NEET Mock Test - 2


180 Questions MCQ Test NEET Mock Test Series 2020 & Past Year Papers | NEET Mock Test - 2


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This mock test of NEET Mock Test - 2 for NEET helps you for every NEET entrance exam. This contains 180 Multiple Choice Questions for NEET NEET Mock Test - 2 (mcq) to study with solutions a complete question bank. The solved questions answers in this NEET Mock Test - 2 quiz give you a good mix of easy questions and tough questions. NEET students definitely take this NEET Mock Test - 2 exercise for a better result in the exam. You can find other NEET Mock Test - 2 extra questions, long questions & short questions for NEET on EduRev as well by searching above.
QUESTION: 1

Figure shows an L–R circuit. When the switch S is closed, the current through resistor R1, R2 and R3 are i1, i2 and i3 respectively. The value of i1, i2 and i3 at t = 0 s is 

Solution:

At t = 0, inductor in replaced by an open circuit. 

QUESTION: 2

A uniform electric field and a uniform magnetic field are acting along the same direction in a certain region. If an electron is projected along the direction of the fields with a certain velocity then

Solution:


Since electron is moving along the field, force qE is repulsive and hence it will slow down. 

QUESTION: 3

In a coaxial, straight cable, the central conductor and the outer conductor carry equal currents in opposite directions. The magnetic field is zero

Solution:

Construct an Ampere’s loop in the concerned regions to see the value of ip.

At a pont outside the co-axial tube at a diatance R from the central axis consider a circular path.
Then from Amperes Law closed integral of B.dl=μ0I(net)
the net current is zero as the current in the two cables flow in opposite directions.

therefore Magnetic field is zero outside

QUESTION: 4

An electron in the beam of a TV picture tube is accelerated by a potential difference 2 kV. Then it passes through a transverse magnetic field to produce a circular arc of radius 0.18m. Find the magnetic field

Solution:

Velocity v acquired  by the electron and accelerating  potential V are related by,
where e and m are charge and mass of the electron

Radius of the circular track:


QUESTION: 5

Two ions have equal masses but one is singly-ionised and the other is doubly-ionised. They are projected from the same place in a uniform magnetic field with the same velocity perpendicular to the field.

Solution:

The radius r of the path of charged particle with charge q in a uniform magnetic field(B) is given by



Now, if charge of singly ionized ion is e, then charge of doubly ionized ion is 2e. Hence, the circle described by the singly ionized charge will have a radius double that of the other one. 

QUESTION: 6

A conductor ABCDEF, shaped as shown, carries a current i. It is placed in the xy plane with the ends A and E on the x-axis. A uniform magnetic field of magnitude B exists in the region. Which of the following option is INCORRECT for the force acting on it? 

Solution:

QUESTION: 7

A current-carrying ring is placed in a magnetic field. The direction of the field is perpendicular to the plane of the ring  

Solution:
QUESTION: 8

A particle with mass m and positive charge q starts from rest at the origin shown in figure. There is a uniform electric field direction and a magnetic field B directed outward the plane of the page. It is shown in more advanced books that the path is a cycloid whose radius of curvature at the top points is twice the y-co-ordinates at that level.  The speed at any point (x, y) is 

Solution:

Work done by electric field = Change in K.E. 

QUESTION: 9

The value of Bohr magneton is 

Solution:
QUESTION: 10

Consider a magnetic dipole kept in the north-south direction. Let P1, P2, Q1 and Q2 be four points at the same distance from the dipole towards north, south, east and west of the dipole respectively. The directions of the magnetic field due to the dipole are the same at

Solution:

QUESTION: 11

Angle of dip at a place is 60° and earth's magnetic field is 8 µT. Its horizontal component BH is 

Solution:

QUESTION: 12

The phenomenon of hysteresis represents the incapability of the material to

Solution:
QUESTION: 13

A small circular ring is kept inside a larger loop connected to a switch and a battery as shown. The direction of induced current when the switch is made

(i) ON
(ii) OFF after it was ON for a long time is  

Solution:

Using lenz law, it is evident that the induced current in the ring will oppose the effect caused by the current in the larger loop. When current is flowing in the larger loop a magnetic field will be set and the ring will oppose it, thus, inducing current in clockwise direction. Similarly when the switch is turned off, the induced current will be in anti-clockwise direction.

QUESTION: 14

A rod AB m oves with a uniform velocity v in a uniform magnetic field as shown in figure 

Solution:

Due to electromagnetic induction, emf e is induced across the ends of the rod. This induced emf is given by
e=Bvl
The direction of this induced emf is from A to B, that is, A is at the higher potential and B is at the lower potential. This is because the magnetic field exerts a force equal to qvB on each free electron where q is −-1.6 × 10-16 C. The force is towards AB by Fleming's left-hand rule; hence, negatively charged electrons move towards the end B and get accumulated near it. So, a negative charge appears at B and a positive charge appears at A.

QUESTION: 15

A conducting ring of radius 1m is placed in a varying magnetic field of B = 0.1t T (where t is time in second) with its plane right angle to B. What will be the induced electric field ?  

Solution:


QUESTION: 16

Two conducting circular loops of radii R1 and R2 (R1 > > R2) are placed in the same plane with their centres coinciding. Find the mutual inductance between them. 

Solution:

Assume current i passes  through outer loop. Then  and 

QUESTION: 17

The number of turns in the primary and secondary coils of an ideal transformer are 100 and 300 respectively. If the input power is 60 watt the output power will be

Solution:

No power loss in case of an ideal transformer. 

QUESTION: 18

The current in a series RL circuit decays as I = I0e–t/τ. Obtain the rms current in the interval 0 ≤ t ≤ τ

Solution:

QUESTION: 19

An alternating current having peak value 14 A is used to heat a metal wire. To produce the same heating effect, a constant current i can be used where i is

Solution:

QUESTION: 20

The value of alternating emf in the following circuit will be

Solution:

QUESTION: 21

The values of L, C and R in an L-C-R series circuit are 4mH, 40pF and 100 Ω respectively. The quality factor of the current is  

Solution:

QUESTION: 22

The self inductance of a motor of an electric fan is 10 H. In order to impart maximum power at 50 Hz, it should be connected to a capacitance of  

Solution:

Maximum power is transferred at resonance. 

or 

QUESTION: 23

Consider the following two statements regarding a linearly polarised, plane electromagnetic wave:

(A) The electric field and the magnetic field have equal average values.
(B) The electric energy density and the magnetic energy density have equal average values  

Solution:

QUESTION: 24

Which of the following rays is emitted by a human body?  

Solution:

IR because temperature of the body is 37°C and it emits heat radiation which falls in IR and microwave region.

QUESTION: 25

In a permanent magnet at room temperature  

Solution:

At room temperature permanent magnet behaves as a ferromagnetic substance for a long period of time. At room temperature, the permanent magnet retains ferromagnetic property for a long period of time.

The individual atoms in a ferromagnetic material possess a dipole moment as in a paramagnetic material.

However, they interact with one another in such a way that they spontaneously align themselves in a common direction over a macroscopic volume called domain. Thus, we can say that in a permanent magnet at room temperature, domains are all perfectly aligned.

QUESTION: 26

Consider the two idealized systems:

(i) a parallel plate capacitor with large plates and small separation and
(ii) a long solenoid of length L >> R, radius of cross-section.
In (i) E is ideally treated as a constant between plates and zero outside.
In (ii) magnetic field is constant inside the solenoid and zero outside.

These idealised assumption, however, contradict fundamental laws as below 

Solution:

Key concept: The electrostatic field lines, do not form a continuous closed path (this follows from the conservative nature of electric field) while the magnetic field lines form the closed paths.

Which implies that number of magnetic field lines entering the Gaussian surface is equal to the number of magnetic field lines leaving it. Therefore case (ii) is not possible.

QUESTION: 27

The self inductance L of a solenoid of length l and area of cross-section A, with a fixed number of turns N increases as  

Solution:

Key concept: The self inductance L of a solenoid depends on various factor like geometry and magnetic permeability of the core material.
L = μr μn2 Al
where, n = N/l (no. of turns per unit length)
1.   No. of turns: Larger the number of turns in solenoid, larger is its self inductance.
2.  Area of cross section: Larger the area of cross section of the solenoid, larger is its self inductance.
3.  Permeability of the core material. The self inductance of a solenoid increases μr times if it is wound over an iron core of relative permeability μr.
The long solenoid of cross-sectional area A and length l, having A turns, filled inside of the solenoid with a material of relative permeability (e.g., soft iron, which has a high value of relative permeability) then its self inductance is L = μrμ0 N2 A/l
So, the self inductance L of a solenoid increases as l decreases and A increases because L is directly proportional to area and inversely proportional to length.
Important point: The self and mutual inductance of capacitance and resistance depend on the geometry of the devices as well as permittivity/ permeability of the medium.  

QUESTION: 28

A circular coil is being expanded radially in a region of magnetic field and no electromotive force is produced in the coil. This can be because  

Solution:

Key concept: As we know whenever the number of magnetic lines of force (magnetic flux) passing through a circuit changes an emf is produced in the circuit called induced emf. The induced emf persists only as long as there is a change or cutting of flux.
The induced emf is given by rate of change of magnetic flux linked with the circuit, i.e., e = -dФ/dt
According to the problem there is no electromotive force produced in the coil. Then the various arrangement are to be thought of in such a way that the magnetic flux linked with the coil does not change even if the coil is placed and expanded in magnetic field.
When circular coil expands radially in a region of magnetic field such that the magnetic field is in the same plane as the circular coil or we can say that direction of magnetic field is perpendicular to the direction of area (increasing) so that their dot product is always zero and hence change in magnetic flux is also zero.

Or
The magnetic field has a perpendicular (to the plane of the coil) component whose-magnitude is decreasing suitably in such a way that the dot product of magnetic field and surface area of plane of coil remain constant at every instant.

QUESTION: 29

For an L CR  circuit, the power transferred from the driving source to the driven oscillator is then the INCORRECT option is 

Solution:

QUESTION: 30

Biot-Savart law indicates that the moving electrons (velocity v) produce a magnetic field B such that  

Solution:
QUESTION: 31

A conducting wire AB bent in the form of a parabola y2 = 2x carries a current i = 2 A as shown in figure. This wire is placed in a uniform magnetic field  Tesla. The magnetic force (in newton) on the wire is  

Solution:

QUESTION: 32

A metal ring of radius r = 0.5 m with its plane normal to a uniform magnetic field B of induction 0.2 T carries a current i = 100 A. The tension in newtons developed in the ring is  

Solution:

T = RBI
= (0.5) (0.2) (100) = 10N. 

QUESTION: 33

Consider six wires coming into or out of the page, all with the same current. Rank the line integral of the magnetic field (from most positive to most negative) taken counterclockwise around each loop shown. 

Solution:

QUESTION: 34

An electron is projected with velocity v0 in a uniform electric field E perpendicular to the field. Again it is projected with velocity v0 perpendicular to a uniform magnetic field B. If r1 is initial radius of curvature just after entering in the electric field and r2 is initial radius of curvature just after entering in magnetic field then the ratio r1 / r2 is equal to

Solution:

QUESTION: 35

Electrons moving with different speeds enter a uniform magnetic field in a direction perpendicular to the field. They will move along circular paths.  

Solution:

QUESTION: 36

A particle having charge of 1C, mass 1 kg and speed 1 m/s enters a uniform magnetic field, having magnetic induction of 1T, at an angle θ = 30° between velocity vector and magnetic induction. The pitch of its helical path is (in meters) 

Solution:

QUESTION: 37

A long straight wire carries a current along the x-axis. Consider the points A(0, 1, 0), B(0, 1, 1), C (1, 0, 1) and D(1, 1, 1). Which of the following pairs of points will have same magnetic fields? 

Solution:

QUESTION: 38

Considering magnetic field along the axis of a circular current carrying loop of radius R, approximately at what distance from the centre of the loop the field is  times of its value at the centre? 

Solution:


QUESTION: 39

A particle enters the region of a uniform magnetic field as shown in figure. The path of the particle inside the field is shown by a dark line. The particle is

Solution:
QUESTION: 40

A wire loop carrying current i is placed in the x–y plane. Magnetic induction at P is 

Solution:

QUESTION: 41

A :  when frequency is greater than resonance frequency in a series LCR circuit, it will be an inductive circuit.

R : Resultant voltage will lead the current

Solution:
QUESTION: 42

A : For a charged particle describing uniform circular motion in a magnetic field T2 ∝ r3 (where T is time period of resolution and r is radius of the circle)

R : The rel ation T2 ∝ r3 is valid only when where f is force acting on the particle. 

Solution:
QUESTION: 43

A : A charged particle is moving from origin in the region in which uniform magnetic field exists in zdirection and a uniform electric filed in y-direction. At a point p(x, y, z) speed of the particle is v. then v will depend on y only.

R : In combined electric field and magnetic fields work is done by electric field only.

Solution:
QUESTION: 44

A : A current carrying conductor produces only an electric filed

R : When a observer is put in the same frame as a moving charge, the observer will find no magnetic field associated with the charge. 

Solution:
QUESTION: 45

A : The phenomenon of electromagnetic induction in accordance with Lenz's law represents the conservation of energy

R : Lenz's law give the same result as we define by the sign rules in Faraday's law.

Solution:

Energy can enter or leave, but not instantaneously. Lenz' law is a consequence. As the change begins, the law says induction opposes and, thus, slows the change.The fact that electromagnetic induction in accordance with Lenz's law represents the conservation of energy can be easily explained.

QUESTION: 46

What is major product of following reaction? 

Solution:
QUESTION: 47

Which of the following compounds will have zero dipole moment?  

Solution:
QUESTION: 48

The carbon –carbon bond length in the compounds

follows the order 

Solution:

As the bond strength increases, the bond length decreases. As we know, bond strength is inthe order of triple bond>double bond>single bond. The C-C bond length of benzene is intermediate between that of double bond and single bond due to resonance. Therefore in benzene, each C-C bind has a partial double bond character. Hence the bond is stronger than single bond and weaker than double bond
Hence here the answer comes B option

QUESTION: 49

In the following compounds, 

The order of basicity is 

Solution:

In IV lone pair is involved in resonance. 

QUESTION: 50

Solution:

QUESTION: 51

Consider the reaction given below: 

Which of these is true regarding the reaction shown above?  

Solution:

From just viewing the compound we cannot tell whether the compound is dextro or levo isomers, so c and d options are obviously incorrect. Also since, cyanide group is converted into carboxylic acid group therefore the configuration of the chiral carbon has not changed, so first option is the correct answer

 

QUESTION: 52

Which of these is the correct fischer projection for the wedge and dash formula given below? 

Solution:
QUESTION: 53

Which of the following will be a product of the reaction

Solution:

As redP,HI, converts carbonyl group into CH2 group, therefore, ethane will be formed from ethanoyl chloride.

QUESTION: 54

Solution:
QUESTION: 55

Which of the following happens to be the carbanion with highest stability? 

Solution:

Option C is correct becz there is a strong electron withdrawing group -NO2 is at para position which decrease electron density at CH2(-) and hence increase stability.

QUESTION: 56

The IUPAC name of the following compound 

Solution:
QUESTION: 57

Which of the following molecule is (are) chiral?

 

Solution:

QUESTION: 58

Which of the following indicates the relative percentage the formation of 2-bromopropane on bromination of propane. The relative reactivities of 1° : 2° : 3° hydrogens to bromination is   1 : 82 : 1600

Solution:
QUESTION: 59

Trans-2-butene on bromination using CS2 gives which dibromoderivative?

Solution:

See its anti addition so meso compound will be formed due to presence of POS so option D is correct

QUESTION: 60

R1CH=CHR+ C6H5COOOH → (A) . Product (a) is 

Solution:
QUESTION: 61

Solution:
QUESTION: 62

The most reactive of these compounds towards sulphonation is

Solution:

Yes, m-xylene is more reactive toward sulphonation because it contain two methyl group, meta position to each other. This is the only reason which will increase electron density at ortho and para position and we know that high electron density at ortho and para is favourable for sulphonation. Hence among all these options d is correct.

QUESTION: 63

Product (b) is  

Solution:
QUESTION: 64



 

Gas X is same is 

Solution:
QUESTION: 65

3-phenyl 1-butene is hydrolyzed (water addition) in the presence of small amount of HCl, the major product formed for the reaction should be

Solution:
QUESTION: 66

Oxymercuration of 1-phenyl acetylene leads to formation of

Solution:

Add H2O without rearrangement(for OMDM)

QUESTION: 67

Where the energy barrier for interconversion of staggered eclipsed conformation for ethane is 12.55 kJ/mol, it is 45 kJ/mol for C2Cl6 . Which of these correctly explains this? 

Solution:
QUESTION: 68

The product obtained on heating n-heptane with Cr2O3-Al2O3 at 600°C is

Solution:

CH3CH2CH2CH2CH2CH2CH3 

QUESTION: 69

The ease of dehydrohalogenation of alkyl halide with alcoholic KOH is

Solution:

The more stable intermediate 3° carbocation is formed more easily than 2° and 2° is formed faster than 1°. Hence, the ease of dehydrohalogenation follows the order 3° > 2° > 1°. 

QUESTION: 70

The intermediate during the addition of HCl to propene in the presence of peroxide is 

Solution:

HCl does not follow Anti Markownikoff's rule. So, it gives carbocation as intermediate.

QUESTION: 71

Which of the following carbocations is expected to be most stable? 

Solution:

Here in the given problem, carbocation (b) is stabilized most, as the CH3 (e releasing group) group is attached to the carbon bearing positive charge.

QUESTION: 72

A group which deactivates the benzene ring towards electrophilic substitution but which directs the incoming group principally to the o- and p-position is  

Solution:

Electron accepting groups which make the substitution difficult are known of deactivating groups. the group or substituent already present on the ring also decides the position of incoming group. ortho and para directing groups are as follow 

−CH3,C2H5(−R),−NH2,−OH, halogens, 

(Cl,Br,I)

QUESTION: 73

In the given conformation C2 is rotated about C2 − C3 bond anticlockwise by an angle of 120° then the conformation obtained is 

Solution:
QUESTION: 74

Which of the following is formed by the thermal decomposition of the hydroxide of  

Solution:

QUESTION: 75

Fish die in water bodies polluted by sewage due to  

Solution:

We need air to breathe, aquatic organisms need dissolved oxygen to respire. It is necessary for the survival of fish, invertebrates, bacteria, and underwater plants. DO is also needed for the decomposition of organic matter. That is why fish die in water bodies polluted by sewage due to reduction in dissolved oxygen.

QUESTION: 76

Which order for basic character of amine is correct for following compounds? 

Solution:

Electron releasing group increasing basic character of amine and aromatic amine is less basic than aryl and aliphatic amine 

QUESTION: 77

Which of the following is the most stable carbocation  

Solution:

is the most stable carbocation 

QUESTION: 78

In the following compounds 

The order of acidity is 

Solution:

NO2 is an electron attracting group where as —CH3 is an electron releasing group. An electron attracting substituent tends to disperse the negative charge of phonoxide ion and thus makes it more stable. This in turn increase the acidity of phenol. The substituent in para-position is more effective than in meta-position as the former involves a resonating structure bearing negative charge on the carbon attached to electron with drawing substituent. An electron releasing substituent tends to increase the negative charge of phenoxide ion and thus makes it more unstable. This, in turn, decrease the acid strength of phenol. 

QUESTION: 79

What is the decreasing order of stability of the ions? 

Solution:

-OCH3 group increases the stability of carbocation and >C=O decreases the stability of carbocation. So, order of stability.

II>I>III

QUESTION: 80

Match the following reactions with the corresponding reaction intermediates 



Solution:
QUESTION: 81

In the above compound, arrange the nitrogen according to their decreasing basic strength 

Solution:

The lone pair of e-s N(3) does not participate in resonance. So, it is the most basic and lone of e-s of N(2) participate in the ring resonance. So, it is least basic.

QUESTION: 82

 The product ‘‘A’’

Solution:

2 radicals are formed but intramolecular reaction is always more favored than inter. So option c will be the major product.

QUESTION: 83

The most stable alkene is 

Solution:

The more alpha hydrogens are present i.e 10

QUESTION: 84

DDT is  

Solution:

DDT, plastics, polythene, bags, insecticides, pesticides, mercury, lead, arsenic, metal articles like aluminum cans, synthetic fibres, glass objects, iron products and silver foils are non-biodegradable pollutants.

QUESTION: 85

(major product) The product Y is

Solution:
QUESTION: 86

A : In na phthalene  the electrophilic attack on indicated position 1 is more hindered so less stable intermediate is formed hence it takes place at position 2.

R : The electrophile attacks on the position which gives more stable intermediate 

Solution:

In the electrophilic substitution , position 1 in naphthalene is more reactive than the position 2 because the carbonation formed by attack of electrophile at position 1is more stable than position 2 because of resonance .Hence both the statement are wrong.

QUESTION: 87

A : Enol fo rm of cy clohexane-1,3,5-trione is more stable than its keto form.

R : It co ntains α- hydrogen atoms. 

Solution:

Enol form of cyclohenxane-1, 3, 5-trione is stabilized by resonance energy. 

QUESTION: 88

A :  Tertiary carbonium ions are generally formed more easily than primary carbonium ions.
R : Hyperconjugative as well as inductive effect due to additional alkyl groups stabilize tertiary carbonium ions. 

Solution:

An alkyl group attached to the positively charged carbon of a carbonium ion tends to release electrons towards that carbon; thus the positive charges gets dispersed as the alkyl group becomes some what positively charged itself. More the number of alkyl groups, the greater is the dispersal of positive charge and therefore more easily it will be formed. 

QUESTION: 89

A : Iodination of alkanes is reversible.
R :  Iodination is carried out in presence of iodic acid. 

Solution:

Iodination is reversible since formed HI is a strong reducing agent and reduces the alkyl iodide back to alkane. 

CH4 + I2 ⇌ CH3I + HI

QUESTION: 90

A:  The reaction given below mainly takes place by path I rather than by path II. 

R :  Organo metallic are strong bases, tertiary halides undergo elimination with such strong bases.  

Solution:

Organometallic compounds are bulky groups therefore they favour elimination rather than substitution.

 

QUESTION: 91

The following pedigree depicts. the inheritance of an X-linked recessive trait 

If the mother shows the affected phenotype and the father is phenotypically normal but having an unknown genotype then: 

Solution:

If female is affected by X - linked recessive trait then her son will be affected by this disease because male has only one X - chromosome that comes from mother. Thus the answer of it is only offspring 1 A that would be affected due to criss-cross inheritance. 

QUESTION: 92

The semi-dwarf variety of rice was introduced in year?

Solution:

IR8 is a high-yielding semi-dwarf rice variety developed by the International RiceResearch Institute (IRRI) in the early 1960s. In November 1966, “the variety was first introduced in the Philippines and India.”

QUESTION: 93

In modern times plant breeding is mainly carried out by ?

Solution:

The advancement in tissue culture, genetics, and molecular biology, plant breeding is carried out by using molecular genetic tools. 

QUESTION: 94

Out of the various objectives of plant breeding, main objective is ?

Solution:

Main aim is increase yield and improve quality. 

QUESTION: 95

Which of the following step is most crucial for the success of plant breeding

Solution:

It is selection and testing of superior recombinants which determines overall success of plant breeding. 

QUESTION: 96

Before the release of new variety developed through plant breeding it is tested in farmers field for how many growing seasons ?

Solution:

It is cultivated for three growing seasons in different agroclimatic zones and its performance is compared with best local cultivar. 

QUESTION: 97

How many of the following statements are correct ?

I. In India agriculture accounts for 33% of GDP and 62% of population is dependent upon it.
II. Between 1960 - 2000, wheat production increased from 11 - 75 million ton and rice from 35 - 89.5 million tons.
III. Semi-dwarf variety of wheat is developed at international center for wheat and maize improvement.
IV. Jaya and Ratna are semi-dwarf variety of rice. 

Solution:
QUESTION: 98

Hybrid variety of maize, jowar and bajra which are high in yield and resistance to

Solution:
QUESTION: 99

Conventional breeding for disease resistance is often constrained by?

Solution:
QUESTION: 100

Cowpea variety resistance to bacterial blight is?

Solution:
QUESTION: 101

Which of the following gene in operon expressed constitutively ?

Solution:

The repressor of the operon is synthesized all the time. 

QUESTION: 102

Which of the following feature do not provide resistance to maize against stem borers

Solution:
QUESTION: 103

Which of the following vegetable crop is rich in vitamin A as well as calcium and iron developed by IARI.  

Solution:

Spinach is a superfood. It is loaded with tons of nutrients in a low-calorie package. Dark, leafy greens like spinach are important for skin, hair, and bone health. They also provide protein, iron, vitamins, and minerals.

QUESTION: 104

Inheritance of a character is also affected by?

Solution:

The promoter and regulatory sequences are the part of the structure of a gene. Here, the promoter sequences are present at the beginning of the gene and can regulate the expression of that particular gene. In common, each gene has its own promoter and regulatory sequences and these sequences are non coding DNA sequences which helps in maintaining the expression of a gene. The characters are governed by some particular gene and inherit from one generation to another. So by knowing this, we can say that inheritance of character is also affected by both promoter and regulatory sequences.

QUESTION: 105

Out of the following four statements how many of them are correct?

(I)  Two nucleotides are joined by 3′ - 5′ phosphodiester linkage.
(II) Chromatin which is a beads – on string structure, where beads represents histone  protein.  
(III) Euchromatin is transcriptionally active whereas heterochromatin is inactive.
(IV) Histone proteins has amino acids with positive charge on side chains.  

Solution:

The major difference between heterochromatin and euchromatin is thatheterochromatin is such part of the chromosomes, which is a firmly packed form and are genetically inactive, while euchromatin is an uncoiled (loosely) packed form of chromatin and are genetically active.

QUESTION: 106

Which of the following inference cannot be drawn from Griffith’s experiment of transformation.  

Solution:

Biochemical nature of genetic material was established by Avery, Macleod & Mc carty.  

QUESTION: 107

Which of the following is not the property of a molecule that can act as genetic material.  

Solution:

It should express itself as “Mendelian characters” 

QUESTION: 108

Distribution of newly synthesized DNA in the chromosomes is by semi-conservative means is experimentally shown by Taylor et al by using  

Solution:
QUESTION: 109

During DNA replication, two strands of DNA cannot be separated in its entire length in one step because?  

Solution:

DNA replication is semiconservative, meaning that each strand in the DNA double helix acts as a template for the synthesis of a new, complementary strand. This process takes us from one starting molecule to two "daughter" molecules, with each newly formed double helix containing one new and one old strand.

QUESTION: 110

If due to mutation both the strands of DNA starts transcribing. Which of the following will not be the consequence of this?  

Solution:

Polyploidy results when the genome doubles or triples or quadruples, etc. If both strands start transcribing it will form complement mRNA strands affecting further translation. The ploidy of the given cell will remain constant.

QUESTION: 111

Study the diagram given below, and based on that find out which of the following statement is incorrect. 

Solution:

Regulatory sequences position is with respect to promoter and not terminator. Inheritance of a character is also affected by promoter and regulatory sequences of a structural gene, because of this it is also referred as regulatory genes. 

QUESTION: 112

Study the diagram given below 

A, B, C and D is marked, which correspond to certain functions which of the following is not matched correctly  

Solution:
QUESTION: 113

Which of the following scientist is responsible for the synthesis of protein in cell-free system?

Solution:
QUESTION: 114
  1. Read the following four statements (A−D)

A. The characters blend in heterozygous condition.
B. Change in a single base pair of DNA does not cause mutation.
C. Cancer cells commonly show chromosomal aberrations.
D. In insect, sex chromosomes in male are ZZ and in females are ZW.

How many of the above statement is/are right? 

Solution:

Characters never blend, change in single base causes mutations in insect XX/XO mechanism of sex determination is present, cancer causing cell are called nodule or mass cell.

Option D is correct because insect are female heterogamety in these where male will be having ZZ chromosomes where female will be having ZW chromosomes

QUESTION: 115

A female is haemophilic because mother of female is  

Solution:

Suppose the X chromosome from her mother has the gene for normal blood clotting. Suppose the X chromosome from herfather has the gene for hemophilia. The daughter is called acarrier for hemophilia. She has the gene on one of her X chromosomes and could pass it on to her children.

QUESTION: 116

How many of the following statements is / are correct for polygenic inheritance?

(1) They show uniformity.
(2) Controlled by three or more genes
(3) It is not influenced by environment.
(4) In polygenic inheritance phenotype reflects the contribution of dominant allele only.  

Solution:
QUESTION: 117

Which is not true for haplodiploid sex determination?  

Solution:

Sperm is produced by male through mitosis. 

QUESTION: 118

The deletion in chromosome number 16 results in which of the following genetic disorder?

Solution:
QUESTION: 119

Conditions of a karyotype 2n ± 1 is observed in all except  

Solution:

Phenyl ketonuria is a autosomal disorder due to point mutation. 

QUESTION: 120

Aneuploidy or euploidy results due to mainly  

Solution:
QUESTION: 121

Which of the following is a set of traits with continuous variation?

Solution:

Skin colour is controlled by 3 pairs of polygenes and height is controlled by 10 pairs of polygenes which are non allelic dominant genes for continious variation. 

QUESTION: 122

If t-RNA isolated from E.coil, and m-RNA as well as ribosomes isolated from mouse liver cell are incubated with ATP and free amino acids, what may happen?

Solution:

Protein specific to mouse will be synthesis.Because , when all this t - RNA , m - RNA & ribosomes are provided correct medium & sufficient energy they will go for preparation of protein , by the process of. Translation.In which t - RNA is bind in ribosomes & m - RNA is decoded according to code on m - RNA amino acid are taken by t - RNA & joined to form chain of polypeptide to synthesis protein. So , Protein is formed according to m - RNA. 

QUESTION: 123

Single nucleotide polymorphism which occurs in human genome helps in?

Solution:
QUESTION: 124

Consider the following human disorders:

i. Haemophilia  
ii. Down’s Syndrome
iii. Cystic fibrosis
iv. Colour blindness
v. Night blindness.

Which of these disorders exhibit ‘Mendelian’ pattern of inheritance?

Solution:
QUESTION: 125

A plant with genotype AABbCcDD is self pollinated. Provided that the four genes are independently assorting, what proportion of the progeny will show the genotype AAbbccDD ?

Solution:

1×1/4×1/4×1 so we are getting =1/16

QUESTION: 126

Like sickle cell anaemia, which is the other genetic disorder related to blood pigment ?

Solution:

Sickle cell disease and thalassemia are genetic disorders caused by errors in the genes for hemoglobin, a substance composed of a protein ("globin") plus an iron molecule ("heme") that is responsible for carrying oxygen within the red blood cell.

QUESTION: 127

To initiate transcription, RNA polymerase binds to the :

Solution:

Promoter sequences are DNA sequences that define where transcription of a gene by RNA polymerase begins. Promoter sequences are typically located directly upstream or at the 5' end of the transcription initiation site.

QUESTION: 128

Which of the following sequences of mRNA will not translate completely?  

Solution:

The seq uence 5′  AUG AAC UAA CCA CUC3. has stop codon UAA in the middle, that will prevent further translation. 

QUESTION: 129

The figure illustrates the process of translation in protein synthesis. If the triplet UAU is modified to UAG what will be the consequence? 

Solution:

If UAU is modified to UAG that is stop codon then synthesis of polypeptide chain will stop at f chain. Thus the polypeptide with a, b, c, d, e and f will be released 

QUESTION: 130

Which of the following element is NOT involved in the information transfer from DNA to finished protein?

Solution:

DNA polymerase is involved in Replication of DNA, not in the transcription (synthesis of mRNA) & translation. 

QUESTION: 131

A : During transcription DNA strand with 5’-3’ polarity is referred as coding strand.

R : C odi ng strand codes for RNA

Solution:
QUESTION: 132

A :  Tendency of parental combinations to remain together is called crossing over.  
R : C rossing over provides a proof in favour of linear arrangement of genes on chromosomes.  

Solution:
QUESTION: 133

A : Germplasm collection is the root for any plant breeding programme.
R : W ild relatives of several cultivated plant are used in plant breeding. 

Solution:
QUESTION: 134

A  : DNA is preferred over RNA as genetic material.
R : RNA  has – OH group at 2 – C of pentose sugar. 

Solution:

Presence of extra – OH group makes RNA more reactive.  

QUESTION: 135

A :  Expressed sequence tags (ESTS) which were used in human genome project, it identifies all the genes that are expressed as DNA.
R : Complete DNA sequence is decoded. 

Solution:

In ESTS sequencing of coding part of DNA that express itself in the form of RNA is done but in question it is given that expressed as DNA that's why answer is d

QUESTION: 136

Mark the incorrect statement?

Solution:
QUESTION: 137

From the evolutionary point of view, mammals were originated from  

Solution:
QUESTION: 138

During evolution of human being, first hominids were  

Solution:

The first early hominid from Africa, the Taung child, as it was known, was a juvenile member of Australopithecus africanus, a species that lived one million to two million years ago, though at the time skeptical scientists said the chimpanzee-size braincase was too small for a hominid.

QUESTION: 139

Which of the following have been found to be very effective as emergency contraceptives as they could be used to avoid possible pregnancy due to rape if

A. Administration of progesterone
B. Progesterone estrogen combination
C. IUDs  
D. Condoms

Solution:
QUESTION: 140

Variation due to mutations are

Solution:
QUESTION: 141

With the given diagram, it can be analyzed that 

(1) In Man, Cheetah, Whale and Bat all have pentadactyl limb plan
(2) All explain the mechanism of adaptive radiation
(3) During evolutionary processes, bat evolved from birds
(4) Limbs pattern of whale and bat explain mechanism of convergent evolution 

Solution:
QUESTION: 142

Mark the incorrect statement regarding infection of AIDS virus  

Solution:

To convert RNA into DNA, retroviruses must contain genes that encode the virus-specific enzyme reverse transcriptase, which transcribes an RNA template toDNA. The fact that HIV produces some of its own enzymes not found in the host has allowed researchers to develop drugs that inhibit these enzymes.

QUESTION: 143

Mark the incorrect in the table? 

Solution:
QUESTION: 144

Secondary immune response is always intensified than primary. This is due to  

Solution:

The primary immune response of the body to antigen occurs on the first occasion it is encountered. The secondary response of both B‐ and T cells is observed following subsequent encounter with the same antigen and is more rapid leading to the activation of previously generated memory cells.

QUESTION: 145

Mark the wrong match? 

1. Active immunity →  both specific and non-specific.
2. Passive immun ity → always non-specific
3. Allergy → IgE  mediated exaggeragated immune response
4. Auto- immunity → Problem related to memory based acquired immunity 

Solution:
QUESTION: 146

Vaccination is

(1) based on the property of memory immune system
(2) a prepa ra tion of antig en ic proteins of pathogen
(3) generate memory B and T cells that recognize the pathogen
(4) provides active immunity to the body 

Solution:
QUESTION: 147

Pick Odd one out?  

Solution:

Thymus is primary limphoid organ while other are secandary limphoid organ.

 

QUESTION: 148

Mark the wrong statement regarding the bone marrow?

Solution:
QUESTION: 149

Mark the incorrect statement about ZIFT  

Solution:

The zygote or early embryos (with upto 8 blastomeres) could then be transferred into the fallopian tube (ZIFT–zygote intra fallopian transfer) and embryos with more than 8 blastomeres, into the uterus (IUT – intra uterine transfer), tocomplete its further development.

QUESTION: 150

Mark the wrong statement?

Solution:

The correct answer would be C as MRI uses strong magnetic field but *non ionising* or *non invasive* technique.

QUESTION: 151

Which of the following side effects in male is not related to use of narcotic analgesics?  

Solution:

Some studies have suggested that larger testicles among some mammals are associated with higher testosterone levels, while smaller testicular volume is associated with decreased sperm production. smaller-than-average testicles. breast tissue growth.

QUESTION: 152

‘Vaccines’ are available for  

Solution:

Most vaccines against bacterial infections are effective at preventing disease, reactions can occur after vaccinations. Vaccines are available against tuberculosis, diphtheria, tetanus, pertussis, Haemophilus influenzae type B, cholera, typhoid, and Streptococcus pneumoniae.

QUESTION: 153

Odd one out?  

Solution:
QUESTION: 154

Periodic abstinence is a method in which the couples avoid coitus from  

Solution:

periodic abstinence. A method of birth control in which a couple tries to avoidpregnancy by refraining from sexual intercourse during certain times within the menstrual cycle.

QUESTION: 155

‘Abused Drugs’ are obtained from

(1) Opioids plants
(2) Cannabinoids plants
(3) Coca alkaloids
(4) Fungi

Solution:
QUESTION: 156

Mark the wrong statement regarding ‘crack’?  

Solution:
QUESTION: 157

Syptoms like internal bleeding muscular pain fever anemia and blockage of intestinal passage can be related to  

Solution:
QUESTION: 158

‘Dysentry’ like symptoms occurs due to  

Solution:
QUESTION: 159

Odd one out

Solution:

Polio is odd one from all these options because polio is a viral disease caused by poliovirus while all those three options are bectiral disease. 

QUESTION: 160

All the finches in the Galapagos Island  

Solution:
QUESTION: 161

Mark the correct statements for natural killer cells
(a) Component of innate immunity
(b) A type of monocy te
(c) Belongs to cellular barrier
(d) Secrete interfer on

Solution:
QUESTION: 162

In life cycle of ‘Plasmodium Vivax’ which of the following stages and process occurs in mosquito only

(1) Sporoz oite
(2) Gamet ocyte
(3) Fertilization of male and female gamete of parasite
(4) Asexual reproduction of plasmodium 

Solution:
QUESTION: 163

In what sense AIDS can be distinguished from cancer?

(1) Infectious and non-infectious disease
(2) Bacterial and viral disease
(3) Trigger cell mediated and antibody mediated immune response
(4) Curable and non-curable diseases 

Solution:

Since AIDS is a infectious disease because it can be transmitted from one person to other through saliva, serum, blood , mucus and any body fluid while in case of cancer transmission is not possible. Talking about cure we know that till date AIDS is non curable disease. Nowday there are lot of options available for treatment of cancer like radiotherapy, chemotherapy, e.t.c. AIDS is actually syndrome which is collection of no. Of disease, harming our immune by every mean that's why its treatment is not known.

QUESTION: 164

Among the following methods which has the highest failure rate?  

Solution:

The rhythm method, also called the calendarmethod or the calendar rhythm method, is a form of natural family planning. To use the rhythm method, you track your menstrual history to predict when you'll ovulate. This helps you determine when you're most likely to conceive.

QUESTION: 165

Which of the following are hormone releasing IUD’s?

(1) Multiload -375
(2) LNG-20
(3) Progestasert
(4) CuT 

Solution:
QUESTION: 166

In the finding of population related to industrial melanism 4% of the moths are homozygous dominant, 32% are heterozygous and 64% are homozygous recessive. From this observation, what will be the frequency of dominant allele

Solution:

Homozygous dominant individuals=P^2=4/100....
P=√4/100.
P=2/10.
P=0.2.
thus ,dominant allele( p) is 0.2

QUESTION: 167

From the evolutionary history of vertebrates, mammals were originated from  

Solution:

Synapsids evolved from basal amniotes and are one of the two major groups of the later amniotes; the other is the sauropsids, a group that includes modern reptiles and birds. The distinctive temporal fenestra developed in the ancestral synapsidabout 312 million years ago, during the Late Carboniferous period.

QUESTION: 168

During evolution of plant, which of the following is most recently evolved?

Solution:
QUESTION: 169

Microbial experiments show that pre-existing advantageous mutations when selected will result in observation of new phenotypes this statement is given

(1) In favour of natural selection theory
(2) Genetic basis of adaptation
(3) Hugo de Vries mutation theory
(4) To explain evolution as a stochastic process 

Solution:
QUESTION: 170

It is evident that first life arose in ocean, but as complexity in organisms appears first multicellular forms migrated to land were

Solution:
QUESTION: 171

Mark the correct statement?

(1) Fossil record says that, for thousand of year, Neanderthal man and modern Homo sapiens lived together, in different parts of world.
(2) Homo  sapiens arose in Africa and moved across continents and developed into distinct races.
(3) Comparison of human and Chimpanzee skull, revealed that, skull of baby chimpanzee is more like adult human skull than adult chimpanzee skull
(4) Homo  erectus were the first human with bipedal locomotion 

Solution:
QUESTION: 172

A human ancestor with bipedal locomotion, meat eating habit and brain capacity more than 800cc, can be

(1) Austra lopithecus
(2) H omo habilis
(3) Neanderthal man
(4) Homo erectus

Solution:
QUESTION: 173

It is evident that the greatest number of different mitochondrial DNA sequences occur among modern Africans. This evidence favors that 

Solution:
QUESTION: 174

Which of the following observation is not correct to explain the mechanism of natural selection  

Solution:

In nature, population growth must eventually slow, and population size ceases toincrease. At carrying capacity, because population size is approximately constant, birthrates must equal death rates, and population growth is zero.

QUESTION: 175

Darwin finches is an excellent example of adaptive radiation. As different finches arise from a common stock of birds, and they were probably  

Solution:
QUESTION: 176

A :  Nature selection for fitness.
R : Fitness is the end result of the ability to adapt and get selected by nature. 

Solution:

“Fitness” is the end result of the ability to adapt and get selected by natureand hence called, “Survival of the fittest through natural selection. Fitness is the potentiality of the selection and the organism's response to it through genetic means.

QUESTION: 177

A :  Contraceptive pills Saheli is very effective with very few side effects.
R : It simply alter the quality of cervical mucus to prevent or retard entry of sperms

Solution:

Saheli prevents the implantation but not fertilization

QUESTION: 178

A :  Cell mediated immunity induces humoral immunity
R :  T cells themselves do not secrete antibodies but help B cells produced them 

Solution:
QUESTION: 179

A :  Cocanie producing a sense of euphoria and increased energy.
R : It interfere with the transport of the neuro-transmitter dopamine 

Solution:
QUESTION: 180

The following is a simplified scheme showing the fact of glucose during aerobic and anaerobic respiration. Identify the end products that are formed at stages indicated as A, B, C and D. identify the correct option from those given below.

Solution:

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