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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.
| 1. |
Describe how the following changes are brought about: (i)Pig iron into steel. (ii)Zinc oxide into metallic zinc. (iii)Impure titanium into pure titanium. |
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Answer» (i) Pig iron is converted into steel by adding carbon and some other elements. (ii) Metallic Zinc is obtained from zinc oxide by reduction with coke. `ZNO+C overset("Heat")to Zn+CO` (iii)Impure titanium is heated with Iodine toform volatile complex TiI_(4) which on furtherheating to higher temperature decomposes to give pure titanium. |
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| 2. |
(a) Give an example of zone refining of metals. (b) What is the role of cryolite in the metallurgy of aluminium? |
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Answer» (a) This method is very useful for producing semiconductor and other metals of very high purity like germanium, silicon, boron etc. (b) Role of cryolite in metallurgy of Aluminimm-Cryolite is added to lower the melting point pf mixture and to increase the conductivity of electrolyte. |
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| 3. |
Write two differences between multimolecular colloids and macromolecular colloids ? |
| Answer» See Q. 21, Set-II, Delhi Board, 2009. | |
| 4. |
(a) What is "population" according to you as a biology student ? (b) "The size of a population for any species is not a static parameter." Justify the statement with specific reference to fluctuations in the population density of a region in a given periods of time. OR (a ) What is hydrarch succession ? (b) Compare the pioneer species and climax communities of hydrarch and xerarch succession respectively. (c) List the factors upon which the type of invading pioneer species depend in secondary hydrarch succession. Why is the rate of this succession faster than that of primary succession ? |
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Answer» Single individuals of any species living in a groups in a well defind geographical area share or compete for similar resource, inter-breed and thus, constitute a population. (b) The size of the population tells us a lot about its status in the habitat. Whatever ecological processes are to be investigated in a population, be the outcome of competition with other species the impact of a predator, it is evaluated in terms of change in the population size. Population size is no more technically called population density. Population size of a species is the number of individuals of a species per unit volume Population density = `("Number of individuals ina a region (N)")/("Unit area in a region (S)")` `implies P Delta = (N)/(S)` A population at any given time is composed of individuals of different ages. When the age distribution (percent individuals of given age or age group) is plotted for the population, the resulting structure is called age pyramid. OR (a) The gradual and fairly predicatable change in the species composition of a given area is called ecological succession. Hydrarch susccession takes place in wetter areas and the successful series progresses from hydric to the mesic conditions. (b) The species that invade a base area are called pioneer species. The changes which finally lead to a community That is in near equilibrium with the enviroment is called climx community. In primary Hydrarch succession, the pioneer are the small phytoplanktons, they are replaced with time by free floating angiosperms, then by rooted hydrophytes, sedges, grasses and finally the trees. The climax would be a forest and later water body is converted into land. Xerarch succession takes place in dry areas and series progresses from xeric to mesic conditions. Primary succession on rock initiated by pioneer lichens which secrete acids to dissolve rock, helps in weathering and soil formation. Later, bryophytes grows which are able to hold in small amount of soil. Later bigger plants takes over ultimately a stable climax forest community is formed. (c) In secondary Hydrarch succession, the species that invade depend upon the condition of soil availability of water. since the soil is already there, the rate of succession is much faster and hence, climax is reached more quickly. |
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| 5. |
It is often said that the pyramid of energy is always upright. On the other hand, the pyramid of biomass can be both upright and inverted." Explain with the help of example and sketches . |
| Answer» For pyramid of Energy : See Q- 28 .Set-II Outside Delhi 2012,2013 | |
| 6. |
List the different attributes that a population has and not an individual organism. ( b) What is population density? Explain any three different ways the population density can be measured ,with the help of an example each . |
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Answer» Following are the attributes that a population has·but an individual organism does not have: 1. Per capita birth·rate. 2. Per capita death rate. 3. Sex ratio : ,Ratio of number of males to females in a population Population density means number of individuals present. per unit area, population density can be measured by determining the population: size. The different methods to study population size are as fallows: ( 1) Per cent cover or biomass : In an area with 200 parthenium plant and only one banyan tree with -large conopy, the denstiy of banyan tree is small but does not reflect its important role in the community. Here percent cover or biomass is a more meaningful method of assesseing population density. ( 2) Total Number : It involves the counting of organisms in the given area. ( 3) Relative Density :In this method m there is no need to count the organsms individually Example the nubmber of fishes caught per trap gives the measure of total density in a given water body. ( 4) Indirect Assesment: The tiger cenus is based on pug marks and faecal pellets. |
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| 7. |
Asnwer the following : (a) Name the em waves which are used for the treatment of certain forms of cancer. Write their frequency range. (b) Thin ozone layer on top of stratosphere is crucial for human survival. Why ? (c) Why is the amount of the momentum transferred by the em waves incident on the surface so small? |
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Answer» (a) Gamma `(gamma)` rays are used for the treatment of certain forms of cancer. Their frequency range is `3xx10^(19)Hz" to "5xx10^(24)Hz.` (b) The thin ozone layer on top of stratosphere absorbs most of the harmful ultraviolet rays coming from the sun towards the Earth. They include UVA, UVB and UVC radiaations, which can destroy the life system on the Earth. Hence, this layer is crucial for human survival. (c) Momentum transferred = (Energy) (Speed of light) `hvc ~~ 10^(-22)(fpr v~ 1020 Hz).` Thus, the amount of the momentum transferred by the em waves incident on the surface is very small. |
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| 8. |
(i) Arrange the following compounds in an increasing order of basic strength : `C_(6)H_(5)NH_(2), C_(6)H_(5)N(CN_(3))_(2), (C_(2)H_(5))_(2)NH and CH_(3)NH_(2)` (ii) Arrange the following compounds in a decreasing order of `pK_(b)` values: `C_(2)H_(5)NH_(2), C_(6)H_(5)NHCH_(3), (C_(2)H_(5))_(2) NH and C_(6)H_(5)NH_(2)` |
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Answer» (i) `C_(6)H_(5)NH_(2) lt C_(6)H_(5)N(CH_(3))_(2) lt CH_(3)NH_(2) lt (C_(2)H_(5))_(2)NH` (ii) `C_(6)H_(5)NH_(2) gt C_(6)H_(5)NHCH_(3) gt C_(2)H_(5)NH_(2) gt (C_(2)H_(5))_(2)NH` |
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| 9. |
Rerrange the following in an increasing order of their basic strengths: `C_(6)H_(5)NH_(2),C_(6)H_(5)N(CH_(3))_(2),(C_(6)H_(5))_(2)` NH and `CH_(3)NH_(2)`. |
| Answer» `C_(5)H_(5)NH_(2) gt C_(6)H_(5)N(CH_(3))_(2) gt CH_(3)NH_(2) gt (C_(6)H_(5))_(2)NH` | |
| 10. |
(b) Estimate the minimum potential difference needed to reduce `AI_(2)O_(3)` at `500^(@)C` The gibbs energy change for the decomposition reaction `2/3AI_(2)O_(3)rarr4/3AI+O_(2)`is 960 kJ (F=96500 C `mol^(-1))` |
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Answer» `AI_(2)O_(3)(2AI^(3+)+3O^(2-))rarr2aI+3/2O_(2)n=6e` `2/3AI_(2)O_(3)rarrr4/3AI+O_(2)n=2/3xx6e=4e` `triangleG=960xx1000=960000` `triangleG=nFE_(cell)` Now, `E_(cell)=(triangleG)/(nF)=(-960000)/(4xx96500)` `E_(cell)=-2.487V` |
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| 11. |
The decomposition of NH, on platinum surface is zero order reaction. If rate constant (k) is `4xx10^(-3) Ms^(-1)`, how long will it take to reduce the initial concentration of `NH_(3)` from 0.1 M to 0.064 M. |
| Answer» `t=([R]_(0)-[R])/K=(0.1-0.064)/(4xx10^(-3))= 9` sec | |
| 12. |
A capacitor, made of two parallel plates each of plate area A and separation d, is being charged by an external ac socurce. Show that the displacement current inside the same as the current charging the capacitor. |
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Answer» Let the alternating emf charging the plates of the capacitor `epsilon=epsilon_(0) sin omegat.` Charge on capacitor `q=C_(epsilon)=C_(epsilon_(0))sin omegat` Let Instantaneous current be I `So," "I=(dq)/(dt)=(d)/(dt)C_(epsilon_(0)) sin omega t` `=C_(epsilon_(0))omega cos t` `=I_(0) cos omega t` `"where, "I_(0)=C_(epsilon_(0))omega` `I_(d)=in_(0)(dphi_(E))/(dt){phi_(E)=EA.` `I_(d)=in_(0) A(DE)/(dt){E=(sigma)/(in_(0))=(q)/(A_(in_(0)))` `I_(d)=in_(0)A(d)/(dt)((q)/(A_(in_(0))))` `I_(d)=(d)/(dt)C_(epsilon_(0))sin omega t = C_(epsilon_(0))omega cos omegat= I_(0) cos omega t` Hence, conduction current is equal to the displacement current. |
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| 13. |
Describe the principle involved in each of the following processes : (i) Zone refining of a metal (ii) Vapour phase refining of metals |
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Answer» (i) Zone refining : This method is based on the principle that the impurities are more soluble in the melt than in the solid state of the metal. |
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| 14. |
Describe the principle controlling each of the following processes: (i) Vapour phase refining of titanium metal. (ii) Froth floatation method of concetration of a sulphide ore |
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Answer» (i) Vapour phase refining of titanium metal. (ii) Froth floatation method of concetration of a sulphide ore |
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| 15. |
Describe the principle behind each of the following processes : (i) Vapour phase refining of a metal. (ii) Electrolytic refining of a metal. (iii) Recovery of silver after silver ore was leached with NaCN. |
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Answer» (i) In this method crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at a lower ten:perature a nd then decomposing the volatile compound at some higher temperature to give the pure metal. Thus, the two requirements are : - the metal should from a volatile compound with a suitable reagent. - the volatile compound should be easily decomposable so that the recovery is easy. (ii) See Q. 21 (or part) (ii), Set-I, Outside Delhi, 2010. (iii) Role of NaCN in the extraction of silver is to do the Leaching of silver ore in the presence of air from which the silver is obtained Later by replacement `4Ag(s)+8CN^(-)(aq)+2H_2O(aq)+O_2(g) to4[Ag(CN)_2]^(-)(aq)+4OH^(-)(aq)` `2[Ag(CN)_2]^(-)(aq)+Zn^(-) (s)to2Ag(s) [Zn(CN_4)]^(2-)(aq)`. |
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| 16. |
(a) Write the principle involved in the vapour phase refining of metals. (b) Write the name of the metal refined by each of the following processes : (i) Mond process (ii) van Arkel method (c ) What is the role of depresent in froth floatation process ? |
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Answer» (a) Vapour phase refining : In this method, crude metal is freed from impurities by first converting it into a suitable volatile compound by heating it with a specific reagent at a lower temperature and then decomposing the volatile compound at some higher temperature to give pure metal. (c ) Depressant is to prevent one type of sulphide ore particles from forming the froth with air bubbles. Example : NaCN is used as a depressant to separate lead sulphide (PbS) ore from Zinc Sulphide (ZnS). `4NaCN + ZnS rarr underset("Prevent from forming the forth)")underset("(Complex formed and")(NO_(2)[Zn(CN_(4))]+Na_(2)S)` |
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| 17. |
An element with density 2.8 `cm^(3)` forms a f.c.c unit cell with edge length `4xx10^(-8)` cm. calculate the molar mass of the element. Given: `(N_(A)=6.022xx10^(23)) mol^(-1)` |
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Answer» `d=2.8 g//cm^(3),z=4,a=4xx10^(-8)cm` `d=(zxxm)/(vxxNA)` `2.8=(4xxM)/(4xx10^(-8^(3))xx6.02xx10^(23))` `2.8=(4xxM)/(4^(3)xx6.02xx10^(-1))` M=26.97 |
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| 18. |
An element crystallises in fcc lattice with cell edge of 400 pm. Calculate its density if 250 g of this element contain `2.5 xx 10^(24)` atoms. |
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Answer» Z = 4 `a = 400 "pm" = 400 xx 10^(-10) cm` `6.022 xx 10^(23)` atoms has a mass = molecular mass (M) `2.5 xx 10^(24)` atoms has mass = 250 g `6.022 xx 10^(23)` atoms has mass = `(250)/(2.5 xx 10^(24)) xx 6.022 xx 10^(23)` Molecular mass (M) = 60.22g `d=(Z xx M)/((a)^(3)xxNa)` `d = (4 xx cancel60.22)/((400xx10^(-10))^(3)xxcancel(6.022)xx10^(23))` `d = (cancel(400))/(cancel(4)xx4xx4) = d = (10)/(16) = 6.25 g//cm^(3)` |
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| 19. |
An element with density `11.2 g cm^(-3)` forms a f. c. c. lattice with edge length of `4xx10^(-8)` cm. Calculate the atomic mass of the element. (Given : `N_(A)= 6.022xx10^(23) mol^(-1)` |
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Answer» Given, Density, `d = 11.2 g cm ^(-3)`, Edge length `a = 4xx 10^(-8) cm,` Avogafro number, `NA = 6.022xx10^(23)` Number of atoms present per unit cell, Z (fcc) = 4 We know for a crystal system = `(ZxxM)/(a^(3) xxNA)` `rArr =(dxxa^(3)xxNa)/(Z) = (11.2xx64xx10^(-24) xx6.022xx10^(23))/(4) = 107.91 g` Thus, atomic mass of the element is `107.91` g. |
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| 20. |
(a) List any four characteristics of an ideal contraceptive. (b) Name two intrauterine contraceptive devices that affect the motility of sperms. |
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Answer» (a) An ideal contraceptive has the following characteristics : 1. It is user - friendly. 2. It should not interfere with the sexual drive desire or the sexual act of the user. 3. It should be easily available. 4. It should be effective and reversible with no or little side effects. ( b) The two IUDS that affect the motility of the sperms are: 1.`C_(u) T` 2. `C_(u) 7` |
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| 21. |
List the post-fertilization events in angiosperms. |
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Answer» The various post-fertilisation events occuring in angiosperms are: i) Endosperms is formed before the development of embryo, as it provides nourishment to the developing embryo. ii) Zygote at the micropylar end undergoes successive divisions to form pro-embryo. It is followed by the formation of globular, heart-shaped and mature embryo. iii) A mature dicotyledonous embryo consists of two cotyledons, epicotyl and hypocotyl. On the other hand, a mature monocotyledonous embryo possesses only one cotyledon,a coleoptile and coleorrhiza. |
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| 22. |
Identify the following paris of Homologous or Analogous organs: i) Sweet potato and potato ii) Eye of octopus and eye of mammals iii) Thorns of Bougainvillea and tendrils of Cucurbits. iv) Fore limbs of Bat and Whale. |
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Answer» i) Analogous organs. ii) Analogous organs. iii) Homologous organs. iv) Homologous organs. |
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| 23. |
Giving three reasons, write how Hardy-Weinberg equilibrium can be affected. |
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Answer» Hardy-Weinberg principle predicts that five conditions can genetic equilibrium and cause evolution to occur. (1) gene flow. (2) genetic drift (3) genetic recombination (3) mutations, or (4) mutations, or (5) natural selection. |
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| 24. |
Name the part of the electromagnetic spectrum of wavelength `10^(-2)m` and mention its one application. |
| Answer» Microwave (Range 0.1 m to `10^(-3) m`). | |
| 25. |
The susceptibility of a magnetic material is `-4*2xx10^-6`. What type of material does it represent? |
| Answer» Diamagnetic substances are present there the form of Silicon. | |
| 26. |
The susceptibility of a magnetic material is `1*9xx10^-5`. What type of material does it represent? |
| Answer» Paramagnetic substances. | |
| 27. |
If `chi` stands for the magnetic susceptibility of a given material, identify the class of material for which (i)`-1 ge chi lt 0` (ii) `0 chi lt in (in "stands for a small positive number")` |
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Answer» `(i)"For "-1 ge x lt 0.` material is diamagnetic (ii) `"For "0 lt chi lt in`, material is paramagnetic. |
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| 28. |
A glass lens of refractive index 1.5 is placed in a trough of liquid. What must be the refractive index of the liquid in order to mark the lens disappear? |
| Answer» The refractive index of the liquid must be greater than or equal to 1.5. | |
| 29. |
The susceptibility of magnetic material is `2.6xx10^(-5)`. Identify the type of magnetic material and state its two properties. |
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Answer» Paramagnetic, as susceptibility x is +ve and not so large. Paramagnetic material (i) Pull most of the field lines inside the material and leave some outside. (ii) Moves towards stronger field in a non-uniform field. |
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| 30. |
The oscillating magnetic field in a plane electromagnetic wave is given by `B_(y)=(8xx10^(-6))sin [2xx10^(11)t+300 pi x ] T` (i) Calculate the wavelength of the electromagnetic wave. (ii) Write down the expression for the oscillating electric field. |
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Answer» Given equation is `B_(y)=(8xx10^(-6))sin [2xx10^(11)t+300 pi x]T` Comparing the given equation with the equation of magnetic field varying sinusoidally with x and t `B_(y)=B_(0)sin ((2pi x)/(lambda)+(2pi t)/(T))` We get, `(2 pi)/(lambda)=300 pi :. lambda=(2)/(300)=0.0067 m. and B_(0)=8xx10^(-6)T` (i) Wavelength of the electromagnetic wave `lambda=0.0067m ` or 0.67 cm (ii)`E_(0)=CB_(0)=3xx10^(8)xx8xx10^(-6)=24xx10^(2)=2400 Vm^(-1)` `:.` The required expression for the oscillating electric field is `E_(z)=E_(0) sin ((2 pi x)/(lambda)+(2pi t)/(T))=2400 sin (300 pi x + 2 xx 10^(11)t)` v/m. |
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| 31. |
State the reason, why heavy water is generally used as a moderator in a nuclear reactor. |
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Answer» (i) Heavy water does not absorb neutron. (ii) It is rich in proton. |
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| 32. |
Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility . What does negative susceptibility signify ? |
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Answer» Magnetic susceptibility : Magnetic susceptibility of a material is defined as the ratio of the intensity of magnetisation (I) induced in the material to the magnetisation force (H) applied on it. Magnetic susceptibility is represented by `X_(m)=(I)/(H)`. `Diamagnetic substances like copper, lead etc. has negative susceptibility. Paramagnetic substances like aluminium, calcium etc. has positive susceptibility. Negative susceptibility of diamagnetic substance does not change with temperature. |
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| 33. |
Nichrome and copper wires of same length and same radius are connected in sereis. Current is `I` passed through them. Which up more ? Justify your answer. |
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Answer» In series combination, I is same for nichrome and copper . As rate heat product = P = ` I^(2)R` or ` P alpha R, , So, R _("nichrome")gt R _("copper")` for given length and diameters. Thus, nichrome will produce more heat. |
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| 34. |
Write any two distinguishing features between conductors, semiconductors and insulators on the basis of energy band diagrams. |
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Answer» Energy bands : A group of very large number of energy levels lying closely in a very small energy range formed an energy band. Formation of energy bands in solids : A free atom has well defined energy levels. Example : Silicon, Si has the electronic configuration `1s^(2),2s^(2),2p^(6),3s^(2),3p^(2)`. The outer s-level has 2N electrons and p-level has 2N electrons. The s-level is completely filled while the p-level is not completely filled and can accommodate 4 N electrons to make 6N electrons. (i) The lower completely filled band at 0 K is the valence band and (ii) The upper unfilled band is the conduction band. The difference between the higher energy in a bond and the lowest energy in the next higher band is called forbidden energy gap E. The conductivity depends upon the energy gap `(E_(g)` between the top fo valence band VB and the bottom of the conduction band CB. Smaller the forbidden energy gap In a semiconductor, higher is the conductivity at a particular temperature. For insulator `E_(g)gt3eV`, for semiconductor `E_(g)=0.2eV`. For metal `E_(g)=0`. Classification of solids : (i) Insulator : In insulator, the valence band is completely filled. The conduction band is empty and forbidden energy gap os quite large. So, no electron is able to go from valence band to conduction band even if electric field is applied. Hence, electrical conduction is impossible. The solid/substance is an insulator. (ii) Metals (conductors) : In metals, either the solution band is partially filled or the conduction and valence bond partly overlap each other. If small electric field is applied across the metal, the free electrons start moving in a direction opposite to the direction of electric field. Hence, metal behaves as a conductor. (iii) Semiconductors : At absolute zero kelvin, the conduction band is empty and teh valence band is filled. The material is insulator at low temperature. However, the energy gap between valence band and conduction band is small. At room temperature, some valence electrons acquire thermal energy and jump to conduction band where they can conduct electricity. The holes left behind in valence band act as a positive charge carrier. |
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| 35. |
For the circuit would the balancing length increase, decrease or remain the same if (i) `R_(1)` is decreased (ii) `R_(2)` is increased, without any change (in each case) in the rest of the circuit ? Justify your answer in each case. |
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Answer» (a) (i) When `R_(1)` is decreased, the balancing length AJ will decrease, we will have to move the Jockey J towards A. as by decreasing `R__(1)` potential gradiant across wire AB will increase `l=(V)/(K)` hence 1 (AJ) will decrease. (ii) When `R_(2)` is increased. Potential drop across `R_(2)` will increas. Hence the blanacing length AJ will increase. We will have to move the Jockey J towards B. (b) A potentiometer being a null device, does not draw any current from the balance point therefore potentiometer measures the actual e.m.f. of the cell. Whereas, voltmeter always draw current from the cell and measures the terminal voltage of the cell instead of the actual emf of the cell. |
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| 36. |
Out of Lyophilic and Lyophobic sols, which can be easily coagulated on the addition of a small amount of electroyte ? |
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Answer» Depending upon the nature of interaction between the dispersed phase and the dispersion medium, colloids are divided into two categories : (i) Lyophilic sols : The colloids in which the particles of the dispersed phase have a strong affinity for the dispersion medium are called lyophilic sols. These collidal solutions, even if precipitated, change back to the colloidal form simly by adding dispersion medium, So lyophilic sols are reversible in nahtre e.g., glue, starch, rubber, etc. (ii) Lyophobic sols : The colloids in which particles of the dispersed phase have no or very little affinity for dispersion medium are called lyophobic sols. These are irreversible in nature i.e., once precipitated, they have little tendency to get back into the colloidal form on simply adding dispersion medium e.g., `As_2S_3 sol`. Lyophobic sols need stabilising agents for teir preservation. |
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| 37. |
Estimate the distance for which ray optics is good approximation for an aperture of `4 mm` and wavelength `400 nm`. |
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Answer» Here, `a=4`mm`=4xx10^(-3)`m. `lambda=400` nm `=400xx10^(-9)m=4xx10^(-7)m`. `:.` Fresnel distance, `Z_(F)=a^(2)/lambda=((4xx10^(-3))^(2))/(4xx10^(-7))=(16xx10^(-6))/(4xx10^(-7))=40 M`. |
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| 38. |
Name two diseases whosepread can be controlled by the eardicatoion of Aedes mosquitoes |
| Answer» Denue and chikungunya | |
| 39. |
Deduce the expression, `N=N_(0)^(-lambdat)` for the law of radioactive decay. (b) Is the nucleus formed in the decay of the nucleus `._(11)^(12)Na`, an isotope or isobar? |
| Answer» Since new nucleus has the same mass number, hence it would be an Isobar. | |
| 40. |
(a) Write the scientific name of the organism Thomas hunt morgan and his colleagues worked with for their experiment s .Explain the correction between linkage and recombination with respect to genes as studied by them (b) How did sturtevant expain gene mapping while working wiht morgan? |
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Answer» (a) Drosphila melanogaster: morgan carried out several dihybrid crosses in drosophila to study gens that were sex lined Morgan and his group kenew that the ngens were locayed on he x chomosome and saw quickly that when cominations were much higher than non parental type. Morgan attributed this due ot the physicla sassociation chromosome and the term recombination to describe the generations of non parental gene combination morgan and his group also found that even when genes wer grouped on the same chromosome some gens were very tightly linked while others were loosely linked (b) Morning student alfred sturteveant used the frequaency of recombination between gene paris on the same chromosome as a measure of the distance between genes and mapped their position on the chromosome |
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| 41. |
Mention the evolutionary significance of the following organisms : (i)Shrews (ii)Lobefins (iii)Homo habilis (iv)Homo erectus |
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Answer» a) Shrews : The first mammals were like shrews. Their fossils are small sized. b) Lobefins: In 1938, a fish caught in South Africa happened to be coelacanth which was thougth to be extinct. These animals called lobefins evolved into the first amphibians that lived on both hand and water. There are no speciments of these left with us. c) Homo habilis: Two mya, Australopithecines probably lived in East African grasslands. Evidence shows, they haunted like being hominid and was called homo habilis. d) Homo erectus: They had larged brain around 900 cubic centimeters. They probably ate meat. They used to hides to protect their body and buried their dead. |
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| 42. |
Explain how does : (a) a primary succession start on a bare rock and reach a climax community? (b) the algal bloom eventually choke the water body in an industrial area? |
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Answer» Such a succession is called Xerarch succession. In this succession - 1. Pioneer species is lichens which secrete acids to dissolve rock causing weathering and soil formation. 2. Next seral stage will be bryophytes which can hold in the small amount of soil. 3. Bryophytes are then succeeded by grasses. 4. Grasses eventually will pave way for bigger trees which will form stable climax community. This remains stable as long as the environment remain unchanged. (b) Effiuent from industries contains large amounts of nutrients. This causes excessive growth of free floating algae causing algal bloom. Alga starts consuming oxygen for its own consumption. This will decrease BOD of water body. This reduced BOD eventually causes death of all aquatic life thus, choking the water body. |
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| 43. |
Which of the following is a free living bacteria that can fix nitrogen in the soil? Spirulina, Azospirillum, Sonalika |
| Answer» Correct Answer - Azospirillum | |
| 44. |
How many kinds of phenotypes would you expect in `F_(2)` generation in a monohybrid cross exhibiting co-dominance ? |
| Answer» There are four kinds of phenotypes expected in a monohybrid cross in `F_(2)` generation showing co-dominance. | |
| 45. |
(b) Name the source used to produce hepatitis B vaccine rDNA technology. |
| Answer» Recombinant DNA technology has allowed the production of antigenic polypeptides of pathogen in bacteria or yeast. Vaccines produced using this approach allow large scale production and hence greater availability for immunisation, e.g., hepatitis B vaccine producced from yeast. | |
| 46. |
Work out the monohybrid cross upto `F_(1)` by taking a suitable example so as to show the following phenotypes. (i)`F_(1)` represents only one of the parents. (ii) `F_(1)` with both the parental characteristics. |
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Answer» Assuming person X with Blood group A : `A_(i)`. Assuming person T with Blood group B : `B_(i)`. `F_(1)` cross `{:(,A,i),(B,AB,Bi),(i,Ai,ii):}` AB : Phenotype with both the parental characteristics. Bi andAi : phenotype with only one of the parents. |
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| 47. |
Name the negatively charged and positively charged components of a nucleosome. |
| Answer» A Histone Octamer is the positively charged component and the DNA helix is the negatively charged component. | |
| 48. |
In a typical nucleus, some regions of chromatin are stained light and others dark. Explain why is it so and what is its significane ? |
| Answer» In a typical nucleus, some regions of chromatin are stained light because of loose packing of chromatin, and some regions of chromatin are stained dark because herein, chromatin is densely packed. Euchromatin is transcriptionally active chromatin and heterochromatin is inactive. | |
| 49. |
How do methanagens help in Producing biogas? |
| Answer» Methanogens such as Methnobacterium act on excreta of cattle and grow anaerobically, producing large amount of methane along with `CO_(2) and H_(2)`. Methane is the predominant gas of bio gas. | |
| 50. |
How are the structural genes inactivated in lac operon in E. Coli? Explain. |
| Answer» In the absence of lactose, no inactivation ofrepressor occurs and hence the repressor binds to the operator region of the operon, thus preventing RNA polymerase from transcribing the operon thus, inactivation of the production of structural genes in E. coli occurs. | |