NEET-UG 2017 Question Paper Solutions

Let dimensions of length is related as,

[L] = [c]ˣ [G]ʸ [ e²/4πϵ₀ ]ᶻ

⇒ e²/4πϵ₀ = [ML³T⁻²]

[L] = [LT⁻¹]ˣ [M⁻¹L³T⁻²]ʸ [ML³T⁻²]ᶻ

[L] = [Lˣ⁺³ʸ⁺³ᶻ M⁻ʸ⁺ᶻ T⁻ˣ⁻²ʸ⁻²ᶻ]

Comparing both sides

−y + z = 0 ⇒ Z ...(i)

x + 3y + 3z = 1 ...(ii)

−x − 4z = 0 (∵ y = z) ...(iii)

From (i), (ii) and (iii)

z = y = −1/2 ; x = −2

Hence,

[L] = c⁻² [ G · e²/4πϵ₀ ]¹ᐟ²

Let the distance be ‘d’ time taken by Preeti to travel up
the stationary escalator = t₁

Velocity of Preeti w.r.t. elevator

v₁ = d/t₁

Since the distance is same let the time taken when Preeti
stands on the moving escalator = t₂.

Velocity of elevator w.r.t. ground

v₂ = d/t₂

The net velocity of Preeti w.r.t. ground

v = v₁ + v₂

d/t = d/t₁ + d/t₂

1/t = 1/t₁ + 1/t₂

t = t₁t₂/(t₁ + t₂)

(time taken by Preeti to walk up the
moving escalator)

Given:

x = 5t − 2t² y = 10t

vₓ = dx/dt = 5 − 4t
vᵧ = dy/dt = 10

aₓ = dvₓ/dt = −4 aᵧ = dvᵧ/dt = 0

𝐚⃗ = aₓî + aᵧĵ

𝐚⃗ = −4î m/s²

Hence, acceleration of particle at (t = 2 s) = −4 m/s²

Before cutting the string

kx = T + 3mg ...(i)

T = mg ..(ii)

Force on spring,

⇒ kx = 4mg

After cutting the string T = 0

For block A:

3ma_A = 4mg − 3mg

a_A= (4mg − 3mg)/3m = g/3

a_A = g/3

For block B:

ma_B= mg

a_B = g

Net force on particle in uniform circular motion is centripetal force (mv²/ℓ) which is provided by tension in string so the net force will be equal to tension i.e., T.

Here, Iω₁ + |ω₂ = 2Iω

⇒ ω = (ω₁ + ω₂)/2

(K.E.)ᵢ = 1/2 Iω₁² + 1/2 Iω₂²

(K.E.)𝒇 = 1/2 × 2Iω² = I[(ω₁ + ω₂)/2]²

Loss in K.E. = (K.E.)𝒇 ᵢ − (K.E.) ᵢ = 1/4 I(ω₁ − ω₂)²

Given, mass of cylinder m = 3 kg

R = 40 cm = 0.4 m

F = 30 N ; α = ?

As we know, torque τ = Iα

F × R = MR²α

α = (F × R)/(MR²)

α = (30 × 0.4)/(3 × (0.4)²)

or, α = 25 rad/s²

Both the astronauts are in the condition of ‘weightlessness. Gravitational force between them pulls towards each other. Hence Astronauts move towards each other under mutual gravitional force.

Above earth surface Below earth surface
gₕ = g (1 − 2h/Rₑ) g_d = g (1 − d/Rₑ)
According to question, gₕ = g_d
g (1 − 2h/Rₑ) = g (1 − d/Rₑ)
Clearly,
d = 2h = 2 km

Bulk modulus is given by
B = p/(ΔV/V)
or
ΔV/V = p/B
3 ΔR/R = p/B (here, ΔR/R = fractional decrease in radius)

⇒ ΔR/R = p/3B

Here, hₒᵢₗ × ρₒᵢₗ × g = h(water)subscript × ρ(water)subscript × g
ρ₀g × 140 × 10⁻³ = ρ_w g × 130 × 10⁻³
ρₒᵢₗ = 130/140 × 10³ ≈ 928 kg/m³

[∵ ρ_w = 1 kgm⁻³]

quivalent thermal conductivity of the composite rod in parallel combination will be,
K = K₁A(T₁ − T₂)/d + K₂A(T₁ − T₂)/d
K_EQ 2A(T₁ − T₂)/d = A(T₁ − T₂)/d [K₁ + K₂]
Hence equivalent thermal conductivities for two rods of
equal area is given by
K_EQ = (K₁ + K₂)/2


 

Given r₁ = 12 cm, r₂ = 6 cm
T₁ = 500 K and T₂ = 2 × 500 = 1000 K
P₁ = 450 watt
Rate of power loss P ∝ r²T⁴
P₁/P₂ = r₁²T₁⁴/r₂²T₂⁴
P₂ = P₁(r₂²T₂⁴)/(r₁²T₁⁴)
Solving we get,
P₂ = 1800 watt

Process I volume is constant hence, it is isochoric.
In process IV, pressure is constant hence, it is isobaric

Given, efficiency of engine,

η = 1/10

work done on system W = 10 J

Coefficient of performance of refrigerator,

β = Q₂/W = (1 − η)/η

= (1 − 1/10)/(1/10)

= (9/10)/(1/10)

= 9

Energy absorbed from reservoir,

Q₂ = βW

Q₂ = 9 × 10 = 90 J

Internal energy of the system is given by
U = f/2 nRT
Degree of freedom
F_diatomic = 5
F_monatomic₎ = 3
and, number of moles
n(O₂) = 2
n(Ar) = 4
U_total = 5/2 (2)RT + 3/2 (4)RT = 11RT

Given, Amplitude A = 3 cm
When particle is at x = 2 cm
According to question, magnitude of velocity = acceleration
ω√(A² − x²) = xω²
√(3)² − (2)² = 2(2π/T)
√5 = 4π/T
⇒ T = 4π/√5 


 

Let ℓ be the complete length of spring.
Length when cut in ratio 1 : 2 : 3 are
ℓ/6, ℓ/3 and ℓ/2
Spring constant (k) ∝ 1/length (ℓ)
Spring constant for given segments
k₁, 6k, k₂ =3k and  k₃ =2k
When they are connected in series
1/k′ = 1/6k + 1/3k + 1/2k
⇒ 1/k′ = 6/6k
∴ Force constant k′ = k
And when they are connected in parallel
k″ = 6k + 3k + 2k
⇒ k″ = 11k
Then the ratios
k /k″ =1/11 i.e.,  1 : 11

Difference in two successive frequencies of closed pipe, 2v/4l ⇒ 260 - 220= 40 Hz

or
2v/4ℓ = 40 Hz
⇒ v/4ℓ = 20 Hz
Which is the fundamental frequency of system of closed organ pipe.

As we known from Doppler’s Effect
fₐₚₚₐᵣₑₙₜ = f₀ [(v + v₀)/(v − vₛ)] = 400 [(340 + 16.5)/(340 − 22)]
fₐₚₚₐᵣₑₙₜ = 448 Hz

Net charge on one H atom = −e + (e + Δe) = Δe
According to question, the net electrostatic force (F_E) = gravitational force (F_G)
F_E = F_G
or
1/4πϵ₀ · Δe²/d² = Gm²/d²
⇒ Δe = m√(G/k) [∵ 1/4πϵ₀ = k = 9 × 10⁹]
= 1.67 × 10⁻²⁷ √(6.67 × 10⁻¹¹ / 9 × 10⁹)
Δe ≈ 1.436 × 10⁻³⁷ C

As the regions are of equipotentials, so Work done

W = qΔV

ΔV is same in all the cases hence work done will also be same in all the cases.

When battery is replaced by another uncharged

capacitor

As uncharged capacitor is connected parallel

So, C′ = 2C

and V_c = (q₁ + q₂)/(C₁ + C₂)

V_c = (q + 0)/(C + C) = CV/(C + C) [∵ q = CV]

⇒ V_c = V/2

Initial Energy of system, Ui = 1/2 CV² ...(i)

Final energy of system, U_f = 1/2 (2C)(V/2)²

= 1/2 CV² (1/2) ...(ii)

From equation (i) and (ii)

U_f = 1/2 Ui

i.e., Total electrostatic energy of resulting system decreases by a factor of 2.

We know that,

R = ρℓ/A

or

R = ρℓ²/Volume ∝ ℓ²

According to question ℓ₂ = nℓ₁

R₂/R₁ = n²ℓ₁²/ℓ₁²

or,

R₂/R₁ = n²

⇒ R₂ = n²R₁

Reading of potentiometer is accurate because during taking reading it does not draw any current from the circuit .

Work done,

W = MB(cosθ₁ − cosθ₂)

When it is rotated by angle 180°

W = MB(cos0° − cos180°) = MB(1 + 1)

W = 2MIA

W =2(NIA) B

= 2 × 250 × 85 × 10⁻⁶ (1.25 × 2.1 × 10⁻⁴) × 85 × 10⁻²

≈ 9.1 μJ

Force per unit length between two parallel current carrying conductors,

F = μ₀i₁i₂/2πd

Since same current flowing through both the wires

i₁ = i₂ = i

so F₁ = μ₀i²/2πd = F₂

∴ Magnitude of force per unit length on the middle wire 'B'

F_net = √(F₁² + F₂²)

= μ₀i²/√2πd

If θ₁ and θ₂ are apparent angles of dip
Let α be the angle which one of the plane make with the magnetic meridian.
tan θ₁ = V/(H cos α)
i.e.,
cos α = V/(H tan θ₁) ...(i)

tan θ₂ = V/(H sin α)

i.e.,

sin α = V/(H tan θ₂) ...(ii)

Squaring and adding (i) and (ii), we get

cos² α + sin² α = (V/H)² (1/tan² θ₁ + 1/tan² θ₂)

i.e.,

1 = V²/H² [cot² θ₁ + cot² θ₂]

or

H²/V² = cot² θ₁ + cot² θ₂

i.e.,

cot² θ = cot² θ₁ + cot² θ₂

Given, no. of turns N = 100
radius, r = 0.01 m
resistance, R = 10π² Ω, n = 2 × 10⁴

As we know,

ε = −N dϕ/dt

⇒ ε/R = −N/R dϕ/dt

ΔI = −N/R dϕ/dt
= Δq/Δt

= −N/R Δϕ/Δt

Δq = −[N/R(Δϕ/Δt)] Δt

'−' ve sign shows that induced emf opposes the change of flux.

Δq = [μ₀nNπr²(Δi/Δt)] (1/R) Δt
= μ₀nNπr²Δi/R

Δq = (4π × 10⁻⁷) × 100 × 4 × π × (0.01)² × 2 × 10⁴/
10π ²
Δq = 32 μC

At t = 0, no current flows through R₁ and R₃.

Current through battery just after the switch closed is

i =ε/ R₂ = 18/9 = 2 A

Given, Eₘₛ = 6 V/m
Eₘₛ/Bₘₛ = c

⇒ Bᵣₘₛ = Eᵣₘₛ/c ...(i)
Bᵣₘₛ = B₀/√2⇒ B₀ = √2 Bᵣₘₛ
B₀ = √2 × Eᵣₘₛ/c From
equation (i)
= √2 × 6/3 × 10⁸
= 2.83 × 10⁻⁸ T

When mirror is rotated by angle θ reflected ray will be rotated by 2θ.
According to the condition,
y/ x = 2θ
⇒ θ = y/2x

For dispersion without deviation

(μ − 1)A₁ + (μ′ − 1)A₂ = 0

|(μ − 1)A₁| = |(μ′ − 1)A₂|

(1.42 − 1) × 10° = (1.7 − 1)A₂

4.2 = 0.7A₂

A₂ = 6°

According to question

8ᵀʰ bright fringe in medium = 5ᵀʰ dark fringe in air

Y₈ₜₕ bright = 8 (λD/μd)

Y₅ₜₕ dark = (2 × 5 − 1) λD/2d = 9λD/2d

⇒ 9 /2 (λD/d ) = 8 (λD/μd)

or, refractive index

μ = 16/9 = 1.78

According to Malus law, I = I₀cos²θ

I₁ = I₀/2

I₂ = (I₀/2) cos²45°
= (I₀/2) × 1/2
= I₀/4

I₃ = (I₀/4) cos²45°
= I₀/8

Resolving power of a microscope = 2μ sin θ/λ

i.e., R ∝ 1/λ, or R₁/R₂ = λ₂/λ₁

Given that the two wavelengths,

λ₁ = 4000 Å and λ₂ = 6000 Å

R₁/R₂ = 6000 Å /4000 Å = 3/2

We know that, de-Broglie wavelength λ = h/√(2mKE)
K.E. of thermal neutron = 3/2 kT
λ = h/√ 2m (3/2 kT)
λ = h/√3mkT

For last line of Balmer series:
n₁ = 2 and n₂ = ∞
1/λB = RZ² [1/n₁² − 1/n₂²]

     = R₁² [1/2² − 1/∞²]
    1/λ_B   = 4/R   ...(i)

     
For last line of Lyman series:
n₁ = 1 and n₂ = ∞
1/λ_L = RZ² [1/n₁² − 1/n₂²]
     = RZ² [1/1² − 1/∞²]
    
λ_L = 1/R                      ...(ii)
Dividing equation (i) by (ii)
λ_B/λ_L = (4/R)/(1/R)
Ratio of wavelengths is
λ_B/λ_ = 4

In forward bias, V₁ > V₂ i.e., in figure (d) p-side is at higher potential w.r.t. n-side of p-n junction diode.

Given, current gain β = 100, R_C = 3kΩ, R_b = 2kΩ

Voltage gain (A_v) = β R_C/R_b = 100 (3/2) = 150

Power gain = A_v β = 150 × 100 = 15000

Y₁ = A̅ +  B̅

Y₂ = Y₁ + Y̅₁ = Y̅₁ = A + B

Y = Y̅₂ = A̅ +  B̅ i.e. NOR gate.

For hydrogen like atoms energy of 2s orbital and 2p-orbital is equal.

After ₈₆Rn (Group 18), elements from atomic number 89 to 103 (actinoids) are in group 3 114 = 103 + 11. Thus, element with atomic number 114 will be in group (3 + 11) or group 14 (carbon family).

The outer shell configuration of group 14 is ns²np².

Hence, only option (a) is correct.

IBr₂⁻, XeF₂
Total number of valence electrons are equal in both the species and both the species exhibit linear shape.

BCl₃ is trigonal planar and hence the bond angle is 120°.

CN⁻ and CO have same no. of electrons and have same bond order equal to 3.

The system is in isolated state.

∴ For an adiabatic process, q = 0

ΔU = q + w

∴ ΔU = w

= −PΔV

= −2.5 atm × (4.5 − 2.5) L

= −2.5 × 2 L-atm.

= −5 × 101.3 J

= −506.5 J ≈ −505 J

Given ΔH = 35.5 kJ mol⁻¹
ΔS = 83.6 J K⁻¹ mol⁻¹
∴ ΔG = ΔH − TΔS
For a reaction to be spontaneous, ΔG = −ve
i.e., ΔH < TΔS

∵ T > ΔH /ΔS = 35.5*10³ Jmol⁻¹/83.6Jk⁻¹

So, the given reaction will be spontaneous at
T > 425 K

Max. pressure of CO₂ = Pressure of CO₂ at equilibrium
SrCO₃(s) ⇌ SrO(s) + CO₂(g)
Kₚ = P_CO₂ = 1.6 atm = maximum pressure of CO₂.
volume of container at this stage,
V = nRT/P                                  ...(i)
Since container is sealed and reaction was not earlier at equilibrium,
n = PV/RT = (0.4 × 20)/(RT)                  ...(ii)
Put equation (ii) in equation (i)
V = [(0.4 × 20)/(RT)]RT / 1.6
V = 5 L

Due to inert pair effect, Pb(II) is more stable than Pb(IV).
Sn(IV) is more stable than Sn(II).

Pb(IV) is easily reduced to Pb(II) and can acts as an oxidising agent whereas Sn(II) is easily oxidised to Sn(IV) and can acts as a reducing agent.

3-keto-2-methylhex-4-enal
Aldehydes get higher priority over ketone and alkene in numbering of principal carbon chain.

The o-isomer is steam volatile due to intra-molecular H-bonding. The p-isomer is not steam volatile due to intermolecular H-bonding or association of molecules.

Thus, both can be separated by steam distillation.

Electrophile is a electron deficient species and can accept pair of electrons from nucleophile.

There is no change in bond angles and bond lengths in the conformation of ethane.

Correct order is

H-C≡C -H > H₃c – C ≡ C-H > H₂c
(Two acidine hydrogensc) (One acidichydrogen)

= CH₂> CH₃ - CH₃

Microorganisms present in the soil is a sink for CO.

K_f (molal depression constant) only depends on the nature of the solvent and is independent of the Concentration of the solution.

Molarity depends on the volume of a solution which can be changed with change in temperature.

Li⁺ being smallest, has maximum charge density.
∴ Li⁺ is most heavily hydrated among all alkali metal ions. Effective size of Li⁺ in aqueous solution is therefore, largest. So, moves slowest under electric field.

For cell,
Zn | ZnSO₄ (0.01 M) || CuSO₄ (1.0 M) | Cu

Ecell = E°cell − (2.303RT/ /nF log ([Zn²⁺] / [Cu²⁺]

∴ E₁ = E°cell − (2.303RT/2F) log(0.01/0.1)

When concentrations are changed for ZnSO₄ and CuSO₄, we can write

E₂ = E°cell − (2.303RT/2F) * log(1/0.01)

∴ E₁ > E₂

Overall rate = k[X][Y₂]      ...(1)
k = rate constant
Assuming step (i) to be reversible, its equilibrium constant,
k(eq)subscript = [X]²/[X₂]
⇒[X]² = k(eq)subscript [X₂]
⇒[X] = k(eq)subscript ½[X₂]½     ...(2)
From (eq)subscript  (1) and (2)
Rate = kk(eq)subscript ½[X₂]½[Y₂]
Overall order = 1/2 + 1 = 3/2 = 1.5

Half life for a first order reaction,
t₁/₂ = 0.693/K
So,
t₁/₂ = 0.693/10⁻² sec
Also, for the reduction of 20 g of reactant to 5 g, two half lives will be required.
∴ For 20 g of the reactant to reduce to 5 g, time taken,
t = 2 × (0.693/10⁻²) sec = 138.6 sec.

A catalyst speeds up both forward and backward reaction with the same rate.
So, equilibrium constant is not affected by the presence of a catalyst at any given temperature.

Zn being more reactive than Ag and Au, displaces them.

Zn being more reactive than Ag and Au, displaces them.

Potassium permanganate has a purple colour. When sulphur dioxide reacts with potassium permanganate, the solution decolourises.

5SO₂ + 2KMnO₄ + 2H₂O → 2H₂SO₄ + 2MnSO₄ + K₂SO₄

Minimum comparable energy gap between 5f, 6d and 7s subshell makes electron excitation easier, hence there is a greater range of oxidation states in actinoids.

In the given complex, the CN of Co is 6, and the complex has octahedral geometry.

In a solution containing HgCl₂, I₂ and I⁻, both HgCl₂ and I₂ compete for I⁻.

Since formation constant of [HgI₄]²⁻ is very large (1.9 × 10³⁰) as compared with I₃⁻ (Kf = 700).

∴ I⁻ will preferentially combine with HgCl₂.

HgCl₂ + 2I⁻ → HgI₂ (Red ppt ) ↓ + 2Cl⁻

HgI₂ + 2I⁻ → [HgI₄]²⁻ soluble

The order of strength of ligand is

en > NH₃ > H₂O

For stronger ligand, the value of Δ₀ for the complex is higher.

Hence, the increasing order of wavelengths of absorption is:

[Co(en)₃]³⁺ < [Co(NH₃)₆]³⁺ < [Co(H₂O)₆]³⁺

Grignard's reagent (RMgX) is a σ-bonded organometallic compound.

Electron withdrawing –NO₂ group has very strong –I and –R effects so, compound 3 will be most acidic

When Ar–O–R ethers are reacted with HI, they are cleaved at weaker O–R bond to give phenol and alkyl iodide.

Since 'A' gives positive silver mirror test therefore, it must be an aldehyde or α-Hydroxyketone. Also, reaction with OH⁻ i.e., aldol condensation (by assuming alkali to be dilute) indicates that 'A' is aldehyde as aldol reaction of ketone is reversible and carried out in special apparatus. It indicates that 'A' is an aldehyde.

–NO₂ group has strong –R effect and –CH₃ shows +R effect.

∴ Order of basic strength is

It is called Hoffmann Bromamide reaction.

Due to denaturation of proteins, helix get uncoiled and protein loses its biological activity.

Dettol is a mixture of chloroxylenol and terpineol which is a very commonly known antiseptic.

When Ar–O–R ethers are reacted with HI, they are cleaved at weaker O–R bond to give phenol and alkyl iodide.

Archaebacteria are able to survive in harsh conditions due to the presence of branched lipid chain in cell membrane that reduces fluidity of cell membrane. It includes halophiles which are exclusively found in saline habitats.

Sticky character of the bacterial wall is due to glycocalyx which is rich in glycoproteins.

Mycoplasmas are smallest, prokaryotes lacking cell wall and are pleomorphic in nature. These are pathogenic to both plants and animals.

Viroids in nature are sub-viral agents as infectious

RNA particles, without protein coat.

Volvox is a motile colonial fresh water green alga. It forms spherical colonies. It was first observed with a light microscope by Van Leeuwenhoek (1700) within the volvocales.

Chlamydomonas has haplontic life cycle hence shows

zygotic meiosis.

Ectocarpus exhibits haplodiplontic life cycle while

Fucus has diplontic life cycle

Pinus is monoecious plant comprising of both male and female cones on same plant.

Double fertilisation is a unique feature exhibited only by angiosperms. It involves both syngamy and triple fusion.

In poriferans (sponges) choanocytes (collar cells)

form lining of spongocoel. Flagella present in collar cells

provide circulation to water in water canal system.

Pharyngeal gflI slns are present in hemlchordates and in chordates. Notochord is present in chordates only. Ventral tubular nerve cord is present in non-chordates.

Sharks and Trygon (sting ray) are the members of cartilag- inous fish while Whale, Dolphin and Seals are aquatic mammals belong to class mammalia.

Halophytes growing in saline soils show vivipary:

seed germination and have pneumatophores for gaseq

exchange.

Thorns are hard, pointed straight structures for protection against grazing animal. These are modified stem.

Coconut fruit is a drupe . A drupe is a fleshy fruit with thin skin and central stone containing the seed

The edible part of coconut is its endosperm. Coconut has double endosperm, liquid endosperm and cellular.

In roots, the root hairs develops from zone of mat ration. This zone is differentiated zone thus bearing hairs.

Cork cambium undergoes periclinal division and Cuts off thick walled suberised dead cells towards outside i.e phellem (cork) and it cuts off thin walled living cells i.e., phelloderm inner side.

Cellulose microfibrils are arranged radially rather than longitudinally which makes easy for the stomata to open.

During secondary growth in plants vascular cambium gives rise to secondary xylem and secondary phloem. Phelloderm is formed by cork cambium. Periderm is the corky outer layer formed in secondary thickening.

The vertebrates process myogenic heart which is self contractile system or auto excitable; it will thus keep working outside the body for some time.

The site of aerobic oxidation of carbohydrates in cells to generate ATP are mitochondria.

Calcium and phosphorus are the remaining elements.

Holoenzyme is conjugated enzyme which consists of a protein part called apoenzyme and a non-protein called cofactor. Coenzyme are also organic compounds but their association with apoenzyme is only transient and acts as cofactors.

The correct sequence of events occurred during mitosis would be as follows:

(i) DNA condensation occurs so that chromosomes become visible during early to mid-prophase.
(ii) Disassembly of nuclear membrane begins at late prophase or transition to metaphase.
(iii) Chromosomes align at equator during metaphase.
(iv) Centromere division occurs during anaphase forming daughter chromosomes.
(v) During anaphase segregation also occurs in which daughter chromosomes separate and move to opposite poles.
(vi) Telophase finally leads to formation of two daughter nuclei.

– Nucleic acids are polymers of nucleotides.
– Proteins are polymers of amino acids.
– Polysaccharides are polymers of monosaccharides.
– Lipids are tri-esters of fatty acids with glycerol.

In bacteria DNA replication occurs in cytoplasm prior to fission. Prokaryotes due to their primitive nature do not show well marked S-phase

During anaphase, Anaphase Promoting Complex (APC) is a protein necessary for separation of daughter chromosomes. A defective APC will cause the chromosomes fail to segregate during anaphase.

By convention, the water potential of pure water at standard temperature, which is not under any pressure, is taken to be zero. (Ψw = 0)

In C₃ plants photosynthesis decreases at higher temperature due to increased photorespiration.

C₄ plants have higher temperature optimum because of the presence of enzyme called pyruvate phosphate dikinase, which is sensitive to low temperature.

In the mesophyll cells cytoplasm of C₄ plants like sugarcane, maize, sorghum etc. PEP is 3C compound which serves as primary CO₂ acceptor.

Krebs cycle begins with condensation of acetyl CoA (2C) with oxaloacetic acid (4C) to form citric acid (6C).

Auxins helps to prevent premature leaf and fruit fall.

– Kupffer-cells are phagocytic cells present in liver.
– Zymogen cells are enzyme producing cells which are secreted by pancreas.
– Paneth cell secrete lysozyme which acts as anti-bacterial agent, into the lumen of intestine.
– Argentaffin cells are hormone producing cells.

In hepatic portal system, hepatic portal vein drains blood to liver from intestine.

Total number of teeth in a human child is 20. In primary dentition premolars are absent.

Rennin and Pepsin enzymes are present in the gastric juice whereas Maltase is present in the intestinal juice.

Volume of air present in lungs after forceful expiration is residual volume prevents the collapsing of alveoli.

In human RBCs, nucleus get degenerates during maturation and it provide more space for oxygen carrying pigment (haemoglobin). It lacks many cell organelles including mitochondria so respires anaerobically.

Descending limb of loop of Henle is permeable to swater but impermeable to electrolytes whereas ascending limb is impermeable to water but permeable to electrolytes.

Pivot joint is a type of synovial joint which provide freely movement between atlas and axis vertebrae of vertebral column.

In human, 12 pairs of ribs are present out of which 7 pairs of ribs 1st to 7th pair) are dorsally attached to vertebral column and ventrally to the sternum.

Miyelin sheath wrapped around the nerve axon. Oligodendrocytes are neuroglial cells which produce myelin sheath in central nervous system while Schwann cell produces myelin sheath in peripheral nervous system.

Pre-synaptic membrane is involved in the release of neurotransmitter in the chemical synapse. The receptors sites for neurotransmitters are present on post-synaptic membrane of neuron.

Carotene is the source of retinal which is involved in formation of thodopsin of rod cells. Retinal, a derivative of vitamin A, is the light-absorbing part of all visual photopigments. Photopigments are occur entirely on the surface of membrane disc.

A decrease in blood pressure / volume stimulates the release of renin, aldosterone, and ADH while increase in blood pressure / volume stimulates the release of Atrial Natriuretic Factor (ANF) secreted by atria of heart, which causes vasodilation and also inhibits RAAS (Renin Angiotensin Aldosterone System) mechanism that decreases the blood volume/pressure.

Insect pollinated plants provide rewards as edible pollen grain and nectar as usual rewards. In order to mate Hialize and maximize pollination flowers have developed a set of attributes which are aimed at attracting the pollinators called attractants.

Wind pollination or anemophily occurs in flowers which are having a single ovule in each ovary, and numerous flowers packed in an inflorescence. It is a non-directional pollination.

Autogamy occurs in bisexual flowers. Geitonogamous flowers are unisexual but present in the same plant. Dioecious condition is observed when unisexual male and female flowers are present on different plants and it prevents both autogamy and geitonogamy.

The first cell of female gametophytic generation in angiosperm is megaspore. It undergoes three successive mitotic division to form 8-mucleated and 7-celled embryo sac.

Capacitation is increase in fertilising capacity of sperms which occurs in female reproductive tract. It is required to render sperm to fertilize an oocyte.

Cu²⁺interfere in the sperm movement, ‘which suppress the sperm motility and fertilising capacity of sperms.

Infertility cases due to inability of the male partner to inseminate the female or due to very low sperm count in the ejaculates, could be corrected by using artificial insemination (AI) technique. In this procedure, semen is injected directly into the vagina or uterus.

During his experiments Mendel have taken seven characters in a pea plant. Among these, nature of trichomes i.e. glandular or non-glandular was not considered by Mendel.

According to NCERT, Mendel conducted hybridisation experiments for 7 years on Pea plant between 1856 to 1863 and his data was published in 1865.

Thalassemia is a quantitative problem of synthesising very few globin molecules while sickle cell anaemia is a qualitative problem of synthesising an incorrectly functioning globin.

Down’s syndrome is caused by non-disjunction of 21ˢᵗ chromosome i.e. Trisomy.

Hershey and Chase proved that DNA as genetic material. They used bacteriophage for their experimental.

DNA fragments are negatively charged because presence of phosphate group.

Ribosomal RNA (rRNA) is most abundant in animal cell. It constitutes 80% of total RNA of the cell.

The association of H1 protein indicates the complete formation of nucleosome which requires DNA condensa tion. The DNA is therefore in condensed form.

If deletion happens at 901st position then the remaining 98 bases specifying for 33 codons of amino acids will be altered

Two DNA polymerase molecules simultaneously work at the DNA fork, one on the leading strand and the other on the lagging strand. DNA polymerase synthesizes each Okazaki fragment a laggIing strand in 5'-3' direction. As the replication fork opens further, new Okazaki fragments appear. The first Okazaki fragment appears away from the replication fork and thus the direction of elongation would be away from replication fork.

In eukaryotes, spliceosomes are used in removal of introns during post-transcriptional processing of hnRNA. they are absent in prokaryotes.

Horse belongs to

Orders - Perissodactyla

Family - Equidae

Genus - Equus

Species - ferus

‘Subspecies - caballus

Cell mediated immune response causes non-acceptance o rejection of graft or transplanted tissues/organs.Transplant rejection is caused primarily by a cell-mediated jmmune response to HLA antigens expressed on donor . antigen-presenting cells (APCs) transferred along with the transplanted organ. Recognition of donor HLA antigens on the cells of the graft induces vigorous T cell proliferation

in the recipient.

MAL Mucosa Associated Lymphoid Tissue constitutes about 50 percent of the lymphoid tissue in human body. It is scattered along mucosal lining in the human body

Inbreeding increases homozygosity. So, mating ofthe related indi viduals of same breed will give homozygous

purelines.

Primary treatment is a physical process which involves two process, i.e. filtration and sedimentation of big solid waste.

Saccharomyces cerevisiae commonly know as Brewer’s yeast, causes fermentation of carbohydrates and produces ethanol.

Ethidium bromide (EtBr) is used to stain the DNA fragments and will appear as orange coloured ‘bands when kept under UV light.

Selectable markers in recombinant DNA technology,

helps in identification and elimination of ‘non-transformants

and selectively permits the growth of the transformants.

DNA fragments during gel electrophoresis, separate

(resolve) according to their size due to sieving effect provided by agarose gel.

The various stages of processing that occur after the completion of fermentation or biosynthetic stage which include separation and purification of product called downstream processing.

The tropical rain forest have five vertical strata on the basis of plants height i.e., ground vegetation, shrubs, short canopy trees, tall canopy trees and tall emergent trees.

In logistic growth curve, the curve has an upper asymptote known as carrying capacity (K) is obtained when the maximum population size is at dN/dt = 0. A population growing in a habitat with limited resources shows logistic growth curve.
For logistic growth,
dN/dt = rN[(K − N)/K]
If K = N then
(K − N)/K = 0

∴ dN/dt =0.

The population reaches asymptote.

Mycorrhizaeis a symbiotic association between fungi and roots of higher plants.

Forest ecosystem has the maximum biomass.

Some very high productive ecosystem are

— Tropical rain forest

— Coral reef

— Estuaries

— Sugarcane fields

Alexander Von Humboldt noticed that within a region species richness increases with the increase in area.

Aerosols through its direct or indirect effects on plants can cause various problems in agriculture. However, continuous increase in air pollution may represent a threat to agriculture in the future that is persistent and largely irreversible.

Artificial selection to obtain cow yielding higher milk output will shift the peak to one direction, so this represent an example of Directional selection. In stabilizing selection, the organisms with the mean value of the trait are selected. In disruptive selection, both extremes get selected.

Rhodospirillum is facultative anaerobe and free living nitrogen fixer. Mycorrhiza show symbiotic relationship between fungi and roots of higher plants.

Ex-situ conservation is offsite strategy for conservation of biological diversity in zoological park and botanical gardens respectively.

Corpus luteum is the temporary endocrine gland formed in the ovary after ovulation. It release hormones like progesterone, oestrogen etc.

Gonorrhoea – Neisseria (Bacteria)
Syphilis – Treponema (Bacteria)
Genital Warts – Human papilloma virus
AIDS – HIV (Virus)

Primary treatment is a physical process which involves two process, i.e. filtration and sedimentation of big solid waste.

In male frog the sperms passage is:
Testes → Vasa efferentia → Kidney → Bidder's canal → Urinogenital duct → Cloaca.

GnRH is secreted by hypothalamus which stimulates anterior pituitary gland for the secretion of gonadotropins (FSH and LH).

Heartwood is inactive physiologically due to deposition of organic compounds and formation of tyloses so it will not conduct water and minerals.

From work-energy theorem,

W_g + W_a = ΔK.E.

or, mgh + W_a = 1/2 mv² − 0

10⁻³ × 10 × 10³ + W_a = 1/2 × 10⁻³ × (50)²

⇒ W_a = −8.75 J

which is the work done due to air resistance

Work done due to gravity = mgh
= 10⁻³ × 10 × 10³ = 10 J

Mechanical advantage, M.A = Load/Effort

If M.A. > 1 ⇒ Load > Effort

Hydration of alkynes give ketones.

XX' → Linear (e.g. ClF, BrF)

XX₃' → T-Shape (e.g. ClF₃, BrF₃)

XX₅' → Square pyramidal (e.g. BrF₅, IF₅)

XX₇' → Pentagonal bipyramidal (e.g. IF₇)

Electron withdrawing—NO₂, group has very strong -I

and —R effects so, compound 3 will be most acidic.

More stable as —ve charge is close to electron withdrawing group.

Also, incoming nucleophile gets attached on same 'C'on

which 'Br' (Leaving group) was present.

. nota cine substitution reaction.

More stable as —ve charge is close to electron withdrawing group.

Also, incoming nucleophile gets attached on same 'C'on

which 'Br' (Leaving group) was present.

. nota cine substitution reaction.