Miller indices It is defined as by considerin.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..

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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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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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homedesigninsertanimationsdevelopmentSolutionhomedes.pdf

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eSolutione.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..

C is correct. DHCP is a statefull method of configuring IPv6 address.pdf

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B) The energy required to remove an electron from an atomSolutio.pdf

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Answer: The following molecules would not require carrier transporters: a) carbon dioxide c) progesterone The following molecules would require carrier transporters: b) fructose d) arganine e) adenosine diphosphate Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in the form of transmembrane channels. These channels are gated, meaning that they open and close, and thus deregulate the flow of ions or small polar molecules across membranes, sometimes against the osmotic gradient. Larger molecules are transported by transmembrane carrier proteins, such as permeases, that change their conformation as the molecules are carried across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly soluble in water. They are transported through aqueous compartments of cells or through extracellular space by water-soluble carriers (e.g. retinol binding protein). ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner mitochondrial membrane. Solution Answer: The following molecules would not require carrier transporters: a) carbon dioxide c) progesterone The following molecules would require carrier transporters: b) fructose d) arganine e) adenosine diphosphate Polar molecules and large ions dissolved in water cannot diffuse freely across the plasma membrane due to the hydrophobic nature of the fatty acid tails of the phospholipids that make up the lipid bilayer. Only small, non-polar molecules, such as oxygen and carbon dioxide, can diffuse easily across the membrane. Hence, no nonpolar molecules are transported by proteins in the form of transmembrane channels. These channels are gated, meaning that they open and close, and thus deregulate the flow of ions or small polar molecules across membranes, sometimes against the osmotic gradient. Larger molecules are transported by transmembrane carrier proteins, such as permeases, that change their conformation as the molecules are carried across (e.g. glucose or amino acids). Non-polar molecules, such as retinol or lipids, are poorly soluble in water. They are transported through aqueous compartments of cells or through extracellular space by water-soluble carriers (e.g. retinol binding protein). ADP/ATP translocases, also known as adenine nucleotide translocases (ANT) and ADP/ATP carrier proteins (AAC), are transporter proteins that enable the exchange of cytosolic adenosine diphosphate(ADP) and mitochondrial adenosine triphosphate (ATP) across the inner mitoc.

AnswerThe following molecules would not require carrier transport.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..

AnswerCyanide is an irreversible enzyme inhibitor of Cytochrome C.pdf

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a.There are two variables in labor market model. One is wage rate .pdf

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A respiratory therapy professional diagnoses and treats those suffering from chronic and other respiratory problems. Some of their patients with respiratory problems would be: • Premature infants • Asthma patients • Patients with lung infections • Heart patients A respiratory therapist (RT) has a variety of job responsibilities. These include talking to patients, gathering information regarding their medical condition, diagnosing breathing conditions and associated treatments, informing the family about the respiratory therapy treatments, etc. Respiratory therapist must update their professional respiratory knowledge periodically and need to be well versed with the latest available techniques, treatments and diseases in respiratory therapy. Care Respiratory therapists often work in medical facilities such as hospitals and nursing homes. They administer oxygen, perform CPR, use ventilators, and give medicines as needed. The ability to provide this care and build relationships with patients is an advantage for people who want to work in the medical field without becoming a medical doctor. Therapists who work in extended care facilities like nursing homes have an even longer time to develop relationships with their patients. Misconception among nurses about a respiration therapist: STRESS: Nurses do think that, being a respiration therapist could be strenuous and stressful, as they spend the majority of the day on their feet. The care that respiratory therapists provide deals with one of the most basic and important life activities -- breathing. Helping people through breathing struggles is stressful under normal conditions. For those therapists who work in hospitals, the stress is more constant. Administering CPR and assisting with treatment in emergency rooms is high pressure. Another factor that can add to the stress is working with children who have breathing issues, or who have been in a drowning accident or other type of accident that has compromised their respiration. WORK CONDITION and ENVIRONMENT: Nurses do believe that, RTs may have to work on weekends, holidays, and various shifts and may also have to work extended hours in case of emergency. A career in respiratory therapy also means being exposed to infections and micro-organisms. Another misconception is that, as being a registered nurse, they do have more opportunities of different places to work like hospitals, clinics, doctor\'s offices, jail\'s, factories, home health, and of course as a school nurse. whereas, as RTs, well not so many as they could work on just hospitals and home health. These are the major misconceptions among nurses about a respiratory therapist. Solution A respiratory therapy professional diagnoses and treats those suffering from chronic and other respiratory problems. Some of their patients with respiratory problems would be: • Premature infants • Asthma patients • Patients with lung infections • Heart patients A respiratory therapist (RT) has a variety of .

A respiratory therapy professional diagnoses and treats those suffer.pdf

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a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the visual symbols. b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex. Broca’s area: It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the lower part of prefrontal cortex. Broca’s aphasia: c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in speech. Lesions in this area results in Wernicke\'s aphasia. d) Damage to left temporal lobe leads to Wernicke\'s aphasia. Therefore, option b is correct. Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral cortex. Solution a) Angular gyrus is present in parietal cortex. It is concerned with storage of memories of the visual symbols. b) Broca\'s aphasia occurs due to damage of left inferior frontal region of cerebral cortex. Broca’s area: It is area 44. Area 44 is the motor speech area. It is also called speech center. It is situated in the lower part of prefrontal cortex. Broca’s aphasia: c) Wernicke\'s area is situated in upper part of temporal lobe. It plays an important role in speech. Lesions in this area results in Wernicke\'s aphasia. d) Damage to left temporal lobe leads to Wernicke\'s aphasia. Therefore, option b is correct. Broca’s aphasia is usually associated with the lesions of left inferior frontal region of cerebral cortex..

a) Angular gyrus is present in parietal cortex. It is concerned with.pdf

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Since youre diluting the solution to 2.5 times .pdf

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wo (-2) above the O3 means there are 2 extra electrons available for bonding. Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except that they have a charge. The only difference between the process for drawing Lewis diagrams for molecules and for ions is in determining the number of available electrons. Count the valence as before then add an electron for each unit of negative charge or subtract one for each unit of positive charge. For example, the sulfite ion has the formula SO32-. The number of valence electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule or you have used all that are available. Solution wo (-2) above the O3 means there are 2 extra electrons available for bonding. Polyatomic ions are groups of atoms joined together by covalent bonds, like molecules, except that they have a charge. The only difference between the process for drawing Lewis diagrams for molecules and for ions is in determining the number of available electrons. Count the valence as before then add an electron for each unit of negative charge or subtract one for each unit of positive charge. For example, the sulfite ion has the formula SO32-. The number of valence electrons is 6 + 3*6 + 2 = 26. The atom with the highest electronegativity in the sulfite ion is the sulfur, so it belongs in the center of the diagram with oxygen atoms bonded to it. Fill in enough bonds to hold the ion together then add non-bonding electrons until all atoms obey the octet rule or you have used all that are available..

wo (-2) above the O3 means there are 2 extra elec.pdf

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the alkali metals are found in the Group I A of p.pdf

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