i 7 (cid:1) ENCADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 QUESTIONS REPEAT IN AIPMT - 2015 FROM iBOOKS PTS QUESTION PAPER AND WORKBOOK IMPORTANT NOTE : (cid:2)(cid:2)(cid:2)(cid:2) 66 Direct (90% - 100%) (cid:2)(cid:2)(cid:2)(cid:2) 24 Very Similar (80% Similar) (cid:2)(cid:2)(cid:2)(cid:2) 86 Similar (50% - 60% Similar & Covered in our Workbooks) PHYSICS Sl. iBOOKS WORKBOOK & PTS QUESTION FROM AIPMT QUESTION 2015 No. 1. A steady current flows in a metallic conductor of non-uniform cross- Across metallic conductor of non-uniform cross section a constant section. The quantity / quantities constant along the length of potential difference is applied. The quantity which remains conductor is/are (a) current, electric field and drift speed (b) drift constant along the conductor is (1) current (2) drift velocity (3) speed only (c) current and drift speed (d) current only. electric field (4) current density. [Refer iBOOKS WORKBOOK - I, 2015 MEP01, Qn.25] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 143, E = 30, G = 134, H = 44] 2. A long wire bent as shown in figure carries A wire carrying current I has the shape as Z Z current I. If the radius of the semicircular shown in adjoining figure. Linear parts of the portion is a, the magnetic field at the centre wire are very long and parallel to X-axis while I C is : semicircular portion of radius R is lying in Y – R (a) µ40aI (b) 4µµµµππππ0al ππππ2++++4 O I aC I Y Z(1 )p la→Bne=. −Mµa0gnIe(πtiˆci −fi2lekˆd) a(2t )p o→→→→Bint==== O−−−− µµiµµs0 : I((((ππππî++++2kˆ)))) I O I Y 4πR 4ππππR (c) µ0I+ µ0I (d) µ0I π2−4 X (3) →B =µ0 I (πî−2kˆ) (4) →B =µ0 I (πî+2kˆ). X 4a 4πa 4πa 4πR 4πR [Refer iBOOKS WORKBOOK - I, 2015 MEP02, Qn.120] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F=147, E=31, G=110, H=15] 3. Which of the following figure represents the variation of particle Which of the following figures represent the variation of particle momentum (p) and associated de-Broglie wavelength (λ) ? momentum and the associated de-Broglie wavelength ? p p p p P P P P (a) (b) (c) (d) (1) (2) (3) (4) . λ λ λ λ λ λ λ λ [Refer iBOOKS WORKBOOK - II, 2015 MEP03, Pg.31, Qn.134] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 148, E = 41, G =102, H = 3] 4. The refracting angle of a prism is ‘A’ and the refractive index of the The refracting angle of a prism is A, and refractive index of the prism is cot A/2. The angle of minimum deviation is (a) 180° - 3A material of the prism is cot (A/2). The angle of minimum deviation (b) 180° + 2A (c) 90° -A (d) 180° -2A is (1) 180° - 2A (2) 90° - A (3) 180° + 2A (4) 180° - 3A. [Refer iBOOKS WORKBOOK - VII, 2015 MEP24, Qn.10] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F=151, E=39, G=106, H=33] 5. The logic circuit shown below has the input waveforms A and B as Which logic gate is represented by the following combination of shown. Pick out the correct output waveform. logic gates ? (1) NAND (2) AND (3) NOR (4) OR. A (a) Y1 Y A B (b) Input A Y (c) B Output B (d) Y2 [Refer iBOOKS WORKBOOK - II, 2015 MEP04, Qn.251] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 152, E = 45, G = 127, H = 6] 6. When a certain metal is illuminated with monochromatic light of A certain metallic surface is illuminated with monochromatic light of wavelength, λ. The stopping potential for photo – electric current wavelength λ, the stopping potential for photo-electric current is 3 for this light is 3V. If the same surface is illuminated with light of 0 V0. For a light of wavelength 2λ, the stopping potential is V0. The wavelength 2λ, the stopping potential is V0. The threshold wavelength for this surface for photo-electric effect is : (1) 4λλλλ (2) threshold wavelength for this metal is (a) 4λλλλ (b) 3λ (c) 2λ (d) λ. λ/4 (3) λ/6 (4) 6λ. [Refer iBOOKS WORKBOOK - II, 2015 MEP03, Qn.97] (PTS) 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 154, E = 40, G = 129, H = 24] STUDENTS' SUCCESS FOR PROSPERITY 8 i (cid:1) ENC ADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 7. Two spherical bodies of mass M and 5M and radii R and 2R Two spherical bodies of mass M and 5M and radii R and 2R are respectively are released in free space with initial separation released in free space with initial separation between their centres between their centres equal to 12R. If they attract each other due equal to 12 R. If they attract each other due to gravitational force to gravitational force only, then the distance covered by the smaller only, then the distance covered by the smaller body before body just before collision is (a) 1.5R (b) 2.5R (c) 4.5 R (d) 7.5 R. collision is (1) 4.5 R (2) 7.5 R (3) 1.5 R (4) 2.5 R. [Refer iBOOKS WORKBOOK – VI, 2015 MEP19, Pg.114, Qn.07] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 167, E = 14, G = 132, H = 36] 8. If the degree of freedom of a gas are f, then the ratio of two specific The ratio of the specific heats Cp =γin terms of degrees of 2 2 1 1 Cv heats CP / CV is given by (a) f ++++1(b) 1− f (c) 1+f (d) 1+f . freedom (n) is given by (1)1+n(2)1++++2(3)1+n (4) 1+1.  3  n  2  n [Refer iBOOKS WORKBOOK – VIII, 2015 MEP28, Qn.38] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 169, E = 22, G = 121, H = 22] 9. If energy E, velocity v and time T are chosen as fundamental units, If energy (E), velocity (V) and time (T) are chosen as the the dimensions of surface tension will be (a) [Ev-2T-2] (b) [Ev-1 T-2] fundamental quantities, the dimensional formula of surface tension (c) [Ev-2T-1] (d) E2v-1 T-1]. will be (1) [EV-1 T-2] (2) [EV-2 T-2] (3) [E-2V-1T-3] (4) [EV-2T-1] [Refer iBOOKS WORKBOOK – VI, 2015 MEP16, Qn.26] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 171, E = 1, G = 116, H = 43] 10. If in a p-n junction diode, a square input signal of 10V is applied as If in a p-n junction, a square input signal of 10 V is applied, as shown then the output signal across R will be shown, L 5v +5 V R RL L -5 V −5v then the output across R will be L (a) (b) +5V (c) 10V (d) (1) 10 V (2) (3) 5 V (4). −5V - 5 V - 10 V −10V [Refer iBOOKS WORKBOOK – II, 2015 MEP04, Qn.242] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 172, E = 44, G = 100, H = 23] 11. An α -particle moving in a circular orbit of radius a makes n An electron moving in a circular orbit of radius r makes n rotations revolutions per second. Then answer the following questions. The per second. The magnetic field produced at the centre has magnetic field produced at the centre has a value B (a) B = 0 (b) µ n2e µµµµ ne µ ne B=µ0ne (c) µ0n2e (d) µµµµ0ne. magnitude : (1) Zero (2) 0r (3) 02r (4) 20πr . 2aπ a a [Refer iBOOKS WORKBOOK - I, 2015 MEP02, Qn.142] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 136, E = 32, G = 113, H = 9] 12. The ratio of the radii of the nuclei 13Al27 and 52Te125 is If radius of the 27Al nucleus is taken to be RAl, then the radius of 13 (a) 13: 52 (b) 3 13:3 52 (c)3 3:5 5(d) 3 : 5. 125 Te nucleus is nearly : 53 (1) 53RAl (2) 35RAl (3) 15331/3RAl (4) 15331/3RAl. [Refer iBOOKS WORKBOOK - V, 2015 MEP12, Qn.21] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 140, E = 43, G = 122, H = 2] 13. Three rings each of mass M and radius R are arranged as shown Three identical spherical shells, each of mass m and radius r are in figure. The moment of inertia of the system about YY' will be placed as shown in figure. Consider an axis XX’ which is touching 3 7 to two shells and passing through diameter of third shell. Moment (a) MR2 (b) 3 M R2 (c) MR2 (d) 5 MR2. of inertia of the system consisting of these three spherical shells 2 2 about XX’ axis is (1) 3 mr2 (2) 16/5 mr2 (3) 4 mr2 (4) 11/5 mr2. Y X A B C Y’ X’ [Refer iBOOKS WORKBOOK – IV, 2015 MEP11, Pg.163, Qn.43] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 175, E = 12, G = 126, H = 20] STUDENTS' SUCCESS FOR PROSPERITY i 9 (cid:1) ENCADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 14. A beam of light of wavelength 600 nm from a distance source falls For a parallel beam of monochromatic light of wavelength ‘λ’ on a single slit 1mm wide and resulting diffraction pattern is diffraction is produced by a single slit whose width ‘a’ is of the observed on a screen 2m away. The distance between 1st dark order of the wavelength of the light. If ‘D’ is the distance of the fringes on either side of central bright fringe is (a) 1.2 cm (b) 1.2 screen from the slit, the width of the central maxima will be (1) mm (c) 2.4 cm (d) 2.4 mm. Dλ/a (2) Da/λ (3) 2Da/λ (4) 2Dλλλλ/a. [Refer iBOOKS WORKBOOK - V, 2015 MEP 13, Qn.40] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 142, E = 37, G = 119, H = 5] 15. A body executing linear simple harmonic motion has a velocity of 3 A particle is executing SHM along a straight lien. Its velocities at cms-1 when its displacement is 4 cm and a velocity of 4 cms-1 when distances x1 and x2 from the mean position are V1 and V2, respectively. Its time period is : its displacement is 3 cm. What is the amplitude of oscillation ? (1) 5 cm (2) 7.5 cm (3) 10 cm (4) 12.5 cm. (1) 2ππππ Vx1222−−−−−−−−xV1222 (2) 2π Vx1122++Vx2222 (3) 2π Vx1122−−Vx2222 (4)2π Vx1122++xV2222 . Refer iBOOKS PTS_STEP_20/Q.46] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 159, E = 24, G = 107, H = 7] 16. The displacement y in cms is given in terms of time t second by the When two displacements represented by y = a sin (ωt) and y = b 1 2 cos (ωt) are superimposed the motion is : (1) simple harmonic with equation y = 3 sin 314 t + 4 cos 314 t, then the amplitude of the amplitude a/b (2) simple harmonic with amplitude a2+b2 (3) S.H.M. is (a) 7 cm (b) 3 cm (c) 4 cm (d) 5 cm. (a+b) simple harmonic with amplitude (4) not a simple harmonic. 2 [Refer iBOOKS WORKBOOK - V, 2015 MEP14, Qn.42] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 138, E = 23, G = 130, H = 42] 17. Two plano concave lenses of glass of refractive index 1.5 have Two identical thin plano-convex glass lenses (refractive indeed 1.5) each having radius of curvature of 20 cm are placed with their radii of curvature of 20 and 30 cm. They are placed in contact with convex surfaces in contact at the centre. The intervening space is curved surfaces towards each other and the space between them filled with oil of refractive. The intervening space is filled with oil of is filled with a liquid of refractive index (4/3). The focal length of the refractive index 1.7. The focal length of the combination is (1) -25 system is (a) 72cm (b) 36cm (c) 48cm (d) 80cm. cm (2) -50 cm (3) 50 cm (4) -20 cm. [Refer iBOOKS WORKBOOK – VII, 2015 MEP23, Qn.121] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 157, E = 36, G = 114, H = 8] 18. When a force is applied on 3 kg block towards right. The ratio of Three blocks A, B and C, of masses 4 kg, 2 kg and 1 kg the contact force between 3kg, 2kg and 2kg, 1kg is (a) 3 : 1 (b) 5 : respectively, are in contact on a frictionless surface, as shown. If a 1 (c) 1 : 3 (d) 1 : 2. force of 14 N is applied on the 4 kg block, then the contact force between A and B is (1) 6 N (2) 8 N (3) 18 N (4) 2 N. F 3 kg 2 kg 1 kg A B C [Refer iBOOKS WORKBOOK – VI, 2015 MEP18, Qn.19] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 173, E = 4, G = 93, H = 1] 19. Two springs have their force constant as k1 and k2 (k1 > k2). when Two similar springs P and Q have spring constants KP and KQ, such that K > K . They are stretched, first by the same amount they are stretched by the same force: (a) no work is done in case of P Q (case A), then by the same force (case B). The work done by the both the springs (b) equal work is done in case of both the springs springs W and W are related as, in case (1) and case (2), P Q (c) more work is done in case of second spring (d) more work is respectively : (1) W = W ; W = W (2) W > W ; W > W (3) W P Q P Q P Q Q P P done in case of first spring. < W ; W < W (4) W = W ; W > W . Q Q P P Q P Q [Refer iBOOKS WORKBOOK – IV, 2015 MEP09, Qn.25] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 178, E = 6, G = 108, H = 37] 20. Power dissipated in an LCR series circuit connected to an a.c. A resistance ‘R’ draws power ‘P’ when connected to an AC source. source of emf ε is : If an inductance is now placed in series with the resistance, such ε2R2+Lω− 1 2 that the impedance of the circuit becomes ‘Z’, the power drawn will (a) ε2 R2+Lω−C1ω2 R (b)  R Cω  be (1) P RZ (2) PRZ (3) P (4) PRZ2. (c) ε2R R2+Lω− 1 2 (d) εεεε2R R2++++Lωωωω−−−− 1 2  Cω   Cωωωω  [Refer iBOOKS WORKBOOK – III, 2015 MEP05, Qn.176] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 168, E = 34, G = 133, H = 31] STUDENTS' SUCCESS FOR PROSPERITY 10 i (cid:1) ENC ADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 BOTANY Sl. iBOOKS WORKBOOK & PTS QUESTION FROM AIPMT QUESTION 2015 No. 21. In Opuntia, the spines are modified (a) stipule (b) bud scale (c) Leaves become modified into spines in (1) Pea (2) Onion (3) Silk axillary bud (d) leaves. Cotton (4) Opuntia. [Refer iBOOKS WORKBOOK - III, MEB07, PAGE 121, Qn 39] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 54, E = 102, G = 1, H = 133] 22. Which one gives the most valid explanation for guard cell Which one gives the most valid and recent explanation for stomatal movements ? (a) Guard cell photosynthesis (b) Starch hydrolysis movements ? (1) Potassium influx and efflux (2) Starch theory (c) Potassium influx and efflux (d) Transpiration. hydrolysis (3) Guard cell photosynthesis (4) Transpiration. [Refer iBOOKS WORKBOOK - IV, MEB09, PAGE 103, Qn 35] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 55, E = 113, G = 51, H = 109] 23. BOD is ______ in polluted water and _____ in potable water. (a) High value of BOD (Biochemical Oxygen Demand) indicates that (1) more, less (b) less, more (c) less in both (d) medium in both. water is highly polluted (2) water is less polluted (3) consumption of organic matter in the water is higher by the microbes (4) water is pure. [Refer iBOOKS WORKBOOK - III, MEB06, PAGE 45, Qn 122] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 57, E = 175, G = 5, H = 131] 24. Cytochromes occur in (a) Matrix of mitochondria (b) Lysosomes (c) Cytochromes are found in : (1) Outer wall of mitochondria (2) Outer membrane of mitochondria (d) Cristae of mitochondria. Cristae of mitochondria (3) Lysosomes (4) Matrix of mitochndria. [Refer iBOOKS WORKBOOK - I, MEB01, PAGE 15, Qn 87] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 70, E = 143, G = 52, H = 76] 25. 1. Which structure are extensive and continuous with the outer Nuclear envelope is a derivative of : (1) Membrane of Golgi complex membrane of nucleus ? (a) RER (b) SER (c) Both RER and SER (d) Golgi body. (2) Microtubutes (3) Rough endoplasmic reticulum (4) Smooth [Refer iBOOKS WORKBOOK - I, MEB01, PAGE 20, Qn 125] 2. Karyotheca is derived from (a) Envelope remains (b) E.R. (c) endoplasmic reticulum. Golgi apparatus (d) Both (a) and (b). [Refer iBOOKS WORKBOOK - I, MEB03, PAGE 54, Qn 5] [Reference of two questions given above] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 87, E = 142, G = 10, H = 89] 26. Keel is characteristic or the flowers of : (a) Gulmohur (b) Cassia (c) Keel is the characteristic feature of flower of (1) Indigofera (2) Aloe Calotropis (d) Bean. (3) Tomato (4) Tulip. [Refer iBOOKS WORKBOOK - VI, MEB17, PAGE 95, Qn 6] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 89, E = 100, G = 32, H = 124] 27. Perigynous type of ovary is found in (a) plum (b) rose (c) pearch (d) Perigynous flowers are found in (1) Cucumber (2) China rose (3) all of these Rose (4) Guava. [Refer iBOOKS WORKBOOK - III, MEB07, PAGE 150, Qn 133] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 90, E = 101, G = 67, H = 72] 28. Closed vascular bundles lack : (a) Conjunctive tissue (b) Cambium Vascular bundles in monocotyledons are considered closed because : (1) Cambium is absent (2) There are no vessels with (c) Pith (d) Ground tissue. perforations (3) Xylem is surrounded all around by phloem (4) A bundle sheath surrounds each bundle. [Refer iBOOKS WORKBOOK - VI, MEB16, PAGE 63, Qn 26] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 111, E = 97, G = 12, H = 112] 29. 1. Which organelle helps in the synthesis of lipids, lipoproteins, Select the correct matching in the following pairs : (1) Smooth ER – cholesterol, steroids and visual pigments in epithelial cells of Synthesis of lipids (2) Rough ER – Synthesis of glycogen (3) retina? (a) SER (b) RER (c) Golgi bodies (d) Sphaerosome. [Refer iBOOKS WORKBOOK – I, MEB01, PAGE 21, Qn 141] Rough ER – Oxidation of fatty acids (4) Smooth ER – Oxidation of 2. S.E.R. takes part in synthesis of (a) Lipids and steroids (b) Vitamins phospholipids. (c) Carbohydrates (d) All of these. [Refer iBOOKS WORKBOOK - I, MEB01, PAGE 19, Qn 110] [Reference of two questions given above] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 118, E = 105, G = 45, H = 102] 30. The archaebacteria occurring in marshes, swamps, rumens of The guts of cow and buffalo possess : (1) Chlorella spp. (2) cattles, gobar gas plants are (a) methanogens (b) ammonifying Methanogens (3) Cyanobacteria (4) Fucus spp. bacteria (c) thermoacidophiles (d) denitrifying bacteria [Refer iBOOKS WORKBOOK - V, MEB12, PAGE 47, Qn 153] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 127, E = 95, G = 20, H = 71] 31. 1. The point where funicle joins with ovule is known as (a) Chalaza The hilum is a scar on the : (1) Fruit, where it was attached to (b) Hilum (c) Integument (d) Micropyle. pedicel (2) Fruit, where style was present (3) Seed, where [Refer iBOOKS WORKBOOK – IV, MEB08, PAGE 31, Qn 126] 2. The ovule is attached to the placenta by (a) hilum (b) funicle (c) micropyle was present (4) Seed, where funicle was attached. petiole (d) pedicel [Refer iBOOKS WORKBOOK - IV, MEB08, PAGE 34, Qn 144] [Reference of two questions given above] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 128, E = 114, G = 18, H = 115] STUDENTS' SUCCESS FOR PROSPERITY i 11 (cid:1) ENCADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 32. Which of the following statements is incorrect about the class Which one of the following matches is correct ? Deuteromycetes ? (a) They reproduce only by asexual spores (1) Alternaria Sexual reproduction absent Deuteromycetes (conidia) (b) Mycelium in these fungi is branched and septate (c) (2) Mucor Reproduction by Conjugation Ascomycetes They have only parasitic forms (d) Examples of these fungi are Alternaria, Colletotrichum and Trichoderma. (3) Agaricus Parasitic fungus Basidiomycetes (4) Phytophthora Aseptate mycelium Basidiomycetes [Refer iBOOKS PTS STEP-2, 2015 Qn 90] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 48, E = 91, G = 6, H = 77] 33. In an experiment, DNA was found to have 31% adenine and 19% In sea urchin DNA, which is double stranded, 17% of the bases were shown to be cytosine. The percentage of the other three guanine. The quantity of cytosine shall be (a) 38% (b) 31% (c) 19% bases expected to be present in this DNA are : (1) G17%, A 16.5%, (d) 62%. T 32.5% (2) G17%, A 33%, T 33% (3) G8.5%, A 50%, T 24.5% (4) G34%, A 24.5%, T 24.5% [Refer iBOOKS WORKBOOK - I, MEB03, PAGE 65, Qn 38] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 68, E = 169, G = 22, H = 116] 34. 1. What is true of Bt toxin ? (a) The concerned bacillus has In Bt cotton, the Bt toxin present in plant tissue as pro-toxin is antitoxins (b) the inactive protoxin gets converted into converted into active toxin due to : (1) acidic pH of the insect gut (2) active form in the insect gut (c) Bt protein exists as active toxin in the bacillus (d) The activated toxin enters the ovaries of action of gut micro-organisms (3) presence of conversation factors the pest to sterilize it and thus prevents its multiplication. in insect gut (4) alkaline pH of the insect gut. [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 98, Qn 203] 2. What is true about Bt toxin ? (a) Bt protein exists as active toxin in the Bacillus (b) The activated toxin enters the ovaries of the pest to sterilize it and thus prevent its multiplication (c) The concerned Bacillus has antitoxins (d) The inactive protoxin gets converted into active form in the insect gut. [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 136, Qn 374] [Reference of two questions given above] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 69, E = 126, G = 28, H = 47] 35. Match the columns and select correct option. Select the correct option : Column – I Column – II Column – I Column – II A. Initiation of spindle fibres p. Anaphase I (A) Synapsis aligns homologous chromosomes (i) Anaphase – II B. Synthesis of RNA and protein q. Zygotene (B) Synthesis of RNA and protein (ii) Zygotene C. Action of endonuclease r. G-phase (C) Action of enzyme recombinase (iii) G – phase 1 2 D. Movement of chromatids towards opposite s. Pachytene (D) Centromeres do not separate but (iv) Anaphase – I poles chromatids move towards opposite poles t. Anaphase II (v) Pachytene (1) (A)–(ii), (B)–(iii), (C)–(v), (D)–(iv) (2) (A)–(i), (B)–(ii), (C)–(v), (a) A – q, B – r, C – s, D – t (b) A – r, B – q, C – p, D – t (c) A – p, (D)–(iv) (3) (A)–(ii), (B)–(iii), (C)–(iv), (D)–(v) (4)(A)–(ii), (B)–(i), (C)– B – r, C – t, D – s (d) A – t, B – r, C – p, D – q. (iii), (D)–(iv) [Refer iBOOKS WORKBOOK - II, MEB04, PAGE 32, Qn 126] 99999999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 88, E = 145, G = 59, H = 132] 36. Floral formula of tobacco plant is – (a) K(5) C(5) A(5) G(2) (b) + + K(5) C(5) A5 G( 2 ) is the floral formula of (1) Sesbania (2) Petunia (3) Brassica (4) Allium. K(4) C(5) A(5) G(2) (c) K4-5 C4-5 A5 G(2) (d) K(5) C5 A10 G(2) NOTE: Both petunia and tobaco belong to solanaceae family, hence 95% similar. [Refer iBOOKS WORKBOOK - VI, MEB17, PAGE 98, Qn 9] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 65, E = 98, G = 75, H = 46] 37. 1. GEAC stands for (a) Genome Engineering Action Committee Which body of the Government of India regulates GM research and (b) Ground Environment Action Committee (c) Genetic safety of introducing GM organisms for public services ? (1) Indian Engineering Approval committee (d) Genetic and Environment Approval Committee. Council of Agriculatural Research (2) Genetic Engineering [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 91, Qn 169] Approval Committee (3) Research Committee on Genetic 2. Which of the following statements is correct regarding Genetic Engineering Approval committee (GEAC) ? (a) It makes Manipulation (4) Bio-safety committee. decision regarding the validity of GM research (b) It ensures the safety of introducing GM-organisms for public services (c) Genetic modification of organisms can have unpredictable results when such organisms are introduced into the ecosystem (d) All of these. [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 124, Qn 321] [Reference of two questions given above] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 74, E = 125, G = 77, H = 50] STUDENTS' SUCCESS FOR PROSPERITY 12 i (cid:1) ENC ADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 38. 1. Incipient nucleus occurs in (a) Myxophyceae (b) True nucleus is absent in : (1) Mucor (2) vaucheria (3) Volvux (4) Chlorophyceae (c) Phaeophyceae (d) Rhodophyceae. Anabaena. [Refer iBOOKS WORKBOOK - V, MEB12, PAGE 40, Qn 117] 2. Nuclear membrane is absent in (a) Agaricus (b) Volvox (c) Nostoc (d) Penicillium [Refer iBOOKS WORKBOOK - V, MEB12, PAGE 56, Qn 192] [Reference of two questions given above] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 80, E = 106, G = 14, H = 52] 39. In the exponential phase of geometric growth, the (a) progeny cells Typical growth curve in plants is (1) Linear (2) Stair-steps shaped stops dividing (b) both progeny cells follow mitotic division (c) both (3) Parabolic (4) Sigmoid. (a) and (b) (d) only one progeny cell follows mitotic division graphs. [Refer iBOOKS WORKBOOK -VI, MEB15, PAGE 16, Qn 40] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 61, E = 112, G = 50, H = 99] 40. Micropropagation is (a) Seed germination in which cotyledons come A technique of micro-propagation is (1) Somatic embryogenesis above ground (b) Development of embryo from somatic cells (c) (2) Protoplast fusion (3) Embryo rescue (4) Somatic hybridization. Production of new plants from cells, tissues or plant parts (d) None of the above. [Refer iBOOKS WORKBOOK -III, MEB06, PAGE 6, Qn 6] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 63, E = 122, G = 71, H = 88] 41. Besides proteins, ribosomes contain (a) DNA (b) RNA (c) Both DNA is not present in : (1) Ribosomes (2) Nucleus (3) DNA and RNA (d) Lipids. Mitochondiria (4) Chloroplast [Refer iBOOKS WORKBOOK - I, MEB01, PAGE 19, Qn 115] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 83, E = 128, G = 36, H = 65] 42. Look at the following diagram and identify The structures that are formed by stacking of organized flattened the labeling 1, 2, 3 and 4 membranous sacs in the chloroplasts are : (1) Grana (2) Stromal (a) Outer membrane, inner membrane, granum and thylakoid respectively lamellae (3) Stroma (4) Cristae. (b) Granum, thylakoid, stroma, inner membrane respectively (c) Inner membrane, outer membrane, granum and thylakoid respectively (d) Outer membrane, inner membrane, thylakoid and granum respectively. [Refer iBOOKS WORKBOOK - I, MEB 01, PAGE 16, Qn 92] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 99, E = 103, G = 34, H = 96] 43. A girdled plant will eventually die because of (a) Absence of In a ring girdled plant : (1) The root dies first (2) The shoot and downward movement of water (b) Absence of upward movement of root die together (3) Neither root nor shoot will die (4) The shoot water (c) Absence of upward supply of organic nutrients (d) Absence of downward movement of organic nutrients. dies first. [Refer iBOOKS WORKBOOK -IV, MEB09, PAGE 94, Qn 16] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 101, E = 111, G = 30, H = 57] 44. Which one of the following shows isogamy with non-flagellated Male gametes are flagellated in : (1) Anabaena (2) Ectocarpus (3) gametes ? (a) Sargassum (b) Ectocarpus (c) Ulothrix (d) Spirogyra Spirogyra (4) Polysiphonia. [Refer iBOOKS WORKBOOK - V, MEB13, PAGE 80, Qn 56] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 112, E = 96, G = 84, H = 135] 45. 1. Read the following four statements (A–D) : (A) In transcription, Gene regulation governing lactose operon of E.coli that involves the adenosine pairs with uracil (B) Regulation of lac operon by lac I gene product is : (1) negative and inducible because repressor repressor is referred to as positive regulation. (C) The human genome has approximately 50,000 genes. (D) Haemophilia is a protein prevents transcription (2) negative and repressible sex-linked recessive disease. How many of the above because repressor protein prevents transcription (3) Feedback statements are right ? (a) Two (b) Three (c) Four (d) One [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 64, Qn 80] inhibition because excess of β-galactosidase can switch off 2. In Escherichia coli, the product of i gene combines with (a) operator transcription (4) Positive and inducible because it can be induced by gene to switch off structural genes (b) Inducer gene to switch off structural genes (c) Operator gene to switch on structural genes (d) lactose. Regulator gene to switch off structural genes. [Refer iBOOKS WORKBOOK - II, MEB05, PAGE 60, Qn 63] [Reference of two questions given above] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 84, E = 168, G = 4, H = 73] 46. 1. Which one of the following is an example of ex-situ conservation ? (a) In which of the following both pairs have correct combination ? (1) Wildlife sanctuary (b) Seed bank (c) Sacred groves (d) National park. In situ conservation : Cryopreservation Ex situ conservation : [Refer iBOOKS WORKBOOK - III, MEB06, PAGE 19, Qn 58] 2. Which one of the following is not used for ex situ plant Wildlife Sanctuary (2) In situ conservation : Seed bank Ex situ conservation ? (a) Shifting cultivation (b) Botanical Gardens conservation : National park (3) In situ conservation : Tissue culture (c) Field gene banks (d) Seed banks. [Refer iBOOKS WORKBOOK - III, MEB06, PAGE 16, Qn 40] Ex situ conservation : Sacred groves (4) In situ conservation : 3. In situ conservation is carried out through (a) National parks (b) National Park Ex situ conservation : Botanical Garden. Wildlife sanctuaries (c) Biosphere reserves (d) All the above. [Refer iBOOKS WORKBOOK - III, MEB06, PAGE 14, Qn 31] [Reference of three questions given above] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 92, E = 132, G = 76, H = 85] STUDENTS' SUCCESS FOR PROSPERITY i 13 (cid:1) ENCADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 47. Examine the figures of different types of chromosomes given below The chromosomes in which centromere is situated close to one end and select the option which represents sub-meta centric are : (1) Acrocentric (2) Telocentric (3) Sub-metacentric (4) Metacentric. chromosome (a) (b) (c) (d) . [Refer iBOOKS WORKBOOK – I, MEB03, PAGE 58, Qn 19] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 100, E = 104, G = 70, H = 106] 48. 1. In which one of the following pollination is autogamous (a) Which one of the following may require pollinators, but is genetically Cleistogamy (b) Geitonogamy (c) Xenogamy (d) Chasmogamy similar to autogamy ? (1) Xenogamy (2) Apogamy (3) Cleistogamy [Refer iBOOKS WORKBOOK – IV, MEB08, PAGE 25, Qn 110] (4) Geitonogamy. 2. Transfer of pollen grains from one flower to another flower of same plant is (a) Geitonogamy (b) Autogamy (c) Cleistogamy (d) Chasmogamy. [Refer iBOOKS WORKBOOK – IV, MEB08, PAGE 26, Qn 117] 3. Geitonogamy involves : (a) Fertilization of a flower by the pollen from another flower of the same plant. (b) Fertilization of a flower by the pollen from the same flower (c) Fertilization of a flower by the pollen from a flower of another plant in the same population (d) Fertilization of a flower by the pollen from a flower of another plant belonging to a distant population. [Refer iBOOKS WORKBOOK – IV, MEB08, PAGE 26, Qn 115] 4. Assertion (A) : Xenogamy is defined as the pollen grain transferred to stigma of same flower. Reason (R) : In Geitonogamy, the pollen grains of a flower transferred to stigma of a different flower of a different plant of different species (a) If both Assertion and Reason are true and Reason is a correct explanation of Assertion (b) If both Assertion and Reason are true but Reason is not a correct explanation of Assertion (c) If Assertion is true but Reason is false (d) Both Assertion and Reason are false. [Refer iBOOKS WORKBOOK - IV, MEB08, PAGE 24, Qn 103] [Reference of four questions given above] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 113, E = 115, G = 21, H = 97] 49. Match column – A with column – B and select the correct option. Which one of the following statements is wrong ? (1) Agar-agar is Column – A Column – B obtained from Gelidium and Gracilaria (2) Chlorella and Spirulina (A) Chlorophyceae (i) Chlorophyll a, b are used as space food (3) Mannitol is stored food in (B) Phaeophyceae (ii) Laterally placed flagella Rhodophyceae (4) Algin and carrageen are products of algae. (C) Rhodophyceae (iii) Pyrenoids (D) Xanthophyceae (iv) Fucoxanthin (v) Chlorophyll a, e (vi) Laminarian starch (vii) Algin (viii) r-phycoerythrin (ix) Floridean starch (x) Lack of motility (xi) Mannitol (xii) Coenocytic plant body (xiii) Chlorophyll a, d (a) (A) - (i) (iii), (B) - (ii) (iv)(vi) (vii) (xi), (C) - (viii) (ix) (x) (xiii), (D) - (v) (xii) (b) (A) - (i) (ii), (B) - (iii) (iv) (vi) (vii) (xi), (C) - (viii) (ix) (x) (xiii), (D) - (v) (xii) (c) (A) - (i) (iii), (B) - (ii) (iv)(v) (vii) (x), (C) - (viii) (ix) (x) (xiii), (D) - (v) (xii) (d) (A) - (ii) (iii), (B) - (i) (iv)(vi) (vii) (xi), (C) - (viii) (ix) (xi) (xiii), (D) - (iv) (xii). [Refer iBOOKS WORKBOOK - V, MEB13, PAGE 70, Qn 5] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 130, E = 94, G = 43, H = 51] 50. Find out the wrong option. (a) Root pressure is due to active Transpiration and root pressure cause water to rise in plants by : (1) absorption whereas the transpiration pull is due to passive pulling and pushing it, respectively (2) pushing it upward (3) absorption. (b) In contrast to transpiration pull, root pressure is pushing and pulling it, respectively (4) pulling it upward. dependent on osmotic gradient. (c) Root pressure is a positive hydrostatic pressure whereas transpiration pull is a negative pressure (d) Guttation is an evident for transpiration pull but translocation of minerals is an evident for root pressure. [Refer iBOOKS WORKBOOK - IV, MEB09, PAGE 105, Qn 40] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 135, E = 108, G = 3, H = 86] STUDENTS' SUCCESS FOR PROSPERITY 14 i (cid:1) ENC ADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 ZOOLOGY Sl. iBOOKS WORKBOOK & PTS QUESTION FROM AIPMT QUESTION 2015 No. 51. Parents have blood groups A and B. Blood group of the child can A man with blood group ‘A’ marries a woman with blood group ‘B’. What be (a) A or B (b) O (c) AB (d) All the above. are all the possible blood groups of their offsprings ? (1) A, B and AB only (2) A, B, AB and O (3) O only (4) A and B only. [Refer iBOOKS WORKBOOK – V, MEZ10, PG 31, Qn 107] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 60, E = 167, G = 65, H = 118] 52. Mass of living matter at a trophic level in an area at any time is The mass of living material at a trophic level at a particular time is called (a) Standing state (b) Standing crop (c) Detritus (d) Humus. called : (1) Standing state (2) Net primary productivity (3) Standing crop (4) Gross primary productivity. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 28, Qn 137] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 104, E = 134, G = 58, H = 53] 53. The rate of production of organic matter during photosynthesis in an In an ecosystem the rate of production of organic matter during ecosystem is (a) primary production (b) Gross primary photosynthesis is termed as : (1) Gross primary productivity (2) productivity (c) Net primary productivity (d) Secondary productivity. Secondary productivity (3) Net productivity (4) Net primary productivity. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 22, Qn 98] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 106, E = 135, G = 29, H = 121] 54. Bruner's gland in human present in (a) Submucosa (b) mucosa (c) Which of the following statements is not correct ? (1) Goblet cells circular muscle layer (d) longitudinal muscle layer. are present in the mucosa of intestine and secrete mucus (2) Oxyntic cells are present in the mucosa of stomach and secrete HCl (3) Acini are present in the pancreas and secrete carboxypeptidase (4) Brunner’s glands are present in the submucosa of stomach and secrete pepsinogen. [Refer iBOOKS WORKBOOK – IV, MEZ06, PG35, Qn 49] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 67, E = 147, G = 66, H = 81] 55. Most of the tree dwellers are found in (a) Tropical rain forest (b) Most animals are tree dwellers in a : (1) thorn woodland (2) temperate Tropical deciduous (c) Temperate evergreen (d) Coniferous forest. deciduous forest (3) tropical rain forest (4) coniferous forest. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 54, Qn 262] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 79, E = 176, G = 80, H = 66] 56. Hysterectomy is removal of (a) uterus (b) ovary (c) oviduct (d) Hysterectomy is surgical removal of : (1) Prostate gland (2) Vas- fallopian tube. deference (3) Mammary gland (4) Uterus. [Refer iBOOKS WORKBOOK – I, MEZ01, PG 13, Qn 100] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 94, E = 163, G = 40, H = 91] 57. Which part of the human ear plays no role in hearing as such but is A gymnast is able to balance his body upside down even in the otherwise very much required ? (a) Organ of Corti (b) Vestibular total darkness because of (1) vestibular apparatus (2) Tectorial apparatus (c) Ear ossicles (d) Eustachian tube. membrane (3) Organ of corti (4) Cochlea. [Refer iBOOKS WORKBOOK – V, MEZ12, PG 142, Qn 146] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 59, E = 156, G = 17, H = 55] 58. Due to ozone layer depletion of upper atmosphere UV radiations Which of the following is not one of the prime health risks reaches the earth surface. In humans, the impact of increased UV- associated with greater UV radiation through the atmosphere due radiation (a) increases the incidence of cataract (b) increases skin to depletion of stratospheric ozone ? (1) Reduced Immune System cancer including melanoma (c) diminishes the functioning of (2) Damage to eyes (3) Increased liver cancer (4) Increased skin immune system (d) All of the three choices are true. cancer. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 63, Qn 316] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 77, E = 180, G = 61, H = 128] 59. Frequency of an allele may change in isolated population due to (a) Which is the most common mechanism of genetic variation in the Genetic drift (b) Mutation (c) Natural selection (d) Gene flow. population of a sexually reproducing organism ? (1) Chromosomal aberrations (2) Genetic drift (3) Recombination (4) Transduction. [Refer iBOOKS WORKBOOK – VI, MEZ15, PG 62, Qn 62] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 108, E = 121, G = 62, H = 103] 60. Preparation of sperm before penetration of ovum is (a) Spermation Capacitation refers to changes in the : (1) ovum before fertilisation (b) Coition (c) Insemination (d) Capacitation. (2) ovum after fertilisation (3) sperm after fertilisation (4) sperm before fertilization. [Refer iBOOKS WORKBOOK – I, MEZ01, PG 27, Qn 225] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 78, E = 159, G = 23, H = 70] STUDENTS' SUCCESS FOR PROSPERITY i 15 (cid:1) ENCADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 61. The vertical distribution of different species occupying different Vertical distribution of different species occupying different levels in levels is called (a) Statification (b) Stratification (c) trophic level (d) a biotic community is known as : (1) Stratification (2) Zonation (3) None of these. Pyramid (4) Divergence. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 38, Qn 192] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 102, E = 131, G = 2, H = 58] 62. Match the parts of the human brain listed under Column I with the Which of the following regions of the brain is incorrectly paired with functions given under Column II ; choose the choice which gives the its function ? (1) Cerebellum – language comprehension (2) correct combination of the alphabets of the two columns. Corpus callosum – communication between the left and right Column – I Column – II cerebral cortices (3) Cerebrum – calculation and contemplation (4) (Parts of the brain) (Functions) Medulla oblongata – homestatic control. A. Cerebral hemisphere p. Relaying impulses B. Thalamus q. Posture and balance C. Cerebellum r. Movement of heart, stomach, lungs, etc. D. Medulla oblongata s. Reflex actions t. Voluntary control, intelligence, hearing, speech (a) A = t, B = q, C = p, D = s (b) A = t, B = p, C = q, D = s (c) A = r, B = s, C = q, D = t (d) A = r, B = q, C = P, D = s. [Refer iBOOKS WORKBOOK – V, MEZ12, PG 108, Qn 51] 99999999....9999%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 122, E = 155, G = 54, H = 127] 63. The maximum amount of electrolytes and water (70 -80 per cent) from Removal of proximal convoluted tubule from the nephron will result the glomerular filtrate is reabsorbed in which part of the nephron ? (a) in : (1) More concentrated urine (2) No change in quality and Distal convoluted tubule (b) Proximal convoluted tubule (c) quantity of urine (3) No urine formation (4) More diluted urine. Descending limb of loop of Henle (d) Ascending limb of loop of Henle. [Refer iBOOKS WORKBOOK – IV, MEZ09, PG 199, Qn 112] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 95, E = 152, G = 16, H = 130] 64. In areas that somehow, lost all living organisms that existed there, Secondary Succession takes place on / in : (1) Degraded forest called as (a) climax community (b) ecological succession (c) primary (2) Newly created pond (3) Newly cooled lava (4) Bare rock. succession (d) secondary succession. [Refer iBOOKS WORKBOOK – II, MEZ02, PG 36, Qn 177] 99995555%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 129, E = 133, G = 46, H = 59] 65. The disease Chikunguniya is transmitted by : (a) house flies (b) Which of the following viruses is not transferred through semen of an Aedes mosquitoes (c) Cockroach (d) Female Anopheles. infected male ? (1) Human immunodeficiency virus (2) Chikungunya virus (3) Ebola virus (4) Hepatitis B virus. [Refer iBOOKS WORKBOOK – II, MEZ03, PG 98, Qn 91] 99990000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 125, E = 161, G = 25, H = 122] 66. A venereal or sexually transmitted disease is (a) Diphtheria (b) Which of the following is not a sexually transmitted disease ? (1) Leprosy (c) Syphilis (d) Tetanus. Acquired immuno Deficiency Syndrome (AIDS) (2) Trichomoniasis (3) Encephalitis (4) Syphilis. [Refer iBOOKS WORKBOOK – II, MEZ03, PG 94, Qn 57] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 53, E = 164, G = 47, H = 98] 67. A prehistoric man with cranial capacity almost similar to that of Which of the following had the smallest brain capacity ? (1) Homo modern man was : (a) Homo habilis (b) Homo heidelbergensis (c) sapiens (2) Homo neanderthalensis (3) Homo habilis (4) Homo Homo erectus (d) Homo sapiens neanderthalensis. erectus. [Refer iBOOKS WORKBOOK – VI, MEZ15, PG 73, Qn 90] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 56, E = 170, G = 24, H = 111] 68. Which of the following characters was not chosen by Mendel ? (a) How many pairs of contrasting characters in pea plants were Pod shape (b) Pod colour (c) Location of flower (d) Leaf shape. studied by Mendel in his experiments ? (1) six (2) eight (3) seven (4) five. [Refer iBOOKS WORKBOOK – V, MEZ10, PG 8, Qn 6] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 64, E = 120, G = 90, H = 125] 69. Lymph differs from blood in having (a) more RBC, less WBC and Which one of the following is correct ? (1) Serum = Blood + platelets (b) more RBC and WBC less plate lets (c) More WBC less Fibrinogen (2) Lymph = Plasma + RBC + WBC (3) Blood = Plasma RBC (d) NO RBC, more WBC and no platelets. + RBC + WBC + Platelets (4) Plasma = Blood – Lymphocytes. [Refer iBOOKS WORKBOOK – III, MEZ04, PG 35, Qn 83] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 72, E = 151, G = 19, H = 83] STUDENTS' SUCCESS FOR PROSPERITY 16 i (cid:1) ENC ADIN, VOL.- XVII, NO. 01 (cid:1) 28 MAY 2015 70. Exchange of segments between non-homologous chromosomes is The movement of a gene from one linkage group to another is called (a) Translocation (b) Inversion (c) Crossing over (d) Tetrasomy. : (1) Duplication (2) Translacation (3) Crossing over (4) Inversion. [Refer iBOOKS WORKBOOK – V, MEZ11, PG 50, Qn 02] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 73, E = 123, G = 48, H = 95] 71. A person entering an empty room suddenly finds a snake right in A chemical signal that has both endocrine and neural roles is : (1) front on opening the door. Which one of the following is likely to Calcitonin (2) Epinephrine (3) Cortisol (4) Melatonin. happen in his neuro-hormonal control system ? (a) Neurotransmitters diffuse rapidly across the cleft and transmit a nerve impulse (b) Hypothalamus activates the parasympathetic division of brain (c) Sympathetic nervous system is activated releasing epinephrine and norepinephrin from adrenal cortex (d) Sympathetic nervous system is activated releasing epinephrine and norepinephrin from adrenal medulla. [Refer iBOOKS WORKBOOK – V, MEZ13, PG 178, Qn 128] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 91, E = 157, G = 8, H = 82] 72. The numbers of chromosome in primary spermatocytes are 48(2n), Which of the following cells during gametogenesis is normally what will be the nos. of chromosomes in secondary spermatocytes? diploid ? (1) Spermatid (2) Spermatogonia (3) Secondary polar (a) 24 (b) 48 (c) 96 (d) 12. body (4) Primary polar body. [Refer iBOOKS WORKBOOK – I, MEZ01, PG 21, Qn 186] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 98, E = 162, G = 86, H = 60] 73. Systolic pressure of heart is higher than diastolic pressure because Blood pressure in the mammalian aorta is maximum during : (1) (a) blood is forcefully pumped into arteries by the heart during Diastole of the right ventricle (2) Systole of the left ventricle (3) systole (b) arteries offer resistance to the flowing of blood (c) Diastole of the right atrium (4) Systole of the left atrium. arteries contract during systole only (d) volume of blood in heart is greater during systole. [Refer iBOOKS WORKBOOK – IV, MEZ08, PG 138, Qn 60] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 109, E = 150, G = 27, H = 90] 74. The disease causing amoebic dysentery, parasites lives in the (a) The active form of Entamoeba histolytica feeds upon : (1) mucosa stomach of human (b) large intestine of human (c) small intestine and submucosa of colon only (2) food in intestine (3) blood only (4) of human (d) Anus of human. erythrocytes; mucosa and submucosa of colon. [Refer iBOOKS WORKBOOK – II, MEZ03, PG 97, Qn 76] 88880000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 124, E = 174, G = 89, H = 64] CHEMISTRY Sl. iBOOKS WORKBOOK & PTS QUESTION FROM AIPMT QUESTION 2015 No. 75. What is correct sequence of bond order ? (a) O2+ >O2− >O2 (b) Which of the following options represents the correct bond order ? OOOO++22++22>>>>OOOO2222>>>>OOOO--22--22 (c) O2 >O2− >O2+ (d) O2− >O2+ >O2. (1)OOOO--22--22O2O2+(4)O2−>O2>O+2 [Refer iBOOKS WORKBOOK - II, MEC06, PAGE 172, Qn 227] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 11, E = 54, G = 157, H = 146] 76. What is correct sequence of bond order ? (a) O2+ >O2− >O2 (b) The correct bond order in the following species is : (1) OOOO++22++22>>>>OOOO2222>>>>OOOO--22--22(c) O2>O2− >O+2 (d) O2− >O2+ >O2. O22+ K+ (2) Ca2+ < < Cl- < S2- (4) Ca2+ < K+ < Ar < S2- < Cl-. K+ < Ar (3) K+ < Ar < Ca2+ (4) Ar < K+ < Ca2+. Refer iBOOKS PTS STEP02, 2015 Qn 23] 2. Consider the isoelectronic series : K+, S2–, Cl– and Ca2+, the radii of the ions decrease as : (1) Ca2+ > K+ > Cl– > S2– (2) Cl– > S2– > K+ > Ca2+ (3) S2– > Cl– > K+ > Ca2+ (4) K+ > Ca2+ > S2– > Cl–. [Refer iBOOKS PTS STEP 26, 2015 Qn 47] [Reference of two questions given above] 111100000000%%%% RRRREEEEPPPPEEEEAAAATTTT [Refer AIPMT 2015 Qn F = 7, E = 47, G = 170, H = 177] STUDENTS' SUCCESS FOR PROSPERITY

QUESTIONS REPEAT IN AIPMT - 2015 FROM iBOOKS PTS QUESTION PAPER AND WORKBOOK

Similar QUESTIONS REPEAT IN AIPMT - 2015 FROM iBOOKS PTS QUESTION PAPER AND WORKBOOK