NEET-UG 2012

${\left({\mu }_0{\epsilon }_0\right)}^{-1/2}$ = $\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ = Speed of light in vacuum = c

So the dimension of ${\left({\mu }_0{\epsilon }_0\right)}^{-1/2}$ = [c] =[LT^-1]

When a stone is dropped from a height h, it hits the ground with a momentum

$P=m\sqrt{2gh}$     ....(1)

where m is the mass of the stone.
When the same stone is dropped from a height 2h (i.e. 100% of initial), then its momentum with which it hits the ground becomes

$P^{\prime }=m\sqrt{2g\left(2h\right)}=\sqrt{2}P$     [Using (1)]     ...(2)

% change in momentum = $\frac{P^{\prime }-P}{P}\times 100\%$

$=\frac{\sqrt{2}P-P}{P}\times 100\%=41\%$

Constant power of car P₀ = F.V = ma.v

$P_0=m\frac{dv}{dt}.v$

P₀dt = mvdv Integrating

$P_0.t=\frac{m{v}^2}{2}$

$v=\sqrt{\frac{2P_{0}t}{m}}$

$\because$ P₀ , m and 2 are constant

$\therefore$ $v\propto \sqrt{t}$

The distance of the centre of mass of the system of three masses from the origin O is

$X_{CM}=\frac{m_1{x}_1+m_2{x}_2+m_3{x}_3}{m_1+m_2+m_3}$

$=\frac{300\times 0+500\times 40+400\times 70}{300+500+400}$

$=\frac{500\times 40+400\times 70}{1200}=\frac{400\left[50+70\right]}{1200}$

$=\frac{50+70}{3}=\frac{120}{3}=40cm$

For smooth driving maximum speed of car v then

$\frac{m{v}^2}{R}={\mu }_{s}mg$

$v=\sqrt{{\mu }_{s}Rg}$

According to the theorem of parallel axes,
I = I_CM + Ma²
As a is maximum for point B.
Therefore I is maximum about B.

Using conservation
L_i = 0 (Initial moment)
L_f = mvR – I$\omega$ (Final moment)

According to the conservation of momentum
L_i = L_f

$\Rightarrow$ mvR – I.$\omega$ = 0

mvR = I.$\omega$

$\omega =\left(\frac{1}{2}\right)$

$\left(v+\omega R\right)t=2\pi R$

$t\left(1+\frac{1}{2}\times 2\right)=2\pi \times 2$

t = 2$\pi$ sec.

Escapevelocity, $v_e=\sqrt{\frac{2GM}{R}}$    ...(i)

where M and R where M and R be the mass and radius of the earth respectively. The orbital velocity of a satellite close to the earth’s surface is

$v_0=\sqrt{\frac{GM}{R}}$   ...(ii)

From (i) and (ii), we get

$v_e=\sqrt{2}{v}_0$

Gravitational field due to the thin spherical shell Inside the shell, i.e. (For r < R)

F = 0

On the surface of the shell, i.e. (For r = R)

$F=\frac{GM}{R^2}$

Outside the shell, i.e. (For r > R)

$F=\frac{GM}{r^2}$

The variation of F with distance r from the centre is as shown in the adjacent figure.

Rate of heat given by steam = Rate of heat taken by ice

where K = Thermal conductivity of the slab
m = Mass of the ice
L = Latent heat of melting/fusion
A = Area of the slab

$\frac{dQ}{dt}=\frac{KA\left(100-0\right)}{1}=m\frac{dL}{dt}$,

$\frac{K\times 100\times 0.36}{0.1}=\frac{4.8\times 3.36\times {10}^5}{60\times 60}$

K =1.24 J/m/s/°C

Initial and final condition is same for all process

$\Delta U_1=\Delta U_2=\Delta U_3$

from first law of thermodynamics

$\Delta Q=\Delta U+\Delta W$

Work done

$\Delta W_1>\Delta W_2>\Delta W_3$   (Area of P.V. graph)

So $\Delta Q_1>\Delta Q_2>\Delta Q_3$

Here, Speed of the train, v_T = 220 ms^–1
Speed of sound in air, v = 330 ms^–1
The frequency of the echo detected by the driver of the train is

${\upsilon }^{\prime }=\upsilon \left(\frac{v+v_T}{v-v_T}\right)=1000\left(\frac{330+220}{330-220}\right)$

$=1000\times \frac{550}{110}=5000Hz$

$y=3\sin \frac{\pi }{2}\left(50t-x\right)$

$y=3\sin \left(25\pi t-\frac{\pi }{2}x\right)$ on comparing with the standard wave equation

$y=a\sin \left(\omega t-kx\right)$

Wave velocity $v=\frac{\omega }{k}=\frac{25\pi }{\pi /2}=50m/\sec$

The velocity of particle

$v_p=\frac{\partial y}{\partial t}=75\pi \cos \left(25\pi t-\frac{\pi }{2}x\right)$

$v_{p_{\max }}=75\pi$

then $\frac{v_{p_{\max }}}{v}=\frac{75\pi }{50}=\frac{3\pi }{2}$

At equilibrium potential of both sphere becomes same if charge of sphere one x and other sphere Q – x
then where Q = 4 × 10^–2 C

v₁ = v₂

$\frac{kx}{1cm}=\frac{k\left(Q-x\right)}{3cm}$

3x = Q – x $\Rightarrow$ 4x = Q

$x=\frac{Q}{4}=\frac{4\times {10}^{-2}}{4}C=1\times 10^{-2}$

Q' = Q – x = 3 × 10^–2 C

Current in the circuit, $I=\frac{\epsilon }{R+r}$

Potential difference across R,

$V=IR=\left(\frac{\epsilon }{R+r}\right)R$

$V=\frac{\epsilon }{1+\frac{r}{R}}$

When R = 0, V = 0 and when R = $\infty$, V = $\epsilon$

$\therefore$ option (c) represents the correct graph.

The power dissipated in the circuit

$P=\frac{V^2}{R_{eq}}$    ...(i)

v = 10 volt

$\frac{1}{R_{eq}}=\frac{1}{R}+\frac{1}{5}=\frac{5+R}{5R}$

$R_{eq}=\left(\frac{5R}{5+R}\right)$

P = 30 W
Substituting the values in equation (i)

$30=\frac{{\left(10\right)}^2}{\left(\frac{5R}{5+R}\right)}$

$\frac{15R}{5+R}=10$

15R = 50 + 10R

5R = 50

R = 10 $\Omega$

Capacitance of a parallel plate capacitor is

$C=\frac{{\epsilon }_{0}A}{d}$   ...(i)

Potential difference between the plates is

V = Ed    ...(ii)

The energy stored in the capacitor is

$U=\frac{1}{2}C{V}^2=\frac{1}{2}\left(\frac{{\epsilon }_{0}A}{d}\right){\left(Ed\right)}^2$   (Using (i) and (ii))

$=\frac{1}{2}{\epsilon }_0{E}^{2}Ad$

According to the principal of circular motion in a magnetic field

$F_c=F_m\Rightarrow \frac{m{v}^2}{R}=qVB$

$\Rightarrow R=\frac{mv}{qB}=\frac{P}{qB}=\frac{\sqrt{2m.k}}{qB}$

$R_{\alpha }=\frac{\sqrt{2\left(4m\right)K^{\prime }}}{2qB}$

$\frac{R}{R_{\alpha }}=\sqrt{\frac{K}{K^{\prime }}}$

but $R=R_{\alpha }$ (given)

$\therefore$ K = K' = 1 MeV

According to work energy theorem

W = U_final – U_initial = MB (cos 0 – cos 60°) = $\frac{MB}{2}$ = $\sqrt{3}J$ ....(1)

The torque on the needle is

$\overrightarrow{\tau }=\overrightarrow{M}\times \overrightarrow{B}=MB\sin \theta$

= $MB\sin 60^{\circ }$ = $\frac{MB\sqrt{3}}{2}$ ....(2)

From equation (1) and (2),

$\tau$ = $\frac{2\sqrt{3}\times \sqrt{3}}{2}$ = 3 J

q = Area under i-t graph

= $\frac{1}{2}\times 4\times 0.1$ = 0.2 C

As q = $\frac{\Delta \varphi }{R}$

$\Rightarrow$ $\Delta$$\varphi$ = qR = 0.2$\times$10 = 2 weber

i_rms = $\frac{i_0}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$ A

e_rms = $\frac{e_0}{\sqrt{2}}=\frac{\frac{1}{\sqrt{2}}}{\sqrt{2}}=\frac{1}{2}$ V

Average power consumed in the circuit,

P = i_rmse_rms cos$\varphi$

= $\left(\frac{1}{2}\right)\left(\frac{1}{2}\right)\cos \frac{\pi }{3}$ = $\frac{1}{8}$ W

If the speed of light c then

E₀ = B₀c

$\Rightarrow$ $\frac{B_0}{E_0}=\frac{1}{c}$

Hence ratio of amplitude of magnetic field to amplitude of electric field for electromagnetic wave in a vacuum is reciprocal of speed of light.

$\mu$ = $\frac{\sin \left(\frac{{\delta }_m+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

= $\frac{\sin \left(\frac{A+A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

= $\frac{\sin A}{\sin \left(\frac{A}{2}\right)}$

= $\frac{2\sin \left(\frac{A}{2}\right)\cos \left(\frac{A}{2}\right)}{\sin \left(\frac{A}{2}\right)}$

= $2\cos \left(\frac{A}{2}\right)$

The angle of minimum deviation is given as

$\delta$_min = i + e – A

for minimum deviation

$\delta$_min = A and i = e then

2A = i + i

$\Rightarrow$ i = A

i_min = 0 = A_min $\Rightarrow$ $\mu$_min = 2

i_max = 90⁰ = A_max $\Rightarrow$ $\mu$_max = $\sqrt{2}$

$\therefore$ $\mu$ lies between 2 and $\sqrt{2}$.

Image position of end A

$\frac{1}{v_A}+\frac{1}{-20}=\frac{1}{-10}$

$\Rightarrow$ v_A = - 20 cm

Image position of the end B

$\frac{1}{v_B}+\frac{1}{-30}=\frac{1}{-10}$

$\Rightarrow$ v_B = -15 cm

Length of the image is

L_A'B' = |v_A| - |v_B| = 20 - 15 = 5 cm

The frequency of the transition v $\propto$ $\frac{1}{n^2}$

when n = 1, 2, 3.

At time t₂, $\frac{2}{3}$ of the sample had decayed

$\therefore$ N = $\frac{1}{3}$N₀

$\therefore$ $\frac{1}{3}$N₀ = N₀$e^{-\lambda t_2}$ ......(1)

At time t₁, $\frac{1}{3}$ of the sample had decayed

$\therefore$ N = $\frac{2}{3}$N₀

$\therefore$ $\frac{2}{3}$N₀ = N₀$e^{-\lambda t_1}$ ......(2)

Divide (i) by (ii), we get

$\frac{1}{2}=\frac{e^{-\lambda t_2}}{e^{-\lambda t_1}}$

$\Rightarrow$ $\frac{1}{2}=e^{-\lambda \left(t_2-t_1\right)}$

$\Rightarrow$ $\lambda \left(t_2-t_1\right)$ = ln 2

$\Rightarrow$ $\left(t_2-t_1\right)$ = $\frac{\ln 2}{\lambda }$ = $\frac{\ln 2}{\frac{\ln 2}{T_{1/2}}}$ = T_1/2 = 50 days

According to Einsten’s photoelectric effect, the K.E. of the radiated electrons

K.E_max = E – W

$\frac{1}{2}m{v}_1^2$ = (1 - 0.5) eV = 0.5 eV

$\frac{1}{2}m{v}_2^2$ = (2.5 - 0.5) eV = 2 eV

$\frac{v_1}{v_2}=\sqrt{\frac{0.5}{2}}$ = 1 : 2

de Broglie wavelength associated with an electron is

$\lambda =\frac{h}{P}$

$\Rightarrow$ P = $\frac{h}{\lambda }$

$\frac{\Delta P}{P}=-\frac{\Delta \lambda }{\lambda }$

$\frac{P}{P_{initial}}=\frac{0.5}{100}$

P_initial = 200P

The Boolen expression for the given combination is

output Y = (A + B) · C

So it is clear that Y = 1, when A = 1, B = 0 and C = 1

Current gain ($\beta$) :

$\beta$ = $\frac{\Delta I_C}{\Delta I_B}$

= $\frac{2\times {10}^{-3}}{40\times {10}^{-6}}$ = 50

Voltage gain of the amplifier is

A_V = $\beta \frac{R_L}{R_i}$

= 50 $\times$ $\frac{4\times {10}^3}{100}$ = 2000

Orbital angular momentum (m) = $\sqrt{l(l+1)}\frac{h}{2\pi }$

For p-electrons; l = 1

Thus, m = $\sqrt{l(l+1)}\frac{h}{2\pi }$ = $\frac{\sqrt{2}h}{2\pi }$ = $\frac{h}{\sqrt{2}\pi }$

Van der Waal gas constant '$a$' represent intermolecular force of attraction of gaseous molecules and Van der Waal gas constant 'b' represent effective size of molecules . Therefore order should be

(I) H₂ < He < O₂ < CO₂   (II) CH₄ > O₂ > H₂

According to Graham's law of diffusion

$r\propto \frac{1}{\sqrt{d}}\propto \frac{1}{\sqrt{M}}$

$\Rightarrow \frac{r_1}{r_2}=\sqrt{\frac{M_2}{M_1}}$

Rate of diffusion = $\frac{Volumeofgasdiffused(V)}{Timestaken(t)}$

$\therefore$ $\frac{V_1/t_1}{V_2/t_2}=\sqrt{\frac{M_2}{M_1}}$

If same volume of two gases diffuse then V₁ = V₂

$\Rightarrow$ $\frac{t_1}{t_2}=\sqrt{\frac{M_2}{M_1}}$

Here t₂ = 3t, M₁ = 4 u, M₂ = ?

$\therefore \frac{3t_1}{t_1}=\sqrt{\frac{M_2}{4}}\Rightarrow 3=\sqrt{\frac{M_2}{4}}$

$\Rightarrow 9=\frac{M_2}{4}\Rightarrow M_2=36u$

C_p for monoatomic gas mixture of same volume = $\frac{5}{2}R$

$\therefore$ C_V = $\frac{3}{2}R$

$\Rightarrow \frac{C_P}{C_V}=\frac{\frac{5}{2}R}{\frac{3}{2}R}=\frac{5}{3}=1.67$

A₂(g) + B_2(g) $\rightleftharpoons$ 2AB_(g);

K_c = $\frac{{\left[AB\right]}^2}{\left[A_2\right]\left[B_2\right]}$

= $\frac{{\left(2.8\times {10}^{-3}\right)}^2}{3\times {10}^{-3}\times 4.2\times {10}^{-3}}$

= 0.62

2SO_2(g) + O_2(g) $\rightleftharpoons$ 2SO_3(g),    K

SO_3(g) $\rightleftharpoons$ SO_2(g) + $\frac{1}{2}$ O_2(g),     K' = $\sqrt{\frac{1}{K}}$

$\therefore$ K' = $\sqrt{\frac{1}{278}}=$ 6.0 $\times$ 10^$-$2

n_CHCl3 = $\frac{25.5}{119.5}$ = 0.213

n_CH2Cl2 = $\frac{40}{85}$ = 0.47

P_T = P^o_AX_A + P^o_BX_B

= $200\times \frac{0.213}{0.683}+41.5\times \frac{0.47}{0.683}$

= 62 + 28.55

= 90.63

We know,

$\Delta$G^o = – nFE^o

$\frac{2}{3}$ Al₂O₃ $\to$ $\frac{4}{3}$ Al + O₂

Total number of Al atoms in Al₂O₃

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

Al³⁺ + 3e^– $\to$ Al

As 3e^– change occur for each Al atom

$\therefore$ n = $\frac{4}{3}\times 3=4$

E^o = - $\frac{\Delta G^{\circ }}{nF}$

= - $\frac{960\times 1000}{4\times 96500}$

= - 2.5 V

Higher the value of reduction potential higher will be the oxidising power whereas the lower the value of reduction potential higher will be the reducing power.

$$\Lambda$$^o_CH3COOH = $\Lambda$^o_CH3COONa + $\Lambda$^o_HCl - $\Lambda$^o_NaCl

= = 91 + 425.9 – 126.4 = 390.5

Let $k_1=A{e}^{-\frac{E_a}{R{T}_1}}$

$\ln k_1=\ln A-\frac{E_a}{R{T}_1}$ ......(1)

$k_2=A{e}^{-\frac{E_a}{R{T}_2}}$

$\ln k_2=\ln A-\frac{E_a}{R{T}_2}$ ....(2)

From eq.(1) and (2), we have

$\ln k_2-\ln k_1=\ln A-\frac{E_a}{R{T}_2}-\ln A+\frac{E_a}{R{T}_1}$

$\Rightarrow$ $\ln \frac{k_2}{k_1}=\frac{E_a}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)$

$\Rightarrow$ $\ln \frac{k_2}{k_1}=-\frac{E_a}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$

Number of atoms in cubic close packing = 4 = O^2-

Number of tetrahedral voids = 2 × N = 2 × 4

Number of A²⁺ ions = 8 $\times$ $\frac{1}{4}$ = 2

Number of octahedral voids = Number of B⁺ ions = N = 4

Ratio of ions will be,

O^2– : A²⁺ : B⁺ = 4 : 2 : 4 = 2 : 1 : 2

Formula of oxide = AB₂O₂

Electronic configuration of O₂

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

Thus the incoming electron will enter in ${\pi }^{\ast }2p_x$ to form $O_2^{+}$

Diatomic species	Bond order
NO	2.5
$$O_2^{-}$$	1.5
$$C_2^{2-}$$	3.0
$$H{e}_2^{+}$$	0.5

The increasing order: $H{e}_2^{+}<{O_2}^{-}<NO<{C_2}^{2-}$

If acidic nature is high, K_a is high and pK_a is low.

since pK_a = – log K_a

Hence the order of pK_a will be

H₂O > H₂S > H₂Se > H₂Te

U exhibits + 3, + 4, + 5, + 6

Th exhibits + 3, + 4 ;

Ac exhibits + 3 only;

Pa exhibits + 3, + 4, + 5

The transition metals and their compounds are used as catalysts because of the variable oxidation states. Due to this, they easily absorb and re-emit wide range of energy to provide the necessary activation energy.

$\left(E_{C{o}^{2+}/Co}^o\right)$ = -0.28

$\left(E_{N{i}^{2+}/Ni}^o\right)$ = -0.25

$\left(E_{C{u}^{2+}/Cu}^o\right)$ = +0.34

$\left(E_{F{e}^{2+}/Fe}^o\right)$ = -0.44

The melting points of the transition elements first rise to a maximum and then fall as the atomic number increases, manganese have abnormally low melting points.

Ni⁺² + 2Dmg– → [Ni(Dmg)₂]


Dimethyl glyoxime act as bidentate ligand.

For low spin d⁶ complex electronic configuration

= $t_{2g}^6{e}_g^0$

$\therefore$ x = 6, y = 0, z = 3

C.F.S.E. = (– 0.4 × 6 + 0 × 0.6)$\Delta$₀ + 3P

= $-\frac{12}{5}{\Delta }_0+3P$

Tertiary alkyl halides shows S_N1 mechanism to the greater extent. In the given reaction negative ion will attack on carbocation. Thus, greater the tendency of ionization (greater ionic character in M-F bond) more favourable will be reaction. In the given options Rb-F is most ionic and hence it will be most favourable for S_N1 mechanism.

Glycol is used as an antifreeze in automobiles.

It is Rosenmund’s reaction.

RCHO + NH₂NH₂ $\to$ RCH $=$ N $-$ NH₂

Such reactions take place in slightly acidic medium and involve nucleophilic addition of the ammonia derivative.

As A gives alcohol on treatment with nitrous acid thus it should be primary amine. C₃H₉N has two possible structure with –NH₂ group.



As it gives isopropylmethylamine thus it should be isopropyl amine not n-propyl amine.

Biodegradable polymer is Nylon-2-Nylon-6 which is copolymer of glycine (H₂N – CH₂– COOH) and amino caproic acid (H₂N–(CH₂)₅ – COOH).

Chloramphenicol is an antibiotic broad spectrum.

Ribosome are small naked (non membrane bound) particles made of r-RNA and proteins. Ribosomes are also seen in the organelles like mitochondria and chloroplasts. They are the cell’s protein factories and are found on RER and scattered in the cytoplasm as polyribosomes. Ribosomes are the sites at which information carried in the genetic code is converted into protein molecules.

Thylakoids are the flattened sac-like membranous structures that are stacked on top of one another to form the grana of plant chloroplast. Chlorophyll and other photosynthetic pigments are situated in the thylakoid membranes, which are the site for the light-dependent reactions of photosynthesis.

Palmitic acid is one of the most common, saturated fatty acids found in animals and plants. It has 16 carbons including the carboxyl carbon. Adenylic acid is a nucleotide consisting of adenine, ribose or deoxyribose, and a phosphate group. It is a constituent of DNA or RNA. It is also called adenosine monophosphate. Amino acids are organic acids (with carboxylic group COOH) having amino group (–NH₂) generally attached to Carbon or carbon next to carboxylic group. The carbon also bears a variable alkyl group (R) or hydrogen or hydrocarbon. In alanine (R) is represented by methyl group.

During anaphase I, from each tetrad two chromatids of a chromosome move as a unit (dyad) to one pole of a spindle and the remaining two chromatids of its homologue migrate to the opposite pole. Thus, the homologous chromosomes of each pair, rather than the chromatids of a chromosome, are separated. As a result, half of the chromosomes, which appear in early prophase, go to each pole. Thus the paternal and maternal chromosomes of each homologous pair segregate during anaphase I independently of the other chromosomes.

Anemophily is an abiotic means of pollination by wind and being non-directional, a wasteful process as the pollen would reach the stigma through wind is a hit-or-miss affair. During the transit of pollen through wind, a considerable amount of pollen is lost because it never reaches a proper stigma. To stand this loss, anemophilous plants have to produce enormous quantities of pollen.

Anemophily is also associated with reduction in the number of ovules per ovary. Some models predict that plants benefit from numerous inexpensive flowers distributed throughout the inflorescence, each with a single ovule or a few ovules. In grasses there is just one ovule per ovary. This is to increase the probability of successful pollination of each ovule.

In over 60 per cent of angiosperms, pollen grains are shed at cell 2-celled stage. In the remaining species the generative cell divides mitotically to give rise to the two male gametes before pollen grains are shed (3-celled stage.)

In a pollen grain, exine is thin or absent at certain places. These areas may have thickened intine or deposition of callose. They are called germ pores (if rounded) or germinal furrows (if elongated). After pollination, the pollen grain on the stigma absorbs water and nutrients from the stigmatic secretion through its germ pores. The tube or vegetative cell enlarges and comes out of pollen grains through germ pore to form a pollen tube.

Spirulina is rich in protein, vitamins & minerals. 250 gram biomass of Methylophilus methylotrophus produces 25 tonne protein/day while cow of 250 Kg produces only 200 gm. protein/day.
Common button mushrooms are a very rich source of vitamin D. A rice variety has been developed which is very rich in iron content.

Sewage treatment is the process of removing contaminants from wastewater and household sewage, both runoff (effluents) and domestic. It includes physical, chemical, and biological processes to remove physical, chemical and biological contaminants. Sewage treatment generally involves three stages, called primary, secondary and tertiary treatment. It is processed by aerobic and then anaerobic bacteria in the secondary treatment in Sewage Treatment Plants (STPs).

In gobar gas the maximum amount of methane which is produced by methanogenic bacteria. Gobar gas is a gas mixture which is generated when organic compounds are fermented in the absence of air (anaerobic fermentation). Biogas contains 50% – 70% methane, 30% – 40% carbon dioxide, and traces of hydrogen, hydrogen sulphide and nitrogen. Methane is a combustible gas, which means it can be burned. It can be used as a fuel for cooking and lighting.

The vascular bundles are arranged in a loose circle inside the endodermis of a monocot root. In a monocot root, more than six vascular bundles are present. It shows polyarch condition.

Nitrogenase enzyme is present in prokaryotic nitrogen fixers. The enzyme nitrogenase requires a high input of energy to carry out biological nitrogen fixation. This can be illustrated by the following equation.

PCR is a technique for enzymatically replicating DNA without using a living organism such as E. coli or yeast. It is commonly used in medical and biological research labs for a variety of tasks like detection of hereditary diseases, identification of genetic fingerprints etc.
The correct steps shown in the above figure are:
A – Denaturation at a temperature of about 94° to 98°C. During the denaturation, the double strand melts open to single stranded DNA, and all enzymatic reactions stop.
B – Annealing (binding of DNA primer to the separated strands. Occurs at 50° to 65°Celsius, which is lower than the optimal temperature of the DNA polymerases)
C – Extension or elongation of the strands using the DNA primer with heat-stable DNA polymerases, most frequently Taq (Thermus aquaticus) at 72ºC.

A palindromic sequence is a nucleic acid sequence (DNA or RNA) that is the same whether read 5' (five-prime) to 3' (three prime) on one strand or 5' to 3' on the complementary strand with which it forms a double helix.
5' - GAATTC - 3'
3' - CTTAAG - 5'
It is a palindromic sequence of DNA cut by restriction enzyme Eco RI.

Biolistic it is direct gene transferred method for constructing recombinant DNA. The gene gun was invented by John C. Sanford with Edward Wolf. A gene gun can be used to genetically infect cells or whole organisms with foreign DNA by aiming the barrel of the gun and firing. The microshot projectiles in the biolistic gene gun are made of microscopic (or nano) sized gold or platinum powders. These expensive powders are soaked in DNA or RNA (in raw or plasmid form) that are engineered for insertion into the genome of the cells or organisms under the gun.

Antibiotics are powerful medicines that fight bacterial infections. They either kill bacteria or keep them from reproducing. In genetic engineering, the antibiotics are used as selectable markers.

Western ghat is biodiversity rich zone along with endemism. The forests of the western ghats are some of the best representatives of non-equatorial tropical evergreen forests in the world. The Western Ghats have evolved into one of the richest centers of endemism owing to their isolation from other moist areas.

An area with particular types of trees dedicated to local deities or ancestral spirits that are protected by local communities through social traditions and taboos incorporating spiritual and ecological values are called as sacred groves.
Sacred groves act as an ideal centre for biodiversity conservation. Several plants and animals that are threatened in the forest are still well conserved in some of the sacred groves. It has been observed that several medicinal plants that are not to be found in the forest are abundant in the sacred groves. Further, rare, endangered, threatened and endemic species are often concentrated in sacred groves.

Plasmodium falciparum – A protozoan pathogen causes the most serious type of malaria that is falciparum malaria.
Musca domestica – House fly, an insect belongs to phylum arthropoda.
Felis tigris - The Bengal tiger, is well protected in Sundarbans (tiger reserve)
E.coli - Escherichia coli, a commonly occurring bacterium in human intestine

In the five kingdom classification system proposed by R.H. Whittaker, the kingdom Protista includes all unicellular eukaryotic organisms. Both Chlamydomonas and Chlorella are unicellular green algae, which are eukaryotic and photosynthetic organisms. They are placed under Protista because of their unicellular organization and eukaryotic cell structure.

Although they are photosynthetic like plants, they are not included in the kingdom Plantae because Plantae is reserved for multicellular, predominantly terrestrial organisms with specialized tissues and organs.

Answer: Option A – protista

Marginal placentation is a placentation with ovules borne on the wall along the ventral suture of a simple ovary. Gram, Arhar, Sun hemp, Moong, Pea & Lupin belong to fabaceae family that bear marginal placentation.

Onion - Bulb - Underground stems -Imbricate aestivation -Axile placentation - Member of Liliaceae.

Polyandrous condition (having large and indefinite number of stamens) is present in Gloriosa (family liliaceae) and Petunia (family solanaceae). Nitrogen fixation is the conversion of inert atmospheric nitrogen into utilisable compounds of nitrogen like nitrate, ammonia, amino acids, etc. Biological nitrogen fixation is performed by free living and symbiotic bacteria and cyanobacteria.

Symbiotic nitrogen fixers occur in association with roots of higher plants. For example Rhizobium is nitrogen fixing bacterial symbiont of papilionaceous roots and Frankia is symbiont in root nodules of several non-leguminous plants like Casuarina. Both Rhizobium and Frankia live free as aerobes in the soil and develop the ability to fix nitrogen only as symbionts when they become anaerobic.

Test cross is performed to determine the genotype of F₂ plant. In a typical test cross an organism showing dominant phenotype and whose genotype is to be determined is crossed with one that is homozygous recessive for the allele being investigated, instead of self-crossing. The progenies of such a cross can easily be analysed to predict the genotype of the test organism.

The inheritance pattern of a particular trait shown in the picture results in haemophilia. Haemophilia is a group of inherited blood disorders in which the blood does not clot properly. It is caused by a fault in one of the genes that determine how the body makes blood clotting factor VIII or IX. These genes are located on the X chromosome.

Change in single base pair of DNA is also a type of mutations called point mutations. It is a type of mutation that causes the replacement of a single base nucleotide with another nucleotide of the genetic material, DNA or RNA. For example, a point mutation is the cause of sickle cell disease.

Let's evaluate each of the statements:

(A) In transcription, adenosine pairs with uracil

This statement is correct. During transcription, adenosine (A) in the DNA pairs with uracil (U) in the RNA.

(B) Regulation of lac operon by repressor is referred to as positive regulation

This statement is incorrect. The regulation of the lac operon by the repressor is an example of negative regulation, where the repressor protein inhibits the expression of the lac genes when lactose is absent.

(C) The human genome has approximately 50,000 genes

This statement is incorrect. The human genome is estimated to have about 20,000-25,000 genes.

(D) Haemophilia is a sex-linked recessive disease

This statement is correct. Haemophilia is indeed a sex-linked recessive disorder, typically associated with mutations on the X chromosome.

Therefore, the correct statements are (A) and (D).

Hence, the number of correct statements is: two.

Option B: Two

Green revolution is the introduction of high-yielding varieties of seeds and the increased use of fertilizers and irrigation, which provided the increase in production needed to make India self-sufficient in food grains, thus improving agriculture in India. Green revolution in India occurred in 1960.

RNA interference technique, sense & antisense RNA fused to form dsRNA that silent the expression of m- RNA of nematode. RNA interference is a novel strategy adopted to prevent infestation of nematode, Meloidegyne incognitia in roots of tobacco plants.

DNA fingerprinting is a test to identify and evaluate the genetic information-called DNA (deoxyribonucleic acid)-in a person’s cells. DNA fingerprinting is a form of identification based on sequencing specific non-coding portions of DNA that are known to have a high degree of variability from person to person. These sections are known as Tandem repeats. The test is used to determine whether a family relationship exists between two people, to identify organisms causing a disease, and to solve crimes.

Gene therapy is an experimental technique that uses genes to treat or prevent disease. The first clinical gene therapy was given for treating adenosine deaminase deficiency. A four-year old girl became the first gene therapy patient on September 14, 1990 at the NIH Clinical Center. Adenosine deaminase deficiency, also called ADA deficiency or ADA-SCID is an autosomal recessive metabolic disorder that causes immunodeficiency. ADA deficiency is due to a lack of the enzyme adenosine deaminase.

Cuscuta, or Dodder plant, is a parasitic plant that wraps around other plants for nourishment. Cuscuta is found on outer side of the host and is total stem parasite. Cuscuta a parasite of Acacia.

Viroids: T. O. Diener discovered that potato tuber spindle disease is caused by a new infectious agent which was smaller than virus. It was found to be free RNA (low molecular weight) later and given the term viroid.

Mustard: An angiosperm, therefore contain synergids (part of ovule).

Ginkgo: A gymnosperm, therefore possess archegonia.

Autotrophs are those organisms that are able to make energy-containing organic molecules from inorganic raw material by using basic sunlight. Nostoc, Chara, Porphyra and Wolffia are photoautotrophs while Nitrosomonas and Nitrobacter are chemoautotrophs.