Sample NEET Full Mock Test


180 Questions MCQ Test NEET Mock Test Series & Past Year Papers | Sample NEET Full Mock Test


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This mock test of Sample NEET Full Mock Test for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET Sample NEET Full Mock Test (mcq) to study with solutions a complete question bank. The solved questions answers in this Sample NEET Full Mock Test quiz give you a good mix of easy questions and tough questions. NEET students definitely take this Sample NEET Full Mock Test exercise for a better result in the exam. You can find other Sample NEET Full Mock Test extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

A cubical block of side 10 cm floats at the interface of an oil and water as shown in the figure. The density of oil is 0.6 g cm-3 and the lower face of ice cube is 2 cm below the interface. The pressure above that of the atmosphere at the lower face of the block is 

Solution:

QUESTION: 2

Initial pressure and volume of a gas are P and V respectively. First it is expanded isothermally to volume 4V and then compressed adiabatically to volume V. The final pressure of gas will be (γ =1.5)

Solution:

In an isothermal process P1V1 = P2V2

or PV = P2 x 4V

In an adiabatic process

QUESTION: 3

A gas is enclosed in a container which is then placed on a fast moving train. The temperature of the gas 

Solution:

According to ideal gas equation , temperature is not related to the velocity of the system

QUESTION: 4

In a thermodynamic process, the pressure of a fixed mass of a gas is changed in such a manner that the gas releases 20 J of heat and 8J of work is done on the gas. If the initial internal energy of the gas was 30 J, then the final internal energy will be

Solution:

Heat energy = work done + internal energy
ΔQ = ΔW + Δu……..(1)
given : heat = 20J = ΔQ
Work done = 8J = Δw
From (1), Δu = 20 – 8 = 12J
i.e. change in energy = 12
∴ 30 – final energy = 12
∴  final energy = 18J   

QUESTION: 5

Five identical rods are joined as shown in figure. Point A and C are maintained at temperature 120°C and 20°C respectively. The temperature of junction B will be

Solution:

If thermal resistance of each rod is considered R then, the given combination can be redrawn as follows


QUESTION: 6

Four massless springs whose force constants are 2k, 2k, k and 2k respectively are attached to a mass M kept on a frictionless plane (as shown in figure). If the mass M is displaced in the horizontal direction, then the frequency of oscillation of the system is

Solution:

QUESTION: 7

A wave travelling along positive x- axis is given by y = A sin(ωt − kx) . If it is reflected from rigid boundary at x = 0, such that 80% amplitude is reflected, then equation of reflected wave is

Solution:

On getting reflected from a rigid boundary the wave suffers a phase difference of π the boundary.
Hence if yincident = A sin(ωt − kx)
then yreflected = (0.8A) sin {ωt − k (−x) + π}
= −0.8A sin(ωt + kx) an additional phase change of π .

QUESTION: 8

A point charge is surrounded symmetrically by six identical charges at distance r as shown in the figure. How much work is done by the forces of electrostatic repulsion when the point charge q at the centre is removed and placed at infinity

Solution:

Total potential at the centre 
Required work done 

QUESTION: 9

A conducting sphere of radius R, and carrying a charge q is joined to a conducting sphere of radius 2R, and carrying a charge – 2q. The charge flowing between them will be

Solution:

Initial charge on sphere of radius R = q
Charge on this sphere after joining 
Now charge flowing between them 

QUESTION: 10

The figure here shows a portion of a circuit. What are the magnitude and direction of the current i in the lower right-hand wire

Solution:

By using Kirchoff's junction law as shown below.

QUESTION: 11

In the following circuit, 18Ω resistor develops a power of 2 J/sec due to current flowing through it. The power developed across 10Ω resistance is

Solution:

The given circuit can be redrawn as follows



QUESTION: 12

Find magnetic field at O

Solution:



QUESTION: 13

A magnet is parallel to a uniform magnetic field. If it is rotated by 60°, the work done is 0.8 J. How much work is done in moving it 30° further

Solution:

W = MB(cos θ1 − cos θ2)
When the magnet is rotated from 0° to 60°, then work done is 0.8 J

In order to rotate the magnet through an angle of 30°, i.e., from 60° to 90°, the work done is
 

QUESTION: 14

A straight wire of length L is bent into a semicircle. It is moved in a uniform magnetic field with speed v with diameter perpendicular to the field. The induced emf between the ends of the wire is

Solution:

QUESTION: 15

If A and B are identical bulbs which bulbs glows brighter (angular frequency of A.C. source = 100π)

Solution:

Inductive reactance=XL=WL=50 * 100*!)^-3=5*10^-2

capacitive reactance=XC=1/WC=1/50*10*10^-12=2*10^9

since XC>XL so resistance provided by capacitor is greater than that of inductor .so bulb which is in series with inductor will glow brighter.so A will glow brighter

QUESTION: 16

A radioactive nucleus undergoes α - emission to form a stable element. What will be the recoil velocity of the daughter nucleus if V is the velocity of α - emission and A is the atomic mass of radioactive nucleus

Solution:


According to conservation of momentum

QUESTION: 17

In an electrical cable there is a single wire of radius 9 mm of Copper. Its resistance is 5 Ω . The cable is replaced by 6 different insulated Copper wires, the radius of each wire is 3 mm. Now the total resistance of the cable will be (l = 1)

Solution:

Initially : Resistance of given cable
 .....(i)
Finally : Resistance of each insulated copper wire is
Hence equivalent resistance of cable
On solving equation (i) and (ii) we get Req = 7.5 Ω

QUESTION: 18

An optical fibre communication system works on a wavelength of 1.3 μm. The number of subscribers it can feed if a channel requires 20 kHz are

Solution:

Optical source frequency 
∴ Number of channels or subscribers 

QUESTION: 19

A ray of light strikes a plane mirror M at an angle of 45° as shown in the figure. After reflection, the ray passes through a prism of refractive index 1.5 whose apex angle is 4°. The total angle through which the ray is deviated is

Solution:


QUESTION: 20

In Young’s double slit experiment, we get 60 fringes in the field of view of monochromatic light of wavelength 4000Å. If we use monochromatic light of wavelength 6000Å, then the number of fringes obtained in the same field of view is

Solution:

n1λ1 = n2λ2 ⇒ 60 × 4000 = n2 × 6000 ⇒ n2 = 40

QUESTION: 21

Two discs of moment of inertia I1 and I2 and angular speeds ω1 and ω2 are rotating along collinear axes passing through their centre of mass and perpendicular to their plane. If the two are made to rotate together along the same axis the rotational KE of system will be

Solution:

By the law of conservation of angular momentum 
Angular velocity of system 

Rotational kinetic energy 

QUESTION: 22

The bob of a simple pendulum is displaced from its equilibrium position O to a position Q which is at height h above O and the bob is then released. Assuming the mass of the bob to be m and time period of oscillations to be 2.0 sec, the tension in the string when the bob passes through O is

Solution:

Tension in the string when bob passes through lowest point 

QUESTION: 23

The acceleration due to gravity about the earth's surface would be half of its value on the surface of the earth at an altitude of (R = 4000 mile)

Solution:

By solving we get h = 1656.85 mile ≈ 1600 mile

QUESTION: 24

A thermodynamic system is taken through the cycle PQRSP. The net work done by the system is

Solution:

QR and SP are two isochoric processes, so, work done for these two is zero.
Workdone in PQ path,
W1=100×103×(300−100)×10−6
20J
Work done in RS path,
W= 200×103×(100−300)×10−6
−40J
∴ Net workdone= W1+W2
−20J
−VE sign implies that work is done by the system.
∴ Workdone by the system is 20J

QUESTION: 25

A sound wave of wavelength 32 cm enters the tube at S as shown in the figure. Then the smallest radius r so that a minimum of sound is heard at detector D is

Solution:

Path difference (πr - 2r) = 

QUESTION: 26

In an LCR circuit R = 100 ohm. When capacitance C is removed, the current lags behind the voltage by π / 3. When inductance L is removed, the current leads the voltage by π / 3. The impedance of the circuit is

Solution:

When C is removed circuit becomes RL circuit hence  .....(i)
When L is removed circuit becomes RC circuit hence  ....(ii)
From equation (i) and (ii) we obtain XL = XC. This is the condition of resonance and in resonance Z = R = 100Ω.

QUESTION: 27

If the magnetic dipole moment of an atom of diamagnetic material, paramagnetic material and ferromagnetic material is denoted by μdpf respectively then

Solution:

Materials with no unpaired, or isolated electrons are considered diamagnetic. Diamagnetic substances do not have magnetic dipole moments and have negative susceptibilities. However, materials having unpaired electrons whose spins do not cancel each other are called paramagnetic. These substances have positive magnetic moments and susceptibilities. μd = 0, μp ≠ 0.

QUESTION: 28

Moment of inertia of a uniform circular disc about a diameter is I. Its moment of inertia about an axis perpendicular to its plane and passing through a point on its rim will be

Solution:

Moment of inertia of disc about a diameter 
Now moment of inertia of disc about an axis perpendicular to its plane and passing through a point on its rim

QUESTION: 29

A body at 1500 K emits maximum energy at a wavelength 20,000 Å . If the Sun emits maximum energy at wavelength 5500 Å , then the temperature of Sun is

Solution:

According to Wien's displacement law λm T = constant or λm T = λm 'T' 

QUESTION: 30

One mole of a monoatomic ideal gas is mixed with one mole of a diatomic ideal gas. The molar specific heat of the mixture at constant volume is

Solution:


QUESTION: 31

If  then projection  of  will be

Solution:


The projection of 

QUESTION: 32

The dimensions of  in the equation , where P is pressure, x is distance and t is time, are

Solution:


QUESTION: 33

A car A is travelling on a straight level road with a uniform speed of 60 km / h. It is followed by another car B which is moving with a speed of 70 km / h. When the distance between them is 2.5 km, the car B is given a deceleration of 20 km / h2. After how much time will B catch up with A

Solution:

Let car B catches, car A after 't' sec, then 

QUESTION: 34

A car has linear velocity v at an instant  on a circular road of radius r. If it is increasing its speed at the rate of 'a' meter / sec2 , then the resultant instantaneous acceleration at this instant will be 

Solution:

QUESTION: 35

A body of mass 2 kg is moving on the ground comes to rest after some time. The coefficient of kinetic friction between the body and the ground is 0.2. The retardation in the body is 

Solution:

We know that

a=μg

=0.2×9.8

=1.96m/s2

QUESTION: 36

A body of mass 2 kg slides down a curved track which is quadrant of a circle of radius 1 metre. All the surfaces are frictionless. If the body starts from rest, its speed at the bottom of the track is (g = 9.8 m/s2)

Solution:

By conservation of energy, 

QUESTION: 37

A ball moving with speed v hits another identical ball at rest. The two balls stick together after collision. If specific heat of the material of the balls is S, the temperature rise resulting from the collision is

Solution:

Initial momentum = mv
Final momentum = 2mV
By the conservation of momentum, mv = 2mV 

QUESTION: 38

If a planet consists of a satellite whose mass and radius were both half that of the earth, the acceleration due to gravity at its surface would be (g on earth = 9.8 m/sec2)

Solution:


According to problem 

= 2 × 9.8 = 19.6 m/s2

QUESTION: 39

Steel and Copper wires of same length and radius are stretched by the same weight one after the other. Young's modulus of Steel and Copper are 2 ×1011 N / m2 and 1.2 ×1011 N / m2 . The ratio of increase in length

Solution:

QUESTION: 40

A soap film of surface tension 3 ×10−2 Nm−1 formed in rectangular frame, can support a straw. The length of the film is 10 cm. Mass of the straw which the film can support is (g = 10 m/s2)

Solution:

The weight of straw will be balanced by the force of surface tension
∴ mg = 2Tl

QUESTION: 41

A : The e.m.f. of the driver cell in the potentiometer experiment should be greater than the e.m.f. of the cell to be determined.
R : The fall of potential across the potentiometer wire should not be less than the e.m.f. of the cell to be determined.

Solution:

If either the e.m.f. of the driver cell or potential difference across the whole potentiometer wire is lesser than the e.m.f. of the experimental cell, then balance point will not obtained.

QUESTION: 42

A : Acceleration of a magnet falling through a long solenoid decreases.
R : The induced current produced in a circuit always flow in such direction that it opposes the change or the cause the produced it.

Solution:

The induced current in the ring opposes the motion of falling magnet. Therefore, the acceleration of the falling magnet will be less than that due to gravity.

QUESTION: 43

A : The power of a pump which raises 100 kg of water in 10 sec to a height of 100 m is 10 KW.
R : The practical unit of power is horse power.

Solution:

The power of the pump is the work done by it per sec.

= 104 W = 10kW
Also 1 Horse power (hp) = 746 W.

QUESTION: 44

A : Two particles of different mass, projected with same velocity at same angles. The maximum height attained by both the particle will be same.
R : The maximum height of projectile is independent of particle mass.

Solution:

it is independent of mass of projectile.

QUESTION: 45

A : If the distance between parallel plates of a capacitor is halved and dielectric constant is made three times, then the capacitance becomes 6 times.
R : Capacitance of the capacitor does not depend upon the nature of the plate’s material.

Solution:

By the formula capacitance of a capacitor

Again for capacity of a capacitor
Therefore, capacity of a capacitor does not depend upon the nature of the material of the capacitor.

QUESTION: 46

The root mean square speeds at STP for the gases H2, N2, O2 and HBr are in the order

Solution:


QUESTION: 47

The rate of diffusion of SO2 and O2 are in the ratio

Solution:

QUESTION: 48

  

Product is:

Solution:

Benzylic rearrangement.

QUESTION: 49

120 gm of urea are present in 5 litre solution, the active mass of urea is

Solution:

QUESTION: 50

For the reaction is equivalent to

Solution:

QUESTION: 51

Which of the following solutions will have pH close to 1.0?

Solution:


QUESTION: 52

HClO is a weak acid. The concentration of H+ ions in 0.1 M solution of HClO(Ka = 5 × 10-8) will be equal to

Solution:

QUESTION: 53

The enthalpy change (∆H) for the process N2 H4 (g) → 2N (g) + 4H (g) is 1724 KJ mol–1. If the bond energy of N–H bond in ammonia is 391 KJ mol–1. What is the bond energy of N–N bond in N2H4

Solution:

The structure of H2−N-N−H2 (So, 4N-H bond present) means their energy
=391×4=1564 so the bond energy of N-N in N2H4
=1724−1564=160KJ/mol.

QUESTION: 54

For the equilibrium at 1 atm and 298 K

Solution:

QUESTION: 55

The time of completion of 90% of a first order reaction is approximately

Solution:

For a first order reaction

QUESTION: 56

The minimum energy required for molecules to enter into the reaction is called

Solution:

The energy necessary for molecules to undergo chemical reaction is known as Activation energy.

QUESTION: 57

Which of the following reaction is possible at anode

Solution:

Oxidation always occurs at anode.

QUESTION: 58

The specific conductivity of N/10 KCl solution at 20ºC is 0.0212 ohm–1cm–1 and the resistance of cell containing this solution at 20ºC is 55 ohm. The cell constant is

Solution:

QUESTION: 59

Which of the following is NOT a redox reaction

Solution:

In the reaction  change in oxidation state is not taking place.

QUESTION: 60

A catalyst

Solution:

A catalyst can increase the rate or reaction and hence increases the frequency of collision of reacting species.

QUESTION: 61

The CORRECT order of second ionization potential of carbon, nitrogen, oxygen and fluorine is

Solution:



According to ionic size and effective nuclear charge, their order of decreasing second ionization potential should be
F > O > N > C
However, O+(g) has half-filled electronic configuration which is more stable than partially-filled electronic configuration in F+(g). Therefore, second ionization potential of O is more than that of F. Hence, the correct order is 

O > F > N > C

QUESTION: 62

In the froth floatation process for the purification of ores, the ore particles float because

Solution:

Particles that can be easily wetted by water are called hydrophilic, while particles that are not easily wetted by water are called hydrophobic. Hydrophobic particles have a tendency to form a separate phase in aqueous media. In froth flotation the effectiveness of an air bubble to adhere to a particle is based on how hydrophobic the particle is. Hydrophobic particles have an affinity to air bubbles, leading to adsorption. The bubble-particle combinations are elevated to the froth zone driven by buoyancy forces

QUESTION: 63

To an aqueous solution of AgNO3 some NaOH(aq) is added, till a brown ppt. is obtained. To this H2O2 is added dropwise. The ppt. turns black with the evolution of O2. The black ppt. is

Solution:


The finely divided Ag is black in colour.

QUESTION: 64

Nitrogen can be obtained from air by removing

Solution:

Air is around 68 percent Nitrogen, 21 percent Oxygen, 0.8 percent Carbon Dioxide and remaining are the residual gases. Therefore by removing Oxygen and Carbon Dioxide, Nitrogen can be collected.

QUESTION: 65

AgCl precipitate dissolves in ammonia due to the formation of

Solution:

QUESTION: 66

50 ml 10N H2SO4, 25 ml 12N HCl and 40 ml 5 N HNO3 were mixed together and the volume of the mixture was made 1000 ml by adding water. The normality of the resultant solution will be

Solution:



QUESTION: 67

The IUPAC name for 

Solution:

QUESTION: 68

The decreasing order of nucleophilicity among the nucleophiles



is

Solution:

Strong bases are generally good nucleophile. If the nucleophilic atom or the centre is the same, nucleophilicity parallels basicity, i.e., more basic the species, stronger is the nucleophile. Hence basicity as well as nucleophilicity order is

Now CN– is a better nucleophile than CH3O–. Hence decreasing order of nucleophilicity is

QUESTION: 69

Which of the following exhibits optical isomerism?

Solution:


Because it has chiral carbon atom.

QUESTION: 70

Which of the following structures correspond to the product expected, when excess of C6H6 reacts with CH2Cl2 in presence of anhydrous AlCl3?

Solution:

It is friedel crafts alkylation

QUESTION: 71

A salt solution is treated with chloroform drops. Then it is shaken with chlorine water. Chloroform layer becomes violet. Solution contains.

Solution:

Cl2 water being aqueous and CH Cl3 being organic form two layers.Cl2 water oxidizes I- → I2 which shows its colour (violet) in organic layer (CHCl3). This is layer test which is used for detection of I- and Br in a solution. Br gives brown layer. 

QUESTION: 72

Which of the following compounds will be most easily attacked by an electrophile

Solution:

Phenol is most easily attacked by an electrophile because presence of –OH group increases electron density at o- and p- positions.

QUESTION: 73

In the following compounds

 

The order of acidity is

Solution:

IV > III > I > II.
−NO2 group is electron withdrawing group while −CH3 group is electron releasing group.

QUESTION: 74

Which of the following compounds will give positive test with Tollen's reagent?

Solution:

Acetaldehyde reduces Tollens' reagent to silver mirror.

QUESTION: 75

A colourless organic compound gives brisk effervescences with a mixture of sodium nitrite and dil. HCl. It could be

Solution:


CO2 evolve with brisk effervescence.

QUESTION: 76

The molarity of orthophosphoric acid having purity of 70% by weight and specific gravity 1.54 would be

Solution:

70% by weight 70gm H3PO4 → 100gm solution / sample

QUESTION: 77

The maximum amount of BaSO4 precipitated on mixing equal volumes of BaCl2 (0.5 M) with H2SO4 (1M) will correspond to

Solution:


One mole of BaCl2 reacts with one mole of H2SO4 . Hence 0.5 mole will react with 0.5 mole of H2SO4 i.e. BaCl2 is the limiting reagent.

QUESTION: 78

The orbital angular momentum of an electron in s orbital is

Solution:

For 2s orbital, l = 0; azimuthal quantum number is 0, this shows angular momentum for the 2s orbitals.

QUESTION: 79

The position of both an electron and a helium atom is known within 1.0 nm and the momentum of the electron is known within 50 × 10–26 kg ms–1. The minimum uncertainty in the measurement of the momentum of the helium atom is

Solution:

The product of uncertainties in the position and the momentum of a sub atomic particle = h/4π . Since ∆x is same for electron and helium so ∆p must be same for both the particle 

QUESTION: 80

In which of the following reactions, there is no change in the oxidation state of any atom?

Solution:

QUESTION: 81

The hybridization in sulphur dioxide is

Solution:

SO2 molecule has sp2 hybridisation.

QUESTION: 82

What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N NaOH solution?

Solution:

For complete neutralization equivalent of oxalic acid = equivalent of NaOH 

QUESTION: 83

The boiling point of benzene is 353.23 K. When 1.80 gm of a nonvolatile solute was dissolved in 90 gm of benzene, the boiling point is raised to 354.11 K. the molar mass of the solute is [Kb for benzene = 2.53 K mol–1]

Solution:

The elevation (ΔTb) in the boiling point = 354.11K-353.23K=0.88K
Substituting these values in expression 

Where, w = weight of solute, w = weight of solute

Hence , molar mass of the slute = 58 gm mol-1

QUESTION: 84

The second order Bragg's diffraction of X–rays with λ = 1Å from a set of parallel planes in a metal occurs at an angle of 60º. The distance between the scattering planes in the crystalis

Solution:


or 

QUESTION: 85

Crystals of covalent compounds always have

Solution:

Constituent particles of covalent compounds are atoms.

QUESTION: 86

A : The solubility of AgCl in water decreases if NaCl is added to it
R : NaCl is highly soluble in water where as AgCl is sparingly soluble.

Solution:

In presence of NaCl, [Cl] increases very much. Hence [Ag+] decreases to keep Ksp = [Ag+] [Cl] constant

QUESTION: 87

A : Aldol condensation can be catalysed both by acids and bases.
R : β - Hydroxy aldehydes or ketones readily undergo acid catalysed dehydration.

Solution:

Both carbanions (formed in presence of base) and enol form (formed in presence of an acid) act as nucleophiles and hence add on the carbonyl group of aldehydes and ketones to give aldols.

QUESTION: 88

A : HClO4 is a stronger acid than HClO3 .
R : Oxidation state of Cl in HClO4 is +7 and in HClO3 is +5.

Solution:

Both assertion and reason are true but reason is not the correct explanation of assertion. Greater the number of negative atoms present in the oxy-acid make the acid stronger. In general, the strengths of acids that have general formula (HO)m ZOn can be related to the value of n. As the value of n increases, acidic character also increases. The negative atoms draw electrons away from the Z-atom and make it more positive. The Z-atom, therefore, becomes more effective in with drawing electron density away from the  oxygen atom that is bonded to hydrogen. in turn, the electrons of H—O bond are drawn more strongly away from the H-atom. The net effect makes it easier from the proton release and increases the acid strength.

QUESTION: 89

A : Borax bead test is not suitable for Al(III).
R : Al2O3 is insoluble in water.

Solution:

Borax bead test is not suitable for Al(III) because its oxidising as well as reducing flame is colourless in both hot as well as cold.
Alumina is insoluble in water as they exist in hydrated form like Al2O3.2H2O, Al2O3.H2O etc.

QUESTION: 90

A : Hydroxyketones are not directly used in Grignard reaction.
R : Grignard reagents react with hydroxyl group.

Solution:

Hydroxy ketones are not directly used in Grignard reagent. Grignard reagents are very reactive. Therefore, they react with hydroxyl group. Here both are correct.

QUESTION: 91

An enzyme that can stimulate germination of barley seeds is

Solution:

The embryo of a seed is a rich source of gibberellins. After water is imbibed, the release of gibberellins from the embryo signals the seed to break dormancy and germinate. Some seeds that require particular environmental conditions to germinate, such as exposure to light or low temperatures, break dormancy if they are treated with gibberellins. Gibberellins support the growth of cereal seedlings (example Barley) by stimulating the synthesis of digestive enzymes such as alpha-amylase that mobilize stored nutrients.
So, the correct answer is 'α - amylase'

QUESTION: 92

DNA is made up of building blocks called    

Solution:

DNA is made of chemical building blocks called nucleotides. These building blocks are made of three parts: a phosphate group, a sugar group and one of four types of nitrogen bases. To form a strand of DNA, nucleotides are linked into chains, with the phosphate and sugar groups alternating.

QUESTION: 93

The primary electron acceptor of PSII during Non- cyclic photophosphorylation in plants is:

Solution:

During Non-cyclic photophosphorylation first electron acceptor from PS-।। Is phaeophytin.

QUESTION: 94

The power of water pump is 2 kW. If g = 10 m / s2, the amount of water it can raise in 1 minute to a height of 10 m is

Solution:

Given,

Power = 2kw = 2000w , Height H = 10m , Time taken t = 60 seconds.

Now ,

Power = Energy/time = mgh/t

2ooo = m*10*10/60

We get m = 1200 kg

Therefore , the pump lifts 1200 kg or 1200 l of water to a height of 10m every minute.

QUESTION: 95

The first biosphere reserve established in India for conserving the gene pool of flora and fauna and the life style of tribals is

Solution:

The Nilgiri Biosphere Reserve was the first Biosphere Reserve in India.

QUESTION: 96

Allogamy is best favoured by:

Solution:

In plants allogamy is used to mean specifically the use of pollen from one plant to fertilize the other and usually synonymous with the term cross fertilization. Dicliny refers to separation of male and female reproductive parts into different flowers.

QUESTION: 97

The population of a culture of E.coli was found to increase from 1.5 x 105cells/ml to 12 x 105 cells/ml in 42min. The generation time of the organism present in the culture is:

Solution:

14 min. Because they need three generations to reach a population of 12 × 10^5..from 1.5 ×10^5 (1.5 -TO 3 TO - 6 -TO 12)×10^5 takes the generation 14 minutes.)

QUESTION: 98

Which of the following are CORRECT statements?

i. M- phase is the shortest phase.
ii. Histones are formed during the S- phase.
iii. Chiasmata begins to appear during the pachytene stage.
iv. During gamete formation the enzyme recombinase participates during Prophase- I.

Solution:

A cell cycle consists of two phases : Interphase and M phase of which Interphase takes most of the time ,therefore M phase is considered as the shortest phase. Also, under M phase Anaphase is the shortest one. So, (i) is correct.
It is also correct that Histones are formed during the S phase as we know DNA is replicated in the S phase and therefore Histones need to be formed in the S phase in order to do their work of wrapping the copied DNA.
For (iv), it is correct as enzyme recombinase is formed during prophase 1 which help in forming recombinant nodule and hence in crossing over.

QUESTION: 99

The largest organelle in animals is __________ and in plants is _____

Solution:

Mitochondria, Chloroplasts produce food for their cells. Chloroplasts help turn sunlight into food that can be used by the cell, a process known as photosynthesis. Like the mitochondria, the chloroplast has an inner and outer membrane.

QUESTION: 100

mRNA is associated with:

Solution:

Polyribosomes or polysome is a complex of mRNA that strung along strand of mRNA that translate the genetic information coded in mRNA.

QUESTION: 101

The cell within cell organelle is:

Solution:

Correct Answer :- b

Explanation : Mitochondria are known as the powerhouses of the cell. They are organelles that act like a digestive system which takes in nutrients, breaks them down, and creates energy rich molecules for the cell. The biochemical processes of the cell are known as cellular respiration.

QUESTION: 102

Select the option with correct labelling of given structure of Golgi apparatus.

Solution:

Golgi apparatus also called Golgi complex and found universally in both plant and animal cell. It is comprised of a series of membrane-covered sacs called cisternae that look something like a stack. The cis face of a Golgi stack is also known as forming a face which has a convex face. The trans face is known as the maturing face which has a concave face. From the forming face, several vesicles are pinched off forming.
So, the correct answer is " A - Cisternae, B - Vesicle, C - Trans face, D - Cis face ".

QUESTION: 103

In sweet pea, genes C and P are needed for flower colour. What would be the percentage of coloured flowers in the offspring of the cross of ccPp XCcpp?

Solution:

probabilities of the gamete formed

Ccpp will form Cp and cp

ccPp will form cP and cp

So the cross will result in 4 type of progenies

CcPp

Ccpp

ccPp

ccpp

hence 25% of the flowers will be coloured.

QUESTION: 104

Given below is a pedigree chart of a family with five children. It shows the inheritance of attached ear-lobes as opposed to the free ones. The squares represent the male individuals and circles represent the female individuals
Which one of the following conclusions drawn is correct?

Solution:

The parents are heterozygous. lt is an autosomal inheritance trait. It is controlled by a rare dominant allele say 'E' (free ear lobe) Here the recessive homozygotes which do not have a dominant 'E' have attached ear lobes.

QUESTION: 105

How many of the following families have been correctly matched with their characteristic inflorescence?
i. Asteraceae –Capitulum
ii. Labiateae – Verticillaster
iii. Euphorbiaceae- Cyathium
iv. Cruciferae – Raceme

Solution:
QUESTION: 106

Pick the odd one out:

Solution:

Option b,c,d belongs to green algae but option a belongs to brown algae.

QUESTION: 107

The flower shown in the following diagram is

Solution:

In the given flower, the following details are evident:
1. the whorl of calyx can be seen at the base and the whorl of corolla can be seen surrounding the inner whorls of androecium and gynoecium. When the calyx and corolla are distinct, the condition is called dichlamydeous.
2. Inner to the corolla, the whorl of androecium can be seen as stamens are present. The gynoecium is visible at the very centre of the flower. When both the male and female sex organs are present in the same flower, it is called bisexual.
3. It is seen that the thalamus is swollen and the ovary is superiorly located than the other floral organs. This is called a hypogynous condition.
Hence, the correct answer is 'Dichlamydeous, bisexual, hypogynous'

QUESTION: 108

Hypohydrophily occurs in:

Solution:

Pollination which occurs by means of water is called hydrophily. When pollination occurs in submerged plants, it is called hypohydrophily. For example, Ceratophyllum. In Ceratophyllum, the male flowers break and reach the surface of water and dehisce there. The pollen grains then sink in the water and come in contact with the stigma and pollination occurs under water.

QUESTION: 109

Which one of the following is INCORRECT?

Solution:

RNA polymerase 1- rRNARNA polymerase 2- tRNA, snRNARNA polymerase 3- rRNA, hnRNA.

QUESTION: 110

If one strand of a DNA molecule has the base sequence ATTGCAT, its complementary strand will have the sequence    

Solution:

Complementarity means if one strand has "A"then, the other strand will have "T" & viceversa And if one strand has "G" then, the other strand will have "C" & viceversa Given , ATTGCAT so,it's complementary strand will have TAACGTA Hence,Option B is correct And remember that for RNA for "A" complementary is "U"(uracil)& viceversa.

QUESTION: 111

Match the following:

Solution:

Mutualism is shown by ophyrs and bees for pollination.
commensalism is shown by epiphyte and mango tree because plants like orchid grow on the branches of a tree.
Parasitism is shown by ticks on a dog because the ticks feed on dogs of their food
Competition is shown by balanus and chathamalas.
Predation is shown by sparrow and gram plants because sparrow's prey is gram seeds

QUESTION: 112

Triticale is a man made cereal produced by the hybridization of:

Solution:

Correct Answer :- a

Explanation : Triticale is a man-made crop developed by crossing wheat and cereal rye. It is probably the only new man-made crop in commercial production. 

Triticale was found to be more efficient than wheat in utilizing and absorbing nitrogen from the soil. 

It also produces a 30% higher yield on acid soils and is superior to wheat on copper-deficient soil.

QUESTION: 113

What would happen if oxygen availability to activated sludge flocs is reduced?

Solution:

If oxygen availability to activated sludge floes is reduced the center of floes will become anoxic, which would cause death of bacteria and eventually breakage of floes.

QUESTION: 114

Pick out the INCORRECT pair:

Solution:

Alec Jeffreys suggested the process of DNA fingerprinting.

QUESTION: 115

The deepest photosynthetic organism in ocean is:

Solution:

Most varieties of algae live near the surface of the water in order to get enough sunlight to live. Since they can absorb blue light, red algae can live in much deeper water where light of long wavelengths -- like red -- can't reach. Red algae have been found living in depths of over 500 feet.

QUESTION: 116

Match column I with the disease resistant variety in column II

Solution:

-Pusa komal is a variety of cowpea which is resistant to the disease bacterial blight. Bacterial blight is a disease of barley caused by the bacterial pathogen Xanthomonas campestris.

-Himgiri is a variety of wheat which is resistant to leaf and stripe rust, and hill bunt. Leaf rust is a fungal disease that affects wheat, barley and rye stems, leaves and grains. Stripe rust can develop on triticale, barley, barley grass, brome grass and some other grasses, but wheat is the main host.

-Pusa swarnim variety of Brassica is resistant to the disease White rust. White rust is a disease in plants caused by the Albugo candida.

-Parbhani kranti variety of lady finger or Okra is resistant to the Yellow mosaic virus which causes yellow vein mosaic disease.

QUESTION: 117

Which of the following is NOT a major characteristic feature of biodiversity hot spots?

Solution:

Hot spot is a place where species are protected not the destruction of habitat.

QUESTION: 118

Which of the following is used in the recovery of healthy plants from diseased plants?

Solution:

Meristem culture is used for shoot apical meristem culture in vitro. When virus elimination is the objective, to obtain disease-free plants, shoot tips of up to 10 mm are used. For rapid clonal propagation, a shoot-tip culture is followed in which (5–10 mm) explants are used.

QUESTION: 119

Which of the following statements describes natural extinction?

i. Extinction abetted by human activity
ii. Quick replacement or loss of existing species
iii. Also known as background extinction
iv. A small population is most likely to be extinct

Solution:

Natural or background extinction is a slow process of replacement of existing species with the better adapted species due to alternate evolution, changes in environmental conditions, predators & diseases.A small population due to inbreeding depression (reduces genetic variability) and normal population fluctuations during unfavorable periods like dtought,harsh winter or summer.

QUESTION: 120

Which of the following can cause the damage of DNA in humans?

Solution:

UVB ranges from 290 to 320 nm. With even shorter rays, most UVC is absorbed by the ozone layer and does not reach the earth. Both UVA and UVB, however, penetrate the atmosphere and play an important role in conditions such as premature skin aging, eye damage (including cataracts), and skin cancers.

QUESTION: 121

Pick the INCORRECT statements:

i. Dikaryon is observed in Ascomycetes and Phycomycetes.
ii. Fungi imperfecti reproduce by Asexual reproduction only.
iii. Basidiocarp is the edible part of mushroom visible above the ground
iv. Blakeslee discovered the phenomenon of heterothallism in Rhizopus.

Solution:

Only their asexual form of reproduction is known, meaning that these fungi produce their spores asexually, in the process called sporogenesis. There are about 25,000 species that have been classified in the deuteromycota and many are basidiomycota or ascomycota anamorphs.
The term heterothallism was first used by Blakeslee in 1904 for the condition of sexual reproduction which he found in certain species of Mucorales, such that 'conjugation is possible only through the interaction of two differing thalli'.
Hence, (i) and (iii) are the two incorrect statements.

QUESTION: 122

Pick the correct match:

Solution:

Diatoms belong to chrysophyta . They are photosynthetic and aquatic.

Sexual reproduction occurs by oogamy in them.

QUESTION: 123

Pick the odd one out:

Solution:

Sapindales belong to order. whereas others are families.

QUESTION: 124

Nitrogenase is sensitive to oxygen. Hence an anaerobic condition is maintained by:

Solution:

Leg haemoglobin is also called oxygen scavenger and it creates anaerobic condition.

QUESTION: 125

Match the following:

Solution:

Minamata disease, sometimes referred to as Chisso-Minamata disease, is a neurological syndrome caused by severe mercury poisoning. Symptoms include ataxia, numbness in the hands and feet, general muscle weakness, narrowing of the field of vision and damage to hearing and speech.
Nitrate in water is almost completely absorbed into the blood. Our bodies convert a portion of that nitrate into nitrite. Nitrite reacts with blood to create methemoglobin. The more methemoglobin in the blood, the worse that blood is at carrying oxygen where it is needed.
Blackfoot disease is a characteristic vascular disease associated with long-term exposure to inorganic arsenic. It occurred in areas of Taiwan with elevated arsenic in drinking water. The patients suffer severe systemic arteriosclerosis with black, mummified dry foot-gangrene in severe cases.
Itai-itai disease is caused by cadmium (Cd) exposure, produced as a result of human activities related to industrialisation, and this condition was first recognised in Japan in the 1960s. Itai-itai disease is characterised by osteomalaecia with severe bone pain and is associated with renal tubular dysfunction.

QUESTION: 126

An element that is immobile and whose deficiency results in necrosis is:

Solution:

Calcium is used to build cell wall structure and because cell walls form around every cell, calcium is an essential element for maintaining cellular integrity.
It is found in large quantities in cells that are dividing (meristematic cells) and this is the reason why calcium deficiency often manifests in the growing roots, shoots, buds and young leaves. The leaves of some plants hook downward and exhibit marginal necrosis.
Calcium is absorbed through the growing root tips and is transported to the growing parts of the plants via xylem vascular system. Once deposited in cell walls it cannot be translocated to other parts of the plant

QUESTION: 127

In a cell:
i. DPE= φw = 0 for a turgid cell
ii.φp is negative in the phloem and positive in the xylem
iii. TP = OP = φs
iv. φw indicates the free energy of the water molecules in a system 

Solution:

When a cell is fully turgid, its OP is equal to TP and DPD is zero. Turgid cells cannot absorb any more water. Thus, with reference to plant cells, the DPD can be described as the actual thrust of a cell for water and can be expressed as DPD=OP-TP. When DPD is zero, entry of water will stop.
In a turgid plant cell, pressure potential is usually positive but, negative in case of xylem, the potential pressure of xylem in a transpiring plant, which is under considerable tension, will be negative.
Gibbs Free Energy. Gibbs Free Energy (G) - The energy associated with a chemical reaction that can be used to do work. The free energy of a system is the sum of its enthalpy (H) plus the product of the temperature (Kelvin) and the entropy (S) of the system: Free energy of reaction (G)

QUESTION: 128

Which of the following is CORRECTLY matched:

Solution:

While movement of water and minerals through the xylem is driven by negative pressures (tension) most of the time, movement through the phloem is driven by positive hydrostatic pressure. This process is termed translocation, and is accomplished by a process called phloem loading and unloading. 

Hence, the correct answer would be Option A.

QUESTION: 129

The net gain of ATP produced during aerobic, anaerobic and HMP are___________ respectively:

Solution:

Aerobic : 
Location - Cytoplasm (glycolysis) and mitochondria
Stages - Glycolysis (anaerobic), Krebs cycle, oxidative phosphorylation
ATP produced - Large amount (36 ATP)

QUESTION: 130

The coenzyme TTP and cofactor Zn2+ are required for:

Solution:

TPP works as a coenzyme in many enzymatic reactions, such as:

-Pyruvate dehydrogenase complex

-Pyruvate decarboxylase in ethanol fermentation

-Alpha-ketoglutarate dehydrogenase complex

-Branched-chain amino acid dehydrogenase complex

-2-hydroxyphytanoyl-CoA lyase

-Transketolase

QUESTION: 131

A : The physical distance between two genes determines the frequency of cross- over.
R : One cross- over reduces the occurence of another cross- over in its vicinity

Solution:

Because crossover is directly proportional to distance. So, frequency of crossover is determined by distance. Closest distance links together the genes that's why reduces the chance of crossover either same chromosome /vicinity region. So, option B is correct.

QUESTION: 132

A : In an aquatic ecosystem, pyramid of biomass is inverted
R : Biomass depends upon the reproductive potential and number of phytoplanktons.

Solution:

In aquatic ecosystem, the pyramid of biomass may be inverted. EXAMPLE : Biomass of zooplanktons is higher than that of phytoplanktons as the life span of former is longer and latter multiply much faster though having shorter lifespan. A number of generations of phytoplanktons may thus be consumed by single generation of zooplanktons . Biomass of fish may be larger as fishes are large in size with longer lifespan and a number of generations of zooplanktons can be consumed by fishes. However, during transfer, only 10 % of the biomass of one generation is passed on to next trophic level . Sum total of biomass of benthic animals and brown algae exceeds the other producers and consumers in aquatic ecosystem

QUESTION: 133

A : Photorespiration decreases the rate of photosynthesis.
R : RUBISCO can also behave as an oxygenase.

Solution:

Photorespiration is a process that reduce the yield of photosynthesis, because the active site of RuBISCO accepts oxygen in place of carbon dioxide and generate no ATP.

QUESTION: 134

A : During photophosphorylation light energy is utilized to produce the proton gradient that is required for ATP synthesis
R : Oxidative phosphorylation results as the energy of oxidation- reduction is utilized for phosphorylation.

Solution:

As electrons travel towards NADP+, they generate a proton gradient across the thylakoid membrane, which is used to drive synthesis of ATP. Thus NADPH, ATP, and oxygen are the products of the first phase of photosynthesis called the light reactions.
Photosynthesis is the process by which plants, some bacteria and some protistans use the energy from sunlight to produce glucose from carbon dioxide and water. This glucose can be converted into pyruvate which releases adenosine triphosphate (ATP) by cellular respiration. Oxygen is also formed.

QUESTION: 135

A : Honey bees often pollinate red coloured flowers.
R : Red coloured flowers have a very strong fragrance.

Solution:

Honeybee often pollinate Purple, violet & blue colour flower & flower which have more fragrance are pink white & purple.

QUESTION: 136

Although both seal and penguin have streamlined, fish- like bodies with a layer of insulating fat, they are not closely related. This similarity results from:

Solution:

Convergent evolution as they originated from anywhere else but their habitat is common.i.e. having common characteristic in species of different lineages.

QUESTION: 137

Which of the following would NOT result from the release of adrenaline (epinephrine)?

Solution:

Intestinal ischemia occurs when the blood flow through the major arteries that supply blood to your intestines slows or stops. The condition has many potential causes, including a blockage in an artery caused by a blood clot, or a narrowing of an artery due to buildup of deposits, such as cholesterol.

QUESTION: 138

Which of the following observations was NOT important in helping Darwin and Wallace develop their theory of natural selection?

Solution:

Darwinism is not based on DNA and it doesn't explain information about DNA and gene .

QUESTION: 139

Which of the following is INCORRECTLY paired with its function?

Solution:

Prostate gland is male accessory gland.

*Helps in sperm activation.

*It forms 30% of semen.

*It COAGULATES SEMEN and reduces sperm mobility so their energy is conserved.

*Prostate gland has no role in secretion of testosterone and male sex hormone.

QUESTION: 140

Suppose you were a neuroscientist and were given a sample of a new snake venom. You test its effect on action at a synapse, and find that it increases the magnitude of the normal depolarizing excitatory response.The most likely explanation for this is that the venom is:

Solution:

When the neurotoxin reaches the Neuromuscular junction then it induces the release of more amount of Ca ions in the pre syanaptic membrane due to which more amount of neurotransmitter is released (from vesicles) at synapse which increases the magnitude of depolarisation at synapse or it can also act as a duplicate of neurotransmitter and can cause increase of depolarisation which depends upon the similarity in their molecular function and structure

QUESTION: 141

Which statement amongst the following is NOT true?

Solution:

Hypothalamus is a region of forebrain not hindbrain.

QUESTION: 142

The trachea, bronchi, and bronchioles of humans have all of the following functions EXCEPT:

Solution:

Increase the surface are for gas exchange because it's done by alveoli.

QUESTION: 143

Which statement is FALSE?

Solution:

Blood that returns to the heart has picked up lots of oxygen from the lungs. At each body part, a network of tiny blood vessels called capillaries connects the very small artery branches to very small veins. The capillaries have very thin walls, and through them, nutrients and oxygen are delivered to the cells.

QUESTION: 144

The mammalian heart beat:

Solution:

Vagal innervation of the heart. At the same time, the two branches of the autonomic nervous system act in a complementary way increasing or slowing the heart rate. In this context, the vagus nerve acts on sinoatrial node slowing its conduction thus actively modulating vagal tone accordingly.

QUESTION: 145

Suppose you are developing a new drug, and have found that when it is administered in humans there is a substantial increase in the volume of urine produced. When you administer antidiuretic hormone (ADH, or vasopressin) at the same time, the volume of urine returns to normal. Which hypothesis best fits these observations? The new drug:

Solution:

Antidiuretic Hormone (ADH). ADH is produced in the hypothalamus and released by the posterior pituitary gland. It causes the kidneys to retain water, constricts arterioles in the peripheral circulation, and affects some social behaviors in mammals. 

QUESTION: 146

Some athletes take “steroids” in an attempt to enhance their physical performance. This can lead to decreased sperm production and even sterility. What is the most likely explanation for this effect?

Solution:

A negative feedback system occurs in the male with rising levels of testosterone acting on the hypothalamus and anterior pituitary to inhibit the release of GnRH, FSH, and LH. The Sertoli cells produce the hormone inhibin, which is released into the blood when the sperm count is too high.

QUESTION: 147

If you were outside for a long time on a hot, dry day, without anything to drink, which of the following would happen?

Solution:

The release of anti-diuretic hormone from the pituitary gland into the bloodstream is controlled by a number of factors. Secretion of anti-diuretic hormone also occurs if the concentration of salts in the bloodstream increases, for example as a result of not drinking enough water on a hot day.

QUESTION: 148

Thyroid stimulating hormone (TSH), luteinizing hormone (LH), and oxytocin are all:

Solution:

The anterior pituitary gland produces the following hormones and releases them into the bloodstream: adrenocorticotropic hormone, which stimulates the adrenal glands to secrete steroid hormones, principally cortisol. growth hormone, which regulates growth, metabolism and body composition.

QUESTION: 149

Which of the following statement about the mammalian heart function is FALSE?

Solution:

The pulmonary artery carries deoxygenated blood from the right ventricle to the lungs. The blood here passes through capillaries adjacent to alveoli and becomesoxygenated as part of the process of respiration. In contrast to the pulmonary arteries, the bronchial arteries supply nutrition to the lungs themselves.

QUESTION: 150

When a doctor gives you an antibiotic when you are sick, he/she always tells you to keep taking the antibiotic until it is all finished. The reason he/she tells you this is because:

Solution:

Telling patients to stop taking antibiotics when they feel better may be preferable to instructing them to finish the course, according to a group of experts who argue that the rule long embedded in the minds of doctors and the public is wrong and should be overturned.

Patients have traditionally been told that they must complete courses of antibiotics, the theory being that taking too few tablets will allow the bacteria causing their disease to mutate and become resistant to the drug.

But Martin Llewelyn, a professor in infectious diseases at Brighton and Sussex medical school, and colleagues claim that this is not the case. In an analysis in the British Medical Journal, the experts say “the idea that stopping antibiotic treatment early encourages antibiotic resistance is not supported by evidence, while taking antibiotics for longer than necessary increases the risk of resistance”.

QUESTION: 151

The secretory phase of the menstrual cycle:

Solution:

Luteal phase. The luteal phase is the final phase of the ovarian cycle and itcorresponds to the secretory phase of the uterine cycle. During the luteal phase, the pituitary hormones FSH and LH cause the remaining parts of the dominant follicle to transform into the corpus luteum, which produces progesterone.

QUESTION: 152

Which statement is FALSE?

Solution:

The islets of Langerhans contain alpha, beta, and delta cells that produce glucagon, insulin, and somatostatin, respectively. The alpha cells of theislets of Langerhans produce an opposing hormone, glucagon, which releases glucose from the liver and fatty acids from fat tissue.

QUESTION: 153

Which one of the following are on the same line of evolution?

Solution:
QUESTION: 154

How are the time of ovulation and the onset of menstruation related in the human menstrual cycle?

Solution:

If an egg is fertilized, progesterone from the corpus luteum supports the early pregnancy (15). If no fertilization occurs, the corpus luteum will start to break down between 9 and 11 days after ovulation (10). This results in a drop in estrogen and progesterone levels, which causes menstruation.

QUESTION: 155

Match the items in column I with those in column II and choose the correct phylum.


 

Solution:

3-(ii) : Pleurobrachia bachei with its oral end down. Pleurobrachia bachei is a member of the phylum Ctenophora and is commonly referred to as a sea gooseberry.
1-(i) : Adamsia belongs to the phylum Coelenterata.
6-(iv) : Taenia is a genus of tapeworms (a type of platyhelminth) that includes some important parasites of livestock.
4-(iii) : Spongilla of phylum Porifera, is a genus of freshwater sponges in the family Spongillidae found in lakes and slow streams.

QUESTION: 156

Which of the following would generally reduce the likelihood of speciation?

Solution:

Immigration and emigration are processes that help mingling between isolated populations. Since genetic isolation is necessary for speciation processes like this decelerate speciation.

QUESTION: 157

Which statement about gas exchange in vertebrates is FALSE?

Solution:

When partial pressure of CO2 is low and partial pressure of O2 is high as in the alveoli, binding of O2 takes place.

QUESTION: 158

Trace a sperm cell from the structure where it is produced to fertilization of the egg:

seminiferous tubules (1);

vas deferens (2);

uterus (3);

fallopian tube (4);

vagina (5);

epididymis (6);

urethra (7).

Solution:

After the process of spermatogenesis (formation of sperm), spermiation takes place i.e. release of sperms from the seminiferous tubules. From there it travels to the rete testis- epididymis-vas deferns-ejaculatory duct-urethra-penis. Then it reaches to vagina by sexual intercourse. From vagina it goes to cervix-uterus and then finally to fallopian tube for further fertilisation

QUESTION: 159

Which statement is FALSE?

Solution:

The membrane at rest is semi-permeable with much higher permeability topotassium than to sodium (it contains potassium "leak channels"). This potential is usually somewhere between -70 and -90 millivolts, with the inside of thecell negative because positive potassium ions have left the cell.

QUESTION: 160

Oogenesis is a process in the ovary that results in the production of female gametes. Which statement about oogenesis in humans is FALSE?

Solution:

Oogenesis is initiated during the embryonic development stage.

QUESTION: 161

How does a noncompetitive inhibitor inhibit binding of a substrate to an enzyme?

Solution:

In biology, the active site is the region of an enzyme where substrate molecules bind and undergo a chemical reaction. The active site consists of residues that form temporary bonds with the substrate (binding site) and residues that catalyse a reaction of that substrate (catalytic site).

QUESTION: 162

Match the following columns :


Solution:

A - Prawn
B - Locust
C - Aurelia
D - Scorpion

QUESTION: 163

Bt corn has been made resistant from corn borer disease by the introduction of the gene

Solution:

Cry1Ab-------Corn borer...
Cry1Ac & Cry 2 Ab---------Cotton bollworms.

QUESTION: 164

If an alien DNA is inserted at Pst I of the given vector, then the transformants will grow in which of the following?