NEET-UG 2019

Given,

X = $\frac{A^2{B}^{1/2}}{C^{1/3}{D}^3}$

$\frac{\Delta X}{X}=\frac{2\Delta A}{A}+\frac{1}{2}\frac{\Delta B}{B}+\frac{1}{3}\frac{\Delta C}{C}+3\frac{\Delta D}{D}$

$\Rightarrow$ $\frac{\Delta X}{X}\times 100=2\left(1\%\right)+\frac{1}{2}\left(2\%\right)+\frac{1}{3}\left(3\%\right)+3\left(4\%\right)$ = 16 %

V_SR = 20 m/s

V_RG = 10 m/s

${\overrightarrow{V}}_{SG}={\overrightarrow{V}}_{SR}+{\overrightarrow{V}}_{RG}$

sin $\theta$ = $\left|\frac{{\overrightarrow{V}}_{RG}}{{\overrightarrow{V}}_{SR}}\right|$ = $\frac{10}{20}$ = $\frac{1}{2}$

$\Rightarrow$ θ = 30° west-side

Assume initial velocity = u


In first case, Moved distance, (x₁) = $\frac{u^2}{2g\sin {60}^o}$

In second case, Moved distance, (x₂) = $\frac{u^2}{2g\sin {30}^o}$

$\therefore$ $\frac{x_1}{x_2}=\frac{\sin {30}^o}{\sin {60}^o}$ = $\frac{1\times 2}{2\times \sqrt{3}}$ = $\frac{1}{\sqrt{3}}$

From the given diagram, the forces are forming closed loop in same order.

$\therefore$ F_net = $\overrightarrow{0}$

$\Rightarrow$ $m\frac{dv}{dt}$ = $\overrightarrow{0}$

$\Rightarrow$ $\overrightarrow{v}$ = constant.

Apply Newtons's 2nd law for equation of motion

T - mg cos $\theta$ = $\frac{m{v}^2}{R}$

T will be maximum when $\theta$ = 0°,

When mass is at lowest point the chance of breaking is maximum.

f_L = $\mu$N = $\mu$mr${\omega }^2$

f_s = mg

As, f_s $\le$ f_L

$\Rightarrow$ mg $\le$ $\mu$mr${\omega }^2$

$\Rightarrow$ $\omega \ge \sqrt{\frac{g}{\mu r}}$

$\Rightarrow$ ${\omega }_{\min }\ge \sqrt{\frac{10}{0.1\times 1}}$ = 10

Work done under the given variable force is :

W = $\underset{y_1}{\overset{y_1}{\int }}Fdy$

Here, y₁ = 0, y₂ = 1 m

$\therefore$ W = $\underset{0}{\overset{1}{\int }}\left(20+10y\right)dy$

= ${\left[20y+\frac{10y^2}{2}\right]}_0^1$ = 25 J

As linear momentum is conserved before and after the collision,

$\therefore$ 4m $\times$ u = 4m $\times$ v₁ + 2m $\times$ v₂

$\Rightarrow$ 2u = 2v₁ + v₂ ......(1)

As collision is elastic so e = 1

We know, e = $\frac{v_2-v_1}{u_1-u_2}$

$\Rightarrow$ 1 = $\frac{v_2-v_1}{u-0}$

$\Rightarrow$ u = v₂ - v₁ ....(2)

Subtractiong (2) from (1), we get

u = 3v₁

$\Rightarrow$ v₁ = $\frac{u}{3}$

After the collision the fraction of energy lost by the colliding body A is

= $\frac{\frac{1}{2}\left(4m\right)u^2-\frac{1}{2}\left(4m\right){\left(\frac{u}{3}\right)}^2}{\frac{1}{2}\left(4m\right)u^2}$

= $\frac{1-\frac{1}{9}}{1}$ = $\frac{8}{9}$

Since, the time period for both the particle in same So, T_A = T_B = T

Angular velocity for A, (${\omega }_A$) = $\frac{2\pi }{T_A}$

Angular velocity for B, (${\omega }_B$) = $\frac{2\pi }{T_B}$

Now, the required ratio is

$\frac{{\omega }_A}{{\omega }_B}=\frac{T_B}{T_A}=\frac{T}{T}=1$

$\omega =\frac{V}{R}=\frac{0.2}{2}=$ 0.1 rad/sec

Apply the law of conservation of energy,

Work required = change in kinetic energy

Since, final KE = 0

And, initial KE = $\frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2$

= $\frac{1}{2}m{v}^2+\frac{1}{2}\frac{M{l}^2}{2}{\omega }^2$

= $\frac{1}{2}\left(100\right){\left(0.2\right)}^2+\frac{1}{2}\frac{100\times {\left(2\right)}^2}{2}{\left(0.1\right)}^2$

= 1 + $\frac{1}{2}\times 4$ = 3 J

Acccording to work energy theorem,

W = $\frac{1}{2}\left({\omega }_f^2-{\omega }_i^2\right)$

Given that,

$\theta$ = 2$\pi$ revolution/minute

$\theta$ = 2$\pi$ $\times$ 2$\pi$ = 4$\pi$² rad

${\omega }_i=3\times \frac{2\pi }{60}$ rad/s

${\omega }_f=0$ rad/s

By putting the value of ${\omega }_f$ and ${\omega }_i$, we get

- $\tau$$\theta$ = $\frac{1}{2}\times \frac{1}{2}m{r}^2\left(0-{\omega }_i^2\right)$

$\Rightarrow$ $\tau$ = $-\frac{\frac{1}{2}\times \frac{1}{2}\times 2\times \left(4\times {10}^{-2}\right)\times {\left(-3\times \frac{2\pi }{60}\right)}^2}{4{\pi }^2}$

= 2 $\times$ 10^-6 Nm

Initial potential energy at Earths surface is

U_i = $-\frac{GMm}{R}$

Final potential energy at height h = R :

U_f = $-\frac{GMm}{2R}$

work done = change in PE

w = U_f - U_i

= $\frac{GMm}{2R}$

= $\frac{g{R}^{2}m}{2R}$

= $\frac{1}{2}$mgR

Accelerattion due to gravity at a depth d from surface of Earth

$g$' = $g\left(1-\frac{d}{R}\right)$ ....(1)

Multiplying by mass 'm' on both sides of equation (1)

m$g$' = m$g\left(1-\frac{d}{R}\right)$

= $200\left(1-\frac{R}{2R}\right)$    ($\because$ $d=\frac{R}{2}$)

= $\frac{200}{2}=100$ N

At any temperature

${\left(\Delta l\right)}_{Cu}={\left(\Delta l\right)}_{Al}$

$l_{Cu}{\alpha }_{Cu}\Delta T=l_{Al}{\alpha }_{Al}\Delta T$

$\Rightarrow$ 88 $\times$ 1.7 $\times$ 10^-5 = $l_{Al}$ $\times$ 2.2 × 10^–5

$\Rightarrow$ $l_{Al}$ = 68 cm

P = P₀ + $\rho$gZ₀ .........(i)

Also, P = P₀ + $\frac{4T}{R}$ .....(ii)

From (i) and (ii),

$\rho$gZ₀ = $\frac{4T}{R}$

$\therefore$ Z₀ = $\frac{4T}{\rho gR}$

= $\frac{4\times 2.5\times {10}^{-2}}{{10}^3\times 10\times {10}^{-3}}$

= 10^-2 m = 1 cm

Elastic potential energy

U = $\frac{1}{2}$ (work done due to gravity)

= $\frac{1}{2}$Mgl

volume flow rate = Av = A$\sqrt{2gh}$

= 2 $\times$ 10^-6 $\times$ $\sqrt{2\times 10\times 2}$

= 2 $\times$ 2 $\times$ 3.14 $\times$ 10^-6 m³/s

= 12.56 $\times$ 10^-6 m³/s

= 12.6 $\times$ 10^-6 m³/s

We know,

Heat transfer (Q) = KA$\left(\frac{\Delta T}{\Delta x}\right)$$\times$t

Here, K is the coefficient of thermal conductivity

J = (K)m²$\frac{kelvin}{m}$$\times$ s

$\therefore$ Unit of K = Wm^–1 K^–1

From the characteristic property of adiabatic process, there is no exchange in heat.

Since, the increase of temperature will increase the kinetic energy for the given gas.

For the given ideal gas,

U = $\frac{F}{2}nRT$

Angular velocity, ($\omega$) = $\frac{2\pi }{T}=\frac{2\pi }{4}=\frac{\pi }{2}$ rad/s

Since, at t = 0, displacement (y) is maximum, so equation will be cosine function.

y = a cos $\omega$t

$\Rightarrow$ y = 3 cos $\left(\frac{\pi t}{2}\right)$

Since, net displacement in complete cycle ∆y = 0

So, Average velocity

=

Displacement

Time travel

= $\frac{y_f-y_i}{T}$ = $\frac{\Delta y}{T}=\frac{0}{T}=0$

From the given displacement

y = A₀ + A sin$\omega$t + B cos$\omega$t.

Let assume, y - A₀ = $\gamma$

$\gamma$ = A sin$\omega$t + B cos$\omega$t

= $\sqrt{A^2+B^2}\sin \left(\omega t+\varphi \right)$

which is S.H.M

$\therefore$ Resultant amplitude of the particle,

= $\sqrt{A^2+B^2}$

For inside (r $<$ R)

Apply, Gauss law,

$\oint E_{in}.\overrightarrow{dS}$ = $\frac{q_{en}}{{\epsilon }_0}$

$\Rightarrow$ E_in = 0 [ As q_en = 0 ]

For outside (r $>$ R)

$\oint {\overrightarrow{E}}_{out}.\overrightarrow{dS}=\frac{q_{en}}{{\epsilon }_0}$

Here, q_en = Q

So, E_out.4$\pi$r² = $\frac{Q}{{\epsilon }_0}$

$\therefore$ E_out $\propto$ $\frac{1}{r^2}$

${\overrightarrow{E}}_{net}={\overrightarrow{E}}_1+{\overrightarrow{E}}_2$

= $\frac{\lambda }{2\pi {\epsilon }_{0}R}$ + $\frac{\lambda }{2\pi {\epsilon }_{0}R}$

= $\frac{\lambda }{\pi {\epsilon }_{0}R}$ N/C

Initially F = $-\frac{K{Q}^2}{r^2}$

If 25% of charge of A transferred to B then after the transformation of charge.

q_A = $Q-\frac{Q}{4}$ = $\frac{3Q}{4}$

and q_b = -Q + $\frac{Q}{4}$ = $-\frac{3Q}{4}$

New force, F₁ = $\frac{K\left(\frac{3Q}{4}\right)\left(-\frac{3Q}{4}\right)}{r^2}$

$=-\frac{9}{16}\frac{K{Q}^2}{r^2}$

= $\frac{9F}{16}$

Resistance for ideal voltmeter = $\infty$

Resistance for ideal ammeter = 0

For 1^st circuit,

V₁ = i₁ $\times$ 10 = $\frac{10}{10}\times 10$ = 10 volt

For 2^nd circuit,

10 $\Omega$ is in series with ideal voltmeter. Therefore it will not affect the circuit.

So, V₂ = i₂ $\times$ 10 = $\frac{10}{10}\times 10$ = 10 volt

$\therefore$ V₁ = V₂

Hence, i₁ = i₂ = $\frac{10V}{10\Omega }$ = 1 A

(i) When all bulbs are glowing, from the given figure.
Equivalent resistance, R_eq = $\frac{R}{3}+\frac{R}{3}=\frac{2R}{3}$

Initial Power, (P_i) = $\frac{3E^2}{2R}$ ......(1)

(ii) When two section A and one from section B are glowing.

R_eq = $\frac{R}{2}+R$ = $\frac{3R}{2}$

Final Power, (P_f) = $\frac{2E^2}{3R}$

$\therefore$ $\frac{P_i}{P_f}=\frac{3E^2}{2R}\times \frac{3R}{2E^2}$ = 9 : 4

Fuse is the protective device because of low melting point of the wire that can save other device on heating.

As, q = CV

Differentiating above equation w.r.t. t, we get

$\frac{dq}{dt}=C\frac{dV}{dt}$

$\Rightarrow$ i_c = $C\frac{dV}{dt}$ [As i = $\frac{dq}{dt}$]

= 20 $\times$ 10^-6 $\times$ 3

= 60 $\times$ 10^-6 A = 60$\mu$A

Also, conduction current in wires is equal to displacement current between the plates of capacitor.

$\therefore$ i_d = i_c = 60 $\mu$A

Radius of the path (r) = $\frac{mv}{qB}$ = $\frac{p}{qB}$

So, the radius for H-atom r_H = $\frac{p}{eB}$

The radius for $\alpha$ particle r_$$\alpha$$ = $\frac{p}{2eB}$

$\therefore$ $\frac{r_H}{r_{\alpha }}=\frac{\frac{p}{eB}}{\frac{p}{2eB}}$ = $\frac{2}{1}$

The magnetic field inside conductor is given as

B = $\frac{{\mu }_0}{2\pi }\frac{i}{R^2}d$

$\therefore$ B $\propto$ d

If, straight line passing through origin, at surface (d = R)

B = $\frac{{\mu }_0}{2\pi }\frac{i}{R}$

For the maximum B at outside surface (d > R)

B = $\frac{{\mu }_0}{2\pi }\frac{i}{d}$

$\Rightarrow$ B $\propto$ $\frac{1}{d}$

In northern hemisphere dip is +ve and in southern hemisphere dip is –ve. Angle of dip is zero at equator.

Given N = 800, A = 0.05 m² , B = 5 × 10^–5 T

$\Delta$t = 0.15 s

We know, e = $-\frac{\left({\varphi }_f-{\varphi }_i\right)}{\Delta t}$

= $-\frac{\left(0-NBA\right)}{\Delta t}$

= $\frac{800\times 5\times {10}^{-5}\times 5\times {10}^{-2}}{0.1}$

= 0.02 V

Eddy current effect is not used in case of electric heater because it works on Joule's heating effect.

Since, Red colour has least frequency. So, red has the longest wavelength among the given colour.

As, $\nu =v\lambda$

When observer faces infront of Sun. He will not observe rainbow.

Let i_c be the critical angle.

Hence, at i = i_c, refracted ray grazes with the surface.

So, angle of refraction is 90°.

Equivalent focal length (in air) $\frac{1}{F_1}=\frac{1}{f}+\frac{1}{f}=\frac{2}{f}$

When glycerin is filled inside, glycerin lens behaves like a diverging lens of focal length (–f)

$\frac{1}{F_2}=\frac{1}{f}+\frac{1}{f}-\frac{1}{f}=\frac{1}{f}$

$\therefore$ $\frac{F_1}{F_2}=\frac{1}{2}$

Angular fringe width (in air) $\theta$_air = $\frac{\beta }{D}$

Angular Fringe width (in water)

$\theta$_w = $\frac{\beta }{\mu D}=\frac{{\theta }_{air}}{\mu }$

= $\frac{0.2^{\circ }}{\frac{4}{3}}$ = 0.15^o

Since, $\alpha$-particle is equivalent to He-atom nucleus, hence it has two protons and two neutrons only.

Apply Bohr's Atomic model for H-atom

KE = –T.E and PE = 2T.E

Given, T.E = –3.4 eV

$\therefore$ KE = +3.4 eV and PE = -6.8 eV

Expression for an electron accelerated through a potential V is given as

$\lambda$ = $\frac{12.27}{\sqrt{V}}\stackrel{o}{A}$

= $\frac{12.27\times {10}^{-10}}{\sqrt{10000}}$

= 12.2 × 10^–12 m

In p-type semiconductor, an intrinsic semiconductor is doped with trivalent impurities, that creates deficiencies of valence electrons called holes which are majority charge carriers.

From the given logic circuit LED will glow, when voltage across LED in high.

This can be verified by the given truth table.
Boolean expression Y = $\overline{A.B}$

This is output of NAND gate.

According to Haber's process,

N₂(g) + 3H₂(g) ⇌ 2NH₃(g)

2 moles of NH₃ uses 3 moles of H₂

So, 20 moles of NH₃ will use $\frac{3}{2}$ $\times$ 20 moles of H₂

= 30 moles of H₂

Balmer series lies in visible region.
Paschen series lies in infra red region.
Brackett series lies in infra red region.
Lyman series lies in ultraviolet region.

As per Aufbau principle (n + l rule), higher is the value of (n + l), greater will be the energy. When (n + l) values are same then which orbital have higher value of n will have more energy.
4d = 4 + 2 = 6
5p = 5 + 1 = 6
5f = 5 + 3 = 8
6p = 6 + 1 = 7
Hence, the correct order of decreasing energy will
be : 5f > 6p > 5p > 4d.

In a disproportionation reaction, same substance undergoes oxidation (increase in oxidation number) and reduction (decrease in oxidation number forming two different products.

Compressibility factor (Z ) = $\frac{V_{real}}{V_{ideal}}$

Here, V_real < V_ideal. Therefore, Z < 1.

In gaseous molecules attractive forces are dominant.

Conjugate base of H₂O is OH^–

Conjugate base of HF is F^–

Ca(OH)2	⇌	Ca+2(aq)	+	2OH-(aq)
		S		2S

pH = 9 ; pOH = 5 ; [OH^–] = 10^–5 = 2S

$\therefore$ S = $\frac{{10}^{-5}}{2}$

K_sp = [Ca⁺²] [OH^–]²

$\Rightarrow$ K_sp = S × (2S)²

$\Rightarrow$ K_sp = 4S³

$\Rightarrow$ K_sp = $4\times {\left(\frac{{10}^{-5}}{2}\right)}^3$ = 0.5 × 10^–15

Basic buffer is mixture of weak base and salt of weak base with strong acid.

milli mole of HCl = 100 × 0.1 = 10 milli mole

milli mole of NH₄OH = 200 × 0.1 = 20 milli mole

HCl	+	NH4OH	$$\to$$	NH4Cl	+	H2O
10		20		0
0		20 - 10 = 10		10

Here in the final solution 10 milli mole weak base NH₄OH and 10 milli mole salt of weak base and strong acid NH₄Cl present. So it is a basic buffer.

For an ideal solution at constant T and P,

$\Delta$_mixH = 0, $\Delta$_mixS > 0, $\Delta$_mixG < 0 and $\Delta$_mixV = 0.

Maximum boiling azeotrope is shown by solution which shows negative deviation from Raoult's law. Except water + Nitric acid, all other mixtures show negative deviation.

2H(g) $\to$ H₂(g)

Due to bond formation, entropy decreases.

W = –P_ext (V₂–V₁)

P_ext = 2 bar

V₁ = 0.1 L

V₂ = 0.25 L

W = –2 bar[0.25 – 0.1] L

W = –2 × 0.15 bar L

W = –0.30 bar L

W = (–0.30) × 100 = –30 J

Here, n = 2

$\Delta$G^o = -nFE^o

= – 2 × 96500 × 0.24

= – 46320 J mol^–1

= – 46.32 KJ mol^–1

We know,

E_cell = $E_{cell}^{\Theta }$ - $\frac{2.303RT}{nF}\log Q$

$\Rightarrow$ E_cell = $E_{cell}^{\Theta }$ - $\frac{0.059}{n}\log Q$

(At equilibrium, E_cell = 0 and Q = K_eq)

$\Rightarrow$ 0 = $E_{cell}^{\Theta }$ - $\frac{0.059}{1}\log K_{eq}$

$\Rightarrow$ $\log K_{eq}=\frac{E_{cell}^{\Theta }}{0.059}$ = $\frac{0.59}{0.059}$ = 10

$\Rightarrow$ K_eq = 10¹⁰

N₂(g) + 3H₂(g) ⇌ 2NH₃(g

Rate of the reaction can be written as :

$-\frac{d\left[N_2\right]}{dt}=-\frac{1}{3}\frac{d\left[H_2\right]}{dt}=+\frac{1}{2}\frac{d\left[N{H}_3\right]}{dt}$

For a first order reaction,

t = $\frac{2.303}{k}\log \frac{\left[A_0\right]}{\left[A\right]}$

for 99% completion of the reaction,

$t=\frac{2.303}{k}\log \frac{100}{100-99}$

$\Rightarrow$ $t=\frac{4.606}{k}$

As anions are in hcp, the number of anions A = 6

Number of cations = 6 $\times$ $\frac{75}{100}$

= 6 $\times$ $\frac{3}{4}$ = $\frac{9}{2}$

So, formula of compound is $C_{\frac{9}{2}}{A}_6$.

$\Rightarrow$ C₉A₁₂

$\Rightarrow$ C₃A₄

AgNO₃ + KI $\to$ AgI + KNO₃

If KI is in excess then negatively charged colloid, [AgI]I^– is formed. Whereas, if AgNO₃ is in excess positively charged colloid AgI/Ag⁺ will form.

Ionisation enthalpy increases in a period from left to right, because of increased nuclear charge and decrease in atomic radii. But because of half filled orbitals of Be, N and fully filled orbitals of Ne, stability of those atoms inceases.

So, the correct increasing order of first ionisation enthalpy will be :
Li < B < Be < C < O < N < F < Ne

$C_2$ has 12 electrons.

Moleculer orbital configuration of $C_2$

= ${\sigma }_{1s^2}{\sigma }_{1s^2}^{\ast }{\sigma }_{2s^2}{\sigma }_{2s^2}^{\ast }{\pi }_{2p_x^2}={\pi }_{2p_y^2}$

Here no unpaired electron present, so it is diamagnetic.

Here, Bond order = $\frac{8-4}{2}$ = 2. So here two bonds are present. For this two bond formation 4 electrons are required which is present in ${\pi }_{2p_x^2}$ and ${\pi }_{2p_y^2}$

So, in C₂ molecule, only $\pi$ electron occupied the bonding molecular orbital.

H–C $\equiv$ C – CH = CH – CH₃

Here, 10 $\sigma$ bonds and 3 $\pi$ bonds are presents.

Br₃O₈

Malachite = CuCO₃.Cu(OH)₂

Be(OH)₂ is an amphoteric hydroxide as it reacts with acid as well as base while others are basic hydroxide.

Be(OH)₂ + 2HCl $\to$ BeCl₂ + 2H₂O

Be(OH)₂ + 2NaOH $\to$ Na₂[Be(OH)₄]

Mg is acts as a cofactor for all the enzymes involved in ATP phosphate transfer.

Clark's method used to remove temporary hardness of water

Ca(HCO₃)₂ + Ca(OH)₂ $\to$ 2CaCO₃ + 2H₂O

The thermal stability of hydrides decreases down the group due to decrease in bond dissociation energy down the group.

Therefore, thermal stability order will be :

H₂Po < H₂Te < H₂Se < H₂S < H₂O

Three equatorial bonds and two axial bonds.
Due to unsymmetry PCl₅ is reactive.

Pure N₂ : BaN₃ $\to$ Ba + N₂

Haber process : N₂ + 3H₂ $\to$ 2NH₃

Contact process : 2SO₂ + O₂ $\to$ 2SO₃

Deacon’s process : HCl + O₂ $\frac{\text{buildrel}CuC{l}_2}{⟶}$ H₂O + Cl₂

PbF₄ is an ionic compound.

[SiCl₆]^2– does not exist because six large chloride ions cannot be accommodated around Si⁺⁴ due to its small size.

Fe : [Ar]3d⁶4s²

Fe²⁺ : 3d⁶

Sulphurdioxide (SO₂) is not a green house gas.

In Ph - OH, the lone pair of electrons on O atom causes delocalization which makes it difficult to protonate due to less availability of electrons.

The basic strength of methyl substituted amines in aqueous form is together affected by inductive effect, solvation effect and steric hindrance. So, the correct order of basic strength will be :

(CH₃)₂NH > CH₃NH₂ > (CH₃)₃N

Biodegradable polymer is Nylon-2-Nylon-6.

Non Essential amino acid is Alanine.

Penicillin – G is Narrow Spectrum antibiotics.

Lysosome and vacuoles do not contain DNA.

The enzymes required for the ETS are present in the inner matrix only as the elementary particles are embedded in the inner membrane of the mitochondrial matrix.

The statement that is not correct is:

Option B: Lysosomes are formed by the process of packaging in the endoplasmic reticulum.

Lysosomes are indeed membrane-bound structures (Option A) that contain numerous hydrolytic enzymes (Option C) which are active under acidic conditions (Option D). However, lysosomes are not formed by the endoplasmic reticulum. Instead, they are formed from vesicles that bud off from the Golgi apparatus. The endoplasmic reticulum is involved in the synthesis of proteins, some of which are destined for lysosomes, but the actual formation and packaging of lysosomes occur at the Golgi apparatus.

Rudolf Carl Virchow in 1858, gave the concept of ''Omnis cellula-e cellula'' regarding cell division. This means that all cells arise from pre-existing cells by cell-division.

Coenzyme or metal ion that is tightly bound to enzyme protein is called prosthetic group while a complete catalytic active enzyme with its bound prosthetic group is called holoenzyme.

Concanavalin A is an example of lection.

G₀ or quiescent phase is the stage wherein cells remain metabolically active, but do not proliferate unless called to do so. These cells that do not divide further exit G₁ phase to enter an inactive stage. Such cells are used for replacing the cells lost during injury.

The correct sequence of phases of cell cycle is:
G₁ $\to$ S $\to$ G₂ $\to$ M

G₁ (Gap 1) phase : It is the first phase of the growth. It is the stage during which the cell grows and prepares its DNA for replication.

S phase (Synthetic phase) : It is the stage during which DNA synthesis occurs.

G₂ phase : In this phase, the cell continues to grow and prepares itself for division.

M Phase : A cell reproduces during the mitotic phase. The mitotic phase has two stages : Karyokinesis and Cytokinesis.

Central cell develops into Endopserm. Endosperm developes from the fusion of secondary diploid nucleus with one male nucleus formina primary endopsermm nucleus (PEN).

Parthenogenesis is the process, whereby the female gamete develops into new organism without fertilisation. Seedless fruits are formed by parthenogenesis.

The pinus seeds have a very hard seed coat so without fungal association the seed cannot absorb water. The mycorrhizae penetrate deep inside the seeds to moisten them.

In parietal placentation the ovules develop on the inner wall of ovary or in peripheral part. E.g. Mustard, Argemone, etc.

Perisperm is persistent in some seeds like castor, black pepper and beetroot.

In flowering plants,

Out of the two male gametes, one gametes fuses with the nucleus of the nucleus of the egg cell and forms the zygote. The process is known as syngamy.

The other male gamete fuses with the two polar nuclei located in the central cell to form a triploid primary endosperm nucleus (PEN). Since, the process involves the fusion of three haploid nuclei, it is known as triple fusion.

Since two kinds of fusions (syngamy and triple fusion) take place in an embryo sac it is known as double fertilisation

Biocontrol refers to the use of biological methods for controlling plant diseases and pests. Bacillus thuriengiensis (Bt), Trichoderma sps and Baculoviruses are used as biocontrol agents.

Rhizobium, Nostoc, Azospirillum and Oscillatoria are used as biofertilisers, which increases the fertility of soil while Tobacco Mosaic virus is a pathogen that infects wide range of plants and aphids are pest that harms crop plants.

Protoxin are converted into active toxins in the alkaline pH of gut.

Lactobacillus is used for production of curd, Saccharomyces cerevisiae is used for making bread, Aspergillus niger is used in citric acid production while Acetobacter aceti is used in acetic acid production.

A biological control being developed for use in the treatment of plant disease is the fungus is trichoderma trichoderma species are free living fungi that are very common in the root ecosystems. They are effective biocontrol agents of several plant pathogens.

Streptokinase is used as a ‘clot buster’ to remove clots from the blood vessels of patients who have myocardial infarction.

Lipases are used in detergent formulations. It helps to remove oily stains from the laundry.

Cyclosporin A is produced by Trichoderma polysporum (fungus) and used as an immunosuppressive agent in organ transplant patients.

Statins are produced by Monascus purpureus yeast. It is used as blood-cholesterol lowering agents. It inhibits the enzymes responsible for synthesis of cholesterol.

Thiobacillus denitrificans cause denitrification. Denitrification is the process of conversion of oxides of nitrogen to free nitrogen by bacteria present in the soil.

In gymnosperms only sieve cells and albuminous cells are present but they lack sieve tubes and companion cells.

The direction of movement of sugars in phloem is Bi-directional.