NEET-UG 2011

In CGS system,

Density = 4$\frac{g}{c{m}^3}$

When unit of mass is 100g and unit of length is 10 cm, then

Density = x $\frac{100g}{{\left(10cm\right)}^3}$

$\therefore$ 4$\frac{g}{c{m}^3}$ = x $\frac{100g}{{\left(10cm\right)}^3}$

$\Rightarrow$ x = 40 unit

Let total distance = 2S

Let particle take t₁ time to cover first S distance,

$\therefore$ t₁ = $\frac{S}{v_1}$

Let particle take t₂ time to cover last S distance,

$\therefore$ t₂ = $\frac{S}{v_2}$

$\therefore$ Average speed = $\frac{2S}{\frac{S}{v_1}+\frac{S}{v_2}}$ = $\frac{2v_1{v}_2}{v_1+v_2}$

Let $\varphi$ be elevation angle of the projectile at its highest point as seen from the point of projection O and $\theta$ be angle of projection with the horizontal.

From figure, tan$\varphi$ = $\frac{H}{R/2}$      ....(1)

In case of projectile motion

Maximum height H = $\frac{u^2{\sin }^2\theta }{2g}$

Horizontal range, R = $\frac{u^2\sin \theta }{g}$

Substituting these values of H and R in (1), we get

$\tan \varphi =\frac{\frac{u^2{\sin }^2\theta }{2g}}{\frac{u^2\sin 2\theta }{2g}}$

$\tan \varphi =\frac{{\sin }^2\theta }{\sin 2\theta }=\frac{{\sin }^2\theta }{2\sin \theta \cos \theta }$

= $\frac{1}{2}\tan \theta$

$\tan \varphi =\frac{1}{2}\tan 45^{\circ }=\frac{1}{2}$

Here, $\theta$ = 45°

$\therefore$ $\tan \varphi =\frac{1}{2}\tan 45^{\circ }=\frac{1}{2}$    $\left(\because \tan 45^{\circ }=1\right)$

$\varphi ={\tan }^{-1}\left(\frac{1}{2}\right)$

Force of friction, f = $\mu$mg

$\therefore$ $a=\frac{f}{m}=\frac{\mu mg}{m}=\mu g$

$=0.5\times 10=5$ ms^-2

Using v² – u² = 2aS

0² – 2² = 2(–5) $\times$ S

$\Rightarrow$ S = 0.4 m

According to conservation of momentum, we get
$mv\hat{i}+\left(3m\right)2v\hat{j}=\left(m+3m\right)\overrightarrow{v^{\prime }}$

where $\overrightarrow{v^{\prime }}$ is the final velocity after collision
$\overrightarrow{v^{\prime }}=\frac{1}{4}v\hat{i}+\frac{6}{4}v\hat{j}=\frac{1}{4}v\hat{i}+\frac{3}{2}v\hat{j}$

According to law of conservation of angular momentum

mvr = mv'r'

$vr=v^{\prime }\left(\frac{r}{2}\right)$

$v^{\prime }=2v$    ...(i)

$\therefore$ $\frac{K}{K^{\prime }}=\frac{\frac{1}{2}m{v}^2}{\frac{1}{2}mv{}^{\prime 2}}={\left(\frac{v}{v^{\prime }}\right)}^2$

$\Rightarrow \frac{K^{\prime }}{K}={\left(\frac{v^{\prime }}{v}\right)}^2={\left(2\right)}^2$    (Using (i))

$\Rightarrow K^{\prime }=4K$

Here, Mass of a particle = M
Mass of a spherical shell = M
Radius of a spherical shell = a
Let O be centre of a spherical shell.
Gravitational potential at point P due to particle at O is

$V_1=-\frac{GM}{a/2}$

Gravitational potential at point P due to spherical shell is

$V_2=-\frac{GM}{a}$

Hence, total gravitational potential at point P is V = V₁ + V₂

$=-\frac{GM}{a/2}+\left(-\frac{GM}{a}\right)=-\frac{2GM}{a}-\frac{GM}{a}=-\frac{3GM}{a}$

$\left|V\right|=\frac{3GM}{a}$

According to law of conservation of mechanical energy

$\frac{1}{2}m{u}^2-\frac{GMm}{R}=0$

$\Rightarrow$ $u^2=\frac{2GM}{R}$

$\therefore$ $u=\sqrt{\frac{2GM}{R}}=\sqrt{2gR}$   $\left(\because g=\frac{GM}{R^2}\right)$

T₁ = 273 + 27 = 300K

T₂ = 273 + 927 = 1200K

For adiabatic process,

$P^{1-\gamma }{T}^{\gamma }$ = constant

$\Rightarrow {P_1}^{1-\gamma }{T_1}^{\gamma }={P_2}^{1-\gamma }{T_2}^{\gamma }$

$\Rightarrow {\left(\frac{P_2}{P_1}\right)}^{1-\gamma }={\left(\frac{T_1}{T_2}\right)}^{\gamma }$

$\Rightarrow {\left(\frac{P_1}{T_2}\right)}^{1-\gamma }={\left(\frac{T_2}{T_1}\right)}^{\gamma }$

${\left(\frac{P_1}{P_2}\right)}^{1-1.4}={\left(\frac{1200}{300}\right)}^{1.4}$

${\left(\frac{P_1}{P_2}\right)}^{-0.4}={\left(4\right)}^{1.4}$

${\left(\frac{P_2}{P_1}\right)}^{0.4}=4^{1.4}$

$P_2=P_1{4}^{\left(\frac{1.4}{0.4}\right)}=P_1{4}^{\left(\frac{7}{2}\right)}$

= P₁ (2⁷) = 2 × 128 = 256 atm

Equation of SHM is given by $x=A\sin \left(\omega t+\delta \right)$
$\left(\omega t+\delta \right)$ is called phase.

When x = $\frac{A}{2}$, then

$\sin \left(\omega t+\delta \right)=\frac{1}{2}$

$\Rightarrow \omega t+\delta =\frac{\pi }{6}$

$\Rightarrow {\varphi }_1=\frac{\pi }{6}$

For second particle,


${\varphi }_2=\pi -\frac{\pi }{6}=\frac{5\pi }{6}$

$\therefore$ $\varphi ={\varphi }_2-{\varphi }_1$

$=\frac{4\pi }{6}=\frac{2\pi }{3}$

As $\upsilon =\frac{1}{2L}\sqrt{\frac{T}{\mu }}$

$\therefore \frac{\Delta \upsilon }{\upsilon }=\frac{1}{2}\frac{\Delta T}{T}$

$\frac{\Delta T}{T}=2\frac{\Delta \upsilon }{\upsilon }=2\times \frac{6}{600}=0.02$

$\overrightarrow{E}=-\overrightarrow{\nabla }V$

where $\overrightarrow{\nabla }=\hat{i}\frac{\partial }{\partial x}+\hat{j}\frac{\partial }{\partial y}+\frac{\partial }{\partial z}$

$\overrightarrow{E}=-\left[\hat{i}\frac{\partial V}{\partial x}+\hat{j}\frac{\partial V}{\partial y}+\hat{k}\frac{\partial V}{\partial z}\right]$

Here, V = 4x²
$\therefore \overrightarrow{E}=-8x\hat{i}$

The electric field at point (1, 0, 2) is

${\overrightarrow{E}}_{\left(1,0,2\right)}=-8\hat{i}V{m}^{-1}$

So electric field is along the negative X-axis.

AC = BC

V_D = V_E

We have, W = Q (V_E – V_D)

$\Rightarrow$ W = 0

Current from D to C = 1A

$\therefore$ V_D – V_C = 2 × 1 = 2V

V_A = 0 $\therefore$ V_C = 1V,

$\therefore$ V_D – V_C = 2

$\Rightarrow$ V_D – 1 = 2   $\therefore$ V_D = 3V

$\therefore$ V_D – V_B = 2 $\therefore$ 3 – V_B = 2 $\therefore$ V_B = 1V

For minimum deflection of 1 division required current = 1 µA

$\Rightarrow$ Voltage required = IR = (1µA) (10) = 10 µV

$\therefore$ 40 µV ≡ 1ºC

$\Rightarrow$ 10 µV ≡ $\frac{1}{4}$ ºC = 0.25ºC

$F_1>F_2$ as $F\propto \frac{1}{d}$, and $F_3$ and $F_4$ are equal and opposite. Hence, the net attraction force will be towards the conductor.

When the ring rotates about its axis with a uniform frequency f Hz, the current flowing in the ring is

I = $\frac{q}{T}$ = qf

Magnetic field at the centre of the ring is

B = $\frac{{\mu }_{0}I}{2R}$ = $\frac{{\mu }_{0}qf}{2R}$

To keep the main current in the circuit unchanged, the resistance of the galvanometer should be equal to the net resistance.

$\therefore$ $G=\left(\frac{GS}{G+S}\right)+S^{\prime }$

$\Rightarrow G-\frac{GS}{G+S}=S^{\prime }$

$\therefore S^{\prime }=\frac{G^2}{G+S}$

When a bar magnet of magnetic moment is placed in a uniform magnetic field , its potential energy is

U = $-\overrightarrow{M}.\overrightarrow{B}=MB\cos \theta$

For stable equilibrium, $\theta$ = 0^o

U = –MB = – (0.4) (0.16) = – 0.064 J

Current flowing in the coil is

I = $\frac{200}{Z}=\frac{200}{\sqrt{R^2+{\left(\omega L\right)}^2}}=\frac{200}{\sqrt{{\left(30\right)}^2+{\left(40\right)}^2}}$ = 4 A

V_rms = $\sqrt{\frac{\frac{T}{2}{\left(V_0\right)}^2+0}{T}}=\frac{V_0}{\sqrt{2}}$

Here, v = +15 cm, u = +(15 – 5) = +10 cm

According to lens formula

$\frac{1}{v}-\frac{1}{u}=\frac{1}{f}$

$\Rightarrow$ $\frac{1}{15}-\frac{1}{10}=\frac{1}{f}$

$\Rightarrow$ f = -30 cm

For without deviation

$\frac{A}{A^{\prime }}=\frac{{\mu }^{\prime }-1}{\mu -1}$

$\Rightarrow$ $\frac{15}{A^{\prime }}=\frac{1.75-1}{1.50-1}$

$\Rightarrow$ A' = 10^o

KE_max = 10eV

$\varphi$ = 2.75 eV

Total incident energy

E = $\varphi$ + KE_max = 12.75 eV

$\therefore$ Energy is released when electron jumps from the excited state n to the ground state.

$\therefore$ E₄ – E₁ = {– 0.85 – (–13.6) ev} = 12.75eV

$\therefore$ value of n = 4

Initially P = 4N₀

Q = N₀

Half life T_P = 1 min.

T_Q = 2 min.

Let after time t number of nuclei of P and Q are equal, that is

N_P = N_Q

$\Rightarrow$ $4N_0{\left(\frac{1}{2}\right)}^{t/1}$ = $N_0{\left(\frac{1}{2}\right)}^{t/2}$

$\Rightarrow$ $2^{t/1}=4.2^{t/2}$

$\Rightarrow$ t = 4 min

After 4 minutes, both P and Q have equal number of nuclei.

$\therefore$ Number of nuclei of R

= $\left(4N_0-\frac{N_0}{4}\right)$ + $\left(N_0-\frac{N_0}{4}\right)$

= $\frac{9N_0}{2}$

E_n = $-\frac{13.6}{n^2}$

$\therefore$ E₁ = –13.6 eV

E₂ = –3.4 eV

E₃ = –1.5 eV

E₄ = –0.85 eV

$\therefore$ E₃ – E₂ = –1.5 – (–3.4) = 1.9 eV

E₄ – E₃ = –0.85 – (–1.5) = 0.65 eV

Obviously, difference of 11.1 eV is not possible.

K.E. = h$\nu$ – h$\nu$_th = eV₀ (V0 = cut off voltage)

$\Rightarrow$ V₀ = $\frac{h}{e}\left(8.2\times {10}^{14}-3.3\times {10}^{14}\right)$

= $\frac{6.6\times {10}^{-34}}{1.6\times {10}^{-19}}\left(8.2\times {10}^{14}-3.3\times {10}^{14}\right)$

$\approx$ 2 V

(n_i)² = n_e × n_h

(1.5 × 10¹⁶)² = n_e (4.5 × 10²²)

So n_e = 5 × 10⁹

Now n_h = 4.5 × 10²²

$\Rightarrow$ n_h $>>$ n_e

Hence, semiconductor is p-type

and n_e = 5 × 10⁹ m^–3

p-n junction is said to be forward biased when p side is at high potential than n side. It is for circuit (A) and (C).

Voltage across 250 Ω resistance = 20 V – 15 V = 5 V

Now current through 250 $\Omega$ resistance: 5/250 = 20 mA

If voltage across load resistance 1 k$\Omega$ is 15 V, then

current through 1 kΩ is 15/1000 = 15 mA

The current through the zener diode is

= Current through 250 $\Omega$ resistance – Current through 1 k$\Omega$ resistance.

= 20 - 15

= 5 mA

8 g H₂ has 4 moles while the others has 1 mole each.

We know that

$\Delta E\propto \left[\frac{1}{n_1^2}-\frac{1}{n_2^2}\right]$, where n₂ > n₁

$\therefore$ n = 6 and n = 5 will give least energetic photon.

Given

P₁ = 1.5 bar, T₁ = 273 + 15 = 288 K, V₁ = V

P₂ = 1.0 bar, T₁ = 273 + 25 = 298K, V₂ = ?

$\frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}$

$\Rightarrow$ $\frac{1.5\times V}{288}=\frac{1\times V_2}{298}$

$\Rightarrow$ V₂ = 1.55V

$\therefore$ Volume of bubble will be almost 1.6 time to initial volume of bubble.

K_sp[AgCl] = [Ag⁺][Cl^-]

$\Rightarrow$ [Ag⁺] = $\frac{1.8\times {10}^{-10}}{{10}^{-1}}$ = 1.8 $\times$ 10^$-$9 M

K_sp[PbCl₂] = [Pb²⁺][Cl^-]²

$\Rightarrow$ [Pb²⁺] = $\frac{1.7\times {10}^{-5}}{{10}^{-1}\times {10}^{-1}}$ = 1.7 $\times$ 10^$-$3 M

Osmotic pressure, $\pi$ = CRT

$\Rightarrow$ $\pi$ = $\frac{n}{V}$RT

$\Rightarrow$ $\pi$V = $\frac{w}{M}$RT

M = $\frac{wRT}{\pi V}$

= $\frac{1.26\times 0.083\times 300}{2.57\times {10}^{-3}\times \frac{200}{1000}}$

= $\frac{1.26\times 0.083\times 300}{2.57\times {10}^{-3}\times 0.2}$

= 61038 g mol^–1

We know that $\Delta$T_f = i × K_f × m

Here i is van’t Hoff’s factor.

i for weak acid is 1 + $\alpha$.

Here $\alpha$ is degree of dissociation i.e., 30/100 = 0.3

$\therefore$ i = 1 + $\alpha$ = 1 + 0.3 = 1.3

$\therefore$ T_f = i × K_f × m = 1.3 × 1.86 × 0.1 = 0.24

$\therefore$ Freezing point = – 0.24

	H (kJ/mol)
1/2A $\to$ B	+150
3B  $\to$ 2C + D	-125
E + A  $\to$ 2D	+350

___________________________________
   B + D $\to$ E + 2C;

$\Delta$H = (300 - 125 - 350) = - 175 kJ/mol

Since the reduction potential of Fe³⁺/Fe²⁺ is greater than that of I₂ /I^– , Fe³⁺ will be reduced and I^– will be oxidised.

2Fe³⁺ + 2I^– $\to$ 2Fe²⁺ + I₂

Rate of disappearance of reactants = Rate of appearance of products

$-\frac{1}{2}$$\frac{-d\left[N_2{O}_5\right]}{dt}$ = $\frac{1}{4}$$\frac{d\left[N{O}_2\right]}{dt}$ = $\frac{d\left[O_2\right]}{dt}$

$\Rightarrow$ $\frac{1}{2}$$k\left[N_2{O}_5\right]$ = $\frac{1}{4}$$k^{\prime }\left[N_2{O}_5\right]$ = $k^{\prime \prime }\left[N_2{O}_5\right]$

$\Rightarrow$ $\frac{k}{2}=\frac{k^{\prime }}{4}=k^{\prime \prime }$

$\Rightarrow$ $k^{\prime }=2k;k^{\prime \prime }=k/2$

Rate = K[A]⁰

Unit of k = mol L^–1 sec^–1

We know, $N=N_0{\left(\frac{1}{2}\right)}^n$

Give N₀(original amount) = 1.28 mg L^–1

N (amount of substance left after time T) = 0.04 mg L^–1

$\therefore$ $\frac{0.04}{1.28}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ $\frac{1}{32}={\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ n = 5

$\therefore$ Time required = 5 × t_1/2 = 5 × 138 = 690 s

For NaCl crystal, $\frac{r_{+}}{r_{-}}=0.414$

Given, r₊ = 100 pm

$\therefore$ $\frac{100}{r_{-}}=0.414$

$\Rightarrow$ $r_{-}=\frac{100}{0.414}$ = 241.5 pm

Na $\to$ Na⁺ + e^-; $\Delta$H = 5.1 eV

Na⁺ + e^- $\to$ Na; $\Delta$H = -5.1 eV

has maximum number of covalent bonds.

O₂⁺ = KKσ2s² σ*2s² σ2p_z² (π2p_x² = π2p_y²) (π*2p_x¹)

O₂ = KKσ2s² σ*2s² σ2p_z² (π2p_x² = π2p_y²) (π*2p_x ¹ = π*2p_y¹)

O₂ and O₂⁺ contain unpaired electron in π* ABMO so paramagnetic.

Slag is formed by the reaction

CaO + SiO₂ $\to$ CaSiO₃

(A) Plaster of paris = CaSO₄.$\frac{1}{2}$H₂O

(B) Epsomite = MgSO₄.7H₂O

(C) Kieserite = MgSO₄.H₂O

(D) Gypsum = CaSO₄.2H₂O

Al reacts with NaOH to give sodium metaaluminate.

2Al(s) + 2NaOH (aq) + 2H₂O(l) $\to$ 2NaAlO₂ + 3H₂

SnO₂ reacts with acid as well as base. So SnO₂ is an amphoteric.

SnO₂ + 4HCl $\to$ SnCl₂ + 2H₂O

SnO₂ + 2NaOH $\to$ Na₂SnO₃ + H₂O

CaO is basic in nature while SiO₂ and CO₂ are acidic in nature.

[Cr(NH₃)₆]⁺³ [Ar] 3d³ 4s⁰

three unpaired electron are present in t_2g orbited

Due to positive oxidation state of Mn back donation in π* ABMO of CO is minimum therefore C–O bond is strongest.

$\therefore$ IUPAC name is trans-2-chloro-3-iodo-2-pentene.

Electron withdrawing group like -NO₂ facilitates nucleophilic substitution reaction in chlorobenzene.

These reactions are purely S_N2 reactions as in reaction (i) and (ii) there is no rearrangement takes place (rearrangement occurs in S_N1 mechanism). Simple substitution of nucleophile takes place.

(A) Benzaldehyde          (ii) Benzoin condensation

(B) Phthalic anhydride    (i) Phenolphthalein

(C) Phenyl benzoate       (iv) Fries rearrangement

(D) methyl salicylate      (iii) Oil of wintergreen

The reactivity of the carbonyl group towards the nucleophilic addition reactions depend upon the magnitude of the positive charge on the carbonyl carbon atom (electronic factor) and also on the crowding around the carboxyl carbon atom in the transition state (steric factor). Both these factors predict the following order
I > II > III

In benzylamine the electron pair present on the nitrogen is not delocalised with the benzene ring.

When the proteins are subjected to the action of heat, mineral acids or alkali, the water soluble form of globular protein changes to water insoluble fibrous protein. This is called denaturation of proteins. During denaturation secondary and tertiary structures of protein destroyed but primary structures remains intact.

Vitamin B complex is not a fat soluble vitamin. It is a water soluble vitamin.

Each mitochondrion is a double membranebound structure with the outer membrane and the inner membrane dividing its lumen distinctly into two aqueous compartments, i.e., the outer compartment and the inner compartment. The inner compartment is called the matrix.

The outer membrane forms the continuous limiting boundary of the organelle. The inner membrane forms a number of infoldings called the cristae towards the matrix. The cristae increase the surface area. The two membranes have their own specific enzymes associated with the mitochondrial function.

Except peroxisome the remaining three and ER are the parts of endomembrane system.

Kinetochores are large protein complexes that bind the centromeres of chromosomes to the microtubules of mitotic spindle fibres, during metaphase in the cell cycle.

Apomixis is a reproductive process in plants that superficially resembles normal sexual reproduction but in which there is no fusion of gametes. The embryos develop simply by division of a diploid cell the ovule. So, the progenies produced are identical to the parent. In vegetative reproduction the progenies produced are also identical to the parent.

During megagametogenesis functional megaspore (mostly chalazal) gives rise to embryo sac. This is the mature female gametophyte generation.

Organic farming is the form of agriculture that relies on the techniques like crop rotation, green manure, compost and biological pest control.

Trichoderma is a species of filamentous fungi. Cyclosporin A is immunosuppressive drug obtained from Trichoderma and used in organ transplantation.

Microbes are used for commercial and industrial production of certain chemicals like organic acids, alcohols and enzymes. Examples of acid producers are Aspergillus niger (a fungus) of citric acid, Acetobacter aceti (a bacterium) of acetic acid; Clostridium butylicum (a bacterium) of butyric acid and Lactobacillus (a bacterium) of lactic acid.

In dicot stem, cambium is present between xylem and phloem, such vascular bundles are called open. In monocot stem, the cambium is absent, such vascular bundles are called closed. Cambium are the meristematic cells which produces secondary xylem and phloem.

Companion cells move sugar and amino acids into and out of the sieve elements. In “source” tissue such as leaf companion cells use transmembrane proteins to take up sugar and amino acids by active transport. Movement of sugars in the phloem begins at the source, where sugars are loaded (actively transported) into a seive tube. Loading sets up a water potential gradient that facilitates movement of sugar.

The main cause of guttation in plants is root pressure. During night when root pressure is high, water drops ooze out with the assistance of special structures called hydathodes which help in guttation.

Cadmium is not an essential element for plants. Phosphorus is a macronutrient and iron and manganese are micronutrients for plants.

In respiration, protons accumulate in the intermembrane space of the mitochondria when electrons move through the ETS.

Recombinant DNA technology is the process joining together two DNA molecules from two different species that are inserted into a host organism to produce new genetic combination.

Biodiversity is the number of variety of organism found within a specified geographic region.

All living things have an ability to respond to their environment, that is also called stimulation.

Microsporum is a member of Deuteromycetes of fungi & Rhizopus is also fungi and member of Zygomycetes.

Whorled phyllotaxy is a feature of Nerium and Alstonia. In Alstonia five leaves are present in a whorl while in Nerium three leaves are present in a whorl.

Water hyacinth is a free floating perennial plant, which can grow to a height of 3 feet.

The figure given in option (d) represents the free central placentation. In free central placentation, ovary is unilocular and ovules are borne on the axis in the center of the ovary and septa are absent. It is seen in Dianthus and Primrose.

Sweet potato is homologous to turnip as both are having same origin i.e., both are root but modified for different functions. Sweet potato is a modified root for storage and vegetative propagation while turnip is modified for storage only.

Manganese (Mn²⁺) is used for photolysis of water to produce oxygen and electrons during light reaction of photosynthesis. It is the phenomenon of breaking up of water into hydrogen and oxygen in the illuminated chloroplast. It acts as an essential cofactor.

In Kranz anatomy, the bundle sheath cells have thick wall, no intracellular spaces and large number of chloroplasts.

Two X chromosomes would lead to the birth of normal human female child.

In test cross, genotype of an organism showing dominant phenotype is determined by crossing it with homozygous recessive genotype.

The unequivocal proof that DNA is the genetic material came from the experiments of Alfred Hershey and Martha Chase (1952). They worked with viruses that infect bacteria called bacteriophages.

In this technique nematode specific genes are introduced in the host plant in such a way that it produces both sense and antisense RNA. The two RNA’s being complementary to each other from a double stranded RNA (dsRNA) which is also called interfering RNA responsible for initiating RNA interference (RNA i). This (dsRNA) bind to and prevent translation of specific mRNA of nematode (gene silencing). Thus transgenic plants based on RNAi technology are resistant to nematode.

In 1997, the first transgenic cow, Rosie, produced human protein enriched milk. The milk contained the human alpha-lactalbumin and was nutritionally a more balanced product for human babies than natural cow-milk. Isolation of DNA from other macromolecule is achieved by treating the bacterial cells/plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cells), chitinase (fungus).

Bacillus thuringiensis produces a large amount of crystalline protein during sporulation. In the cell toxins are formed along with the spore and are referred to as parasporal body. The bacteria are capable of entering the insect’s blood and using the host insect to reproduce. The proteins from ingested spores are activated by gut, high pH and the polypeptide toxins destroy gut epithelial cells and kill the pest.

The logistic population growth is expressed by the equation $\frac{dN}{dt}=rN\left(\frac{\left(K-N\right)}{K}\right)$ where N is population density at time t, r is the Malthusian parameter (rate of maximum population growth) and K is the so called carrying capacity (i.e. maximum sustainable population). It is a type of population growth when resources are limiting.

Bad ozone is formed in the lower atmosphere (troposphere) that harms plants and animals. Good ozone is found in the upper part of the atmosphere called the stratosphere and it acts as a shield absorbing ultraviolet radiation from the sun. UV rays are highly injurious to living organisms since DNA and proteins of living organisms preferentially absorb UV rays and its high energy breaks the chemical bonds within these molecules.

The given figure is of female thallus of Marchantia (bryophyte) in which A, B, C and D are archegoniophore, female thallus, gemmacup and rhizoids respectively.

The sporophyte in mosses is more elaborate than that in liverworts. The male and female cones or strobili is borne on same tree in (Pinus). In Cycas male cones and megasporophylls are borne on different trees.

Selaginella and Salvinia are advanced pteridophytes.

The break down of detritus into smaller particles by earthworm is known as fragmentation.

Hydrarch succession takes place in wetter areas and the successional series progress from hydric to the mesic condition . Xerarch succession takes place in dry areas and the series progress from xeric to mesic condition. Hence, both hydrarch and xerarch succession leads to medium water conditions (mesic).

Sparrow can be herbivorous (eating seeds and fruits) or carnivorous (eating insects).

Pneumonia disease is spread by the organism Streptococcus pneumoniae and the mode of infection is by droplet infection.

Common cold is due to rhinovirus. Recent studies found that antibiotics do not kill this virus.

By counting the number of QRS complexes that occur in a given time period, one can determine the heart beat rate (pulse) of an individual. The QRS complex represents the depolarisation of the ventricles, which initiates the ventricular contraction.

Cardiac muscles are found in the wall of the heart. It is involuntary and slightly striated. Smooth muscles are found in gastrointestinal tract. These are non-striated and involuntary. Striated (or skeletal) muscles are found in the limbs and body walls. These muscles are voluntary (under the control of animals’ will) and show dark and light bands thus are striated.

Gamete Intra Fallopian Transfer (GIFT) is transfer of an ovum collected from a donor into the Fallopian tube of another female who cannot produce ova but can provide proper environment for fertilization and further development.

Clitellum - secretes cocoon during breading season of earthworm. Gizzard is used for- grinding of food particles. Setae help in locomotion. Typhlosole increases the absorption area in intestine.

Human possesses enucleated RBCs in mature state. But frog blood has both white and red blood cells which are nucleated. Frog cells do not lack platelets.

Simple squamous epithelium is composed of large flat cells whose edges fit closely together like the tiles in a floor, hence it is also called pavement epithelium. The nuclei of the cells are flattened and often lie at the centre of the cells and cause bulgings of cells surface. The epithelium lines the blood vessels, lymph vessels, heart, terminal bronchioles, alveoli of the lungs, walls of the Bowman’s capsules, descending limbs of loop of Henle. In the blood vessels and heart it is called endothelium.

The adult dental formula of human is Incisor $\frac{2}{2}$, Canine $\frac{1}{1}$, Premolar $\frac{2}{2}$, Molar $\frac{3}{3}$..

Duodenum follows the stomach. It is somewhat C-shaped and about 25 cm. long. It receives the hepatopancreatic ampulla of the hepatopancreatic duct formed by the union of bile duct (from liver) and pancreatic duct (from pancreas) and whose opening is guarded by sphincter of Oddi. Pancreatic juice contains proenzymes—trypsinogen, chymotrypsinogen and procarboxypeptidase. In the presence of enterokinase (a protease of intestinal juice), inactive trypsinogen is converted to active trypsin. Trypsin then activates chymotrypsinogen and procarboxypeptidase into chymotrypsin and carboxypeptidase respectively. This enables simultaneous action of all pancreatic proteases for a rapid digestion of proteins. The bile provides alkaline medium for various reactions.

Reptiles, birds, land snails and insects excrete nitrogenous wastes as uric acid in the form of pellet or paste with a minimum loss of water and are called uricotelic animals.

Melatonin is a hormone that regulates the 24 hour rhythm of our body such as the sleep-wake cycle etc. Melatonin is produced by pineal gland in the brain, which is responsible for operation and regulation of the biological clock in mammals.

The process of fusion of a sperm with an ovum is called fertilization. During fertilization, a sperm comes in contact with the zona pellucida layer of the ovum and induces changes in the membrane that block the entry of additional sperms. Thus, it ensures that only one sperm can fertilize an ovum. The secretions of the acrosome help the sperm enter into the cytoplasm of the ovum through the zona pellucida and the plasma membrane. In contact with the surface of egg covering, the acrosome releases its contained hydrolytic enzymes, also called sperm lysins. It is known as acrosomal reaction. Acrosome reaction results in dissolving of corona cells and degeneration of zona pellucida which helps in sperm penetration.