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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.