AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf

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Answer: Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth complex in the electron transport chain. Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which results in blockade of electrons transfer to the final electron acceptor oxygen. This blockade causes build up of proton motive force in the intermembrane space as the protons are not allowed to flow back into the matrix which subsequently halts ATP production. Solution Answer: Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth complex in the electron transport chain. Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which results in blockade of electrons transfer to the final electron acceptor oxygen. This blockade causes build up of proton motive force in the intermembrane space as the protons are not allowed to flow back into the matrix which subsequently halts ATP production..

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A is correct. Public and private community strings are used within SNMP to read and write, respectively. B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for writing configuration data, while public allows for reading MIB information. C and D are incorrect because they do not exist as community strings. Solution A is correct. Public and private community strings are used within SNMP to read and write, respectively. B, C, and D are incorrect. B is incorrect because the choices are backwards. Private allows for writing configuration data, while public allows for reading MIB information. C and D are incorrect because they do not exist as community strings..

A is correct. Public and private community strings are used within S.pdf

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4 ansit is periodic signal andSolution4 ansit is periodic .pdf

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3030Solution3030.pdf

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5Solution5.pdf

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1-They have terrestrial adaptation and dominant sporophytic phase. 2-ovule is produced, but but it is not enclosed within ovary wall. 3- all are heterosporous(producing mega and microspores).microspore termed pollen grains make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte which contain multicellular female sex organ (archegonium). 4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non flagellated sperms. 5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid. 1-Dominant sporophytic phase. 2-ovule develops within ovary of a carpel, and enclosed by ovary wall. 3-All are heterosporous(producing mega and microspores).micro or male gametophytes develops within microspore (pollengrain) and without antheridium.megaspore bear mega or female gametophyes and without archegonium. the gametophytes of angiosperms are much reduced in size and cell number. the mature microgametophytes within the pollen grain consist of only three cells, and the mature megagametophytes within the ovule , in most angiosperm. consists of only seven to eight nuclei. 4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on the ovules . 5-double fertilization occurs , to produce embryo as well as endosperm . the endosperm, nutritive tissue for the future embryo , is not haploid but triploid and forms after fertilization.gymnospermsAngiosperm 1-They have terrestrial adaptation and dominant sporophytic phase. 2-ovule is produced, but but it is not enclosed within ovary wall. 3- all are heterosporous(producing mega and microspores).microspore termed pollen grains make direct contact with the ovules . microgametophytes develops within the pollenwall)i.e endosporic) and without male sex organ, antheridium. megaspors bears megagametophyte which contain multicellular female sex organ (archegonium). 4-they dont require water for sperm to swim in to reach the egg, all gymnosperm have non flagellated sperms. 5-no double fertilization to produce endosperm . the endosperm nutritive tissue is haploid. 1-Dominant sporophytic phase. 2-ovule develops within ovary of a carpel, and enclosed by ovary wall. 3-All are heterosporous(producing mega and microspores).micro or male gametophytes develops within microspore (pollengrain) and without antheridium.megaspore bear mega or female gametophyes and without archegonium. the gametophytes of angiosperms are much reduced in size and cell number. the mature microgametophytes within the pollen grain consist of only three cells, and the mature megagametophytes within the ovule , in most angiosperm. consists of only seven to eight nuclei. 4-pollination is indirect in which pollen falls on the stigma of the carpel rather than directly on the ovules . 5-double fertilization occurs , to produce embryo as well as endosperm . th.

1-They have terrestrial adaptation and dominant sporophytic phase..pdf

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(1)Solution(1).pdf

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What does absense of MHC Class I indicate? Almost all nucleated cells will have MHC Class I molecules on their surface and their main role is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will not express MHC Class I molecules on their surface and thus there will not be any expression of tumor antigens to Tc cells and these will survive and will go on producing new cells and form the tumor. Like this MHC class I negative cells present within the primary tumor will develop into unnoticed micrometastases and will have an immunological equilibrium and succeeds in surviving in a dormant stage for long periods of time What does the presentation of tyrosinase indicate? The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell surface as peptides and overexpression of these antigens in tumor cells will result in an abnormally high density of specific peptide:MHC complexes on the cell surface making them immunogenic and provides them tolerance and is seen as tumor rejection in most of the cases. What dos decreased CTL infiltratioin indicate? It is the CD8+ cytotoxic T lymphocytes (CTLs) which will reject the tumor by infiltrating the solid tumors by recognizing the tumor antigens, and will kill those tumor cells. And to do this CTLs will try to infiltrate tumors in depth and it is done only when tumor cells express the cognate CTL antigen. If the tumor will not express the cognate antigen, CTL infiltration will not occur and the tumor cells cannot be killed and so the expression of tumor antigen on the cells will attract CTL infiltration, at the same time when the tumor antigens are not presented on these cells, it will decrease the CTL infiltration. Solution What does absense of MHC Class I indicate? Almost all nucleated cells will have MHC Class I molecules on their surface and their main role is to present foreign protein to cytotoxic T cells (Tc Cells) so that the cells which got infected can be cleared by these Tc cells. But in some cases, it has been observed that some tumor cells will escape killing or elimination by Tc cells by becoming MHC Class I negative, here they will not express MHC Class I molecules on their surface and thus there will not be any expression of tumor antigens to Tc cells and these will survive and will go on producing new cells and form the tumor. Like this MHC class I negative cells present within the primary tumor will develop into unnoticed micrometastases and will have an immunological equilibrium and succeeds in surviving in a dormant stage for long periods of time What does the presentation of tyrosinase indicate? The tyrosinase will cleave the tumor proteins and will express the melanoma antigens on the cell surface as peptides and overexpres.

What does absense of MHC Class I indicateAlmost all nucleated cel.pdf

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The probabilities must add up to one. Add the ones you have and su.pdf

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The most likely F1 will be the one that is made from 4 and 5. This is because see the map between c and d gene. In map 1 the distance is 15 and in map 2 the distance is 20. What we know is , the larger the distance between two genes there are more recombinants between those 2 genes. Here in map 1 the distance between c and d is 10 which means the 2 genes are more likely to be in a shorter dstance ie they are linked. Linked genes fail to separate during meiosis and results in lesser number of recombinants and more of parents. So here in 2x3 F1s we cannot say that the heterozygous will replace the homozygous recessive allele. But in map2 ie F1s of 4x5, the distance between c and d is 20, and there will be less linkage compared to F1s of 2x3 and hence there is a chance for deletion of homozygous recessive alleles in them. Solution The most likely F1 will be the one that is made from 4 and 5. This is because see the map between c and d gene. In map 1 the distance is 15 and in map 2 the distance is 20. What we know is , the larger the distance between two genes there are more recombinants between those 2 genes. Here in map 1 the distance between c and d is 10 which means the 2 genes are more likely to be in a shorter dstance ie they are linked. Linked genes fail to separate during meiosis and results in lesser number of recombinants and more of parents. So here in 2x3 F1s we cannot say that the heterozygous will replace the homozygous recessive allele. But in map2 ie F1s of 4x5, the distance between c and d is 20, and there will be less linkage compared to F1s of 2x3 and hence there is a chance for deletion of homozygous recessive alleles in them..

The most likely F1 will be the one that is made from 4 and 5. This i.pdf

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Subculturing or passage is the method of preserving or continuing the culture for longer periods Subculturing can be done in three procedures based on the cells we need to passage or subculture. 1. Bacterial subculture. 2.Monolayer sub culture. 3. Suspension sub culture. Requirements :- For any of the sub culture it needs the following criteria The adequate conditions like subculture should be done periodically, need of adding nutrients or shifting of the innoculum both are considered as subculture. 1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are grown on the roux flask or other surface which are cultured by removing by trypsinization procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles medium. So that the cells will grow vigorously and differentiate. 2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium components so that cells grow in high nutrient concentration, 3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium components. Solution Subculturing or passage is the method of preserving or continuing the culture for longer periods Subculturing can be done in three procedures based on the cells we need to passage or subculture. 1. Bacterial subculture. 2.Monolayer sub culture. 3. Suspension sub culture. Requirements :- For any of the sub culture it needs the following criteria The adequate conditions like subculture should be done periodically, need of adding nutrients or shifting of the innoculum both are considered as subculture. 1. MONOLAYER CULTURES when the continous cell lines or other adhered cells which are grown on the roux flask or other surface which are cultured by removing by trypsinization procedures and grown in the fresh medium eg DMEM medium Dulbeco modified Eagles medium. So that the cells will grow vigorously and differentiate. 2. Suspension cultures when cells of the well grown medium is transferred to the fresh medium components so that cells grow in high nutrient concentration, 3. Bacterial sub culturing : Transfer of innoculum(colony) from solid agar surface to broth medium, solid surface to solid surface, liquid broth to liquid broth, liquid broth to solid agar medium surface, sub culturing can be done by streaking, pouring, spreading onto fresh medium components..

Subculturing or passage is the method of preserving or continuing th.pdf

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public class Interest{ double interest(double rate, double amount){ return amount+(amount*rate); } double totalSaving(int months,double monthly,double rate){ double yearly=rate/12; for(int i=1;i<=months;i++){ System.out.print(\"After Month \"+i+\" Amount \"); monthly=interest(yearly,monthly); System.out.println(monthly); } } public static void main(String [] args){ Interest i=new Interest(); i.totalSaving(12,100,0.5) } } Solution public class Interest{ double interest(double rate, double amount){ return amount+(amount*rate); } double totalSaving(int months,double monthly,double rate){ double yearly=rate/12; for(int i=1;i<=months;i++){ System.out.print(\"After Month \"+i+\" Amount \"); monthly=interest(yearly,monthly); System.out.println(monthly); } } public static void main(String [] args){ Interest i=new Interest(); i.totalSaving(12,100,0.5) } }.

public class Interest{double interest(double rate, double amount){.pdf

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Phototaxis the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward or away from light is termed as phototaxis movement. This is advantageous for phototrophic organisms as they can orient themselves most efficiently to receive light for photosynthesis. Phototaxis is called positive if the movement is in the direction of light and negative if the direction is opposite. Phototropism, Growth with respect to light, expressed in all shoots and some roots. Phototropism is one of the many plant tropisms or movements which respond to external stimuli Charles Darwin studied plant growth phenomena involving tropisms. He found the bending of plants toward light. This phenomenon, which is caused by differential growth, is called phototropism. In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are sheathed in a protective organ called the coleoptile. Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an hour. Simillarity between Phototaxis and phototropism Both are related with abiotic factor i.e light. ___________________. . Solution Phototaxis the Phenomenon of movement of motile unicellular organisms such as algae and bacteria toward or away from light is termed as phototaxis movement. This is advantageous for phototrophic organisms as they can orient themselves most efficiently to receive light for photosynthesis. Phototaxis is called positive if the movement is in the direction of light and negative if the direction is opposite. Phototropism, Growth with respect to light, expressed in all shoots and some roots. Phototropism is one of the many plant tropisms or movements which respond to external stimuli Charles Darwin studied plant growth phenomena involving tropisms. He found the bending of plants toward light. This phenomenon, which is caused by differential growth, is called phototropism. In experiments he used seedlings of canary grass (Phalaris canariensis), the youngest leaves are sheathed in a protective organ called the coleoptile. Coleoptiles are very sensitive to light, especially to blue light . If illuminated on one side with a short pulse of dim blue light, they will bend (grow) toward the source of the light pulse within an hour. Simillarity between Phototaxis and phototropism Both are related with abiotic factor i.e light. ___________________. ..

Phototaxisthe Phenomenon of movement of motile unicellular organis.pdf

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Miller indices: It is defined as by considering the plane in which it intersects the main crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may be used to identify the plane or the surface. Solution Miller indices: It is defined as by considering the plane in which it intersects the main crystallographic axes of the solid. A set of numbers which explains the intercepts and thus it may be used to identify the plane or the surface..

Miller indices It is defined as by considerin.pdf

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Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a psychoactive drug or medicine. Characteristics of Cannabis were studied to determine the suitable and efficient environmental conditions for its indoor mass cultivation for pharmaceutical studies. Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on gas and water vapour exchange characteristics of the Cannabis. The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic photon flux densities at lower temperatures. Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the plant makes the cellulose which gives the fibrous structures to the plant. The osmosis process helps in the absorbption of water and minerals from soil. The osmotic pressure mainly helpful in intake and transport of minerals and the soil condition also plays an important role. Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich environment. Solution Marijuana is also called as cannabis and it is a preparation of cannabis plant and it is used as a psychoactive drug or medicine. Characteristics of Cannabis were studied to determine the suitable and efficient environmental conditions for its indoor mass cultivation for pharmaceutical studies. Effect of different photosynthetic photon flux densities,temperature and CO2 concentrations on gas and water vapour exchange characteristics of the Cannabis. The rate of photosynthesis and water use efficiency of Cannabis increases with photosynthetic photon flux densities at lower temperatures. Genrally the photosynthetic process will occur by taking the carbondioxide and water and in the presence of sunlight and chlorophyll it forms the glucose and oxygen and from the glucose the plant makes the cellulose which gives the fibrous structures to the plant. The osmosis process helps in the absorbption of water and minerals from soil. The osmotic pressure mainly helpful in intake and transport of minerals and the soil condition also plays an important role. Cannabis gives its potential for better survival,growth and productivity in drier and CO2 rich environment..

Marijuana is also called as cannabis and it is a preparation of cann.pdf

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IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets means all the hydrogens are isolated from one another, but there are two types in the molecule giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2- dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen contributing another signal. All non-equivalent hydrogens are too far apart to split the signal, resulting in singlets) Solution IHD = 1 Therefore there can be only 1 -bond or ring. The two signals composed of singlets means all the hydrogens are isolated from one another, but there are two types in the molecule giving differing chemical shifts. The structure must be symmetrical. Likely compound is 2,2- dimethylpropanal (three methyl groups contributing one signal, and one aldehyde hydrogen contributing another signal. All non-equivalent hydrogens are too far apart to split the signal, resulting in singlets).

IHD = 1 Therefore there can be only 1 -bond or ring. The two signals.pdf

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I have the answer ready in paint but i am not able to upload. 1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0) Hence the and moves in a circle of radius 1 unit. 2) F\'(t)= (-2sin2t, 2cos2t) F\'(pi/2) = (0, -2) This represents the Speed at t= pi/2 3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1) Solution I have the answer ready in paint but i am not able to upload. 1) F(0)=(1,0), F(pi/4)=(0,1), F(pi/2) =(-1,0), F(3pi/4)=(0,-1) & F(pi/2)=(1,0) Hence the and moves in a circle of radius 1 unit. 2) F\'(t)= (-2sin2t, 2cos2t) F\'(pi/2) = (0, -2) This represents the Speed at t= pi/2 3) Integral 0 to 1 F(t) dt = Area of circle = pir^2 = pi (as r = 1).

I have the answer ready in paint but i am not able to upload. 1) F.pdf

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C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address states of each device. A, B, and D are incorrect. Each is a stateless method with no device maintaining the address states. Solution C is correct. DHCP is a statefull method of configuring IPv6 addresses. It maintains the address states of each device. A, B, and D are incorrect. Each is a stateless method with no device maintaining the address states..

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AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf

1. Answer: Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth complex in the electron transport chain. Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which results in blockade of electrons transfer to the final electron acceptor oxygen. This blockade causes build up of proton motive force in the intermembrane space as the protons are not allowed to flow back into the matrix which subsequently halts ATP production. Solution Answer: Cyanide is an irreversible enzyme inhibitor of Cytochrome C Oxidase (Complex IV) the fourth complex in the electron transport chain. Cyanide ions bind to the iron atom of the Complex IV in the mitochondrial membrane, which results in blockade of electrons transfer to the final electron acceptor oxygen. This blockade causes build up of proton motive force in the intermembrane space as the protons are not allowed to flow back into the matrix which subsequently halts ATP production.

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