NEET-UG 2004

F = $\frac{G{m}_1{m}_2}{r^2}$

$\Rightarrow$ G = $\frac{F\times r^2}{m_1{m}_2}$

$\therefore$ Dimension of G = $\frac{\left[ML{T}^{-2}\right]\left[L^2\right]}{\left[M\right]\left[M\right]}$ = $\left[M^{-1}{L}^3{T}^{-2}\right]$

F = $\frac{1}{4\pi {\epsilon }_0}\frac{q^2}{r^2}$

$\Rightarrow$ ${\epsilon }_0=\frac{q^2}{4\pi F{r}^2}$

$\therefore$ Unit of ${\epsilon }_0$ = coulomb²/newton-metre²

$\left|\overrightarrow{A}\times \overrightarrow{B}\right|=AB\sin \theta$

$\overrightarrow{A}\overrightarrow{B}=AB\cos \theta$

$\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{3}\overrightarrow{A}\overrightarrow{B}$

$\Rightarrow AB\sin \theta =\sqrt{3}AB\cos \theta$

$\Rightarrow \tan \theta =\sqrt{3}$    $\therefore \theta$ = 60^o

$\therefore$ $\left|\overrightarrow{A}\times \overrightarrow{B}\right|=\sqrt{A^2+B^2+2AB\cos 60^{\circ }}$

$=\sqrt{A^2+B^2+AB}$ = (A² + b² + AB)^1/2

Free body diagram of two masses is



We get equations
$T+ma=f_{\mu }$ or $T=\mu N_A(fora=0)$

and T = ma + mg or T = m_Bg (for a = 0)

$\therefore$ $\mu N_A=m_{B}g$

$\Rightarrow m_B=\mu m_A=0.2\times 2=0.4kg$

The wedge is given an acceleration to the left.

$\therefore$ The block has a pseudo acceleration to the right, pressing against the wedge because of which the block is not moving.

$\therefore$ mgsin$\theta$ = macos$\theta$

or $a=\frac{g\sin \theta }{\cos \theta }$

Total reaction of the wedge on the block is N = mgcos$\theta$ + masin$\theta$.

$\Rightarrow N=mg\cos \theta +\frac{mg\sin \theta \sin \theta }{\cos \theta }$

$\Rightarrow N=\frac{mg\left({\cos }^2\theta +{\sin }^2\theta \right)}{\cos \theta }=\frac{mg}{\cos \theta }$

It is observed that if mass colloids with spring in such case, spring will be compressed, so kinetic energy of mass will be equal to that of springs potential energy.

Further, if y is the compression, then

$\frac{1}{2}m{v}^2=\frac{1}{2}k{y}^2$

$\Rightarrow \frac{1}{2}\left(0.5\right)\left(1.5\right)=\frac{1}{2}\left(50\right)y^2$

$\therefore$ y = 0.15 m

Kinetic energy = $\frac{p^2}{2m}$

$\frac{E_1}{E_2}=\frac{p_1^2/2m_1}{p_2^2/2m_2}\Rightarrow \frac{E_1}{E_2}=\frac{m_2}{m_1}$ as m₁ > m₂

$\therefore$ E₁ < E₂

From the question, as height is same for both balls, so their velocities on reaching the ground will be same.

Hence in such case kinetic energy ∝ mass

So, $\frac{K.E_1}{K.E_2}=\frac{m_1}{m_2}=\frac{1}{2}$

$C.M=\frac{m_1{x}_1+m_2{x}_2}{m_1+m_2}$   ...(i)

After changing position of m₁ and to keep the position of C.M. same

$C.M=\frac{m_1\left(x_1-d\right)+m_2\left(x_2+d_2\right)}{m_1+m_2}$

$\Rightarrow 0=\frac{m_{1}d+m_2{d}_2}{m_1+m_2}$
   [Substituting value of C.M. from (i)]

$\Rightarrow d_2=\frac{m_1}{m_2}d$

Radius of gyration of disc about a tangential axis in the plane of disc is $\frac{\sqrt{5}}{2}R=K_1$, radius of gyration of circular ring of same radius about a tangential axis in the plane of circular ring is

$K_2=\sqrt{\frac{3}{2}}R$

$\therefore \frac{K_1}{K_2}=\frac{\sqrt{5}}{\sqrt{6}}$

The initial angular momentum of the system is $I_1\omega$ and final angular momentum is $\left(I_1+I_2\right){\omega }^{\prime }$ where

${\omega }^{\prime }$ = final angular velocity of combination of discs
On equating the initial and final angular momentum, we get

${\omega }^{\prime }=\frac{I_1\omega }{\left(I_1+I_2\right)}$

I_AX = m(AB)² + m(OC)²
= m$l$² + m ($l$ cos 60º)²
= m$l$² + m$l$²/4 = 5/4 m$l$²

${\omega }_f={\omega }_i-\alpha t\Rightarrow 0={\omega }_i-\alpha t$

$\therefore$ $\alpha =\frac{{\omega }_i}{t}$ where $\alpha$ is retardation.

The torque on the wheel is given by

$\tau =I\alpha =\frac{I\omega }{t}=\frac{I.2\pi \upsilon }{t}=\frac{2\times 2\times \pi \times 60}{60\times 60}$

$\tau =\frac{\pi }{15}$ N m

This is the torque required to stop the wheel in 1 min. (or 60 sec.).

From equation of acceleration due to gravity.

$g_e=\frac{G{M}_e}{R_e^2}=\frac{G\left(4/3\right)\pi R_e^3}{R_e^2}{\rho }_e$

$g_e\propto R_e{\rho }_e$

Acceleration due to gravity of planet

$g_p\propto R_p{\rho }_p$

$\therefore$ $R_e{\rho }_e=R_p{\rho }_p\Rightarrow R_e{\rho }_e=R_{p}2{\rho }_e$

$\Rightarrow R_e=\frac{1}{2}R$   ($\because$ R_e = R)

From Wein’s displacement law

${\lambda }_{m}T$ = constant

$\Rightarrow {\lambda }_m\propto T^{-1}$

$T_1=T,W=6Rjoules,\gamma =\frac{5}{3}$

$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=\frac{nR{T}_1-nR{T}_2}{\gamma -1}$

$=\frac{nR\left(T_1-T_2\right)}{\gamma -1}$

$n=1,T_1=T\Rightarrow \frac{R\left(T-T_2\right)}{5/3-1}=6R$

$\Rightarrow T_2=\left(T-4\right)K$

As PV = nRT

$n=\frac{m}{molecularmass}=\frac{5}{32}$

$\Rightarrow PV=\left(\frac{5}{32}\right)RT$

In S.H.M, $v_{\max }=A\omega =A\left(2\pi f\right)$

$f=\frac{v_{\max }}{2\pi A}=\frac{31.4}{2\left(3.14\right)\times 5}=1$ Hz per sec.

$\frac{1}{k_{eq}}=\frac{1}{k_1}+\frac{1}{k_2}\Rightarrow \frac{1}{k_{eq}}=\frac{k_1+k_2}{k_1{k}_2}$

$\Rightarrow k_{eq}=\frac{k_1{k}_2}{k_1+k_2}$

y₁ = 10^–6sin[100t + (x/50) + 0.5]

y₂ = 10^–6cos[100t + (x/50)]
[using cosx = sin(x + $\pi$/2)]

= 10^–6sin[100t + (x/50) + $\pi$/2]

= 10^–6sin[100t + (x/50) + 1.57]

The phase difference = 1.57 – 0.5 = 1.07

[or using sinx = cos($\pi$/2 – x). We get the same result.]

Let f' be the frequency of sound heard by cliff.

$\therefore$ $f=\frac{vf}{v-v_c}$   ...(1)

Now, for the reflected wave, cliff acts as a source,

$\therefore$ $2f=\frac{f^{\prime }\left(v+v_c\right)}{v}$   ...(2)

$2f=\frac{f\left(v+v_c\right)}{v-v_c}$

$\Rightarrow 2v-2v_c=v+v_c$

$\Rightarrow \frac{v}{3}=v_c$

When the dipole is in the direction of field then net force is qE + (–qE) = 0


and its potential energy is minimum = – P.E. = –qaE

Using $\frac{1}{2}m{v}^2=qV$

$V=\frac{1}{2}\times \frac{2\times {10}^{-3}\times 10\times 10}{2\times {10}^{-6}}=50kV$

$R\alpha \ell$

For 300 cm, R = 100 $\Omega$

For 50 cm, $R^{\prime }=\frac{100}{300}\times 50=\frac{50}{3}\Omega$

$\therefore$ IR = 6

$\Rightarrow I{R}^{\prime }=\frac{6}{R}\times R^{\prime }=\frac{6}{100}\times \frac{50}{3}=1volt.$

$P=\frac{V^2}{R}$

$\Rightarrow R=\frac{V^2}{P}=\frac{{\left(220\right)}^2}{60}=\frac{4\times {\left(110\right)}^2}{60}$

$R^{\prime }=\frac{{\left(110\right)}^2}{60}=\frac{R}{4}$

A balanced Wheststone’s bridge exists between A & B.

$\therefore$ R_eq = R

Current through circuit = V/R

Current through AFCEB = V/2R

Resistance of wire = $\rho \frac{l}{A}$

$R\propto \frac{l}{A}=\frac{l}{\pi r^2}$

When length and radius are both doubled

$R_1\propto \frac{2l}{\pi {\left(2r\right)}^2}\Rightarrow R_1\propto \frac{1}{2}R$

The specific resistance of wire is independent of geometry of the wire, it only depends on the material of the wire.

$R=\frac{r}{n}\Rightarrow r=nR$

When connected in series, $R_{eq}=nr$

= n (nR) = n²R

The resistance of each bulb

$=\frac{V^2}{P}=\frac{{\left(200\right)}^2}{60}\Omega$

When three bulbs are connected in series their resultant resistance

$=\frac{3\times {\left(200\right)}^2}{60}$

Thus power drawn by bulb when connected across 200 V supply

$P=\frac{V^2}{R_{re}}=\frac{{\left(200\right)}^2}{3\times {\left(200\right)}^2/60}=20W$

For converting galvanometer to voltmeter, a high resistance should be connected in series with galvanometer.

The total current shown by the galvanometer,

I_g = 25 $\times$ 4 $\times$ 10^-4 A = 10^-2 A

The value of resistance connected in series to convert galvanometer into voltmeter of 25 V is

V = I_g(R_e + R_g)

$\Rightarrow$ R_e = $\frac{V}{I_g}-R_g$ = 2450 $\Omega$

$\frac{\Delta \varphi }{\Delta t}$ = E = iR

$\Rightarrow$ $\Delta \varphi =\left(i\Delta t\right)R$ = QR

$\Rightarrow$ Q = $\frac{\Delta \varphi }{R}$

Time constant of LR circuit is $\tau$ = L/R.

$\therefore$ $\tau$ = 40/8 = 5 sec.

Red and green rays emerge from two points, propagating in two different parallel directions.

Using law of triangle, we get $\angle r$ = 30^o

$\mu$ = $\frac{\sin i}{\sin r}$

$\Rightarrow$ $\sqrt{2}\times \sin 30^{\circ }$ = sini

$\Rightarrow$ sini = $\frac{1}{\sqrt{2}}$

$\Rightarrow$ i = 45^o

Here, $\frac{x}{1000}=\frac{1.22\lambda }{D}$

$\Rightarrow$ x = $\frac{1.22\times 5\times {10}^3\times {10}^{-10}\times {10}^3}{10\times {10}^{-2}}$

$\Rightarrow$ x = 1.22 $\times$ 5 $\times$ 10^-3 m = 6.1 mm

$\therefore$ x is of the order of 5 mm.

N = N₀${\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ $\frac{25}{100}$ = ${\left(\frac{1}{2}\right)}^n$

$\Rightarrow$ n = 2

The total time in which radium change to 25 g is

= 2 × 1600 = 3200 yr.

Bohr model of atoms assumes that the angular momentum of electrons is quantised.

In nuclear fusion the mass of end product or resultant is always less than the sum of initial product, the rest is liberated in the form of energy, like in Sun energy is liberated due to fusion of two hydrogen atoms.

As m₁ + m₂ = m₃ + E

Since E = (m₁ + m₂ – m₃) × c²

Z is number of protons and A is the total number of protons and neutrons.

Mass defect = ZM_p + (A –Z)M_n – M(A, Z)

$\Rightarrow$ $\frac{B.E}{c^2}$ = ZM_p + (A –Z)M_n – M(A, Z)

$\therefore$ M(A, Z) = ZM_p + (A $-$ Z)M_n $-$ $\frac{B.E}{c^2}$

The maximum kinetic energy of photoelectron ejected is given by

K.E. = h$\nu$ - W = h$\nu$ – h$\nu$₀

The truth table of OR gate is

$\begin{array}{ccc}A& B& Y\\ 0& 0& 0\\ 0& 1& 1\\ 1& 0& 1\\ 1& 1& 1\end{array}$

It is seen that output of OR gate is Y = A + B.
If both or any one has 1 input then output will be 1.

A diode is said to be reverse biased if p-type semiconductor of p-n junction is at low potential with respect to n-type semiconductor of p-n junction. It is so for circuit (c).

In semiconductors at room temperature the electrons get enough energy so that they are able to over come the forbidden gap. Thus at room temperature the valence band is partially empty and conduction band is partially filled. Conduction band in semiconductor is completely empty only at 0 K.

V = $\frac{V_0}{\pi }$

$\Rightarrow$ V = $\left(\frac{10}{\pi }\right)V$

At STP, 22.4 L H₂

= 6.023 $\times$ 10²³ molecules

15 L H₂ = $\frac{6.023\times {10}^{23}\times 15}{22.4}$ = $4.033\times 10^{23}$

5 L N₂ = $\frac{6.023\times {10}^{23}\times 5}{22.4}$ = 1.344 $\times$ 10²³

2 g of H₂ = 6.023 $\times$ 10²³

0.5 g H₂ = $\frac{6.023\times {10}^{23}\times 0.5}{2}$ = 1.505 $\times$ 10²³

32 g O₂ = 6.023 $\times$ 10²³

10 g O₂ = $\frac{6.023\times {10}^{23}\times 10}{32}$ = 1.882 $\times$ 10²³

E = h$v$ or $v$ = E/h

For H atom, E = $\frac{-21.76\times {10}^{-19}}{n^2}Jat{m}^{-1}$

$\Delta E=-21.76\times 10^{-19}\left(\frac{1}{4^2}-\frac{1}{1^2}\right)$

= 20.40 $\times$ 10^-19 J atm^-1

$v$ = $\frac{20.40\times {10}^{-19}}{6.626\times {10}^{-34}}$ = 3.079 $\times$ 10¹⁵ s^-1

H₂O + $\stackrel{0}{\mathrm{Br}}$₂ $\to$ HO$\stackrel{+1}{\mathrm{Br}}$ + H$\stackrel{-1}{\mathrm{Br}}$

In the above reaction the oxidation number of Br₂ increases from zero (in Br₂ ) to +1 (in HOBr) and decreases from zero (in Br₂ ) to –1 (in HBr). Thus Br₂ is oxidised as well as reduced and hence it is a redox reaction.

For an acid-base indicator

H_In ⇌ H⁺ + In^-

K_in = $\frac{\left[H^{+}\right]\left[I{n}^{-}\right]}{\left[H_{in}\right]}$

$\Rightarrow$ $\left[H^{+}\right]=\frac{K_{in}\left[I{n}^{-}\right]}{\left[H_{in}\right]}$

Take – log on both sides

$-\log \left[H^{+}\right]=-\log \left(\frac{K_{in}\left[I{n}^{-}\right]}{\left[H_{in}\right]}\right)$

$\Rightarrow$ pH = –log K_In + $\log \frac{\left[I{n}^{-}\right]}{\left[H_{in}\right]}$

$\Rightarrow$ pH = pK_In + $\log \frac{\left[I{n}^{-}\right]}{\left[H_{in}\right]}$

$\Rightarrow$ $\log \frac{\left[I{n}^{-}\right]}{\left[H_{in}\right]}$ = pH - pK_In

AX2	⇌	A2+	+	2X-
s		s		2s

K_sp = = [A²⁺] [X^– ]² = s × (2s)² = 4s³

$\Rightarrow$ 3.2 $\times$ 10^$-$11 = 4s³

$\Rightarrow$ s³ = 8 × 10^–12

$\Rightarrow$ s = 2 × 10^–4 mol L^–1

$\Delta$G = $\Delta$H – T$\Delta$S

$\Delta$G = –382.64 × 103 J mol^–1 – (298K) (–145.6 JK^–1 mol^–1)

= –382640 + 43388.8

= – 339251.2 J mol^–1 = – 339.3 kJ mol^–1

H_2(g) + Br_2(g) $\to$ 2HBr_(g),   $\Delta$H^o_f = ?

$\Delta$H^o_f = $\Sigma$(B.E.)_reactants – $\Sigma$(B.E.)_products

= (B.E.)_H–H + (B.E.)_Br–Br –2(B.E)_H-Br

= [433 + 192] – 2(364) kJ mol^–1

= (625 – 728) kJ mol^–1 = –103 kJ mol^–1

For the reaction to be spontaneous, the total entropy of system and universe increases i.e.,

$\Delta S_{system}+\Delta S_{surroundings}>0$

Work done during the expansion, W = – pdV

W = –3 atm (6 dm³ – 4 dm³)

= – 3 atm ( 2 dm³ ) (1 dm³ = 1 L)

= – 3 atm × 2 L

= – 6 L atm

As, 1 L atm = 101.32 J

$\therefore$ W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J

We know, from Nernst Equation

E_cell = E^o_cell - $\frac{2.303RT}{nF}{\log }_{10}K$

At equilibrium E_cell = 0

$\therefore$ 0 = E^o_cell - $\frac{2.303\times 8.314\times 298}{2\times 96500}{\log }_{10}K$

$\Rightarrow$ 0 = 0.295 - $\frac{2.303\times 8.314\times 298}{2\times 96500}{\log }_{10}K$

$\Rightarrow$ 0.295 = $\frac{0.0591}{2}{\log }_{10}K$

$\Rightarrow$ ${\log }_{10}K$ = 10

$\Rightarrow$ K = 1 $\times$ 10¹⁰

For a first order reaction, A $\to$ products

Rate(r) = k[A]

$\Rightarrow$ k = $\frac{r}{\left[A\right]}$

$\Rightarrow$ k = $\frac{1.5\times {10}^{-2}}{0.5}=3\times 10^{-2}$

So, $t_{1/2}=\frac{0.693}{3\times {10}^{-2}}$ = 23.1 min

${}_{28}^{60}$ Ni + ${}_0^1$ n $\to$ ${}_{27}^{60}$Co + ${}_1^1$ n

To determine the formula of the compound formed by elements X and Y based on their positions in a cubic structure, we need to count the number of atoms per unit cell for each type of atom and then simplify the ratio.

For element X:

• Corners of a cube: Each corner atom is shared by 8 adjacent cubes, so each corner contributes 1/8 of an atom to the cube. Since there are 8 corners in a cube, the total contribution from corners is $8\times \frac{1}{8}=1$ atom of X per unit cell.

For element Y:

• Face-centers of a cube: Each face-centered atom is shared by 2 adjacent cubes, so each face-center contributes 1/2 of an atom to the cube. Since there are 6 faces on a cube, the total contribution from face-centers is $6\times \frac{1}{2}=3$ atoms of Y per unit cell.

Therefore, within one unit cell, we have 1 atom of X and 3 atoms of Y, giving us the simplest ratio of $X:Y=1:3$. Hence, the formula of the compound is $X{Y}_3$.

So, the correct answer is:

Option A: XY₃

Triglycerides are lipids, hence these are hydrolysed by lipases to glycerol and fatty acids.

Ionic radius in the n^th orbit is given by

r_n = $\frac{n^2{a}_0}{Z^{\ast }}$ or, $r_n\propto \frac{1}{Z^{\ast }}$

When n is the principal quantum number, a₀ the Bohr's radius of H-atom and Z^*, the effective nuclear charge.

In H₂O the central atom oxygen has two lone pairs of electrons and two bond pairs of electrons. Thus, the shape of H₂O is bent or V-shape. Thus, the two dipoles of H-O bonds have some resultant value as they are not exactly in the opposite direction to cancel out each other. While in BeF₂, there are two bond pairs on central atom Be, hence it is linear in shape. Thus, the two dipoles of bonds are exactly in opposite direction to each other and cancel each other to give net zero dipole moment.

Lone pair - Lone pair repulsion > lone pair - bond pair repulsion > bond pair-bond pair repulsion.

The lone pairs of electrons are arranged on equatorial position as they are at maximum distance at these positions and hence they have minimum repulsion at these positions.

SiF₄ has symmetrical tetrahedral shape which is due to sp³ hybridisation of the central silicon atom in its excited state configuration.

SF₄ has distorted tetrahedral or sea-saw geometry which arises due to sp³d hybridisation of the central sulphur atom and due to the presence of one lone pair of electrons in one of the equatorial hybrid orbital.

The MX₆ molecule with regular octahedral geometry is as follows :
Thus, there are three X—M—X bonds with bond angle 180^o.

The given compound X must be CaCO₃.

A binary compound is one made of two different elements.

Metals which have variable oxidation number can form more than one type of binary compound like Fe shows the oxidation state +2 and +3. Hence it forms two types of binary compounds. e.g., FeCl₂, FeCl₃.

Zeolites are alumino silicates having three dimensional open structure in which four or six membered ring predominates. Thus, due to open chain structure, they have cavities and can take up water and other small molecules.

As sixth period can accommodate only 18 elements in the table, 14 members of 4f series (atomic number 58 to 71) are separately accommodated in a horizontal row below the periodic table. These are called as lanthanides.

Ti⁺² = [Ar]₁₈ 3d²4s⁰

V³⁺ = [Ar]₁₈ 3d²4s⁰

Cr⁺⁴ = [Ar]₁₈ 3d²4s⁰

Mn⁺⁵ = [Ar]₁₈ 3d²4s⁰

Diaminodichloroplatinum (II) commonly known as cis - platin is found to have anticancer property.

tris-(Ethylenediamine) cobalt (III) bromide [Co(en)3]Br₃ exhibits optical isomerism.

In the formation of d²sp³ hybrid orbitals, two (n – 1)d orbitals of e_g set [i.e. (n – 1)d_z² and (n – 1)d_{x² – y²} orbitals)], one ns and three np (np_x, np_y and np_z) orbitals combine together and form six d²sp³ hybrid orbitals.

Al(OC₂H₅)₃ does not have metal-carbon bond. Its structure is

CN^– is a pseudohalide, i.e., it is a stronger coordinating ligand with the ability to form $\sigma$ and $\pi$-bonds.

In [Mn(H₂O)₆]²⁺

Mn⁺² = [Ar] 3d⁵4s⁰

In presence of weak field ligand H₂O, there will be no pairing of electrons. So it will form a high spin complex, i.e. the number of unpaired electrons = 5.

Only four structural isomers are possible for diphenyl methane.

Reaction of HBr with propene in the presence of peroxide gives n-propyl bromide. This addition reaction is an example of anti-Markovnikov’s addition reaction.

With ethene in presence of anhydrous AlCl3, benzene gives ethyl benzene (PhEt).

Chlorine of vinyl chloride (CH₂==CHCl) is non-reactive substitution reaction because it shows the following resonating structure due to +M effect of –Cl atom.

In second Cl–atom has positive charge and partial double bond character with C of vinyl group, so it is more tightly attracted towards the nucleus and it does not get replaced by nucleophile in nucleophilic substitution reaction.

Formation of a yellow precipitate on heating a compound with an alkaline solution of iodine is known as iodoform reaction. Methyl alcohol does not respond to this test. Iodoform test is exhibited by ethyl alcohol, acetaldehyde, acetone, methyl ketones and those alcohols which possess CH₃CH(OH) – group.

Carbonyl compounds (aldehydes and ketones) are obtained by the oxidation of 1° and 2° alcohols respectively. Among the given options, only (a) is 2° alcohol hence it can be oxidized to ketone.

ROH + PCl₅ $\to$ RCl + POCl₃ + HCl

RCOOH + PCl₅ $\to$ RCOCl + POCl₃ + HCl

Polystyrene is a chain growth polymer.

In peptide linkage i.e., (– CONH) group, the carboxyl group of one amino acid molecule forms an amide by combination with the amino group of the next amino acid molecule with the liberation of water molecule.

There are four bases in mRNA, i.e., adenine, cytosine, guanine and uracil. They form triplets and each triplet behave as a code for the synthesis of a particular amino acid.

Haemoglobin acts as an oxygen carrier in the blood since it reacts with oxygen to form unstable oxyhaemoglobin which easily breaks to give back haemoglobin and oxygen.

The $\alpha$-helix structure is formed when the chain of $\alpha$-amino acids coils as a right handed screw (called $\alpha$-helix) because of the formation of hydrogen bonds between amide groups of the same peptide chain, i.e., NH group in one unit is linked to carbonyl oxygen of the third unit by hydrogen bonding. This hydrogen bonding between different units is responsible for holding helix in a position.

Insulin is a hormone secreted by the pancreas that lowers blood glucose level by promoting the uptake of glucose by cells and the conversion of glucose to glycogen by the liver and skeletal muscle.

Cetyl trimethyl ammonium bromide forms cationic micelles above certain concentration.
In the molecules of detergents, the negative ions aggregate to form a micelle of colloidal size. In polar medium like water the negative ion has a long hydrocarbon chain and a polar group –Br^– at one end and on the other hand it has N⁺ ion thus cationic micelle is formed.

When CHCl₃ is treated with conc. HNO₃ its hydrogen is replaced by NO₂ group to form chloropicrin.

CHCl₃ + HONO₂ $\to$ CNO₂Cl₃ + H₂O

The thylakoid membranes possess the chlorophylls. The outer membrane is permeable to a number of solutes. Inner membrane has a number of carrier proteins. Stroma forms the matrix containing the enzyme for Calvin cycle.

Telomeres are non sticky terminal ends of the chromosomes. It has heterochromatin and repetitive DNA.

Copper is associated as an activator with tyrosinase. It is widely distributed in plants, animals and man. It is also known as polyphenol oxidase or catecholase. It oxidizes tyrosine to melanin in mammals and causes the cut surfaces of many fruits and vegetable to darken.

DNA replication is restricted to Sphase of interphase. G₂ phase is followed by the mitotic phase which is shorter than the interphase. In G₂ phase, DNA content is double than the amount present in the original cell.

Metaphase is the best time to count and study the number and morphology of chromosomes. The distinctly visible chromosome arrange themselves at the equatorial or metaphasic plate. The centromeres lie at the equatorial plate while the limbs are placed variously according to their size and spiral arrangement. At prophase the chromosomes appear thin and filamentous, forming a network.

So they are not very clearly visible. At telophase the chromosomes uncoil and lengthen and therefore are not clearly seen. Anaphase also shows chromosomes distinctly and they can be counted. But during anaphase chromatids seperate and start moving towards opposite pole. So for counting metaphase is the best stage.

At the beginning of M phase or mitotic phase, the nuclear membrane and nucleolus disappear. The decondensation of chromosomes occur. But when nuclear envelop start forming then nuclear lamina assemble and chromosomes start condensing.

Endosperm is formed due to fusion of the haploid male gamete with the polar nucleus of the embryo sac. But in this case the male plant is tetraploid so that its gametes would be diploid. When these diploid gametes fuse with two polar nuclei of the embryo sac the resultant endosperm would be tetraploid.

Anthesis is the process of opening floral buds. Reception of pollen by stigma is called pollination. Formation of pollen is called microsporogenesis.

In campylotropous ovule, the body is curved but the embryosac is straight eg. Capsella. In Anatropous, the body of ovule is inverted and gets fused with funiculus along its whole length on one side (most of the angiosperms). In orthotropous condition the body of ovule lies straight and upright over the funicle. e.g. Piperaceae, Polygonaceae.