RE-AIPMT

23 Pages·2015·0.25 MB·English

RE-AIPMT – 2015 TEST PAPER WITH ANSWER & SOLUTION
(HELD ON SATURDAY 25th JULY, 2015)
1. 2,3(cid:150)Dimethyl(cid:150)2(cid:150)butene can be prepared by 4. The number of structural isomers possible from the
heating which of the following compounds with a molecular formula C H N is :
3 9
strong acid ? (1) 2 (2) 3 (3) 4 (4) 5
Ans. (3)
(1) (CH ) C=CH(cid:150)CH (cid:150)CH
32 2 3
(2) (CH ) CH(cid:150)CH (cid:150)CH=CH
32 2 2 Sol. C H N : CH CH CH NH
3 9 3 2 2 2
(3) (CH )CH(cid:150)CH(cid:150)CH=CH CH CH CH
32 (cid:150) 2 3 3 1(cid:176) amine
CH
3 NH
2
(4) (CH ) C(cid:150)CH=CH
33 2 CH (cid:150)CH (cid:150)NH(cid:150)CH } 2(cid:176) amine
Ans. (4) 3 2 3
CH N CH
3 3
Sol. CH3 H+ CH3Å CH3 3(cid:176) amine
H C C CH CH H C C CH CH
3 2 3 3 5. If the equilibrium constant for
CH CH
3 3 N2(g) + O2(g) ‡ˆˆˆˆˆ†ˆ 2NO(g) is K, the equilibrium
~CH
3
1 1
constant for 2 N2(g) + 2 O2(g) ‡ˆˆˆˆˆ†ˆ NO(g) will
H+ Å
H C C C CH H C C CH CH be :-
3 3 3 3
CH CH CHCH
3 3 3 3 COD1E-A
2. Gadolinium belongs to 4f series. It's atomic number (1) K (2) K2 (3) K1/2 (4) K
2
is 64. Which of the following is the correct electronic
Ans. (3)
configuration of gadolinium ?
(1) [Xe] 4f75d16s2 (2) [Xe] 4f65d26s2 Sol. N2(g) + O2(g) ‡ˆˆˆˆˆ†ˆ2NO(g); K
(3) [Xe] 4f86d2 (4) [Xe] 4f95s1 1 1
Ans. (1) 2N2(g)+2O2(g)‡ˆˆˆˆˆ†ˆNO(g); K'
Sol. Gd= [Xe]6s24f75d1 when a reaction is multiplied by 1/2 then
64 54
K' = (K)1/2
3. The formation of the oxide ion, O2(cid:150) (g), from oxygen
6. Which one of the following pairs of solution is not
atom requires first an exothermic and then an
an acidic buffer ?
endothermic step as shown below :
(1) H CO and Na CO
O(g) + e(cid:150) ® O- ; DH = (cid:150)141 kJ mol(cid:150)1 2 3 2 3
(g) f (2) H PO and Na PO
3 4 3 4
O(cid:150)(g) + e(cid:150) ® O2- ; DH = +780 kJ mol(cid:150)1
(g) f (3) HClO and NaClO
4 4
Thus process of formation of O2(cid:150) in gas phase is (4) CH COOH and CH COONa
3 3
unfavourable even thought O2(cid:150) is isoelectronic with Ans. (3)
neon. It is due to the fact that, Sol. HClO and NaClO cannot act as an acidic buffer.
4 4
(1) Oxygen is more electronegative 7. Aqueous solution of which of the following
(2) Addition of electron in oxygen results in larger compounds is the best conductor of electric current?
size of the ion (1) Ammonia, NH
3
(3) Electron repulsion outweighs the stability gained (2) Fructose, C H O
6 12 6
by achieving noble gas configuration (3) Acetic acid, C H O
2 4 2
(4) O(cid:150) ion has comparatively smaller size than (4) Hydrochloric acid, HCl
Ans. (4)
oxygen atom
Sol. Aqueous solution of HCl is the best conductor of
Ans. (3)
electric current because HCl is strong acid, so it
dissociates completely into ions.
1
8. Caprolactam is used for the manufacture of : Sol. Given species : O2, O2(cid:150)1, O2+1, O22(cid:150)
Total number of electrons
(1) Terylene (2) Nylon - 6, 6
(3) Nylon - 6 (4) Teflon O ®16e-
2
Ans. (3)
O-1 ®17e-
2
(CH ) O+1 ®15e-
25 O 2
Sol. z O2- ®18e-
HN C O nH N (CH ) C OH 2
H 25
H OH
O+1 O O-1 O-2
2 2 2 2
Caprolactum
Bond order 2.5 2 1.5 1
O
N (CH ) C Stability (cid:215) B.O.
H 25
n * Stability order éëO+1 >O >O-1 >O2-ùû
2 2 2 2
Nylon-6 12. The number of water molecules is maximum in :-
(1) 18 gram of water
9. On heating which of the following releases CO most
2 (2) 18 moles of water
easily ?
(3) 18 molecules of water
(1) MgCO (2) CaCO (4) 1.8 gram of water
3 3
(3) K CO (4) Na CO Ans. (2)
2 3 2 3 Sol. Q 1 mole water = 6.02 (cid:215) 1023 molecules
Ans. (1)
\ 18 mole water = 18 (cid:215) 6.02 (cid:215) 1023 molecules
Sol. Thermal stability order so, 18 mole water has maximum number of
molecules.
K CO >Na CO >CaCO >MgCO
2 3 2 3 3 3 13. In which of the following pairs, both the species are
Therefore MgCO releases CO most easily
RE-AIPMT-2015 3 2 not isostructural ?
MgCO ¾¾D®MgO+CO (1) NH , PH
3 2 3 3
(2) XeF , XeO
10. Strong reducing behaviour of H PO is due to : 4 4
3 2 +
(3) SiCl , PCl
(1) High oxidation state of phosphorus 4 4
(4) Dimond, silicon carbide
(2) Presence of two (cid:150)OH groups and one P(cid:150)H bond Ans. (2)
(3) Presence of one (cid:150)OH group and two P(cid:150)H bonds Sol. (i) Hybridiation of NH3 [s =3, lp=1]
sp3 geometry : tetrahedral
(4) High electron gain enthalpy of phosphorus
Ans. (3)
Sol. Strong reducing behaviour of H PO N (pyramidal) P (pyramidal)
3 2
All oxy-acid of phosphorus which contain P(cid:150)H bond
H H H H
H H
act as reductant.
(ii) Structures of XeF is square planar.
4
O
F F
P OH
Xe
H sp3 d2 hybridisation
H
F F
presence of one (cid:150)OH group and two P(cid:150)H bonds (square planar)
11. Decreasing order of stability of O , O(cid:150), O+ and O2(cid:150)
2 2 2 2 Structure of XeO is tetrahedral
4
is :-
O
(1) O > O+ > O2(cid:150) > O(cid:150)
2 2 2 2
Xe
(2) O(cid:150) > O2(cid:150) > O+ > O sp3 hybridisation
2 2 2 2 O O
(3) O+ > O > O(cid:150) > O2(cid:150) O
2 2 2 2
(4) O2(cid:150) > O(cid:150) > O > O+ so XeF4 and XeO4 are not isostructural
2 2 2 2
Ans. (3)
2
(iii) Structure of SiCl4 is tetrahedral 16. Reaction of phenol with chloroform in presence of
dilute sodium hydroxide finally introduces which one
Cl
of the following functional group ?
Si
(1) (cid:150)CHCl (2) (cid:150)CHO
sp3 hybridisation 2
Cl Cl (3) (cid:150)CH Cl (4) (cid:150)COOH
Cl 2
Ans. (2)
Structure of PCl + is tetrahedral
4 Reimer Tieman reaction
Cl
OH OH
+
P
sp3 hybridisation CHO
Cl Cl
Cl Sol. + CHCl + NaOH
3
(iv) Diamond & SiC both are isostructural because
both have tetrahedral arrangement and central
atom is sp3 hybridised. 17. The vacant space in bcc lattice unit cell is :
(1) 23% (2) 32%
14. In the reaction with HCl, an alkene reacts in
(3) 26% (4) 48%
accordance with the Markovnikov's rule, to give a
Ans. (2)
product 1(cid:150)chloro(cid:150)1(cid:150)methylcyclohexane. The
Sol. Packing efficiency in bcc lattice = 68%
possible alkene is :-
\ vacant space in bcc lattice = 100 (cid:150) 68 = 32%
18. Which of the statements given below is incorrect ?
CH
2 CH (1) ONF is isoelectronic with O N(cid:150)
3 2
(1) (A) (2) (B) (2) OF is an oxide of fluorine
2
(3) Cl O is an anhydride of perchloric acid
2 7 CODE-A
(4) O molecule is bent
3
CH
3 Ans. (2)
(3) (A) and (B) (4) Sol. (i) No. of electron in ONF = 24
No. of electron in NO (cid:150) = 24
2
both are isoelectronic
Ans. (3)
(ii) OF is a fluoride of oxygen not oxide of fluorine
2
because EN of fluorine is more than oxygen
CH CH
2 3 Cl OF = oxygen difluoride
Å 2
Sol. +
H Cl (iii) Cl O is an anhydride of perchloric acid
2 7
2HClO ¾¾D¾®Cl O
4 -H2O 2 7
CH ..
3
CH CH
3 3 Cl
+ Å (iv) O molecule is bent O bent shape
H Cl 3
O O
19. The name of complex ion, [Fe(CN) ]3(cid:150) is :-
6
15. Assuming complete ionization, same moles of which (1) Tricyanoferrate (III) ion
of the following compounds will require the least (2) Hexacyanidoferrate (III) ion
amount of acidified KMnO for complete oxidation? (3) Hexacyanoiron (III) ion
4
(4) Hexacyanitoferrate (III) ion
(1) FeC O (2) Fe(NO )
2 4 22
Ans. (2)
(3) FeSO (4) FeSO
4 3 Sol. [Fe(CN) ](cid:150)3
6
Ans. (3)
Hexacyanido ferrate (III) ion
3
20. If avogadro number NA, is changed from Ni2+ ®[Ar]183d84s0
6.022 (cid:215) 1023 mol(cid:150)1 to 6.022 (cid:215) 1020 mol(cid:150)1, this
would change :
3d 4s 4P 4d
(1) the ratio of chemical species to each other in a
due to presence of strong field ligand all unpaired
balanced equation
electrons are paired up.
(2) the ratio of elements to each other in a compound
3d 4s 4P 4d
(3) the definition of mass in units of grams
.. .. .. ..
(4) the mass of one mole of carbon
Ans. (4)
CN(cid:151) CN(cid:151)CN(cid:151)CN(cid:151)
Sol. Q mass of 1 mol (6.022 (cid:215) 1023 atoms) of carbon Hybridisation of [Ni(CN) ]2(cid:150) is dsp2
4
= 12g
24. The heat of combustion of carbon to CO is
2
If Avogadro Number (NA) is changed (cid:150)393.5 kJ/mol. The heat released upon formation
than mass of 1 mol (6.022 (cid:215) 1020 atom) of carbon of 35.2 g of CO from carbon and oxygen gas is:
2
(1) (cid:150)630 kJ (2) (cid:150)3.15 kJ
12´6.022´1020
= =12´10-3g
6.022´1023 (3) (cid:150)315 kJ (4) +315 kJ
Ans. (3)
Therefore the mass of 1 mol of carbon is changed
Sol. Formation of CO from carbon and dioxygen gas
21. Which of the following statements is not correct for 2
can be represented as
a nucleophile ?
C(s) + O ®CO ; DH = (cid:150)393.5 kJ mol(cid:150)1
2(g) 2(g) f
(1) Nucleophiles attack low e(cid:150) density sites
(1 mole = 44 g)
(2) Nucleophiles are not electron seeking Heat released on formation of 44 g CO
2
RE-AIPMT-2015
(3) Nucleophile is a Lewis acid = (cid:150)393.5 kJ mol(cid:150)1
(4) Ammonia is a nucleophile
-393.5kJ mol-1
= ´35.2g
Ans. (3) 44g
Sol. Reason : Nucleophiles are electron rich species so = (cid:150)315 kJ
act as Lewis base.
25. 20.0 g of a magnesium carbonate sample
22. A gas such as carbon monoxide would be most likely decomposes on heating to give carbon dioxide and
to obey the ideal gas law at :
8.0g magnesium oxide. What will be the percentage
(1) high temperatures and high pressures purity of magnesium carbonate in the sample ?
(2) low temperatures and low pressures (1) 60 (2) 84
(3) high temperatures and low pressures (3) 75 (4) 96
(4) low temperatures and high pressures (At. Wt. : Mg = 24)
Ans. (3) Ans. (2)
Sol. Real gases show ideal gas behaviour at high Sol. MgCO (s)®MgO(s)+CO (g)
3 2
tempratures and low pressures.
23. The hybridization involved in complex [Ni(CN)4]2(cid:150) moles of MgCO3 = 2804 = 0.238 mol
is (At.No. Ni = 28)
(1) d2sp2 (2) d2sp3 From above equation
1 mole MgCO gives 1 mole MgO
(3) dsp2 (4) sp3 3
\ 0.238 mole MgCO will give 0.238 mole MgO
3
Ans. (3)
= 0.238 (cid:215) 40 g = 9.523 g MgO
Sol. [Ni(CN) ]2(cid:150)
4 Practical yield of MgO = 8 g MgO
oxidation state of Ni is +2
x (cid:150)4 = 2
8
x = +2 \ % purity = (cid:215) 100 = 84%
9.523
4
26. What is the mole fraction of the solute in a 1.00 m 30. The following reaction
aqueous solution ?
H
(1) 0.0354 (2) 0.0177
NH N
2
(3) 0.177 (4) 1.770 +Cl NaOH
O
Ans. (2)
O
Sol. 1.00 m solution means 1 mole solute is present in
1000 g water. is known by the name :
1000 (1) Acetylation reaction
n = =55.5mol H O
H2O 18 2 (2) Schotten-Baumen reaction
n 1 (3) Friedel(cid:150)Craft's reaction
X = solute =
solute n +n 1+55.5 = 0.0177 (4) Perkin's reaction
solute H2O
Ans. (2)
27. The correct statement regarding defects in
Sol. Benzoylation of aniline is an example of Schotten
crystalline solids is :- Bauman reaction.
(1) Frenkel defect is a dislocation defect 31. The sum of coordination number and oxidation
(2) Frenkel defect is found in hallides of alkaline number of the metal M in the complex
metals [M(en) (C O )]Cl (where en is ethylenediamine) is :-
2 2 4
(3) Schottky defects have no effect on the density (1) 7 (2) 8 (3) 9 (4) 6
Ans. (3)
of crystalline solids
Sol. [M(en) (C O )]Cl
2 2 4
(4) Frenkel defects decrease the density of crystalline
oxidation state of M = + 3
solids Coordination number of M = 6
Ans. (1) Sum of oxidation state + coordination number
Sol. Frenkel defect is a dislocation defect =3 + 6 = 9 CODE-A
32. Reaction of carbonyl compound with one of the
28. The stability of +1 oxidation state among Al, Ga,
following reagents involves nucleophilic addition
In and TI increases in the sequence :
followed by elimination of water. The reagent is :
(1) TI < In < Ga < Al
(1) hydrocyanic acid
(2) In < TI < Ga < Al
(2) sodium hydrogen sulphite
(3) Ga < In < Al < TI
(3) a Grignard reagent
(4) Al < Ga < In < TI
(4) hydrazine in presence of feebly acidic solution
Ans. (4)
Ans. (4)
Sol. Stability of +1 oxidation state due to inert pair effect Sol. Reaction of carbonyl compounds with ammonia
Tl > In > Ga > Al derivatives is an example of Nucleophilic addition
29. Two possible stereo-structures of elimination reaction.
33. Which one of the following esters gets hydrolysed
CH CHOH.COOH, which are optically active, are
3
called :- most easily under alkaline conditions ?
(1) Enantiomers OCOCH
3
(1)
(2) Mesomers
(3) Diastereomers
OCOCH
(4) Atropisomers 3
(2)
Ans. (1) Cl
Sol. COOH
COOH OCOCH
3
H OH HO H (3)
O N
CH 2
3 CH
3
OCOCH
Both are enantiomers 3
(4)
HCO
3
5
Ans. (3) Ans. (1)
Sol. EWG (electron withdrawing group) increases Sol. [Co(en)2Cl2]Cl
reactivity towards nucleophilic substitution reaction. Possible isomers -
(i) Geometrical isomers
(cid:150)NO is strong electron withdrawing group.
2
34. In an S 1 reaction on chiral centres, there is : Cl
N
Cl
(1) 100% retention
(2) 100% inversion cis en Co optically active
(3) 100% racemization
(4) inversion more than retention leading to partial en
recemization
Ans. (4) Cl
Sol. S 1 reaction gives racemic mixture with slight
N
predominance of that isomer which corresponds to
inversion because SN1 also depends upon the trans en Co en optically inactive
degree of 'shielding' of the front side of the reacting
carbon.
35. The rate constant of the reaction A ® B is Cl
0.6 (cid:215) 10(cid:150)3 mole per second. If the concentration
(ii) In trans form plane of symmetry present, so trans
of A is 5 M, then concentration of B after
form is optically inactive but cis is optically active.
20 minutes is :-
Total number of stereoisomer = 2+1 =3
(1) 0.36 M (2) 0.72 M
38. The variation of the boiling points of the hydrogen
RE-A(I3P) M1.0T8-2 M015 (4) 3.60 M
halides is in the order HF > HI > HBr > HCl.
Ans. (2)
What explains the higher boiling point of hydrogen
Sol. For zero order reaction :
fluoride ?
x = K.t
(1) The bond energy of HF molecules is greater than
= 0.6 (cid:215) 10(cid:150)3 (cid:215) 20 (cid:215) 60
x = 0.72 M in other hydrogen halides
36. What is the pH of the resulting solution when equal (2) The effect of nuclear shielding is much reduced
volumes of 0.1 M NaOH and 0.01 M HCl are in fluorine which polarises the HF molecule
mixed? (3) The electronegativity of fluorine is much higher
(1) 7.0 (2) 1.04 than for other elements in the group.
(3) 12.65 (4) 2.0 (4) There is strong hydrogen bonding between HF
Ans. (3) molecules
Sol. N1V1 (cid:150) N2V2 = N.V. Ans. (4)
0.1 (cid:215) 1 (cid:150) 0.01 (cid:215) 1 = N (cid:215) 2
Sol. Due to strong H-bonding in HF molecule, boiling
0.09 point is highest for HF
[OH(cid:150)] = N = = 0.045 N
R 2 HF > HI > HBr > HI
pOH = (cid:150) log (0.045) = 1.35
39. What is the mass of the precipitate formed when
\ pH = 14 (cid:150) pOH = 14 (cid:150) 1.35 = 12.65
50 mL of 16.9% solution of AgNO is mixed with
3
37. Number of possible isomers for the complex
50 mL of 5.8% NaCl solution ?
[Co(en) Cl ] Cl will be : (en = ethylenediamine)
2 2 (Ag = 107.8, N = 14, O = 16, Na = 23, Cl =35.5)
(1) 3 (2) 4
(1) 7 g (2) 14 g
(3) 2 (4) 1
(3) 28 g (4) 3.5 g
6
Ans. (1) 42. Method by which Aniline cannot be prepared is :-
Sol. 16.9 g AgNO is present in 100 mL solution.
3 (1) reduction of nitrobenzene with H /Pd in ethanol
\ 8.45 g AgNO is present in 50 mL solution 2
3 (2) potassium salt of phthalimide treated with
5.8 g NaCl is present in 100 mL solution
\ 2.9 g NaCl is present in50 mL solution chlorobenzene followed by hydrolysis with
aqueous NaOH solution
AgNO + NaCl ® AgCl + NaNO
3 3 (3) hydrolysis of phenylisocyanide with acidic
8.45 2.9
mol
170 58.5 solution
=0.049mol =0.049mol ® 0 0 (4) degradation of benzamide with bromine in
after 0 0 ® 0.049 mol 0.049 mol
alkaline solution
reaction
Ans. (2)
mass of AgCl precipitated
= 0.049 (cid:215) 143.5 g ..
= 7g AgCl : C l:
40. The oxidation of benzene by V O in the presence
2 5
Sol.
of air produces :
(1) benzoic acid (2) benzaldehyde
(3) benzoic anhydride (4) maleic anhydride due to resonance C(cid:150)Cl bond acquires double bond
Ans. (4) character.
43. Which of the following reaction(s) can be used for
O
Sol. Air CH C the preparation of alkyl halides ?
O
V O
2 5 CH C (I) CH CH OH + HCl¾¾anh¾.ZnC¾l2¾®
3 2
O
(II) CH CH OH + HCl ¾¾®
3 2
Maleic anhydride
(III) (CH ) COH + HCl ¾¾®
33 CODE-A
41. Which of the following is not the product of
(IV) (CH ) CHOH + HCl ¾¾anh¾.ZnC¾l2¾®
32
(1) (IV) only (2) (III) and (IV) only
(3) (I), (III) and (IV) only (4) (I) and (II) only
dehydration of OH ?
Ans. (3)
Sol. (I) and (IV) can be used due to presence of anhydrous
ZnCl (III) gives alkyl halide due to formation of more
2
stable carbocation.
(1) (2) 44. Which is the correct order of increasing energy of
the listed orbitals in the atom of titanium ?
(At. no. Z = 22)
(1) 3s 3p 3d 4s (2) 3s 3p 4s 3d
(3) (4)
(3) 3s 4s 3p 3d (4) 4s 3s 3p 3d
Ans. (4) Ans. (2)
Sol. Ti(22) = 1s22s22p63s23p64s23d2
order of energy is 3s 3p 4s 3d
45. In the extraction of copper from its sulphide ore,
Å
Sol.
the metal is finally obtained by the reduction of
cuprous oxide with :-
Intermediate carbocation (more stable).
(1) copper(I) sulphide
No rearangement in C+ takes place.
(2) sulphur dioxide
(3) iron(II) sulphide
(4) carbon monoxide
So product is not possible.
Ans. (1)
Sol. Self reduction
Cu S + 2Cu O ®6Cu + SO ­
2 2 2
7
46. Root pressure develops due to : 53. Axile placentation is present in :
(1) Increase in transpiration (1) Argemone (2) Dianthus
(2) Active absorption (3) Lemon (4) Pea
(3) Low osmotic potential in soil Ans. (3)
(4) Passive absorption 54. In which of the following both pairs have correct
Ans. (2) combination :
47. Which one is a wrong statement ?
Gaseous nutrient cycle Sulphur and Phosphorus
(1) Brown algae have chlorophyll a and c, and
(1)
Sedimentary nutrient cycle Carbon and Nitrogen
fucoxanthin
(2) Archegonia are found in Bryophyta, Pteridophyta Gaseous nutrient cycle Carbon and Nitrogen
(2)
and Gymnosperms
Sedimentary nutrient cycle Sulphur and Phosphorus
(3) Mucor has biflagellate zoospores
Gaseous nutrient cycle Carbon and sulphur
(4) Haploid endosperm is typical feature of (3)
Sedimentary nutrient cycle Nitrogen and phosphorus
gymnosperms
Gaseous nutrient cycle Nitrogen and sulphur
Ans. (3)
(4)
48. Which of the following structures is not found in Sedimentary nutrient cycle Carbon and Phosphorus
prokaryotic cells?
Ans. (2)
(1) Plasma membrane
55. In mammalian eye, the 'fovea' is the center of the
(2) Nuclear envelope
visual field, where :
(3) Ribosome
(1) more rods than cones are found.
(4) Mesosome
(2) high density of cones occur, but has no rods
RAEns-.A(I2P)MT-2015
(3) the optic nerve leaves the eye
49. Which one of the following animals has two separate
(4) only rods are present
circulatory pathways ?
Ans. (2)
(1) Shark (2) Frog (3) Lizard (4) Whale
56. Choose the wrong statement :
Ans. (4)
(1) Yeast is unicellular and useful in fermentation
50. Most animals that live in deep oceanic waters are:
(2) Penicillium is multicellular and produces
(1) Detritivores
antibiotics
(2) Primary consumers (3) Neurospora is used in the study of biochemical
(3) Secondary consumers genetics
(4) Tertiary consumers (4) Morels and truffles are poisonous mushrooms
Ans. (1) Ans. (4)
51. An association of individuals of different species 57. Which of the following are not membrane-bound?
living in the same habitat and having functional (1) Mesosomes
interactions is : (2) Vacuoles
(1) Population (2) Ecological niche (3) Ribosomes
(3) Biotic community (4) Ecosystem (4) Lysosomes
Ans. (3) Ans. (3)
52. The oxygen evolved during photosynthesis comes 58. In which of the following interactions both partners
from water molecules. Which one of the following are adversely affected ?
pairs of elements is involved in this reaction? (1) Mutualism
(1) Magnesium and Chlorine (2) Competition
(2) Manganese and Chlorine
(3) Predation
(3) Manganese and Potassium
(4) Parasitism
(4) Magnesium and Molybdenum
Ans. (2)
Ans. (2)
8
59. A colour blind man marries a woman with normal 66. A protoplast is a cell :
sight who has no history of colour blindness in her (1) without cell wall
family. What is the probability of their grandson (2) without plasma membrane
being colour blind ? (3) without nucleus
(1) 0.25 (2) 0.5
(4) undergoing division
(3) 1 (4) Nil
Ans. (1)
Ans. (4) 67. In which group of organisms the cells walls form two
60. Ectopic pregnancies are referred to as : thin overlapping shells which fit together ?
(1) Pregnancies terminated due to hormonal (1) Slime moulds (2) Chrysophytes
imbalance (3) Euglenoids (4) Dinoflagellates
(2) Pregnancies with genetic abnormality. Ans. (2)
(3) Implantation of embryo at site other than uterus. 68. The DNA molecules to which the gene of interest
is integrated for cloning is called :
(4) Implantation of defective embryo in the uterus
(1) Carrier (2) Transformer
Ans. (3)
(3) Vector (4) Template
61. Cellular organelles with membranes are :
Ans. (3)
(1) Lysosomes, Golgi apparatus and mitochondria
69. Male gametophyte in angiosperms produces :
(2) Nuclei, ribosomes and mitochondria
(1) Three sperms
(3) Chromosomes, ribosomes and endoplasmic
(2) Two sperms and a vegetative cell
reticulum
(3) Single sperm and a vegetative cell
(4) Endoplasmic reticulum, ribosomes and nuclei
(4) Single sperm and two vegetative cells
Ans. (1)
Ans. (2)
62. Cell wall is absent in : CODE-A
(1) Nostoc (2) Aspergillus 70. Coconut water from a tender coconut is :
(3) Funaria (4) Mycoplasma (1) Degenerated nucellus
(2) Immature embryo
Ans. (4)
(3) Free nuclear endosperm
63. The term "linkage" was coined by :
(4) Innermost layers of the seed coat
(1) W.Sutton (2) T.H. Morgan
Ans. (3)
(3) T.Boveri (4) G.Mendel
71. The species confined to a particular region and not
Ans. (2)
found elsewhere is termed as :
64. Which of the following biomolecules does have a
(1) Rare (2) Keystone (3) Alien (4) Endemic
phosphodiester bond ?
Ans. (4)
(1) Nucleic acids in a nucleotide
72. Metagenesis refers to :
(2) Fatty acids in a diglyceride
(1) Presence of a segmented body and
(3) Monosaccharides in a polysaccharide parthenogenetic mode of reproduction
(4) Amino acids in a polypeptide (2) Presence of different morphic forms
(3) Alternation of generation between asexual and
Ans. (1)
sexual phases of an organism
65. The primary dentition in human differs from
(4) Occurrence of a drastic change in form during
permanent dentition in not having one of the
post-embryonic development
following type of teeth :
Ans. (3)
(1) Incisors
73. The enzymes that is not present in succus entericus
(2) Canine
is :
(3) Premolars (1) lipase (2) maltase
(4) Molars (3) nucleases (4) nucleosidase
Ans. (3) Ans. (3)
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74. Eutrophication of water bodies leading to killing of 79. During biological nitrogen fixation, inactivation of
fishes is mainly due to non-availability of : nitrogenase by oxygen poisoning prevented by :
(1) oxygen (2) food (1) Cytochrome (2) Leghaemoglobin
(3) light (4) essential minerals (3) Xanthophyll (4) Carotene
Ans. (1) Ans. (2)
75. The function of the gap junction is to : 80. Grafted kidney may be rejected in a patient due to
(1) stop substance from leaking across a tissue
(1) Innate immune response
(2) performing cementing to keep neighbouring
(2) Humoral immune response
cells together
(3) Cell-mediated immune response
(3) Facilitate communication between adjoining cells
(4) Passive immune response
by connecting the cytoplasm for rapid transfer
Ans. (3)
of ions, small molecules and some large
81. The body cells in cockroach discharge their
molecules
(4) separate two cells from each other. nitrogenous waste in the haemolymph mainly in the
Ans. (3) form of :
76. Match the following list of microbes and their (1) Calcium carbonate (2) Ammonia
importance : (3) Potassium urate (4) Urea
(a) Saccharomyces (i) Production of Ans. (3)
cerevisiae immunosuppressive 82. Filiform apparatus is characteristic feature of :
agents
(1) Synergids
(b) Monascus (ii) Ripening of Swiss (2) Generative cell
purpureus cheese (3) Nucellar embryo
(4) Aleurone cell
(c) Trichoderma (iii) Commercial
RE-AIPMT-2015
polysporum production of ethanol Ans. (1)
83. Acid rain is caused by increase in the atmospheric
(d) Propionibacterium (iv) Production of blood
concentration of :
sharmanii cholesterol lowering
(1) O and dust (2) SO and NO
agents 3 2 2
(3) SO and CO (4) CO and CO
(a) (b) (c) (d) 3 2
(1) (iii) (i) (iv) (ii) Ans. (2)
(2) (iii) (iv) (i) (ii) 84. The wheat grain has an embryo with one large,
(3) (iv) (iii) (ii) (i) shield-shaped cotyledon known as :
(4) (iv) (ii) (i) (iii) (1) Coleoptile (2) Epiblast
Ans. (2) (3) Coleorrhiza (4) Scutellum
77. Arrange the following events of meiosis in correct Ans. (4)
sequence : 85. Among china rose, mustard, brinjal, potato, guava,
(a) Crossing over cucumber, onion and tulip, how many plants have
(b) Synapsis superior ovary?
(c) Terminalisation of chaismata (1) Four (2) Five (3) Six (4) Three
(d) Disappearance of nucleolus Ans. (3)
(1) (b), (c), (d), (a) (2) (b), (a), (d), (c) 86. Which of the following is not a function of the
(3) (b), (a), (c), (d) (4) (a), (b), (c), (d) skeletal system?
Ans. (3) (1) Locomotion
78. The cutting of DNA at specific locations became (2) Production of erythrocytes
possible with the discovery of : (3) Storage of minerals
(1) Ligases (2) Restriction enzymes (4) Production of body heat
(3) Probes (4) Selectable markers Ans. (4)
Ans. (2)
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