H denote H atom with 1s1 electron configuration, while H- is an anio.pdf
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H denote H atom with 1s1 electron configuration, while H- is an anion with 1s2 electron configuration. H- holds one more electron than H, thus different density and density potential images. Solution H denote H atom with 1s1 electron configuration, while H- is an anion with 1s2 electron configuration. H- holds one more electron than H, thus different density and density potential images..
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Moles = 0.142138 = 0.00102 moles .pdf
Moles = 0.142/138 = 0.00102 moles
Solution
Moles = 0.142/138 = 0.00102 moles.
The name of As2S5 is diarsenic pentasulfide.TRUEExplanationAs .pdf
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\"
Solution
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\".
The main thing we have to do is correct tap on the system status not.pdf
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over
Solution
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over.
The answer is b. H2SBoth S and O come from the same group of the .pdf
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other.
Solution
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other..
Molarity , M = ( massMolar mass) Volume of sol.pdf
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M
Solution
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M.
Many molecules contain bonds that fall between th.pdf
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment.
Solution
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment..
the 4 examples are1) HI HI = H+ + I-2) HBr HBr = H+ + Br-.pdf
the 4 examples are:
1) HI
HI => H+ + I-
2) HBr
HBr => H+ + Br-
3) HCl
HCl => H+ + Cl-
4) H2SO4
H2SO4 => 2H+ + SO4-2
Hope this will do..:)
Solution
the 4 examples are:
1) HI
HI => H+ + I-
2) HBr
HBr => H+ + Br-
3) HCl
HCl => H+ + Cl-
4) H2SO4
H2SO4 => 2H+ + SO4-2
Hope this will do..:).
Recommended
Moles = 0.142138 = 0.00102 moles .pdf
Moles = 0.142/138 = 0.00102 moles
Solution
Moles = 0.142/138 = 0.00102 moles.
The name of As2S5 is diarsenic pentasulfide.TRUEExplanationAs .pdf
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\"
Solution
The name of As2S5 is diarsenic pentasulfide.
TRUE
Explanation
As there are 2 atoms of Arsenic so we uise word\"di\"
As there are 5 atoms of sulfide so we use word \"penta\".
The main thing we have to do is correct tap on the system status not.pdf
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over
Solution
The main thing we have to do is correct tap on the system status notice symbol and open the
Network and Sharing Center from the setting menu.
At that point tap on the Change connector settings hyperlink on the left-hand side of the Network
and Sharing Center.
Presently right-tap on the system connector you wish to change the DNS settings for and select
properties from the setting menu.
At the point when the properties for your system connector open, you should choose Internet
Protocol Version 4 (TCP/IPv4) from the rundown, then snap properties.
Our illustration will have opened the IPv4 settings for your system connector. You will find in
the base portion of the discourse that your PC is set to get its DNS settings \"consequently\". You
should change that with the goal that we can physically determine the DNS servers we need to
utilize. Presently basically pick an administration underneath and enter the DNS server
addresses.
Google DNS
Favored: 8.8.8.8
Interchange: 8.8.4.4
OpenDNS
Favored: 208.67.222.222
Interchange: 208.67.220.220
We selected to run with Google DNS for the time being — while they could utilize this
information to track your perusing propensities, by and by we\'re not that stressed over.
The answer is b. H2SBoth S and O come from the same group of the .pdf
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other.
Solution
The answer is: b. H2S
Both S and O come from the same group of the periodic table and have 6 valence electrons.
Thus the structures of H2S and H2O are most similar to each other..
Molarity , M = ( massMolar mass) Volume of sol.pdf
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M
Solution
Molarity , M = ( mass/Molar mass) / Volume of solution in L = ( 5.00 g / 15.00
g/mol ) / 0.100 L = 3.3333 M.
Many molecules contain bonds that fall between th.pdf
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment.
Solution
Many molecules contain bonds that fall between the extremes of ionic and covalent
bonds. The difference between the electronegativities of the atoms in these molecules is large
enough that the electrons aren\'t shared equally, and yet small enough that the electrons aren\'t
drawn exclusively to one of the atoms to form positive and negative ions. The bonds in these
molecules are said to be polar, because they have positive and negative ends, or poles, and the
molecules are often said to have a dipole moment. for Eg : HCl molecules, for example, have a
dipole moment because the hydrogen atom has a slight positive charge and the chlorine atom has
a slight negative charge. Because of the force of attraction between oppositely charged particles,
there is a small dipole-dipole force of attraction between adjacent HCl molecules. non-polar
molecule has it\'s electron density shifted by a charged molecule, i.e., if a positively charged
species is brought near a species with an evenly distributed charge, the electrons go towards the
side where the positive species is, and away from the opposite side, creating an artificial dipole.
Eg : a helium atom is perfectly symmetrical. But movement of the electrons around the nuclei of
a pair of neighboring helium atoms can become synchronized so that each atom simultaneously
obtains an induced dipole moment..
the 4 examples are1) HI HI = H+ + I-2) HBr HBr = H+ + Br-.pdf
the 4 examples are:
1) HI
HI => H+ + I-
2) HBr
HBr => H+ + Br-
3) HCl
HCl => H+ + Cl-
4) H2SO4
H2SO4 => 2H+ + SO4-2
Hope this will do..:)
Solution
the 4 examples are:
1) HI
HI => H+ + I-
2) HBr
HBr => H+ + Br-
3) HCl
HCl => H+ + Cl-
4) H2SO4
H2SO4 => 2H+ + SO4-2
Hope this will do..:).
molality = moles of solute mass of solvent = 4..pdf
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987
Solution
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987.
SolutionTrueIn a shared block prevents need another user from p.pdf
Solution
:
True
In a shared block prevents need another user from performing DDL or DML operations on the
table.
The shared lock is automatically obtained when the select...for Update command is executed.
There is a lock known as share row exclusive, when this is lock is used then the whole table can
be seen by the other user. Similarly the other users can see all the rows in the table, but they
cannot perform DDL or DML operations on the table. When this is used in the shared mode then
it will automatically locks from updating of the table..
shares.txt(Save the file in the D drive.Then the path of the file po.pdf
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shares_sold*selling_price;
//Calculating service tax after selling of shares
service_tax_during_sell=amount_received_after_sell*0.015;
//calculating Total Service tax paid
total_tax=service_tax_during_buy+service_tax_during_sell;
//calculating amount profit or lost
amount_profit_lost=(amount_received_after_sell-tot_amt_invested)-(total_tax);
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\\SharesOutput.txt\");
//Writing the contents to the file
myfile
<<\"No_Of_Shares_sold\"<<\"||\"<<\"Purchase_Price\"<<\"||\"<<\"Sell_Price\"<<\"||\"<<\"Amt_i
nvest\";
myfile<<\"||\"<<\"service_charges\"<<\"||\"<<\"Amt_Gained\"<<\"||\"<<\"Amt_received_After_s
ell\ \";
myfile<<\"\\t\"<
Solution
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shar.
Scientific approaches have been defined to depict the methods to stu.pdf
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It is a comprehensive method which includes artificially designed
algorithm based non-human in silico/in vivo animals as models with special relevance for their
corresponding human behaviuor. This implies that a pre-defined set of rules are designed for a
model and certain stimuli are given. The nature of comprehensive behaviour of the organismic
model is then recorded and analysed. This gives a clear idea about comparative psychology of an
animal with respect to humans.
3. Behavioral studies and ecology- These studies comprise of a close integration of various
behavioral aspectso an animal with its specific environment. This field of science has been raised
in a past few decades in order to enhance the psychological information about the animal
behaviour. This helps the investigator to presume the behaviuor of an animals under a stimulus
without putting the animal exactly under the stimulus based upon previous experience,
recordings and integrated information.
Thus, collectively, these set of informations can help an investigator to understand animal
behaviuor scienctifically and psycho-socially.
Solution
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It i.
Ques-1Government funding is limited so that the factors are neede.pdf
Ques-1:
Government funding is limited so that the factors are needed to consider to eliminate an
infectious disease are whether the infectious agent mode of transmission and infectious agent
etiology and disease symptoms in human beings. These factors are going to promote
\"epidemiologists\" to better control \"both nosocomial infections\", pandemic infections of
communicable disease. The possible preventive measures such as vaccination can be
implemented to eliminate the disease.
Assessment of chronic disease has not only been possible because of epidemiology but also acute
disease conditions, causes, outcome of treatment to that disease conditions is also possible by
studying the disease status in the public communities. This epidemiology study used to mainly
for those infectious diseases prevailed to a large number of populations within a geographical
area in a short period of time finally useful to implement public health policies against that
infection based on epidemiological data; for example infectious disease often prevail in a
population in 2 weeks. Example of epidemic diseases is meningococcal infections; Epidemiology
studies also useful to study the nature of pandemic diseases to monitor the health of communities
& populations; These studies are used for those diseases occurring in large number of
populations across the world in almost all geographical areas. Example is flu infections.
Ques-10:
No, we never win war on microbes because microbes can change their genome persistently &
actively through mutations when they reach the host genome. For example, HIV -1 infection
therefore, it is difficult to introduce medication to defeat this microbe
Some diseases causing bacterial spores are dormant for 1000 years also and medication is
ineffective to combat these spores due to the following reason
Bacillus subtilis produce dormant spores by sporulation ate suitable conditions and these spores
are called as endospores. These endopores have the capability to survive at extreme temperatures
followed by insufficient nutrient availability and their spore DNA is completely resistant to
mutagens such as UV light, chemicals. They are also resistant to desiccation and resistant to
extreme freezing and chemical disinfectants.
Factor-1: These endospores possess very complex diversified structure that can withstand all the
possible mutation -inducing changes. These possess structural adaptations such as exosporium,
hard spore coat and these are acts as very hard complex resistant materials, which can eliminate
digesting enzymes such as lysozymes. These endospores possess cortex composed of
peptidoglycan and it has a core wall that is present under the cortex and surrounds protoplast.
These layers potentially protect spore DNA material.
Factor-2: Core material of the spore also possesses calcium dipicolinate that enable stability of
chromosomal DNA by protecting it from oxidizing agents. Inside the core material, spore
possess DNA and it is covered .
Please follow the data and description Impact of absorption on tr.pdf
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This ranges from the level, range, medium and all other factors. These also vary
extensively based on the distances greater than about a hundred meters from the source. This
variation is caused by changes in weather conditions and by topographical features such as
ground cover, hills and other obstacles between the source and the receiver.
There are several important factors which affect the propagation such as geometric spreading,
atmospheric effects, and surface effects as discussed below :
a) Geometric Spreading :
This generally refers to the spreading of energy as a result of the expansion of the wavefronts.
Geometric spreading is merely independent of FREQUENCY and has a major effect in almost
all the propagation situations.
b) Air Absorption :
In general there are two types where the energy is absorbed by the atmosphere. These are
molecular relaxation and viscosity effects. High frequencies are absorbed more than low. The
amount of ABSORPTION depends on the temperature and humidity of the atmosphere.
c) Wind and Temperature Gradients :
The speed that the energy propagates in a gas depends on the temperature of the gas. Higher
temperatures produce higher speeds of sound. Since the temperature of the atmosphere is not
uniform there are local variations in the sound speed. For example, under normal conditions the
atmosphere is cooler at higher altitudes. This results in the waves being \'bent\' upwards. This
will result in the formation of a shadow zone, which is a region in which sound does not
penetrate. In reality some sound will enter this zone due to scattering. Scattering occurs when
sound waves are propagating through the atmosphere and meet a region of inhomogeneity and
some of their energy is re-directed into many other directions.
d) Ground Absorption :
This is to be noted in general as the energy is propagating over the ground, and then the
ATTENUATION will occur due to acoustic energy losses on REFLECTION. These losses will
depend on the surface. Smooth, hard surfaces will produce little ABSORPTION whereas thick
grass may result in sound levels being reduced by up to about 10 dB per 100 meters at 2000 Hz.
High frequencies are generally attenuated more than low frequencies.
These are some of the effects or the types where the absorption impacts the transmission of the
data or the energy from the source to the destination.
Hope this is helpful.
Solution
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This rang.
Part-aLactose is a disaccharide and composed of glucose and galac.pdf
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential for enzymatic
hydrolysis.
C12H22O11 (lactose) + H2O (enzymatic hydrolysis by lactase) C6H12O6 (D-galactose) +
C6H12O6 (D-glucose)+ heat.
In the absence of lactases in the small intestine is leading to lactose digestion by gut bacteria
through \"bacterial fermentation\" finally result in formation of lactic acid and CO2 result in
persistent flatulence and abdominal distension
Part-c:
Osmotic gradient is produced due to lactose intolerance in the large intestine through lactic acid
formation and bicarbonate formation between the \"lacteals of intestine\" and gut result in higher
fluid secretion into the intestine to attain homeostasis finally loose fecal formation or diarrhea.
Lactose intolerance: This is defined as a disorder in which inability of the production of enzyme
lactase to properly digest lactose in both adults and children as well, when they had dietary
intake of lactose containing food products (ex. milk). This condition is considered as congenital
since the gene coding for the enzyme lactose to break lactose sugar polymer
Solution
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential f.
ln(lnx)=4if ln a = c , then a = ec== lnx = e4== x = ee^4.pdf
ln(lnx)=4
if ln a = c , then a = ec
==> lnx = e4
==> x = ee^4
Solution
ln(lnx)=4
if ln a = c , then a = ec
==> lnx = e4
==> x = ee^4.
In the first, the NO2 attaches at the 3 position .pdf
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again
Solution
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again.
hydride ion reacts with water to give hydrogen an.pdf
hydride ion reacts with water to give hydrogen and hydroxyl ion.
Solution
hydride ion reacts with water to give hydrogen and hydroxyl ion..
Labels on the right side of image (when we look at it)1. medial le.pdf
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus
Solution
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus.
joint distribution function of X and Y will be given by int-( i.pdf
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2
Solution
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2.
he had not removed and washed with distilled wate.pdf
he had not removed and washed with distilled water
Solution
he had not removed and washed with distilled water.
H2O = It is a liquid at room conditions, hydrogen.pdf
H2O = It is a liquid at room conditions, hydrogen bonding forces are responsible
for H2O to keep it as a liquid. C2H2 = It is a gas at room conditions, and dispersion forces are
responsible to it CCl4 - liquid at room conditions, and dispersion forces are responsible to it KCl
- This is a solid and ionic forces are responsible to it Pb - Lead is a solid, and meta bonding is
present in it
Solution
H2O = It is a liquid at room conditions, hydrogen bonding forces are responsible
for H2O to keep it as a liquid. C2H2 = It is a gas at room conditions, and dispersion forces are
responsible to it CCl4 - liquid at room conditions, and dispersion forces are responsible to it KCl
- This is a solid and ionic forces are responsible to it Pb - Lead is a solid, and meta bonding is
present in it.
generally, compounds with Na+ or NO3- are soluble in water and will .pdf
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s)
Solution
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s).
A. http signifies that the website uses hypertext transfer protocl t.pdf
A. http signifies that the website uses hypertext transfer protocl to transfer the hypertext context
from on eplace to another place through browser.
B. www.cisco.com is an URL (Universal Resource Locator) to locate the server with unique
name. No one will own same URL\'s. They must change either their extension e.g.,
.in,.net,.org,etc...
C. education.htm is the HTML which contains the hypertext information that can be displayed on
a web browser.
D. www.cisco.com/education.htm is an address which directly goes to education page of cisco
website. Usually when we just enter www.cisco.com we will land on index page or home page. /
indicates that webpages in that website.
E. WWW means World Wide Web which indicates that the website can be accessed from
anywhere in the world.
Solution
A. http signifies that the website uses hypertext transfer protocl to transfer the hypertext context
from on eplace to another place through browser.
B. www.cisco.com is an URL (Universal Resource Locator) to locate the server with unique
name. No one will own same URL\'s. They must change either their extension e.g.,
.in,.net,.org,etc...
C. education.htm is the HTML which contains the hypertext information that can be displayed on
a web browser.
D. www.cisco.com/education.htm is an address which directly goes to education page of cisco
website. Usually when we just enter www.cisco.com we will land on index page or home page. /
indicates that webpages in that website.
E. WWW means World Wide Web which indicates that the website can be accessed from
anywhere in the world..
Energy for primary active transport comes from AT.pdf
Energy for primary active transport comes from ATP molecules. ATP is produced
by the mitochondria. e.g. Na+/K+ ATPases are archetypal primary active transporters and they
transport 3 Na+ ions out of cells and 2 K+ ions into cells by hydrolyzing one molecule of ATP.
Energy for secondary active transport is stored in an established concentration gradient of ions
across the cell membrane and drives transport with the tendency of ions to move from an area of
high concentration to an area of lower concentration. e.g. Na+/glucose secondary transporters
use the Na+ gradient (higher outside cells than inside) established by Na+/K+ ATPases to
transport a molecule of glucose along with a sodium ion into cells.
Solution
Energy for primary active transport comes from ATP molecules. ATP is produced
by the mitochondria. e.g. Na+/K+ ATPases are archetypal primary active transporters and they
transport 3 Na+ ions out of cells and 2 K+ ions into cells by hydrolyzing one molecule of ATP.
Energy for secondary active transport is stored in an established concentration gradient of ions
across the cell membrane and drives transport with the tendency of ions to move from an area of
high concentration to an area of lower concentration. e.g. Na+/glucose secondary transporters
use the Na+ gradient (higher outside cells than inside) established by Na+/K+ ATPases to
transport a molecule of glucose along with a sodium ion into cells..
Cl- has larger radius than Ca2+SolutionCl- has larger radius t.pdf
Cl- has larger radius than Ca2+
Solution
Cl- has larger radius than Ca2+.
Answer- In the windows server uses active directory for its direc.pdf
Answer:-
In the windows server uses active directory for its directory service, Both use tree structures and
both allow an administrator to manipulate individual objects or container objects that represent
collections of device and individuals. Active Directory allows forests,while NDS does not
Solution
Answer:-
In the windows server uses active directory for its directory service, Both use tree structures and
both allow an administrator to manipulate individual objects or container objects that represent
collections of device and individuals. Active Directory allows forests,while NDS does not.
Answer part a)According to the gene map of the wild type and the m.pdf
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males will be
normal both geotypically and phenotypically. Also, all the females will be phenotypically normal
but 50% of them would be heterozygous carriers.
Genotypes: XtX : XtY : XX : XY :: 1:1:1:1
Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY
Solution
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males w.
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
More Related Content
More from appsmobileshoppe
molality = moles of solute mass of solvent = 4..pdf
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987
Solution
molality = moles of solute / mass of solvent = 4.47 mol of CaCl2 / 1 g of water
mole fraction = moles of CaCl2 / (moles of CaCL2 + moles of water) = 4.47 / (4.47 + 1/18) =
0.987.
SolutionTrueIn a shared block prevents need another user from p.pdf
Solution
:
True
In a shared block prevents need another user from performing DDL or DML operations on the
table.
The shared lock is automatically obtained when the select...for Update command is executed.
There is a lock known as share row exclusive, when this is lock is used then the whole table can
be seen by the other user. Similarly the other users can see all the rows in the table, but they
cannot perform DDL or DML operations on the table. When this is used in the shared mode then
it will automatically locks from updating of the table..
shares.txt(Save the file in the D drive.Then the path of the file po.pdf
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shares_sold*selling_price;
//Calculating service tax after selling of shares
service_tax_during_sell=amount_received_after_sell*0.015;
//calculating Total Service tax paid
total_tax=service_tax_during_buy+service_tax_during_sell;
//calculating amount profit or lost
amount_profit_lost=(amount_received_after_sell-tot_amt_invested)-(total_tax);
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\\SharesOutput.txt\");
//Writing the contents to the file
myfile
<<\"No_Of_Shares_sold\"<<\"||\"<<\"Purchase_Price\"<<\"||\"<<\"Sell_Price\"<<\"||\"<<\"Amt_i
nvest\";
myfile<<\"||\"<<\"service_charges\"<<\"||\"<<\"Amt_Gained\"<<\"||\"<<\"Amt_received_After_s
ell\ \";
myfile<<\"\\t\"<
Solution
shares.txt(Save the file in the D drive.Then the path of the file pointing to that file is
D:\\\\shares.txt )
williams
kane
500
15
18
________________________________________________
C++ code
#include
#include
using namespace std;
int main()
{
//Declaring variables
int noOfShares;
double purPrice,sellPrice;
string firstname,lastname,fname,lname;
double
no_of_shares_sold,purchase_price,selling_price,tot_amt_invested,amount_received_after_sell,se
rvice_tax_during_buy,service_tax_during_sell,total_tax,amount_profit_lost;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\\shares.txt\");
//Reading the data from the file
while(dataIn>>fname>>lname>>noOfShares>>purPrice>>sellPrice)
{
firstname=fname;
lastname=lname;
no_of_shares_sold=noOfShares;
purchase_price=purPrice;
selling_price=sellPrice;
}
//closing the input stream
dataIn.close();
//calculating total amount invested
tot_amt_invested=no_of_shares_sold*purchase_price;
//Calculating service tax during buying of shares
service_tax_during_buy=tot_amt_invested*0.015;
//calculating the amount received after selling
amount_received_after_sell=no_of_shar.
Scientific approaches have been defined to depict the methods to stu.pdf
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It is a comprehensive method which includes artificially designed
algorithm based non-human in silico/in vivo animals as models with special relevance for their
corresponding human behaviuor. This implies that a pre-defined set of rules are designed for a
model and certain stimuli are given. The nature of comprehensive behaviour of the organismic
model is then recorded and analysed. This gives a clear idea about comparative psychology of an
animal with respect to humans.
3. Behavioral studies and ecology- These studies comprise of a close integration of various
behavioral aspectso an animal with its specific environment. This field of science has been raised
in a past few decades in order to enhance the psychological information about the animal
behaviour. This helps the investigator to presume the behaviuor of an animals under a stimulus
without putting the animal exactly under the stimulus based upon previous experience,
recordings and integrated information.
Thus, collectively, these set of informations can help an investigator to understand animal
behaviuor scienctifically and psycho-socially.
Solution
Scientific approaches have been defined to depict the methods to study animal behaviour. These
methods help in understanding not only animal behaviour but also enhance the understanding of
an investigator in order to ddevelop a more comprehensive scienctific approach towards
psychological facts associated with animals. These methods are discussed as below:
1. The Ethology-Technically, ethology is the study of animal/human behaviuor in its
geographical location in association with its psycho-social environment with other factors by
means of close observations. This includes observational analysis by field scans i.e. residing in
the field/animal niche by the investigator for close observations. Importantly, ethology also
comprises of a biological context with each and every aspect of human/animal behaviuor. This
represents one of the most improtant tools to study and understand animal behaviour.
2. Comparative psychology- It i.
Ques-1Government funding is limited so that the factors are neede.pdf
Ques-1:
Government funding is limited so that the factors are needed to consider to eliminate an
infectious disease are whether the infectious agent mode of transmission and infectious agent
etiology and disease symptoms in human beings. These factors are going to promote
\"epidemiologists\" to better control \"both nosocomial infections\", pandemic infections of
communicable disease. The possible preventive measures such as vaccination can be
implemented to eliminate the disease.
Assessment of chronic disease has not only been possible because of epidemiology but also acute
disease conditions, causes, outcome of treatment to that disease conditions is also possible by
studying the disease status in the public communities. This epidemiology study used to mainly
for those infectious diseases prevailed to a large number of populations within a geographical
area in a short period of time finally useful to implement public health policies against that
infection based on epidemiological data; for example infectious disease often prevail in a
population in 2 weeks. Example of epidemic diseases is meningococcal infections; Epidemiology
studies also useful to study the nature of pandemic diseases to monitor the health of communities
& populations; These studies are used for those diseases occurring in large number of
populations across the world in almost all geographical areas. Example is flu infections.
Ques-10:
No, we never win war on microbes because microbes can change their genome persistently &
actively through mutations when they reach the host genome. For example, HIV -1 infection
therefore, it is difficult to introduce medication to defeat this microbe
Some diseases causing bacterial spores are dormant for 1000 years also and medication is
ineffective to combat these spores due to the following reason
Bacillus subtilis produce dormant spores by sporulation ate suitable conditions and these spores
are called as endospores. These endopores have the capability to survive at extreme temperatures
followed by insufficient nutrient availability and their spore DNA is completely resistant to
mutagens such as UV light, chemicals. They are also resistant to desiccation and resistant to
extreme freezing and chemical disinfectants.
Factor-1: These endospores possess very complex diversified structure that can withstand all the
possible mutation -inducing changes. These possess structural adaptations such as exosporium,
hard spore coat and these are acts as very hard complex resistant materials, which can eliminate
digesting enzymes such as lysozymes. These endospores possess cortex composed of
peptidoglycan and it has a core wall that is present under the cortex and surrounds protoplast.
These layers potentially protect spore DNA material.
Factor-2: Core material of the spore also possesses calcium dipicolinate that enable stability of
chromosomal DNA by protecting it from oxidizing agents. Inside the core material, spore
possess DNA and it is covered .
Please follow the data and description Impact of absorption on tr.pdf
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This ranges from the level, range, medium and all other factors. These also vary
extensively based on the distances greater than about a hundred meters from the source. This
variation is caused by changes in weather conditions and by topographical features such as
ground cover, hills and other obstacles between the source and the receiver.
There are several important factors which affect the propagation such as geometric spreading,
atmospheric effects, and surface effects as discussed below :
a) Geometric Spreading :
This generally refers to the spreading of energy as a result of the expansion of the wavefronts.
Geometric spreading is merely independent of FREQUENCY and has a major effect in almost
all the propagation situations.
b) Air Absorption :
In general there are two types where the energy is absorbed by the atmosphere. These are
molecular relaxation and viscosity effects. High frequencies are absorbed more than low. The
amount of ABSORPTION depends on the temperature and humidity of the atmosphere.
c) Wind and Temperature Gradients :
The speed that the energy propagates in a gas depends on the temperature of the gas. Higher
temperatures produce higher speeds of sound. Since the temperature of the atmosphere is not
uniform there are local variations in the sound speed. For example, under normal conditions the
atmosphere is cooler at higher altitudes. This results in the waves being \'bent\' upwards. This
will result in the formation of a shadow zone, which is a region in which sound does not
penetrate. In reality some sound will enter this zone due to scattering. Scattering occurs when
sound waves are propagating through the atmosphere and meet a region of inhomogeneity and
some of their energy is re-directed into many other directions.
d) Ground Absorption :
This is to be noted in general as the energy is propagating over the ground, and then the
ATTENUATION will occur due to acoustic energy losses on REFLECTION. These losses will
depend on the surface. Smooth, hard surfaces will produce little ABSORPTION whereas thick
grass may result in sound levels being reduced by up to about 10 dB per 100 meters at 2000 Hz.
High frequencies are generally attenuated more than low frequencies.
These are some of the effects or the types where the absorption impacts the transmission of the
data or the energy from the source to the destination.
Hope this is helpful.
Solution
Please follow the data and description :
Impact of absorption on transmission ranges :
In general the sound, light needs a medium to travel along the desination. Witht the transmission
there might be some level of difference or the deprecation at the destination and source
comparitively. This rang.
Part-aLactose is a disaccharide and composed of glucose and galac.pdf
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential for enzymatic
hydrolysis.
C12H22O11 (lactose) + H2O (enzymatic hydrolysis by lactase) C6H12O6 (D-galactose) +
C6H12O6 (D-glucose)+ heat.
In the absence of lactases in the small intestine is leading to lactose digestion by gut bacteria
through \"bacterial fermentation\" finally result in formation of lactic acid and CO2 result in
persistent flatulence and abdominal distension
Part-c:
Osmotic gradient is produced due to lactose intolerance in the large intestine through lactic acid
formation and bicarbonate formation between the \"lacteals of intestine\" and gut result in higher
fluid secretion into the intestine to attain homeostasis finally loose fecal formation or diarrhea.
Lactose intolerance: This is defined as a disorder in which inability of the production of enzyme
lactase to properly digest lactose in both adults and children as well, when they had dietary
intake of lactose containing food products (ex. milk). This condition is considered as congenital
since the gene coding for the enzyme lactose to break lactose sugar polymer
Solution
Part-a:
Lactose is a disaccharide and composed of glucose and galactose through a glycosidic linkage
Part-b:
Process of lactose digestion & absorption:
Simple digestion of glucose rich carbohydrates by salivary amylase, pancreatic amylase and
intestinal \"maltase, sucrase and lactase\" into simple monomers of glucose units
Lactose digestion without lactose intolerance: It is common disaccharide that is present in the
milk and dairy products as it can be digested internally by lactase at temperature 37 °C and at pH
of 6. This is representing that enzymes specifically exerts their action at a specific temperature
and PH. Beta glycosidic bonds of lactose digested and forms D-galactose and D-glucose
individual units. Thereby substrate modification of the lactose by enzyme lactase revealing that
3’-OH and 2’-OH groups of the substrate galactopyranose component is very crucial for
competitive lactase enzymatic binding in which predominantly 3’-OH is essential f.
ln(lnx)=4if ln a = c , then a = ec== lnx = e4== x = ee^4.pdf
ln(lnx)=4
if ln a = c , then a = ec
==> lnx = e4
==> x = ee^4
Solution
ln(lnx)=4
if ln a = c , then a = ec
==> lnx = e4
==> x = ee^4.
In the first, the NO2 attaches at the 3 position .pdf
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again
Solution
In the first, the NO2 attaches at the 3 position In second, NO2 becomes NH2 In
third it becomes NN In fourth, it becomes the first diol again.
hydride ion reacts with water to give hydrogen an.pdf
hydride ion reacts with water to give hydrogen and hydroxyl ion.
Solution
hydride ion reacts with water to give hydrogen and hydroxyl ion..
Labels on the right side of image (when we look at it)1. medial le.pdf
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus
Solution
Labels on the right side of image (when we look at it)
1. medial leminiscus
2.medulla oblongata
3. dorsal root ganglion
Labels on the left:
1. ventral nuclei
2. nucleus gracilis and nucleus cuneatus
3. fasciculus gracilis and fasciculus cuneatus.
joint distribution function of X and Y will be given by int-( i.pdf
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2
Solution
joint distribution function of X and Y will be given by int->( infite to infinte ) int ->( infite to
infinte)g(x,y)dydx
int->ingration means
but here the values are between 0 to 1
int->( 0 to 1)int ->( 0 to 1)(Y-X)dXdY
int->( 0 to1) (0 to 1) (YX-X^2 / 2)= int ->( 0 to 1) (Y-1/2)dY=y^2/2-y=1/2-1=-1/2.
he had not removed and washed with distilled wate.pdf
he had not removed and washed with distilled water
Solution
he had not removed and washed with distilled water.
H2O = It is a liquid at room conditions, hydrogen.pdf
H2O = It is a liquid at room conditions, hydrogen bonding forces are responsible
for H2O to keep it as a liquid. C2H2 = It is a gas at room conditions, and dispersion forces are
responsible to it CCl4 - liquid at room conditions, and dispersion forces are responsible to it KCl
- This is a solid and ionic forces are responsible to it Pb - Lead is a solid, and meta bonding is
present in it
Solution
H2O = It is a liquid at room conditions, hydrogen bonding forces are responsible
for H2O to keep it as a liquid. C2H2 = It is a gas at room conditions, and dispersion forces are
responsible to it CCl4 - liquid at room conditions, and dispersion forces are responsible to it KCl
- This is a solid and ionic forces are responsible to it Pb - Lead is a solid, and meta bonding is
present in it.
generally, compounds with Na+ or NO3- are soluble in water and will .pdf
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s)
Solution
generally, compounds with Na+ or NO3- are soluble in water and will be ions in solution.
Therefore you\'re left with Ba2+ and SO42- to react together and form a salt. BaSO4 is not a
very soluble salt, so it\'s going to precipitate out of the solution. (Ksp is about 1.1x10-10).
NaNO3 is going to just stay in solution as the ions, so Na+ and NO3- are the spectator ions.
Ba2+(aq) + NO3-(aq) + Na+(aq) + SO42-(aq) ---> BaSO4 (s) + NO3-(aq) + Na+(aq)
(if looking for only 2 products, write the Na+ and NO3- ions as NaNO3 (aq) )
The balanced net ionic equation is Ba2+ (aq) + SO42- (aq) --> BaSO4 (s).
A. http signifies that the website uses hypertext transfer protocl t.pdf
A. http signifies that the website uses hypertext transfer protocl to transfer the hypertext context
from on eplace to another place through browser.
B. www.cisco.com is an URL (Universal Resource Locator) to locate the server with unique
name. No one will own same URL\'s. They must change either their extension e.g.,
.in,.net,.org,etc...
C. education.htm is the HTML which contains the hypertext information that can be displayed on
a web browser.
D. www.cisco.com/education.htm is an address which directly goes to education page of cisco
website. Usually when we just enter www.cisco.com we will land on index page or home page. /
indicates that webpages in that website.
E. WWW means World Wide Web which indicates that the website can be accessed from
anywhere in the world.
Solution
A. http signifies that the website uses hypertext transfer protocl to transfer the hypertext context
from on eplace to another place through browser.
B. www.cisco.com is an URL (Universal Resource Locator) to locate the server with unique
name. No one will own same URL\'s. They must change either their extension e.g.,
.in,.net,.org,etc...
C. education.htm is the HTML which contains the hypertext information that can be displayed on
a web browser.
D. www.cisco.com/education.htm is an address which directly goes to education page of cisco
website. Usually when we just enter www.cisco.com we will land on index page or home page. /
indicates that webpages in that website.
E. WWW means World Wide Web which indicates that the website can be accessed from
anywhere in the world..
Energy for primary active transport comes from AT.pdf
Energy for primary active transport comes from ATP molecules. ATP is produced
by the mitochondria. e.g. Na+/K+ ATPases are archetypal primary active transporters and they
transport 3 Na+ ions out of cells and 2 K+ ions into cells by hydrolyzing one molecule of ATP.
Energy for secondary active transport is stored in an established concentration gradient of ions
across the cell membrane and drives transport with the tendency of ions to move from an area of
high concentration to an area of lower concentration. e.g. Na+/glucose secondary transporters
use the Na+ gradient (higher outside cells than inside) established by Na+/K+ ATPases to
transport a molecule of glucose along with a sodium ion into cells.
Solution
Energy for primary active transport comes from ATP molecules. ATP is produced
by the mitochondria. e.g. Na+/K+ ATPases are archetypal primary active transporters and they
transport 3 Na+ ions out of cells and 2 K+ ions into cells by hydrolyzing one molecule of ATP.
Energy for secondary active transport is stored in an established concentration gradient of ions
across the cell membrane and drives transport with the tendency of ions to move from an area of
high concentration to an area of lower concentration. e.g. Na+/glucose secondary transporters
use the Na+ gradient (higher outside cells than inside) established by Na+/K+ ATPases to
transport a molecule of glucose along with a sodium ion into cells..
Cl- has larger radius than Ca2+SolutionCl- has larger radius t.pdf
Cl- has larger radius than Ca2+
Solution
Cl- has larger radius than Ca2+.
Answer- In the windows server uses active directory for its direc.pdf
Answer:-
In the windows server uses active directory for its directory service, Both use tree structures and
both allow an administrator to manipulate individual objects or container objects that represent
collections of device and individuals. Active Directory allows forests,while NDS does not
Solution
Answer:-
In the windows server uses active directory for its directory service, Both use tree structures and
both allow an administrator to manipulate individual objects or container objects that represent
collections of device and individuals. Active Directory allows forests,while NDS does not.
Answer part a)According to the gene map of the wild type and the m.pdf
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males will be
normal both geotypically and phenotypically. Also, all the females will be phenotypically normal
but 50% of them would be heterozygous carriers.
Genotypes: XtX : XtY : XX : XY :: 1:1:1:1
Phenotypes: Affected : normal : carriers :: 1: 2: 1XYXtXtXXtYXXXXY
Solution
Answer part a)
According to the gene map of the wild type and the mutant gene, it can be seen that the mutant
lacks a site for restriction digestion. This shows that in wild type, the restriction enzyme can
digest the DNA at any three sites and thus atleast two bands will appear after amplification. On
the contrary, the deletion of middle sequence of restriction site will lead to formation of a single
band in mutant gene. The smaller band will represnt the mutant and the larger band will
represent the normal/wild type allele. Thus, according to this information, it can be deciphered
that the mother is a heterozygous in nature and carries a mutant allele for Tay-Sachs disease
while the father does not. Thus, the lower band will be a representative of mutant or \"t\" and the
upper band will be a representative of normal \"T\" allele.
Answer part b)
Since the mother is a carrier, there are 50% chances that the mother will transmit the mutant
allele to her offspring. Thus, the couple can bear a normal heterozygous carrier female offspring
or an affected male offspring. According to the information in the image, the genotype bands of
father and offspring match which clearly suggest that the offspring is an unaffected heterozygous
carrier female.
Answer part c)
There are always 50% changes for a male to be affected by the disease. Rest 50% males w.
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