECE (Medical) l Solved Paper 2013 |
hc
1
= mcv = mv 2
2
λ Ph
EPh 2c
=
Ee
v
∴
or
are input of OR gate and the output of A and
B and as well as C are the inputs of AND
gate.
33. We have, v ∝ r
⇒
v = Hr
where, H is Hubble's" constant.
34. Slope of displacement versus time graph gives
the velocity. This is constant here only during
the bc, as the graph is a straight line in it.
Constant velocity means no acceleration,
35. For a long conductor,
magnetic field at
distance r is
µ 0 2I µ 0I
× =
4π r 2 πr
1
∴
B∝
r
So, graph between B and r should be a
parabola as given in option (c).
I
36. Time period is given by, T = 2π
MH
B=
Hence, it does not depend upon the length of
the suspension.
37. We have, B = µH
B
H
So, the permeability µ is defined as the ratio of
magnetic induction to the magnetising field.
38. As φ = B . A
∴
⇒
ml 2
+ ml 2
4
5
= ml 2
4
 2c 
 
 v
32. Y = 1, if A = 1, B = 0 and C = 1because A and B
µ=
φ = 2 × 0.5 × cos 60 °
1
= 2 × 0.5 × , = 0.5 Wb
2
39. In the diagram (b), the positive terminal is at
lower potential and negative terminal is at
higher potential, hence it is reversed
biased.
40. Moment of inertia of the system
I = I1 + I2 + I3
=
41. Given Rg = 50 Ω,
Ig = 25 × 4 × 10 −4 A = 10 −2 A
Range of
V = 25 volts
V
 25

Resistance, R =
− Rg =  −2 − 50 Ω
 10

Ig
= 2450 Ω
Thus, a resistance of 2450 Ω is connected is
series to convert it into a voltmeter.
v + vs
42. We have, ν =
ν
v − vs
v + vs
or
2f =
f
v − vs
or
or
∴
2(v − v s ) = v + v s
2v − 2v s = v + v s
v = 3v s or v s =
v
3
43. Rise = Real depth-apparent depth
=R.D−
R.D
 1
= R . D 1 − 
 µ
µ
Thus, for least rise µ should be least.
Refractive index is least for red colour.
sin i
c
44. We have, µ =
and also µ =
sin r
v
c sin i
∴
=
v sin r
⇒
 sin r 
v =c .

 sin i 
45. As the angle between v and E is neither zero
nor π, a constant force (− qE ) will act on the
electron. The motion of the electron, therefore
will be along a parabola.
E
46. We know r =  − 1 R
V

hence,
hence,
R = 3.9π, E = 2.0 V, r = 0.1 Ω
V = 195
. V
47. Heat required, Q = 1000 × 1 × (100 − 20 )
2
 l
= 0 + m   + ml 2
 2
17
= 1000 × 80 cal
Heat produced in time t is, H = VIt
18
| BCECE (Medical) l Solved Paper 2013
= 220 × 4 × t J
220 × 4 × t
cal
=
418
.
220 × 4 × t
=
= 1000 × 80
418
.
1000 × 80
t =
= 6.3 min
418
. × 220 × 4
∴
⇒
48. (d) Pair production is just oppsite to mass
annihilation. When suddenly γ-ray energy
disappears to form a pair of a particle and an
anti-particle, the process is called pair
production.
49. (b) Resolving power ∝
1
1
and ∝
λ
v
∴
R.P ∝ v
⇒
(R . P )2 = (R . P ) 1 ×
80kV
20kV
= 2(R . P )1
50. (c) Electric field between the oppositely
charged plates of a capacitor is twice that due
to one plate Hence, when one plate is
removed, the electric force reduces to half of
its earlier value.
Chemistry
1.
CH 3 OH
 
H+
CH 3 C  CH CH 3 →

CH 3
3. Positive inductive effect (+ I effect) is arises
due to electron releasing group ( y ). It
develops negative charge on chain and exerts
a positive inductive effect.
δ−
CH 3 O H 2
− H2O


CH 3 C  C H CH 3 →

CH 3
CH 3
 ⊕
1,2 −methyl
CH 3 C  C H CH 3 →
shift

CH 3
2 °carbocation
CH 3

Br –
CH 3 C  C H CH 3 →

CH 3
⊕
3 ° carbocation
Br CH 3
 
CH 3 C  C H CH 3

CH 3
2.
A β-hydroxy carbonyl compound is formed by
condensation of carbonyl compound with at
least one hydrogen atom at α-carbon.
δ+
C ← y
⊕
Therefore, + I effect tends to decrease positive
charge and tends to increase stability.
∆
4. C 4H 6O 4 →
C 3H 6O 2
A
B
From above equation, it is clear that CO 2 is
lost due to heating. Thus, ‘A’ is a dibasic acid
with two COOH groups on same carbon
atom (gem position). Thus, ‘A’ is
COOH ∆
CH 3COH
→ CH 3CH 2COOH
'A'
'B'
COOH
propanoic acid
methyl malonic acid
Soda lime
→ CH 3CH 3
∆
'C'
ethane
For 1 mole of ‘C’ again, CH 3CH 3 (mol wt = 30)
Q 30 g of C occupies volume = 22.42
22.4
∴1 g of C with occupy volume =
30
~
− 0.75 L
5. When an extra electron is added in nitrogen
atom, added electron will be screened by five
BCECE (Medical) l Solved Paper 2013 |
electrons in second orbit (2 s2 2 p3 ) and two
electrons in first orbit (1s2 ).
Added electron
is screened
1 erg =
(n-1)
(n)
Now screening constant, σ = 5 × 0.35 in nth
orbit + 2 × 0.85 in (n − 1th
) orbit
= 175
. + 170
. = 3.45
∴Effective nuclear charge = 7 − 3.45 = 3.55
6. MS 2 ≡ 2 D
100 40.06
∴
or
or
or
R
mol K
8.314
1 kcal =
R
mol K
0.002
Cu 2 (OH)2 CO 3
11. Mischmetal is an alloy of lanthanide metal
(95%) and iron (about 5%) with traces of
sulphur, carbon, calcium and aluminium. It is
used in the production of bullets, shells and
lighter flint.
M ≈ 160 − 64
M~
− 96
12. [Fe(H 2O)5 NO]2+
s
−u
2
7. Formal charge, F = v −
This is a typical case of complex formed by
electron exchange. NO changes to NO+ by
loss of electron and this electron is gained by
Fe 2+ which changes to Fe+
NO → NO + + e −
Where, v = valency electrons
s = shared electrons
u = unshared electrons
2 shared electrons
Fe 2+ + e − → Fe +
Thus, oxidation state of iron = + 1
C :← 2
≡≡
6 unshared electrons
Fe2+ =Ar 4s 3d6
unshared electrons
v
s
u
F
H
N
C
1
2
0
0
5
8
0
1
4
6
2
–1
8. R = 0.0821 L atm mol −1 K −1
7
= 8.314 × 10 ergs mol
−1
−1 −1
= 8.314 J mol K
= 0.002 kcal mol −1 K −1
∴
Cr2+ (z = 24) = Ar 3d4
2Cu 2 + H 2O + CO 2 + O 2 →
M + 64 ≈ 160
H N
9. Fe2+ (z = 26) = Ar 3d6
Cu 2 (OH)2 SO 4
40.06 (M + 64) = 100 × 64
6400
M + 64 =
40.06
or
Thus, II < III < I < IV
10. 2Cu + H 2O + SO 3 + O 2 →
M + 64 64
100
40.06
=
M + 64
64
or
R
mol K
8.314 × 10 7
1J=
⊕
19
1 L atm =
R
mol K
0.0821
K
−1
Fe+ = Ar 4s 3d7
Three unpaired electrons + 1oxidation state is
conirmed by magnetic moment of iron
= 3 (3 + 2 )
= 15 BM. and also by diamagnetic character
of NO (which is only possible when it has no
unpaired electron).
140 °C
13. 3BCl 3 + 3NH 4Cl → B3N 3H 3Cl 3
C 6H5Cl
NaBH4
→
B3N 3H 6
borazine or
inorganic benzene
20
| BCECE (Medical) l Solved Paper 2013
14. Let atomicity of sulphur is n, then it exists as S n .
∴Molar mass, m = 32 n
1000 w K f
Now from,
∆Tf =
mw
1000 × 2.56 × 0.1
0.010 =
m × 100
1000 × 2.56 × 0.1
or
m=
= 256
0.010 × 100
or
32 n = 256
256
or,
n=
=8
32
∴
K =
+
4
18. Q Salts NH 4 A, NH 4 B, NH 4C ) are of weak acid
and weak base.
pK a pK a
−
2
2
Thus, greater the value of pK a of HA, greater
the pH.
Now
K′ =
K′
=
K
= 0.0072 × 76 × 13.6 cm of mercury column
j
NH+4
+ OH
−
−
–
Also,
∴
or
+
[OH ] [NH 4OH] [H ]
×
[NH 4OH]
[NH+4 ] [H 2O]
= K w = 1 × 10 −14
K ′ = K × 1 × 10 −14
= 1.8 × 10 9 × 10 −14 = 1.8 × 10 −5
17. ph of 0.01 M CH 3COOH
1
pH = [pK a − log C ]
2
1
=
( 474
. − log 0 .01)
2
=
Thus, order of pH va;ies is
= 0.0072 × 76 cm of Hg
= [OH − ] [H+ ] (QH 2O is in excess)
∴
∴pKa (HA, HB, HC ) | HC < HB < HA
8
9
10
19. p = 0.0072 atm
[OH ]
[NH 4OH]
[NH+4 ]
pK b (A− , B − , C − ) | A− < B − < C −
4
5
6
(NH 4 C < NH 4 B < NH 4 A).
+ H 2O
Again,
NH 3 + H 2O → NH 4OH
∴
pH = 7 +
∴
[NH+4 ] [H 2O]
= 1.8 × 10 9
[NH 4OH] [H+ ]
[NH+4 ]
0 .01
= 474
.
0 .01
∴Change in pH = 4.74 − 3.37 = 1.37
moles have been doubled, thus half-life is
also doubled, i . e ., now it is 4 min. Thus,
8 mol change to 4 mol (half) in = 4 min
and 4 mol change to 2 mol (half) in = 2 min.
Thus, total time = 4 + 2 = 6 min.
j NH
[CH 3COO]
[CH 3COOH]
= 474
. + log
15. By set I, half-life is 2 min. In set-II, number of
16. NH 4OH + H+
pH = pK a + log
1
( 474
. + 2 ) = 3.37
2
When 0.01 mol CH 3COONa is added to it, it is
now a buffer and [[CH3COONa]= 0.01] M.
Now from
p = h × 1 cm of water column
h × 1 = 0.0072 × 76 × 13.6 cm
h = 7.4 cm
20. Q ∆H° and ∆S ° remain constant in the given
temperature range.
∆G ° − ∆H °
∴
∆S ° = −
T
 − 66.9 + 41.8
=−



300
= 0.08367 kJ K −1 mol −1
∴
°
∆G330
= ∆H ° − T∆S °
= − 41.8 − 330 × 0.08367
= − 69.4 kJ mol −1
21. From Kirchhoff's equation
∆H2 (1000 K ) = ∆H1 (300 K )
+ ∆Cp (1000 − 300 )
Here, ∆H2 (1000 K) = − 123.77 kJ mol −1
∆H1 (300 K) = ?
∆CP = 2CP (NH 3 ) − [C P(N 2 ),+3C P (H 2 )]
BCECE (Medical) l Solved Paper 2013 |
=− 6 R
27. Rate of oxidation depends upon steric
hinderance. Greater the steric hinderance the
slower is the oxidation. Thus, the correct order
is
= − 6 × 8.314 × 10 −3 kJ
− 123.77 = ∆H1 (300 K)
∴
− 6 × 8.314 × 10 −3 × 700
or
For two moles of NH 3
∆H (300 K)
∴ ∆Hf (NH 3 ) = 1
2
88.85
=–
2
= − 44.42 kJ mol −1
22.
Electrode
General Name
(A)
(B)
(C)
Calomel
Glass
Hydrogen
Reference
Memberane
Gas
(D)
Quinhydrone
Ecell < 1.1 if
+
−
3+
−
24. 2H +2e
+3e
Redox
2+
0.0591
[Zn ]
log
2
[Cu 2 + ]
= 1.1 −
Al
0.0591
[Zn 2 + ]
log
2
[Cu 2 + ]
→ H 2
hv
NH3
— NH2
ZnCl2
Aniline
(Primany amine)
Phenol
O

29. ‘A’ gives iodoform test thus A has  C CH 3
group. ‘C’ eliminates CO 2 on heating hence,
C has two COOH groups at same (gem)
position.
O

Since, only (a) has  C CH 3 group, therefore,
O
O


compound ‘A’ is CH 3 C CH 2 C OC 2H 5 .
∆
CH 3I ↓ +
yellow ppt
B
CH2CH3
→
— OH
(III)
O
O


NaOC CH 2 C ONa
→ Al
w(H 2 ) E(H 2 ) 1
=
=
w(Al) E (Al) 9
→ H + Br2
28.
(II)
A
1 – ethylbicyclo [2, 2, 1] heptane
26.
(I)
O
O


NaOH/I 2
CH 3 C CH 2 C OC 2H 5 →
[Zn 2 + ]
>1
[Cu 2 + ]
1
1
w(H 2 ) = 4.5 × = 0.5 g = mol H 2
9
4
22.4
=
= 5.6 L H 2 at STP
4
25.
CH 2OH CH 3CHOH (CH 3 )2 COH

>

> 
CH 2OH CH 3CHOH (CH 3 )2 COH
∆H1 (300 K) = − 88.85 kJ
°
23. Ecell = Ecell
=−
21
→
(i) Mg /ether
→
(ii) 1. D O
2
Br
→ D
↓H +
O
O
O



∆
CH 3 C OH ← HOC CH 2 C OH
C
O

30. (I) Methyl ketones (CH 3 C R) react rapidly with
halogens, (Cl 2 , Br2 , I2 ) in presence of alkali to
form haloform.
O

CH 3 C CH 3 + 3I2 + 4NaOH →
O

CH 3 C ONa + CHI3 + 3NaI + 3H 2O
22
| BCECE (Medical) l Solved Paper 2013
O
O


∆
+
CH 3 C ONa + H → CH 3 C OH
calcium and magnesium. When such type
of water is used for cleaning clothes with
soap (i . e ., sodium stearate) a curdy
precipitate of calcium and magnesium salts
are formed.
2+
RCOONa
or Mg 2 + →
14243 + Ca
144244
3
−Na +
(II) Stronger oxidising agents oxidise ketones to
carboxylic acids.
O

K2 Cr2 O7 + H2 SO 4
CH 3 C CH 3 
→
soap
in hard water
Cardy precipitate (RCOO)2 Ca or Mg
36.
O
O


CH 3 C H + HC OH
Structure of compound
(III) Tollen’s reagent (ammonical silver nitrate
solution), being weak oxidising agent can not
reduce acetone into acetic acid.
O

CH 3 C CH 3 + [Ag(NH 3 )+2 ] →No reaction
(a)
O
Name of
compound
Ascorbic acid
O
CH (OH) CH2OH
HO
OH
(b)
Coumarin
31. Benzyl carbocation is more stable then alkyl
⊕
(CH 3 CH 2 ). CH 3O  and CH 3 , both
stabilise benzyl carbocation and among them
CH 3O  is more stable then CH 3 . Thus,
correct order of stabilities will be
IV < II < III < I
O
37.
S.No.
32. Haemoglobin is transports protein. It transport
MnO
CH3COOH
(i)
Adenine
A
DNA, RNA
Guanine
G
DNA, RNA
(CH3COO)2Mn +
(iii)
Cytosine
C
DNA, RNA
(HCOO)2Mn
(iv)
Thymine
T
DNA
(v)
Uracil
U
RNA
MnO
HCOOH
Heterocycli
Abbreviatio Occurrenc
c amine
n
e
base
(ii)
oxygen to the cells.
33.
O
CH3CHO
∆
34.
38. Theoretical value of dipole moment of 100%
ionic character = e × d
NH2
= (1.6 × 10 −19 C) (1.41 × 10 −10 m)
NH2
Br2 water
Br
Br
NaNO2 / HCl
Observed value of dipole moment
0°C
= 2.60 × 10 −30 Cm
Br
N2Cl
Br
Br
Br
(i) BF –
4
(ii)
= 2.26 × 10 −29 Cm
∴Percent ionic character
F
Br
=
observed value
× 100
theoretical value
=
2.60 × 10 −30
× 100
2.26 × 10 −29
Br
Br
35. Hardness in water is due to dissolved
sulphates, chlorides and bicarbonates of
= 11.5%
39. Oxygen is more electronegative than chlorine
and cannot be oxidised by it.
BCECE (Medical) l Solved Paper 2013 |
40. In Goldschmidt’s alumino thermite process, is
used when a high temperature is needed for
carbon to reduce an oxide. In this process a
large amount of energy (1675 kJ mol −1) is
liberated on oxidation to Al 2O 3 .
45.
–2
+4
H2O2 + MnO2
+2e–
B2O 3 + 2Al → 2B + Al 2O 3
(Oxidation number two O atoms taken)
On balancing of oxidation number
3H 2O 2 + 2MnO 2 → 2MnO −4 + 6H 2O
Cr2O 3 + 2Al → 2Cr + Al 2O 3
41. In Nelson’s cell, H2 gas is liberated at cathode
wheres Cl 2 gas is liberated at anode, on
electrolysis of solution of NaCl.
+
−
j Na + Cl
H O j H + OH
2
+
−
H + + e − → H
H + H → H 2 ↑
at anode
Cl − → Cl + e −1
H 2O 2 : MnO 2 = 3 : 2
KCl AsNO 3
=
M1V1
M2V2
1 × 20
M1 =
25
= 0.8 M
i = (1 + x ) = 1 + 1 = 2
∴
∆Tf = molality × K f × i
= 0.8 × 2 × 2 = 3.2 °
47. According to Lux- Flood concept, acid is
Cl + Cl → Cl 2 ↑
42. SnCl 2 + 2NaOH → Sn(OH)2 ↓ + 2NaCl
white ppt
Na 2SnO 2
sodium stannite
+ 2H 2O
43. N −3 (aq ) + 3H 2O(l ) → NH 3 (g ) + 3OH −(aq )
N −3
Thus,
46.
at cathode
Sn(OH)2 + 2NaOH →
Mn+7O4 + 2H2O–4
–3e–
3Mn 3O 4 + 8Al → 9Mn + 4Al 2O 3
NaCl
23
accepts proton, hence, it is strong
Bronsted Lowry base and reacts with water to
produce NH 3 and OH − ions.
 1
1
− 2
2
n2 
 n1
44. ν (wave number) = RH Z 2 
1
1
ν1 (He + , Z = 2) = RH (2) 2  2 − 2 
n 1 n 2 
 1
1
ν 2 (Be 3 + ,Z = 4) = RH = (4)2  2 − 2 
n2 
 n1
∴
ν 2 (4)2 16
=
=
=4
ν1 (2 )2
4
∴
ν 2 = 4ν1
ν 2 = 4 x cm −1
defined as a substance which can accept
oxide and base as a substance which can
donate oxide.
CaO + CO 2 → Ca 2 + CO 23 −
base
acid
48. Negatively charged colloidal sol is formed
when AgNO 3 is completely precipitated as
AgI and extra KI is absorbed on AgI.
Ag + + I− → AgI
AgI + I− → [AgI]I−
Thus,
[Ag + ] < [I− ]
x
1
49. log   = log k + log p
 m
n
Given, log k = 0.3010 = intercept on y-axis
1
and
= tan 45° = 1 = slope
n
 x
∴ log   = 0.3010 + 1x log 0.2
 m
= log 2 + log 0.2
= log 0.4
∴
x
= 0.4
m
24
| BCECE (Medical) l Solved Paper 2013
50. We have to calculate ∆H° for the following
reaction in which one mole of CaC 2 (s) is
formed form its elements.
Ca(s) + 2C (s) → CaC 2 (s)
This is obtained when we operate.
(B) + (C ) + z(D ) − (E ) − a
∴∆H = (− 635.1) + (− 65.2) + 2 (− 393.5)
− (− 1299.58) − (− 127.9) kJ mol −1
= − 59.82 kJ mol −1
Biology
1. The fruit of Ananas comosus (pineapple or
5. If a pulse crop possesses premature yellowing
ananas) is sorosis, (a type of multiple fruits),
developing from spike, spadix or catkin. In
this type, the flowers associate by their
succulent petals, the axis bearing them grows
and becomes fleshy or woody, thus, the
whole inflorescence turns into a compact fruit.
of leaves and decrease in yield application of
magnesium and iron to promote synthesis of
chlorophyll may become most beneficial to
overcome the problem and to obtain maximum
seed yield.
2. Parthenocarpy is the development of fruits
without prior fertilisation which results in the
formation of seedless fruits. In some plant
species, parthenocarpic (seedless) fruits may
be produced naturally or they may be induced
by treatment of the unpollinated flowers with
auxin, e.g., parthenocarpic tomato fruits can
be produced by treating the plants with low
concentration of gibberellic acid (promotes
fruit set) and auxin (completes the
development process).
Removal of androecium, before pollen
release is called emasculation which is helpful
in
preventing
unwanted
pollination.
Vernalized seeds are the chill treated seeds
for breaking dormancy. Phenyl mercuric
acetate is an antitranspirant.
3. The disease ‘blue baby syndrome’ or cyanosis
results from methaemoglobin. The main cause
of this disease are the nitrate fertilisers on soil,
which enter the human body through water
and converted to nitrites by microbial flora of
intestine. The nitrites combine with
haemoglobin to form methaemoglobin
causing methaemoglobinaemia in adults and
blue baby syndrome in newly borne babies or
infants.
4. The autonomous nervous system regulates
the secretion of glands whereas the glands do
not regulate the nervous system.
Magnesium is an important part of ring
structure of chlorophyll molecule and its
deficiency causes chlorosis and premature
leaf abscission.
In iron deficiency also, the leaves become
chlorotic because iron is required for the
synthesis of some of the chlorophyll protein
complexes in the chloroplast.
6. Miller and Urey were the two scientists who
recreated the conditions of primitive earth in
laboratory and abiotically synthesised amino
acids and bases. They synthesised glycine,
aspartic acid and alanine in abundant
quantities, while glutamic acid was not
synthesised in their experiment.
7. Monoculture involves the exclusive cultivation of
a single crop over wide areas. It is an efficient
way to use certain kinds of soils but the crop
plants grown in monoculture are highly prone to
pests and thus, it carries the risk of an entire
crop being destroyed with the appearance of a
single pest species or disease.
8. In J-shaped form of population growth, the
density increases in an exponential manner and
then a crash occurs or the increase stops
abruptly as environmental resistance becomes
effective. This is mathematically described by
an equation of exponential or geometric
increase, which is as follows
dN
= rN
dt
BCECE (Medical) l Solved Paper 2013 |
where, d = rate of change
t = time
N = number of females at a particular time
r = biotic potential of each female
9. In photogystem-I, the primary electron
acceptor is probably a Fe-S protein. The
reduced primary acceptor transfers the
electrons to secondary electron acceptor
(most probably P430 ). The sequence of
electron transfer is as follows
e–
e–
A 1 → A 2
Chlrophyll -a Phylloquinone (Fe -s)Protein
not contain carbon di-oxide (CO 2 ).
Photochemcial smog materials causes
damage to plants, human health hazards and
corrosion problems.
13. The members of Myxomycetes are called
(N can also be considered as the total
population and as the biotic potential of each
individual).
P700 →
25
e–
→ A 3
P430
The reduced P430 passes its electrons to
ferridoxin (Fd) present at outer surface of
thylakoid membrane.
10. Drosophila melanogaster is commonly called
as fruitfly and is often used in genetic and
developmental biology researches. The ripe
banana is the most suitable medium for the
culture of this fly.
Moist bread is a culture medium for the
fungus Rhizopus, while agar-agar is used as a
tissue culture medium.
11. Tetracyclin interfere with the attach of tRNA
carrying
the
amino
acid
to
the
m-RNA-ribosome complex preventing the
addition of amino acids to the growing
polypeptide chain. Streptomycin interfere with
the initial steps of protein synthesis by
changing the shape of SOS portion of 70S
prokaryotic ribosome.
Erythromycin reacts with SOS portion of the
70S prokaryotic ribosome.
12. Photochemical smog is highly oxidising
polluted atmosphere comprising largely of
ozone (O 3 ), oxides of nitrogen (NO x ),
hydrogen peroxide (H 2O 2 ), organic peroxides,
Peroxy Aceryl Nitrate (PAN) and Peroxy
Benzyl Nitrate (PBN).
Some sulphates and nitrates can also be
formed in photochemical smog due to
oxidation of sulphur containig components (
SO 2 , H 2S) and NO x (N 2O 5 , NO 2 ) but it does
slime molds because they contain and
secrete slime. They are included in lower
fungi. Their somatic phase is a multinucleate,
diploid holocarpic Plasmodium (a product of
syngamy).
In Plasmodium, propagation occurs through
fission or thick walled cysts or sclerotium like
structures. Reproduction takes place by the
formation of uninucleate, thick walled resting
spores which are produced within minute
fruiting bodies like structures i.e., the
sporangia, however, the true fruiting bodies
are absent in slime molds.
Fruiting bodies and mycelium are absent in
lower fungi. Protonema is not formed in fungi.
14. In
sweet pea (Pisum sativum), the
placentationis marginal, in which, the
placenta develops along the junction of two
carpels, in a unilocular ovary.
In basal placentation, the ovules are few or
reduced to one and are borne at the base of
ovary, e.g., compositae.
In axile placentation, margins of carpels fold
inwards, fusing together in centre of ovary to
form a single central placenta. Ovary is
divided into as many locules, as there are
carpels, e.g.,
Hibiscus, Asphodelus.
Free-central placentation possesses a
placenta arises as a central upgrowth from
ovary base, e.g., Stenaria.
15. In a cob of maize, each ovary has a long silky
(hairy) style, called as corn silk. Collectively
these styles protrude at the end of a young
cob. The grains are formed on the cob, which
remain covered by the leafy bracts.
16. The types of gametes produced by a plant
depend upon the number of heterozygous
pair. Number of types of gametes = 2n
n = number of heterozygous pair
21 = 2
The gametes are - ABC and AbC.
26
| BCECE (Medical) l Solved Paper 2013
17. In mitochondria, the enzymes of electron
transport chain are found in the inner
membrane while outer membrane contains
enzymes involved in mitochondrial lipid
synthesis and those enzymes that convert
lipid substrates into forms that are
subsequently metabolized in the matrix.
The outer membrane resembles a sieve that is
permeable to all molecules of 10,000 daltons
mole, weight or less including small proteins.
The inner membrane is impermeable and highly
convoluted, forming a series of infoldings,
known as cristae, in the matrix space.
18. Coenzyme is an organic non-protein molecule
that associates with an enzyme molecule in
catalysing biochemical reactions. It usually
participates
in
the
substrate-enzyme
interaction by donating or accepting certain
chemical groups.
Holoenzyme is a complex comprising of
enzyme molecule and its cofactor. The
enzyme is catalytically active in this state.
Apoenzyme is an inactive enzyme that must
associate with a specific cofactor molecule in
order to function.
Isoenzyme or isozyme is one of the several
forms of an enzyme that catalyse the same
reaction but differ from each other in such
properties as substrate affinity and
maximum rates of enzyme-substrate
reaction.
19. Bowman’s glands tubular mucus lands
(olfactory glands) occur below the olfactory
epithelia. Their ducts open on the olfactory
epithelial surface. These glands secrete
watery mucous to protect and keep the
epithelium moist.
20. The bacterium Clostridium botulinum, causing
botulism (a form of food poisoning) is an
obligate anaerobic, endospore forming, gram
positive, rod shaped bacterium found in soil
and in many fresh water sediments. This
bacterium produces a toxin, called botulinum
toxin which is the most toxic substance known
(cause fatal food poisoning) but in minute
doses it is used to treat certain conditions
involving muscle dysfunction. Rhinoceros
(Rhinoceros unicornis) is an endangered
animal and conserved in Kaziranga National
Park.
21. Sulphur is constituent of certain amino acids.
The amino acids form the protein by
polymerisation. The pulses are rich in protein.
22. In mosses, the sporophyte developing from
the embryo is a simple structure without
rhizoids and is differentiated into foot, seta
and capsule. It is parasitic (partially or wholly)
on the gametophyte as it is organically
attached and is nutritionally dependent upon
the gametophyte.
23. Curing of tea leaves is brought about by the
activity of bacteria. It is essentially an
oxidation dry fermentation process, during
which, wateris driven off, the green colour is
lost and the leaves assume a tougher texture
and undergo chemical changes. Mycorrhiza
is mutually beneficial association between
fungi and the roots of higher plants. Virsus
and fungi are not involved in the curing
process of tea leaves.
24. Salamander is semi-terrestrial lizard-like tailed
amphibian lives under stones, logs and inside
crevices. They show hibernation.
Salmon are anadromous Le., they spend their
adult lives at sea but return to fresh water to
spawn. The pacific species is legandry, i.e.,
after migrating downstream as a smolt a
sockeye salmon ranges many hundreds of
miles over the pacific for nearly four year and
than returns to spawn in the head waters of its
parent stream. Migration is characteristic
feature of birds. Arctic tern travels about 1100
miles during winter and return back during
summer.
25. A major break through in the studies of cell
came with the development of electron
microscope because the resolution power of
the electron microscope is much higher than
that of the light microscope.
As an average the resolving power of a light
microscope is 0.25 mn— 0.3 nmwhile of
electron microscope, is 2-10Å though the
oritically, it is 0.25Å. The magnification range
of light microscope is 2000-4000 while of
electron microscope is 1,00,000-3,00,000.
BCECE (Medical) l Solved Paper 2013 |
26. Metameric segmentation is the characteristic
of Annelida (e.g., earthworm) and Arthropoda
(e.g., cockroach). In earthworm, body
consists of 100-120 ring like segments or
somites called metameres. It shows true
segmentation, i.e., external segmentation
corresponds with internal segmentation.
Cockroach also shows the metameric
segmentation. Its anterior few segments are
specialised to form head. Such metamerism is
called heteronomous metamerism. Metameric
segmentation is absent in platyhelminthes,
Echinodermata, Mollusca, etc.
27. Trypanosoma, Noctiluca, Monocystis and
Giardia are all unicellular protists.
Trypanosom a gambiense is the single celled,
parasitic
zoo
flagellate
causing
trypanosomiasis or sleeping sickness. Giardia
or the ‘Grand old man of the intestine’ is a
diplomonadid parasitic flagellate occurring
in the intestine of man and other animals and
causes giardiasis or diarrhoea (i.e., very
loose and frequent stool containing large
quantity of fat).
Noctilluca
is
a
marine,
colourless
dinoflagellate. It is a voracious predator and
has a long, motile tentacle, near the base of
which, its single short flagellum emerges.
Monocystis is a microscopic, unicellular
endoparasitic protozoan found in the coelom
and seminal vesicles of earthworm. As it is an
endoparasite, it does not possess any special
structure for locomotion.
28. Phylogeny (Gr, Phylon = tribe or race; geneia
= origin) is the origin and diversification of any
taxon or the evolutionary history of its origin
and diversification. It is usually represented as
a diagrammatic phylogenetic tree (that traces
putative evolutionary relationships), i.e.,
dendrogram.
Palaeontology is the study of fossils.
Ontogeny is the whole course of an
individual’s development and life history.
29. Concept of hot spots was developed by
Norman Meyer. Hot spots are the areas with
high density of diversity or megadiversity
which are also the most threatened ones.
Today, the number of hot spots identified by
27
ecologists are 25 in the world of which two hot
spotsare present in India, i.e., Western ghats
and Eastern Himalayas. Western ghats occur
along the western coast of India for a distance
of about 1600 km in Maharashtra, Karnataka,
Tamil Nadu and Kerala extending over to
Srilanka.
Eastern Himalayas hot spot extends from
Bhutan to Myanmar covering most of
north-east. In India, Indo-Gangetic plain,
Eastern ghats and Aravali hills are mainly not
considered as hot spot of biodiversity.
30. Polygenic
inheritance
involves
the
determination of a particular phenotypic
characteristic by many genes, called
polygenes (i.e., the group of genes
influencing a quantitative characteristic), each
having a small effect individually.
The characteristics controlled in this way
show continuous variation and are called
polygenic characters, e.g., height and stun
colour in humans.
The polygenic inheritance is called
multifactorial inheritance or quantitative
inheritance. The pink flower colour in Mirabilis
jalapais an example of incomplete dominance
white production of male honey bee is an
example of parthenogenesis.
31. Cortisone is a corticosteroid that is itself
biologically inactive and is formed naturally in
the adrenal gland (adrenal cortex) from the
active hormone cortisol. Cortisol promotes
the synthesis and storage of glucose and
important in the normal response to stress,
suppresses inflammation and regulates
deposition of fat in body.
Acetylcholine is a neurotransmitter secreted
from the nerve endings.
Epinephrine and nor-epinephrine are
secreted from the medulla of adrenal gland
and these also act as neurotransmitter.
32. Cortisol or hydrocortisone is the principal
glucocorticoid hormone of many mammals
including humans (corticosterone is more
abundant in some small mammals. It
regulates the glucose metabolism and
promotes gluconeogenesis, especially during
starvation and raises blood pressure.
Cortisone is an inactive form of cortisol.
28
| BCECE (Medical) l Solved Paper 2013
33. Second
messengers are the organic
molecules and sometimes the metal ions,
acting as intracellular signals, whose
production or release usually amplifies a
signal such as a hormone, received at the cell
surface. Sodium (Na) is not a second
messenger in hormone action.
Cyclic Adenosine Monophosphate (cAMP) was
the first ‘second messenger’ to be discovered.
In addition to cyclic AMP, Cyclic Guanosine
Monophosphate (cGMP) functions as a
second messenger in certain cases. Calcium
ions (Ca 2+ ) also act as second messenger in
phospholipase C-Ca + second messenger
system.
34. When a cross (dihybrid) is made between
plants bearing round yellow (RRYY) and
wrinkled green (rryy) seeds, all the plants in F1
generation are with yellow round seeds
(showing the genotype RrYy). The phenotype
in F2 will be as follows
P1 Cross
rryy
Parents RRYY
yellow, round
green, wrinkled
Gamete
formation
RY
RrYy
F1 yellow, round (in both cases)
RrYy × RrYy
RY
RY
Ry
1/16 RRYY
2/16 RRYy
2/16 RrYY
4/16 Rryy
1/16 RRyy
2/16 Rryy
1/16 rryy
2/16 rrYy
1/16 rryy
F2 phenotypic ratio
9/16 yellow,round
3/16 yellow, wrinkled
3/16 green, round
1/16 green, wrinkled
35. B-DNA is helical structure with 20Å diameter
and the distance between the two base pairs
is 3.4 Å and there are 10 base pairs in each
turn or pitch (one round). Hence, one turn of
the helix is approximately 34Å or 3.4 nm
(10Å = 1.0 nm)
36. JD Watson and FHC Crick (1953) showed that
DNA has a double helical structure with two
polynucleotide chains connected by hydrogen
bonds and running in opposite directions
(antiparallel). The antiparallel strands of a DNA
molecule means that the phosphate groups at
the start of two DNA strands are in opposite
position (pole).
37. Histamine is a protein, acting as a vasodilator
ry
Fertilisation
F1
F2 genotypic ratio
rY
ry
RRYY
RRYy
RrYY
RrYy
yellow, round
yellow, round
yellow, round
yellow, round
RRYy
RRyy
RrYy
Rryy
(widening of blood vessels) in inflammatory
and allergic reactions and also increases the
permeability of small vessels. It is secreted
from the mast cells, which widely occur in the
areolar connective tissue.
Two other active substances secreted by
mast cells are heparin, a proteoglycan, which
prevents coagulation of blood vessels and
serotonin, a protein, which acts as
vasoconstrictor to stop bleeding and to
increase blood pressure.
38. Colourblindness is a disease in which a
Ry
yellow, round yellow, wrinkled yellow, round yellow, wrinkled
rY
ry
RrYY
RrYy
rrYY
rrYy
yellow, round
yellow, round
green, round
green, round
Rryy
rrYy
rryy
RrYy
yellow, round
round, green wrinkled, yellow green, wrinkled
F2 generation
person is unable to differentiate between red
and green colour. The gene for this disease is
located on the X-chromosome. So, if a
colourblind woman marries a normal man, it
will produce all the sons colourblind (XC Y). In
case of a carrier woman the probability of a
colourblind and a normal son is 50 : 50.
BCECE (Medical) l Solved Paper 2013 |
(Normal man)
(Carrier woman) Parents
XCX
XY
X
Fertilisation
Y
XC
X
Gametes
29
hormones followed rapidly by involution of the
endometrium itself to about 65% of its
previous thickness.
43. Residual air is the air that remains in lungs after
the most forceful expiration. It is about 1200 mL.
As the residual air remains in the lungs therefore
it has no effect on respiration efficiency.
XCX
XC Y
XX
(Normal
(Carrier
daughter) daughter)
(Colour
blind son)
(Colour blind
woman)
(Normal man) Parents
XY
XCXC
XC
XC
XCX
XY Progenies
(Normal
son)
Fertilisation
XCY
(Colour
(Carrier
daughter) blind son)
X
Y
Gametes
XCX
XCY Progenies
(Carrier
(Colour
daughter) blind son)
39. Antibodies are ‘the proteins (glycoproteins)
called immunoglobulins. These are produced
by the lymphocytes in response to entry of a
foreign substance or antigen into the body.
Lipoproteins are the micellar complex of
protein and lipids. Steroids are a group of
lipids derived from a saturated compound
cyclopentano perhydrophenanthrene, which
has a nucleus of four rings.
Prostaglandin is a group of organic
compounds derived from essential fatty acids
and causing a range of physiological effects
in animals.
40. The
Central Pollution Control Board
prescribed the BOD limit for the discharge of
industrial and municipal waste water as < 10
ppm.
41. Class–Oligochaeta
includes
terrestrial
earthworms and some other species that live
in fresh water. Aquatic oligochaetes excrete
ammonia, while terrestrial oligochaetes
excrete urea but Lumbricus produces both
ammonia and urea.
42. Menstruation is caused by the reduction of
oestrogen and progesterone, especially
progesterone at the end of monthly ovarian
cycle. The first effect is decreased stimulation
of the endometrial cells by these two
44. According to mass flow hypothesis the
transport of organic solutes takes place from
source to sink this transport also depends on
metabolic energy.
According
to
cytoplasmic
streaming
hypothesis (put forward by de Vries 1885) the
transport of organic solutes takes place by
combination of diffusion and cytoplasmic
streaming. Cytoplasmic streaming carry
organic solutes from one end the other end of
sieve tube. Proteins has a role as defence
against phloem feeding insects and sealing of
damaged sieve tubes.
Protein is absent
gymnosperms.
in
monocots
and
45. Barley seeds are rich in carbohydrate (starch).
The starch is hydrolysed by α-amylase to
monosaccharides unit at the time of
germination of seeds.
46. In a cereal grain (e.g., wheat), the single
cotyledon of embryo is represented by the
scutellum. Scutellum is specialised for
nutrient absorption from the endosperm.
Coleoptile is a modified ensheithning leaf that
covers and protects the young primary leaves
of a grass seedling. Coleorhiza is a sheath like
structure found on the radicle which covers
and protects it during the growth into the soil.
47. In our body, the blood transports the CO 2 in
three ways
(i) Majority of carbon dioxide produced
(70%) in our cells is transported in the
form of bicarbonates. In this way first, the
CO 2 that dissolves in blood plasma reacts
with water forming carbonic acid, which
dissociates into hydrogen and bicarbonate
ions.
CO 2 + H 2O → H 2CO 3
Carbonic acid
H 2CO 3 → H + + HCO 3
30
| BCECE (Medical) l Solved Paper 2013
(ii) About 7% of all the CO 2 of transported by
blood from tissues to the lungs is in
dissolved state in plasma.
(iii) About 23% of CO 2 , collected from cells
through tissue fluids, is transported by
blood in the form of carbamino
compound, carbamino -haemoglobin
(CO 2HHb).
48. In meristem culture, the shoot apical meristem
atong with some surrounding tissue is grown
in vitro. It is used for clonal propagation and
recovery of virus free plants. It is also
potentially useful in germ plasm exchange
and long term storage of germ plasm through
freeze preservation.
Protoplast culture involves the culture of
protoplasts (plant cell without cell wall) on
suitable medium.
Embryo rescue is the culture of incised
immature embryo to raise new plants. Anther
culture involves the culture of anthers on
artificial culture medium. It is helpful in
obtaining haploid plants.
49. Sickle-cell aneamia (in which RBCs become
sickle-shaped and stiff) is a genetic disorder
that is autosomal and linked to a recessive
allele. It has not been eliminated from the
African population because it provides
immunity against malaria. People who are
heterozygous for sickle-cell allelic arc much
less susceptible for falciparum malaria, which
is one of the main causes of illness and death
in them. Thus, the sickle-cell allele is
maintained at high levels in populations where
falciparum malaria is common.
50. Both sickle-cell anaemia and Huntingdon’s
chorea are congenital genetic disorders.
Sickle-cell anaemia was first reported by
James Herrick (1904). In this disease the
patient’s haemoglobin level reduced to half of
the normal and the RBCs become
sickle-shaped. A single mutation in a gene
can cause sickle-cell anaemia.
Huntington’s chorea is caused by autosomal
mutation, which is dominant. The gene is
present on chromosome number 4.