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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.
| 1001. |
Name the source organism that possesses Taq polymerase. What is so special about the function of this enzyme? |
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Answer» The source of Taq Polymerase-It isolated from thermophilic bacterium, Thermus aquaticus. Function of this Enzyme-Taq polymerase is a thermostable enzyme which remains active during the high temperature induced denaturation of DNA. |
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| 1002. |
Name the respective pattefn of inheritance where F: phenotype (a) does not resemble either of the two parents and is in between the two. (b) resembles only one of the two parents. |
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Answer» (a) Incomplete dominance, (b)Comp1ete dominance . |
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| 1003. |
How is the entry of only one sperm and not many ensured into an ovum during fertilisation in humans? |
| Answer» During fertilisation in humans, when the ‘sperm comes in contact with the zona pellucida layer of ovum, it induces changes in the ovum membrane that prevents the entry of additional sperms. Thus, only one sperm can fertilise an ovum. | |
| 1004. |
a) What is an age-pyramid? b) Name three representative kinds of age-pyramids for human population and list the characteristics for each one of them. |
| Answer» sacred groves : in india religious and cultural traditions stressed on protection of nature vast stretches of forest land were left untouched and all the animal an plant life within tham worshipped such sarcred goves are seen in khasi and jantia hills in Meghalya and aravilli Hills fo rajasthan In megalye the speed groves are last resvoirs fo rate and threatened plants . | |
| 1005. |
DIFFERENCE BETWEEN EUCHROMATIN & HETEROCHROMATIN |
| Answer» `{:(,"Euchromatin","Heterochromatin"),(1.," It is the region of chromatin that is "," It is the region of chromatin that is densely"),(,"loosely packed","packed."),(2.,"It stains light","It satins dark."):}` | |
| 1006. |
List the events that reduce the biological oxygen demand (Bod) of a promary effluent during sewage treatment . |
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Answer» the events that reduce the BOD of a primary effuent are : 1. the primary effluent is passed into large aeration tanks where it is mechanically agitated and air is pumped into it . 2. Mechanical agitation of primary efuent leads to gowth of aerbic microbes which reduces the BOD. |
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| 1007. |
(a) Explain the following : (i) `NF_(3)` is an exothermic compound whereas `NCl_(3)` is not (ii) `F_(2)` is most reactive of all the four common halogens. (b) Complete the following chemical equations : (i) `C+H_(2)SO_(4) (conc) to` (ii) `P_(4)+NaOH+H_(2)Oto` (iii) `underset("excess")(Cl_(2)+F_(2))to` |
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Answer» (a) (i) See Q. 29 (b) (ii), Delhi Board, 2010. (ii) Because of small size, highest eledtro negativity and low bond dissociation enthalp) of f-f bond, fluorine is most reactive. (b) (i) `C+2H_2SO_4("conc".)to2SO_2+CO_2+2H_2O` (ii) `P_4+3NaOH+3H_2Otounderset("Phosphine")(PH_3)+underset("Sodium hypaphosphite")(3NaH_2PO_2)` (iii) `{:(C_2+3F_2overset("heat")to2ClF_3),(""" (excess)"):}` |
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| 1008. |
(a) What is meant by unidentate, bidentate and ambidentate ligands? Give two examples for each. (b) Calculate the overall complex dissociation equilibrium constant for the `Cu(NH_3)_(4)^(2+)` ion, given that `beta4` for this complex is `2.1xx10^13`. |
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Answer» (a) A molecule or an ion which has on]y one donor atom to form one coordinate bond with the central metal atom is called unidentate ligand, eg., `Cl^-and NH_3`. A molecule or an ion which contains two donor atoms and hence forms two coordinate bonds with the central metal atom is called a didentatc ligands, eg., `NH_2 -CH_2-CH_2 - NH_2 and overset(-)COC -COO` A molecule or an ion which contains two donor atoms but only of them forms a coordinate bond at a time with the centeral metal atom is called ambidentate ligand, ag, : `CN^-or NC :and : NO_2or : ONO^-`. (b) (i) `Cu^(2+)+NH_3hArrCu(NH_3)^(2+)` `k_1=([Cu(NH_3)]^(2+))/([Cu^(2+)][NH_3])` (ii) `Cu(NH_3)^(2+)+NH_3hArrCu(NH_3)_(2^(2+))` `k_2=([Cu(NH_3)_2]^(2+))/([Cu(NH_3)]^(2+)[NH_3])` (iii) `beta_4=([Cu(NH_3)_4]^(2+))/([(NH_3)]^(4)[Cu^(2+)])` `[beta_4=k_1xxk_2xxk_3xxk_4]` |
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| 1009. |
A translucent white waxy solid (A) on heating in an inert atomosphere is converted to its allotropic form (B). The solid (A) on reaction with very dilute aqueous KOH liberates a highly poisonous gas (C ) having rotten fish smell. With excess of chlorine, (A) forms (D) which hydrolyses to compound (E ). Identify compounds (A) to (E ). |
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Answer» (A) White waxy solid (A) is white phosphorus. When white phosphorus is heated in an inert atmosphere at 573K, it changes to red phosphorus. (B) is red phosphorus. (a) On heating with KOH liberates phosphine. (c) Which is poisonous gas with rotten fish smell. `{:(P_4+3KOH+3H_2OtoPH_3+3KH_2PO_2),(""" Phosphine(c)"):}` White phosphorus (`P_4`) burns with excess of `Cl_2` to form phosphorus pentachloride (D). `{:(P_4+10Cl_2overset("heat")to4PCl_5),(" "(D)):}` Hydrolysis of (D) gives phosphoric acid (E) `{:(PCl_5+4H_2OtoH_3PO_4+5HCl),(" "(E)):}` |
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| 1010. |
(a) Define the terms osmosis and osmotic pressure. Is the osmotic pressure of a solution a colligative property? Explain. (b) Calculate the boiling point of a solution prepared by adding `15.00g` of `NaCl` to `250.0g` of water. (`K_(b)` for water `=0.512K kg "mol"^(-1)`, Molar mass of `NaCl=58.44g`) |
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Answer» (a) Osmosis : The spontaneous flow of solvent molecules from a less concentrated solution to a more concentrated solution though semi permeable membrane. Osmotic pressure: The minimum excess pressure that has to be applied on the solution to prevent the entry of solvent into the solution through SPM. It is a colligative property becuase it depends upon, no. of solute particles not on the . nature of particle. e.g., o. IMKCl and Na Cl has same osmotic pressure under same ten1p and pressure condition. (b) (b) Elevation is boiling point `[i=2" as " NaClhArrNa^++Cl]` `DeltaT_b=(K_bxx1000xxW_B)/(W_AxxM_B)=(0.512xx1000xx15xx2)/(250xx25.44)=(512xx2)/974=1.051k` Boiling point of water = 373 k `373+1.051=374.051 k` |
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| 1011. |
An element exists in bcc lattice with a cell edge of 288 pm. Calculate its molar density is `7.2g//cm^(3)` |
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Answer» (ii) BCC lattice. Hence `z=2` `a=288p m=288xx10^(-12)m` `d=7.2g//cm^(3)` `d=(ZxxM)/(a^(3)xxNa)impliesM=(d xx a^(3)xxNa)/(Z)` `M=(7.2xx(288xx10^(-10))xx6.022xx10^(23))/(2)` `M=3.6xx288xx288xx288xx10^(-7)xx6.022` `M=51.8g//mol` |
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| 1012. |
Explain the following observations : (i) Copper atom has completely field d orbitals `(3d^(10))` in its ground state, it is regarded as a transition element. (ii) `Cr^(2+)` is a stronger reducing agent than `Fe^(2+)` in aqueous solutions. |
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Answer» The outer configuratin of copper is `3d^(10)4s^(1).` In addition to + 1, it shows an oxidation state of +2. In +2 state, the configuration is `d^(9)` .i.e., the sub shell is incompletely filled. Hence, it is a transition element. (iii), `Cr^(2+)` strong reducing agent than `Fe^(2+)` because `Cr^(2+)` get oxidised to `Cr^(3+)` easily which has half filled `t_(2)^(3)g` configuration which is more stable is more stable than `3d^(5)` in `Fe^(3+).` |
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| 1013. |
(a) Draw a labelled diagram of AC generator. Derive the expression for the instantaneous value of the emf induced in the coil. |
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Answer» (a) N//A (b) `A = 200 cm^(2)` `= 200 xx 10^(-4)m^(2) = 2 xx 10^(-3) m^(2)` `n= 20` `omega = 50 "Rad" s^(-1)` `B = 3 xx 10^(-2)T` Max. emf `E_(o) = 3 xx 10^(-2) T` `= 20 xx 3 xx 10^(-2) xx 50 = 6000 xx 10^(-4) = 0.` Volt Max. Current `I_(0) = (E_(0))/(R ) = (0.6)/(R )A`. |
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| 1014. |
(a) What is amplitude modulation? Draw a diagram showing an amplitude modulated wave obtained by modulation of a carrier sinusoidal wave on a modulating signal. (b) Define the terms (i) modulation index, and (ii) side bands. Mention the significance of side bands. |
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Answer» (b) (i) Modulation Index : (ii) Side Bands : Side band is the portion of a modulated carrier wave that is either above or below the base band signal. `(w_(1)-w_(m))` is the lower side band `(w_(1)+w_(m))` is called the upper side band. Both side bands are used to carry a message. |
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| 1015. |
(a) Draw a schematic diagram of an AC generator. Explain is working and obtain the expression for the instantaneous value of the emf in terms of the magnetic field B, number of turns N of the coil of area A rotating with angular frequency `omega`. Show how an alternating emf is generated by loop of wire rotating in a magnetic field. (b) A circular coil of radius 10 cm and 20 turns is rotated about its vertical diameter with angular speed of 50 rad `s^(-1).` in a uniform horizontal magnetic field of `3.0xx10^(-2)T`. (i) Calculate the maximum and average emf induced in the coil. (ii) If the coil forms a closed loop of resistance `10 Omega`, calculate the maximum current in the coil and the average power loss due to Joule heating. |
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Answer» (b) `r=(10)/(100)=(1)/(10)m` N=20 `omega=50 rad//sec` `B=3:0xx10^(-2)T` (i) `epsilon_(0)=NBA omega=Nbpir^(2)omega` `=20xx3xx10^(-2)xx3.14xx(1)/(10)xx(1)/(10)xx50` =0.942 volt Average emf over one complete cycle=0 for half cycle `=(2)/(pi)epsilon_(0)=(2)/(3.14)xx0.942=0.6 volt.` (ii) `R=10 Omega` Maximum current `I_(0)=(epsilon_(0))/(sqrt(2))` `=(0.942)/(sqrt(2))=0.66 AmP` Average power `P_(av)=I_(rms)^(2)R=(I_(0)^(2))/(2)R` `=(0.66xx0.66)/(2)xx10=2.718 W.` |
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| 1016. |
Give the IUPAC names of the following compounds : (i) `[Co(NH_3)_(4)(H_(2)O)Cl]CL_2` (ii) `[CrCl_2(en)_2]Cl` |
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Answer» (i)Tetraamineaquachloridocobalt (iii) Chloride. (ii) Dichloridobis (ethane-1,2-diamine) chromium (iii) chloride. |
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| 1017. |
A resistance R is connected across a cell of emf `epsilon` and internal resistance r. A potentiometer now measures the potential difference between the terminals of the cell is V. Write the expression for r in terms of `epsilon`, V and R`. |
| Answer» Expression : `r=(epsi/V-1)xxR`. | |
| 1018. |
Define the activity of a radionuclide. Write its SI unit. Give a plot of the activity of a radioactive species versus time. |
| Answer» The total decay rate of a radioactive sample is called the activity of the sample. The S.I unit of activity is Becquerel (Bq). | |
| 1019. |
Two identical cells each of emf `epsilon`, having negligible internal reistance r, are connercted in parallel with each other across an external resistance R. What is the current through this resistance. |
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Answer» The cells are arranged as shown in the circuit diagram. As the internal resistance is negligible, so total resistance of the circuit = R. So, current through the resistance, `I = (E)/(R)`. |
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| 1020. |
Name the products of hydrolysis of sucrose. Why is sucrose not a reducing sugar ? |
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Answer» Products of hydrolysis of sucrose are glucose and fructose. Sucrose is not a reducing sugar because carbonyl groups of sucrose is not free. |
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| 1021. |
Write IUPAC name of the following compound : `CH_(3)NHCH (CH_(3))_(2)` |
| Answer» `underset("N-methylpropan-2-amine")(CH_(3) - underset(H)underset(|)(N) - underset(CH_(3))underset(|)overset(2)(CH) - overset(3)(C)H_(3))` | |
| 1022. |
Two convex lenses of same focal length but of aperture `A_(1)` and `A_(2)(A_(2) lt A_(1))` are used as the objective lenses in two astronomical telescope having identical eye pieces. What is the ratio of their resolving power ? Which telescope will you prefer and why ? Give reason. |
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Answer» Resolving Power, `R.P.=a/(1.22 lambda):. ((R.P.)_1)/((R.P.)_2)=A_1/A_2` The telescope with objective of aperture A, should be preferred for viewing as this would : (i) Give a better resolution. (ii) Have a higher light gathering power of telescope. |
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| 1023. |
Apart from being part of the food chain, predators play other important roles. Mention any two such roles supported by examples. |
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Answer» Predators play an important role in: i) Maintaing the prey population, this regulates inter-species competition. This lso helps in checking any overgrazing by herbivores, which might ultimately lead to an unbalanced ecosystem. For exmaple, if tigers are removed from a forest, spotted deer will multiply rapidly. This would result in rapid destruction of herbs and grasses in the forest. 2) Indicating ecological disturbance. Usually predators are small in number and highly sensitive to ecological changes, owing to their excessive adaptation. For example, Gharials are extremely adapted to their freshwater habitat. They start dying rapidly and in large numbers as soon as a river starts getting polluted, thus indicating ecological disturbances. |
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| 1024. |
Two large parallel thin plates having uniform charge densities `+ sigma and - sigma` are kept in X-Z plane at a distance d apart. Sketch an equipotential surface due to electric field between the plates. If a particle of mass m and charge `-q` remains stationary between the plates, what is the magnitude and direction of the field ? |
| Answer» Electric field due to two infinite plane parallel sheets of charge : | |
| 1025. |
What is the force between two small charged of `2 xx 10^(-7)` C placed 30 cm apart in air? |
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Answer» `F=(1)/(4piepsi_(0))(q_(1)q_(2))/(r^(2))`. `9xx10^(9)xx(2xx10^(-7)xx2xx10^(-7))/((30xx10^(-2))^(2))` `=4xx10^(-3)`N. |
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| 1026. |
To which part of the electromagnetic spectrum does a wave of frequency `3xx10^(13)`Hz belong? |
| Answer» A wave of frequency `3xx10^(13)` Hz will belong to the microwaves of electromagnetic spectrum. | |
| 1027. |
Write the functions of the following in communication systems: (i). Receiver (ii). Demodulator |
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Answer» (i) Transmitter : It is a device or a set-up which transmits the message to receiver through communication channel. (ii) Modulator : It is a device in which the amplitude frequency (phase) of a high frequency carrier wave is made to change in accordance with message signal through super position. |
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| 1028. |
To which part of the electromagnetic spectrum does a wave of frequency `5xx10^(19)`Hz belong ? |
| Answer» Correct Answer - Infra-red. | |
| 1029. |
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `2.5xx10^(-7)m_(2)` carrying current of 1.8 A. Assume the density of conduction electrons to be `9xx`10^(26)m_(3)`. |
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Answer» We know that `V_(d)=I//nqA`. `(1.8)/((9xx10^(28))xx(1.6xx10^(-19))xx(2.5xx10^(-7)))=5xx10^(-4)ms^(-1)` |
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| 1030. |
Estimate the average drift speed of conduction electrons in a copper wire of cross-sectional area `1.0xx10^(-7)m^(2)` carrying a current of `1.5xx10^(-19)`A. Assume the density of conduction electrons to be `9xx10^(28)m^(-3)`. |
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Answer» We know that `I="neA"" " vd` Where, I is the current, n is charge density, e is charge of electron and A is cross-section area. `V_(d)=I//"neA"` `=1.5xx10^(19)//9xx10^(28)xx1.6xx10^(-19)xx1xx10^(-7)` `V_(d)=10^(-3)ms^(-1)` This is the required average drift velocity. |
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| 1031. |
The instantaneous current and voltage of an a.c. circuit are given by i=0 sin 3000 t A and v=200 sin 300 t V. What is the power dissipation in the circuit? |
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Answer» We have given, `i=10sin 100t A and V = 200 sin 300 t V` `:. i_(0) = 10A and V_(0)=100V` `:.` Average power dissipation `=V_(0)i_(0)` `=100xx10=1000W`. |
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| 1032. |
(a) Indiscriminate human activities such as alien species invasion, fragmentation and habits loss have accelerated the loss of biodiversity. Justify by taking one example of each. (b) State the importance of (i) IUCN Red data list and (ii) Hot spots in conservation of biodiversity. |
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Answer» (a) Alien species invasions : When alien species are introduced unintentionally or deliberatelt for whatever purpose, some of them turn invasive, and cause decline or extinction of indigenous species E.g. The recent illegal introduction of African catfish claries gariepinus for aqua culture purposes is posing a threat to the indigenous catfishes in our rivers. Habit loss and fragmentation : The most important cause driving animals and plants to extinction. When large habitals are broken up into fragments due to various human activities, badly affected, leading to population declines. For example , The Amazon rain forest harbouring probably millions of species is being cut and cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle. (b) The IUCN Red list is set upon precies criteria to evaluta the extinction risk of the thousands of species and sub-species. The role of IUCN (1996) Red data list are : (i) to provide scientifically based information on status of species and subspecies at a global level, (ii) todraw attention to the magnitude and importance of threatened biodiversity. (iii) to conserve biological diversity. Hot spot has been developed in 1988. Thses include the proteced areas rich in species which are under a constant threat of being ober exploited. There are 25 hot spots in the whole world and two are in India. These are 'The Eastern Himalays' and 'The Western Ghats'. |
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| 1033. |
(a) Indiscriminate human activities such as alien species invasion, fragmentation and habits loss have accelerated the lose of biodiversity. Justify by taking one example of each. (b) State the importance of (i) IUCN Red data list and (ii) Hot spots in conservation of biodiversity. |
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Answer» (a) Alien species invasions : When alien species are introduced unintentionally or deliberatelt for whatever purpose, some of them turn invasive, and cause decline or extinction of indigenous species E.g. The recent illegal introduction of African catfish claries gariepinus for aqua culture purposes is posing a threat to the indigenous catfishes in our rivers. Habit loss and fragmentation : The most important cause driving animals and plants to extinction. When large habitals are broken up into fragments due to various human activities, badly affected, leading to population declines. For example , The Amazon rain forest harbouring probably millions of species is being cut and cleared for cultivating soya beans or for conversion to grasslands for raising beef cattle. (b) The IUCN Red list is set upon precies criteria to evaluta the extinction risk of the thousands of species and sub-species. The role of IUCN (1996) Red data list are : (i) to provide scientifically based information on status of species and subspecies at a global level, (ii) todraw attention to the magnitude and importance of threatened biodiversity. (iii) to conserve biological diversity. Hot spot has been developed in 1988. Thses include the proteced areas rich in species which are under a constant threat of being ober exploited. There are 25 hot spots in the whole world and two are in India. These are 'The Eastern Himalays' and 'The Western Ghats'. |
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| 1034. |
Give two reasons for keeping beehives in crop fields during flowering period. |
| Answer» See Q.24, Set-I., Delhi Board-2014. | |
| 1035. |
(a) State the role of DNA ligase in biotechnology. (b) What happens when Meloidegyne incognitia consumes cells with RNAi gene? |
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Answer» Role of DNA ligase in biotechnology: DNA ligase helps in joining of DNA fragments end to end, having same kind of sticky ends. (b) If Meloidegyne incognitia consume cells with RNA igene then silencing of specific mRNA occurs, due to a complementary dsRNA molecule formation. This ds RNA binds to and prevents translation of mRNA (silencing) and thus, causing death of the nematode. |
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| 1036. |
The microscoppiocl len grains of the past are obtained as fossils. Mention the characteristic of the pollen grains that makes it happen. |
| Answer» The Iexine of pollen grains have an outermost hard layer composed of a chemical, sporopollenini It is highly resistant to high temperature, strong acids and alkali. | |
| 1037. |
Name the plant source of the drug popularly called "smack". How does it affect the body of the abuser? |
| Answer» It is obtained from pa paver somniferum. Smack is a stronger analgesic than morphine. It reduces heart beat, blood pressure and increases blood sugar. | |
| 1038. |
(a) State the difference between meiocyte and gamete with respect to chromosome number. (b) Why is a whiptail lizard referred to as parthenogenetic? |
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Answer» (a) Meiocyte (gamete mother cells) is diploid (2n), where as gamete is haploid (n). (b) Whiptail lizard is said to be parthenogenetic because female gamete undergoes development to form new organisms without fertilization. |
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| 1039. |
(a) How does cleistogamy ensure autogamy? (b) State one advantage and one disadvantage of cleistogamy to the plant. |
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Answer» (a) Cleistogamous flowers are those which do not open at all and pollen from other plants cannot land on the stigma of these flowers. Thus, in such plants cross pollination cannot occur and only autogamy occurs. Therefore, cleistogamy ensures autogamy. (b) Cleistogamy has this advantage that the plant produces assured seed set even in the absence of pollinators and disadvantage is that self pollination occur which reduces that chances of variation and evolution of genetically superior progeny. |
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| 1040. |
Why do clown fish and sea anemone pair up? What is this relationship called? |
| Answer» This interaction is called commensalism. Here the clown fish lives in the tentacles of sea-anemone. The fish gets protected from predators which stay away from stinging tentacles. However anemone does not get any benefit from clown fish. | |
| 1041. |
Name the technique and the property of plant cells that can help to grow somaclones of certain desired varitey of apple. Explain how somaclones of apple can be obtained in the lab so as to get the desired variety on a large scale. |
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Answer» Tissue culture or micropropagation is used which utilizes totipotency to grow somaclones of certain desired varieties of apples in short period of time. A specific tissue of desired varieties is chosen and grown invitro under sterile conditions to give callus. Callus is cut into small parts taken in different test tubes added with specific phyto-hormones. Roots and shoots grows after few days. Plants after few weeks of hardening process get ready to be planted in field. |
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| 1042. |
What is cleistogamy ? Write one advantage and one disadvantage of it, to the plant. |
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Answer» Flowers which are never open to ensure self-pollination is called cleistogamy. They remain closed so that cross-pollination does not occur. Cleistogamy has the advantage that the plant produces assured seed set even in the absence of pollinators and disadvantage is that self-pollination occur which reduces chances of variation and evolution of genetically superior progeny. |
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| 1043. |
Which type of population interaction is found between clown fish and sea anemone? |
| Answer» Interaction that is seen between clownfish and sea anemones is called commensalism. The fish gets protected from predators which stay away from stinging tentacles of anemones however, anemones does not get any benefit from clowfish. | |
| 1044. |
Name the kind of diseases/disorders and any two symptoms that are likely to occur in humans if: (a) Mutation in the gene that codes for an enzyme phenylalanine hydroxylase occurs. (b) The karyotype is XXY. |
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Answer» (a) Phenylalanine hydroxylase converts amino acid phenylalanine into another amino acid, tyrosine. It due to mutation, phenylalanine hydroxylase (PAH) does not form, Phenylalanine `overset("PAH")rarr` Tyrosine It leads to buildup of phenylalanine to toxic levels, a condition called phenylketonuria. The disease leads to mental disorders, seizures and intellectual disability. (b) The karyotype XXY is present in a syndrome called Klinefelter syndrome. It results from two or more X-chromosome in males. Symptoms include weaker muscles, greater height,less body hair, breast growth poor coordination and infertility. |
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| 1045. |
Normally one embryo develops in one seed but when an orange seed is squeezed many embryos of different shapes and sizes are seen. Mention how it has happened. |
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Answer» The nuceller cells surrounding the embryo sac start dividing, protrude into the embryo sac and develop into the embryos. In such species each ovule contains many embryosf Occurence of more than one embryo in a seed is referred as polyembryony |
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| 1046. |
Mention the role of the codons AUG and UGA during protein synthesis. |
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Answer» AUG it acts as a initiator condon in translation. UGA it acts as the terminator condon. |
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| 1047. |
why is making cells competent essential for biotechnology experiments? List any two ways by which this can be achieved. |
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Answer» DNA being a hydrophilic molecule, can not pass, through the cell membrances. So the bacteria should be made competent to accept the DNA molecules. Competency is the ability of a cell to take up foreign DNA. Two ways to make cells competent: (i) Treatment with divalent cations like calcuim. (ii) Use of gene gun. |
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| 1048. |
Write the importance of cryopreservation in conservation of biodiversity. Ans |
| Answer» In-Vitro conservation especially cryopreservation is useful technique for preserving vegetatively propagated crops. e.g. potato seeds of plants, and preserving sperms, eggs, cells and embryonic tissues of animals at `196^@`C temperature. | |
| 1049. |
How can bacterial DNA be released from the bacterial cell for biotechnology experiments ? |
| Answer» Since, DNA is enclosed Within the membranes, we have to break the cell Open to release DNA along with other macromolecules such as RNA, proteins, polysaccharides and also lipids. This can be achieved by treating the bacterial cell/plant or animal tissue with enzymes such as lysozyme (bacteria), cellulase (plant cell), chitinase (fungus) | |
| 1050. |
Draw a schematic diagram of a human sperm and label the cellular components. Give the functions of any three parts. Where are the sperm heads found embedded to survive after spermiogenesis? |
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Answer» Head : The acrosome contains hydrolytic enzymes which dissolves the egg membrane for fertilisation. Middle piece : It contains mitochondria that provide energy for the movement of the sperms. Tail: It helps in the motility of sperms that helps for the swimming of sperms in a liquid medium. The sperm head becomes embedded in the Sertoli cells : See Q.19.,Delhi Board, Set-I,2008. |
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