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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 451. |
(i) Name any two causative organisms responsible for ringworm. (ii) State any two symptoms of the disease. |
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Answer» (i) For two causative organisms resposible for ring worm : See Q.24 (a)(ii), Delhi Board, 2014 (Comptt.) (ii) Two symptoms of the disease are : 1. Stool with excess mucus and blood clots. 2. Constipation. Housefy acts as mechanical carrier and transmits the parasits from faces of infected person to the food articles. |
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| 452. |
Explain the process of polliantion in Vallisnaria. |
| Answer» In vallisnaria, the female flower reaches the surface of water by along stalk and the polllen grains are released on surface of water . The pollen grains are carried passively by water currents and reach the stigma eventually. | |
| 453. |
Why does hnRNA to undergo splicing ? Where does splicing occur in the cell ? |
| Answer» hnRNA undergoes splicing to remove introns and jion exons. Splicing occurs in the nucleous of the cell. | |
| 454. |
(a) What is a tropihic level in an ecosystem? What is standing crop with reference to it ? (b) Explain the role of the first troplhic level in an ecosystem . (c ) How is the detritus food chain connected with the grazingfood chain in a natural ecosystem? |
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Answer» (a) tropic level : organisms occupy a plce in the natureal surrounding or ina community accroding to their feeding relationshio with other organisms .Based on the source of their nutrition or gfood organisms occupu a specifice place in the food chain and this is knwo as trophic level (b) First trophic level is formed by prducer theis is the basic unit these organisms can live without feeding on any another level.The only thing that these organisms need to sur vie is sunlight and water which they can turn into energy themselves .all other trophic level depends on thius for energy . (c ) GFC is grazing food chain : it is depicted as below: Grass`rarr`Deer`rarr`Lion DFC is detritus food chain : It begin with dead organic matter .It is made up of decomposers which are hetyerotyrophic organisms like fungi bactgreria etc GFC is the major condut for energy flow .DFC may be connected with GFC at some levels : some of the organisms of DFC are prey to the GFC animlas. |
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| 455. |
Explain the following observation : (i) Transition elements generally form coloured compounds. (ii) Zinc is not regaded as a transition element. |
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Answer» (i) Because the energy of excitation of an electron in d-orbital corresponds to the visible region. (ii) Because Zn has completely filled do orbitals `(3d^(10)). |
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| 456. |
State the functions of Ribozyme and release factor in protein synthesis respectively. |
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Answer» Ribozyme: Ribozyme in bacteria is 235 rRNA, acts as an enzyme for the formation of a peptide bond between two amino acids. Release Factor binds to the stop condon (UAA) terminating translation and releasing the complete polyptide. |
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| 457. |
(a) Expand CFC. (b) How does it reduce ozone to oxygen ? |
| Answer» (a ) CFC- Chlorofluoro carbons. | |
| 458. |
Write the functions of (a) cry IAC gene (b) RNA interference (RNAi) |
| Answer» (a) It produces in active protoxin in the host cell and also produces proteins to control the cotten ball worm. | |
| 459. |
Name the technique by which Gene expression can be controlled with the help of RNAi molecule. |
| Answer» Correct Answer - RNA interference. | |
| 460. |
Define the term standing crop. |
| Answer» Each trophic level has a certain mas of living matrial at a particular time called as the standing crop. | |
| 461. |
Draw the structure of 2-methylbutanal. |
| Answer» `CH_(3)-CH_(2)-underset(CH_(3))underset("| ")"CH"-CHO` | |
| 462. |
What happens when ethyl chloride is treated with aqueous KOH? |
| Answer» `CH_3CH_2Cl+KOH(aq)toCH_3CH_2OH` | |
| 463. |
Baculoviruses are good example of bio-control agents. Justify giving three reasons. |
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Answer» Baculovirus are used as biocontrol agents : 1. They are pathogens that attack insects and other arthropods. The majority of baculoviruses used as biological control agents are in the genus Nucleopolyhedrovirus. 2. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications. 3. They have been shown to have no negative impacts on plants, mammals, birds, fish or even on non-target insects. |
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| 464. |
The emf of a cell is always greater than its terminal voltage. Why? Give reason. |
| Answer» The emf of a cell is greater than its terminal voltage because there is some potential drop across the cell due to its small internal resistance. | |
| 465. |
A single pea plant in your kitchen garden produces pods with viable seeds, but the individual papaya plant does not. Explain. |
| Answer» A pea plant is monoecious and even a single plant produces seeds white papaya is dioecious so an individual pant does not produce viable seeds. | |
| 466. |
Explain the role of baculviruses as biological control agents. Mention their importance in orgenic farming. |
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Answer» Baculoviruses are pathogens that attack insects and other arthropods. The majority of baculoviruses used as biological cantrol agents are in the genus-Nucleopoly-hedrovirus. These viruses are excellent candidates for species-specific, narrow spectrum insecticidal applications. Importance in organic farmic : The use of the chemical fertilisers to meet the ever-increasing demand of agricultural products has contributed significantly to enviromental pollution . The over use of chemical fertilisers is problematic and there is a large pressure to switch to organic farming to use of bio-fertilisers. Baculoviruses have beeb shown to have no negative impact on plant, mammals, birds, fish and even non-target insect. This is useful in overall Integrated Pest Menagement (IPM). |
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| 467. |
Read the graph given above and correlate the uterine events that take place according to the hormonal levels on (i) 6 - 15 days (ii) 16 - 25 days (iii) 26 - 28 days (if the ovum is not fertilised ) (b) Specify the sources of the hormones mentioned in the graph. |
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Answer» (i) It is the follicular phase where the FHS stimulates the overian follicles to secrete estrogen which in ture stimulates the proliferation of the endometrium of the uterine wall . The endometrial lining begins to thicken. (ii) Secretion of progesterone maintains the endometrium which is neccessary for implantation of fertilised ovum and other events of pregnancy. (iii) When fertilisation does not take place, the endometrium, disintegrates, leading to menstruation making a new cycle. (iv) Oestrogen is secreted by the ovarian follicle and progesterone is secreted by the corpus luteum. |
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| 468. |
Describe the mechanism of pattern of inheritance of ABO blood groups in humans. |
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Answer» Human blood group is determined by glycoprotein/ antigen A and glycoprotein/ antigen B The alleles are `I^A,I^B`. Hence referred to as multiple allelism The individual inherits any two of them as given below `I^A,I^A" "Agroup ` `i^BI^B, I^Bi" " B group ` `I^AI^B" "AB group ` `ii " " O group ` In the case of A, B and 0 groups, Law of dominanee is the pattern tifinheritance as `I^A "and" I^B` are dominant over i. In AB group both the alleles `I^A "and " I^B` express themselves. Thus, it is the case of co-dominance. |
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| 469. |
Of physiosorption or chemisorption, which has a higher enthalpy of adsorption? |
| Answer» Correct Answer - Chemisorption. | |
| 470. |
Name the method used for refining of copper metal |
| Answer» Correct Answer - Electrorefining | |
| 471. |
A 10% solution (by mass) of sucrose in water has freezing point of 269.15 K. Calculate freezing point of 10% glucose in water, if freezing point of pure is 273.15 K (Given molar mass of sucrose= 342 g `mol^(-1)`, Molar mass of glucose =180 g `mol^(-1)`). |
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Answer» 10% solution (by mass) of sucrose: `{:(w_(B) = 10 g),(m_(B) = 342 g//mol),(w_(S) = 100g"," w_(A) = 100 - 10 = 90g):}` `[(w_(B) = "Mass of solute"),(w_(S) = "Mass of solution"),(w_(A) = "Mass of solvent")]` `Delta T_(f) = k_(f) xx (w_(B))/(m_(B) xx w_(A) (kg))` `T_(A)^(@) -T_(S)^(@) = k_(f) xx (w_(B))/(m_(B) xxw_(A) (kg))` `273.15 - 269.15 = k_(f) xx (10 xx 1000)/(342 xx 90)` `k_(f) = (4 xx 342 xx 90)/(10000) rArr (342 xx 9)/(250)` 10% solution (by mass) of glucose `{:(w_(B) ",",m_(B) = 180 g//mol),(w_(S) = 100g",",w_(A) = 90g):}` `T_(A)^(@) - T_(S) = k_(f) xx (w_(B))/(m_(B) xx w_(A) (kg))` `273.15 - T_(S) = (342 xx 9)/(250) xx (10 xx 1000)/(180 xx 90)` `273.15 - T_(S) = 7.6 k` `T_(S) = 273.15 - 7.6` `T_(S) = 265.55 k` |
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| 472. |
Write IUPAC name of the complex : `[CoCl_(2)(en)_(2)]^(+).` |
| Answer» dichlorobis (ethane - 1, 2- diamine) cobalt (ii) ion. | |
| 473. |
Calculate the degree of dissociation `(alpha)` of acetic acid if its molar conductivity `(^^_(m))` is 39.05 `S cm^(2) mol^(-1)` Given `lamda^(@) (H^(+)) = 349.6 cm^(2) mol^(-1) and lamda^(2) (CH_(3)COO^(-)) = 40.9 S cm^(2) mol^(-1)` |
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Answer» Given: (i) `^^_(m)^(c) = 39.05 S cm^(2)//mol` (ii) `lamda^(@) (H^(+)) = 349.6 S cm^(2)//mol and lamda^(@) (CH_(3)COO^(-)) + lamda^(@) (H^(+))` `= 40.9 + 349.6 = 390.5 S cm^(2)//mol` Degree of dissociation `(alpha) = (^^_(m)^(c))/(^^_(m)^(@)) = (39.05)/(390.5) = alpha = 0.1` |
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| 474. |
Define the following terms: (i) Colligative properties (ii) Molality (m) |
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Answer» (i) Colligative properties: Those properties of ideal solution which depend only on the number of particles of the solute dissolved in a definite amount of the solvent and do not depend on the nature of solute. For example, (i) Elevation in boiling point (ii) Osmotic pressure (iii) Depression in freezing point. (ii) Molality (m): It is defined as the numbers of moles of the solute dissolved in 1000 grams (1kg) of the solvent `{:("Molality (m)" = (n_(B))/(w_(A) (kg)),n_(B) rarr "Moles of solute"),(,w_(A)rarr"Mass of solvent"):}` |
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| 475. |
(a) Why does presence of excess of lithium makes Li Cl crystals pink? (b) A solid with cubic crystal is made of two elements P and Q. Atoms of Q are at the comers of the cube and Pat the body-centre. What is the formula of the compound? |
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Answer» (a) When a crystal of LiCl is heated in an atmosphere of Li vapour, the Li atoms lose electron to for `Li^(+)` ions. The released electrons diffuse into the crystal and occupy anionic sites. These electrons impact pink colour to the LiCl crystal. (b) It is given that the atoms of Q are present at the corners of the cube. Therefore, number of atoms of Q in one unit cell `=8xx(1)/(8)=1`. It is also given that the atoms of P are present at the body-centre. Therefore, numbers of atoms of P in one unit cell = 1. This means that the ratio of the number of Q atoms P: Q = 1 : 1. Hence, the formula of the compound is PQ. |
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| 476. |
Given the structures of A, B and C in the following reactions : (i) `CH_(3) Br overset (KCN) toAoverset(LilH_(4))to B overset(HNO_(2))underset(273k)toC` (ii) `CH_(3) COOH overset (NH_(3))undersetDelta toAoverset(Br_(2)+KOH)to B overset(CHCL_(3)+NaOH)toC` |
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Answer» (i) `CH_(3) Br overset (KCN) to CH_(3)underset(A)CN overset(LilH_(4))to CH_(3)underset(B)CH_(2)NH_(2) overset(HNO_(2))underset(273k)toCH_(3)underset((C))CH_(2)-OH` (ii) `CH_(3) COOH overset (NH_(3))undersetDelta toCH_(3)underset((A))(CONH_(2))overset(Br_(2)+KOH)to CH_(3)underset((B))(NH_(2)) overset(CHCL_(3)+NaOH)toCH_(3)underset((C))(NC)` |
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| 477. |
complete the following equation : (i) c+conc.`H_(2)SO_(4)rarr` (ii) `XeF_(2)+H_(2)Orarr` |
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Answer» (i) `C+2H_(2)SO_(2)(conc)rarrCO_(2)+2SO_(2)+2H_(2)O` (ii)`2XeF_(2)+2H_(2)Orarr2Xe+O_(2)+4Fhf` |
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| 478. |
Three long straight parallel wires are kept as shown in figure. The wire (3) carries a current I. (i) The direction of flow of current I in wire (3), is such that the net force, on wire (1), due to other two wires, is zero (ii) By reversing the direction of I, the net force, on the wire (2) due to the other two wires, becomes zero. What will be the directions of current I, in the two cases? Also obtain the relation between the magnitudes of current `I_1`, `I_2` and I. |
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Answer» (i) Direction of I is downward. (ii) Direction of I is upward. `(mu_(0))/(4pi)=(2I" "I_(2))/(a)=(mu_(0))/(4pi)(2I_(1)" "I_(2))/(a)` `impliesI=I_(2)` Also, `(mu_(0))/(4pi)=(2I" "I_(2))/(2a)=(mu_(0))/(4pi)(I_(1)" "I_(2))/(a)impliesI=I_(2)`. |
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| 479. |
Define ionization energy. How would the ionization energy change when electron in hydrogen atom is replaced by a particle 200 times heavier than electron, but having the same charge? |
| Answer» It is defined as the energy required to knock an electron completley out of the atom. It will remain same as it depends on charge and not on mass. | |
| 480. |
Ram was a daily wage worker in a factory. He was suffering from Cancer. On hearing this, most of his co-workers, started avoiding him under the impression that it was a contagious disease. When Prof. Srivastava came to know about this, case, he took him to a leading radiologist, who examined him and told that it was at the beginning stage. He advised that it could be easily cured and also certified that it was not a communicable disease. After this, Ram was given proper treatment by the doctor and got cured completely. (1) What moral values did Prof. Srivastava display ? (2) How is mean life of a radioactive element related to its half life? (3) A radioactive sample has activity of 10,000 disintegrations per second after 20 hours . After next 10 hours, its activity reduces to 5,000 dps. Find out its half - life and initial activity. |
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Answer» (1) Awareness and Humanity. (2)`t_(1//2)=(0.6931)/lambda and tau = 1//lambda` `(T/(1//2))/tau=0.6931 and T_(1//2)=0.6931 tau` (3)` R_(0)=10,000s^(-1) R= 5000 s^(-1) t= 10h` ` = 5000/10000 = e^(-10 lambda) rArr e^(-10) lambda=2` taking ` log 10 lambda = log2 rArr 10 lambda = 2.303 xx log _(10^(2)) lambda = (0.6931)/10 per h` `T_(1//2) = (0.6931)/lambda =cancel(6931)/cancel (6931) xx 10 h = 10h` Intial activity = `A _(0)` Using `5000=R_(0)(e^(-0.693))/(36 xx 10^(3)) xx 30 xx 60 xx 60 rArr 5000 = A_(0) e^(-.693 xx3)` `A_(0) = 5 xx 10^(3) xx e^(0.693 xx 3) ` decay/sec. |
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| 481. |
An iron-cored solenoid has self- inductance. 2.8 H. When the core is removed, the self inductance becomes 2 mH. What is the relative permeability of the core used? |
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Answer» If Li initial inductance and La is inductance after removing the core : `(Li = cancel (u_0)u_(r)cancel n^ cancel (2) cancel (Al))/(La =cancel(u_(0)) cancel n ^cancel(2) cancel (Al))` `2.8/(2 xx 10^(-3))=(Li)/(La)=u_(r)` `u_(r) = 1.4 xx 10^(3)=1400` |
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| 482. |
Why are medicines more effective in collidal state ? |
| Answer» Medicines are more effective in colloidal state because they have large surface area. These medicines are used to kill the germs by getting adosorbed on them. | |
| 483. |
Define ambident nucleophile with an example. |
| Answer» See Q.15, Paper-2014 (C), Outside Delhi, Set-I, [Page 270] | |
| 484. |
(i) Can two equpotential surfaces intersect each other ? Give reason. (ii) Two charges `+ q and -q` are located at points A `(0, 0, -2) and B(0, 0, 2)` respectively. How much work will be done in moving a test charge from point P`(4, 0, 0) " to " (-5, 0,0)` ? |
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Answer» (ii) Potential at P (4, 0, 0) is `V_(1) = (-q)/(4 pi epsi_(0)).(1)/(sqrt((4 -0)^(2) + (0 + 2)^(2))) + (q)/(4 pi epsi_(0)).(1)/(sqrt((4 -0)^(2) + (0 -2)^(2)))` `= (-q)/(4 pi epsi_(0)).(1)/(sqrt(16 + 4)) + (q)/(4 pi epsi_(0)) .(1)/(sqrt(16 + 4)) = 0` Potential at `q(-5, 0, 0)` is `V_(2) = (-q)/(4 pi epsi_(0)).(1)/(sqrt((-5 -0)^(2) + (0 + 2)^(2))) + (q)/(4pi epsi_(0)).(1)/(sqrt((-5-0)^(2) + (0-2)^(2)))` `= (-q)/(4pi epsi_(0)).(1)/(sqrt(25 + 4)) + (q)/(4pi epsi_(0)).(1)/(sqrt(25 + 4)) = 0` `:.` Work done `= q (V_(2) - V_(1)) = q (0 -0) = 0` Hence, `w = 0` |
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| 485. |
How would you account for the following ? (i) Many of the transition elements are known to form interstitial compounds. (ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the correponding group members of the second (4d) series. (iii) Lanthanoids from primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds , +4 or even +6 being typical. |
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Answer» Many of the transition elementare known to form interstial compounds. Transition elements form intensitial compounds as they are capable of entrapping small atoms like H,C or N in the interstitial sites in their crystal Lattice. (ii) The metallic radii of the third (5d) series of the transition metals are virtually the same as those of the corresponding group memebers of the second (4d) series. This is due to filling 4f orbitals which have poor shielding effect or due to lanthanoid contraction. (iii) This is attributed to the fact that 5f, 6d and 7s levels are of comparable energies in case of actionoids. Hence all these three subshells can participate on bonding. |
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| 486. |
Define the term polymersation. |
| Answer» Polymerisation : It is process of formation of a high molecular mass polymer from one or more monomers by linking together numbr or repeating structural units through covalent bonds | |
| 487. |
A capacitor C, a variable resistor R and a bulb B are connected in series to the ac mains in circuits as shown if Fig. The bulb glows with some brightness. How will the glow of the bulb change if (i) a dielectric slab is introduced between the plates of the capacitor, keeping resistance T to be the same, (ii) the resistance R to be the same, (ii) the resistance R is increased keeping the same capacitance. |
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Answer» (i) As the dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V = Q/ C). As the result, the potential drop across the bulb will increase (since both are connected in series). So, its brightness will increase. (ii) As the resistance (R) is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease (since both are connected in series). So its brightness will decrease. |
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| 488. |
How is amplitude modulation achieved ? (b) The frequencies of two side bands in an AM weve are 640 kHz and 660 kHz respectively. Find the frequencies of carrier and modulating signal. What is the bandwidth required for amplitude modulation ? |
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Answer» (a) When a modulating AF weve is super imposed on a high frequency carrier weve in a manner that the freaquency of modulated weve is same as that of the carrier weve but its amplitude is made proportional to the instantaneous amplited of the audio frequency modulating voltage process is called Amplitude Modulation (b) `f_(LSB)= f_(c) - f_(m) = 640kHz` `f_(USB)= f_(c) + f_(m) = 640kHz` Adding `2f_(c) = 130 kHz So, f_(c) = 650kHz` Subtracting `2f_(m) = 20 kHz f_(m) = 10 kHz` Band width `2xx` frequency of modulating signal = ` 2 xx 10 kHz = 20 kHz.` |
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| 489. |
Write two properties of ameterial of a meteriable suitable for making (a) a perment magnet, and (b) an eletromagnet. |
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Answer» Permenent magnets are formed with material having (i) hight retentwity and (ii) high (i) high permeability (ii) how retentively . |
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| 490. |
(a) The potential difference applied across a given resistor is altered so that the heat produced per secound increases by a factor of 9 . By what factor does the appliced potential difference change ? (b) In the figure shown, an ammeter A and a resistor of 4 `Omega` are connected to the terminals of the socrce . The emf of the source is 12 V having an internal resisitance of 2 `Omega` Calculate the voltmeter and ammeter readings. |
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Answer» (a) H = Vlt ` H_(1) = (V_(1)^(2) t)/(R ) and H_(2) = (V_(2)^(2) )/(R )t` Taking `H_(2) =9H_(1),(H_(1))/(H_(2))= (H_(1))/(9H_(1) )= ((V_1^2)/cancel Rcancelt)/((V_2^2)/cancelRcancel t)` `rArr (1)/(9) = (V_(1)^(2))/(V_(1)^(2) )=((V_(1))/(V_(2) ))^(2) rArr (V_(1))/(V_(2) )= (1)/(3)` `rArr V_(2) 3V_(1)` Hence, Increase by a factor of 3. (b) Ammeter reading `I = (E)/((r + R)) = (12) /(2+4) = (12)/(6) = 2 Amp` Volmeters reading = V = E - Ir = 12 - (2.2)V = 8V. |
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| 491. |
(a) Why is haemophilia generally observed in human males ? Explain the conditions under which a human female can be haemophilic. (b) Draw the male Reproductive system. |
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Answer» (a) Haemophilia is caused due to the recessive gene on X-chromosome. Y has no ailele for this. If a male is` X^hY`, then he is haemophilic. If male inherits Xh from the mother, he Will be haemophilic (with the genotype `X^hY`) If female inherits `X^hX`", one from the carrier mother and one from her haemophilic féther, then she can be haemophilic (b) For diagram please see Q. No. 29. (a). |
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| 492. |
Write the technical term used in human ABO blood groups fro `I^(A),I^(B)` and i. |
| Answer» The technical term used is allele. | |
| 493. |
Mention one application for each of the following : (a) Passive immunization (b) Antihistamine (c ) Colostrum (d) Cytokinin - barrier |
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Answer» (a) Passive immunization : Even in ceses of snakebites , the injection which is given to the patients contain preformed antibodies against the snake venom . We need to directly inject the preformed antibodies or antitoxin (a preparation containing antibodies to the toxin). (b) Antithistamine : Pharmaceutical drug that opposes the activity of histamine reception in the body. (c ) Colostrum : The yellowish fluid colostrum secreted by mother during the initial days of lactation has abundant antibodies (IgA) to protect the infant. (d) Cytokinin - barrier : Virus infected cells secrete proteins called interferons which protect non - infected cells from further viral indection. |
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| 494. |
How do algal blooms affect the life inwater bodies |
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Answer» Algal bloom pollutes water and deteriorates the water quality by depleting the oxygen content. This turns the water toxic and increase its BOD. (ii) Algal booms lead to death of aquatic organisms due to oxygen depletion |
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| 495. |
Many fresh water. animals can not survive in a marine environment. Explain. |
| Answer» Many freshwater animals can not survive in a marine environment because their bodies are not adapted to the marine environment. They lose body water because of the higher concentration of salt in the surrounding water, which makes survival difficult. | |
| 496. |
The diagram above is that of a typical biogas plant. Explain the sequence of events occurring in a biogas plant. Identify a, b and c. |
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Answer» : Sludge tank : Gas holder Charge kit The sequence of events occuring in a biogas plant is that theglog: |
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| 497. |
Explain the process of artificial hybridisation to get improved crop, variety in (i) plants bearing bisexual flowers (ii) female parent producing unisexual flowers. |
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Answer» Artifical Hybridisation is the major approach of crop improvement programme : (i) Plants bearing bisexual flowers: Here emasculation is done, i.e., removal of anthers from the flower bud before the another splits using forceps. Emasu]ated flo wers have tn be covered with a bag of suitable size, generally made up of butter paper to prevent contamination of its stigma with unwanted pollen. This process is called bagging. When the stigma is bagged flower achieves receptivity, mature pollen grains collected from anther of the male parent are dusted un the stigma, and the flowers are rebagged. (ii) If the female parents produces unisexual flowers there is no need for emasculation. The female flower buds are bagged before the flowers open. When the stigma becomes recepative, pollination is carried out using the desired pollen and the flower rebagged. |
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| 498. |
Enumerate ate four objectives for improving the nutritional quality of different crops for the health benefits of the human population by the-process of 'Biofortification". |
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Answer» Biofortification involves the breeding of crops to increase their nutritional value 1. To improve protien content and quality 2. To improve oil content and quality ( 3) To improve vitamin content ( 4) To improve micro nutrient and mineral content . |
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| 499. |
(a) Explain how to find whether an E. coli bacterium has transformed or not when a recombinant DNA bearing ampicillin resistant gene is transferred into it. (b) What does the ampicillin resistant gene act as in the above case? |
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Answer» (a) The successful transformation of E. coli with a recombinant DNA can be detected by growing the E.coli cell on culture media having ampicillin. The transformed cells will survive as they carry recombinant DNA with ampicillin resistance gene, where as non-transformed cells will die as they are ampicillin sensitive. (b) Antibiotic resitance genes such as ampicillin resistance, serve as selectable markers as in the above case. |
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| 500. |
Mention two objectives of-setting up GEAC by our Government. |
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Answer» The two objectives are : 1. To make decisions regarding the validity of GM-research. 2. To decide regarding the safety of introducing GM-organisms for public services. |
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