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H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4
Solution
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
Computer graphics are pictures and movies produced use computers fre.pdf
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years
Solution
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate
Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
C is correct. Only the host that the unicast message is addressed to.pdf
mol L-1 s-1 is the answer
Solution
mol L-1 s-1 is the answer.
According to the seriousness level, the following organs areThymu.pdf
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14
Solution
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
First substance = Ionic solid Second substance = Covalent solid
Solution
First substance = Ionic solid Second substance = Covalent solid.
AdvantagesThe main objective of business combination is to elimina.pdf
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O
Solution
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
coordination no is the no of ligands in the coordination sphere and in this case it is 6
Solution
coordination no is the no of ligands in the coordination sphere and in this case it is 6.
Recommended
H2SO4 may act as 1) an acidexample 2NaOH + H2SO4 - Na2SO4 +.pdf
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4
Solution
to minimise the length, term with \'m\' should be reduced so first to absolute values
must be opened with negative sign S2 = -4m+1 - 5m+6 + 10m-3 S2 = m+4.
Computer graphics are pictures and movies produced use computers fre.pdf
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years
Solution
no. of sand grains in 1 truck = 10/0.1 x 10 ^-6=10^8 sand grains total time taken=
(10 X 6.02x10^23 x 10 )/10^8= 6.02 x 10^17 min= 6.872 x 10^ 13 years.
Circle.javaimport java.text.DecimalFormat;public class Circle {.pdf
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate
Solution
NaCl - sodium chloride MgO - magnesium oxide AgI - silver iodiode check its not
L CuSO4 - copper sulphate.
C is correct. Only the host that the unicast message is addressed to.pdf
mol L-1 s-1 is the answer
Solution
mol L-1 s-1 is the answer.
According to the seriousness level, the following organs areThymu.pdf
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14
Solution
empirical formula of C2H4 is CH2 empirical formula mass = 12+2 = 14.
Among 40 subjects randomly choose 20 subjects and assign themSol.pdf
First substance = Ionic solid Second substance = Covalent solid
Solution
First substance = Ionic solid Second substance = Covalent solid.
AdvantagesThe main objective of business combination is to elimina.pdf
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O
Solution
ex-1-enol: CH3 - CH2 - CH2 - CH2 - CH2 = CH2-OH hexanal CH3 - CH2 - CH2 -
CH2 - CH2 - CH2=O.
a. Germ cell speciication in Drosophila is primarily a maternlly con.pdf
coordination no is the no of ligands in the coordination sphere and in this case it is 6
Solution
coordination no is the no of ligands in the coordination sphere and in this case it is 6.
A is correct. The packets to be filtered would be heading into the r.pdf
At the beginning of the 11th year, there would be 1,024,000 rabbits.
Solution
At the beginning of the 11th year, there would be 1,024,000 rabbits..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body.
Solution
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body..
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding
genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
The YSTRs are also helpful when a number of males are involved. As regular STRs cause
masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces.
3.The important question while dealing with low DNA quantity is that whether the DNA that
was deposited was deposited during the crime or at some other time.
4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from
fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also
share the same profile.
7. DNA mixtures are very complex.
8.Interpretation is required in many cases.
16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian
rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock
that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by
circadian clocks.
PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present
Solution
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present.
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions.
Solution
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions..
Use Daltons Law of partial pressures. P(Total).pdf
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear.
Solution
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear..
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4
Solution
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4.
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved.
Solution
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it.
Solution
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it..
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08
Solution
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08.
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g)
Solution
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
Solution!DOCTYPE html html head style type=textcss.pdf
Solution
IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ
IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of
pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit
supports cross platformBLUEJFreeyesit suitable for small software projectsJCREATORfreeyesit
is fast and efficient ideJGRASPfreeyesit is super lightweight ideGREENFOOTfreeyesit supports
designing for 2D games CODEENVY$1(month)nocloud based ide.
TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os.
Solution
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os..
The Roman Empire A Historical Colossus.pdf
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
More Related Content
More from ANJALIENTERPRISES1
A is correct. The packets to be filtered would be heading into the r.pdf
At the beginning of the 11th year, there would be 1,024,000 rabbits.
Solution
At the beginning of the 11th year, there would be 1,024,000 rabbits..
1). A) ipsilateralBelow the point of spinal cord, the nerve paths .pdf
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body.
Solution
1). A) ipsilateral
Below the point of spinal cord, the nerve paths belong to the same side (no crossing over occurs).
So, the nociceptive information can be found on the same side of the body..
17-The Y-Chromosome DNA testing helps in the examination of the male.pdf
17-The Y-Chromosome DNA testing helps in the examination of the male specific portion of the
biological evidence. This is useful where small amount of male DNA is found amidst a large
amount of female DNA , as in the sexual assault cases.It can also help in the missing person
investigation as it extends the range of potential reference samples. Sine the Y chromosome is
passed on from the fathers to their sone (unchanged, except in the case of some mutations),
therefore all the males of the same paternal lineage will possess common Y chromosome
haplotype.
A set of Short tandem repeats (STR) are found on the male specific Y chromosome.The coding
genes found on the short arm of the Ychromosome are vital to male sex determination,
spermatogenesis andother male related functions. The Y-STRs are polymorphic among the
nonrelated males.
Sometimes when sepoarating the mal ecomponent and the female component becomes difficult,
then the YSTRs help in the determination. Since The Y-STRs are not present in the females,
therefore in a case where the male and female are involved, it becomes clear that as the Y-STRs
are present the male component is easily detected as only this part will be amplified in the PCR.
The YSTRs are also helpful when a number of males are involved. As regular STRs cause
masking effect the YSTR will help to identify all the males who have contributed to the
evidence.
ADVANTAGES-
1. extremely discriminating.
2. Extremely sensitive.e. low amounts of DNA can be used for determination.
3. Not onlu body fluids, but alsp touched materials or handled surfaces, where skinn cells ahev
been left can be used for DNA profiling.
4. Multiple testing systems ar available, which increases the likelyhood of successfully
generating DNA profiles.
5. Apart from standart=d STR enem mt DNA or YSTRs are used for profiling.
6. With the presence of the current technologies, from amidst mixed DNA samples, the DNA of
a particular individual can be profiled.
LIMITATIONS
1. Random match probability cannot be excluded.
2. Sometimes even when crime is not committed, we do get ,low levels of DNA on surfaces.
3.The important question while dealing with low DNA quantity is that whether the DNA that
was deposited was deposited during the crime or at some other time.
4.Searching evidence is challenging.
5.Sometimes finding the evidence is altogether impossibl.
6. In case of mtDNA /T STRs , as they are passed on from generation to generation either from
fathers to sons or mothers to daughters and son, therefore sometimes unrelated persons are also
share the same profile.
7. DNA mixtures are very complex.
8.Interpretation is required in many cases.
16- from Cyanobacterial to humans, many terristrial organisms have acquired the circadian
rhythms to adapt to sunlight to increase their survival rate.Circadian Rhythm is a biological clock
that displays an endogenous, entrainable oscillation of 24 hours.Thesse oscillations are driven by
circadian clocks.
PULSE CHASE EXPERIME.
#includestdio.h#includestring.h#includestdlib.h#define M.pdf
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=node;
return list;
}
else
{
link_node *prev=list;
while(temp!=NULL&&compare_node(node,temp)>=0)
{
prev=temp;
temp=temp->next;
}
prev->next=node;
node->next=temp;
return list;
}
}
//method definition for displaying the list
void display_list( struct link_node *head )
{
link_node *temp;
while(head)
{
printf(\"%s\ \",head->node_str);
temp=head;
head=head->next;
free(temp);
}
}
Sample output:
Enter the string:
Geneva
America
Solution
#include
#include
#include
#define MAX_STR_LEN 80
//decalring the structure
struct link_node
{
//declaring the variables
char node_str[ MAX_STR_LEN ];
struct link_node *next;
};
//method declaration
typedef struct link_node link_node;
struct link_node *add_node( struct link_node *list,struct link_node *node );
int compare_node( struct link_node *n1, struct link_node *n2 );
void display_list( struct link_node *head );
//implementing the method main
int main()
{
//declaring the main variables
link_node * head=NULL,*temp;
char str[MAX_STR_LEN];
do
{
//getting the STRING input
printf(\"Enter the string : \");
gets(str);
//malloc() is implemented for memory
temp=(link_node*)malloc(sizeof(link_node));
temp->next=NULL;
strcpy(temp->node_str,str);
head=add_node(head,temp);
}while(strlen(str)>1);
display_list(head);
return 0;
}
//method definition for compare node
int compare_node( struct link_node *n1, struct link_node *n2 )
{
if(strcmp(n1->node_str,n2->node_str)==0)
return 0;
if(strcmp(n1->node_str,n2->node_str)<0)
return -1;
else
return 1;
}
//method definition for adding the node
struct link_node *add_node( struct link_node *list,struct link_node *node )
{
link_node *temp=list;
if(list==NULL)
{
return node;
}
if(compare_node(node,list)==-1)
{
node->next=list;
list=no.
The pH and pOH of a solution are defined as pH .pdf
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74
Solution
The pH and pOH of a solution are defined as: pH = -log[H+] pOH = -log[OH-]
and pH + pOH = 14.00 [H+] = 0.055 Conversions: [OH-] = 1.00 x 10^-14 / [H+] = 1.00 x 10-14
/0.055= 1.8x10^-13 pOH = -log[OH-]=12.74.
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, .pdf
goal_state = [1, 8, 7, 2, 0, 6, 3, 4, 5] #goal_state = [1, 0, 7, 2, 8, 6, 3, 4, 5] import sys global
count def display_board( state ): #print \"-------------\" print \" %i %i %i \" %
(state[0], state[3], state[6]) #print \"-------------\" print \" %i %i %i \" % (state[1],
state[4], state[7]) #print \"-------------\" print \" %i %i %i \" % (state[2], state[5],
state[8]) #print \"-------------\" def move_up( state ): new_state = state[:] index =
new_state.index( 0 ) if index not in [0, 3, 6]: temp = new_state[index - 1]
new_state[index - 1] = new_state[index] new_state[index] = temp return
new_state else: return None def move_down( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [2, 5, 8]: temp = new_state[index + 1]
new_state[index + 1] = new_state[index] new_state[index] = temp
return new_state else: return None def move_left( state ): new_state = state[:]
index = new_state.index( 0 ) if index not in [0, 1, 2]: temp = new_state[index
- 3] new_state[index - 3] = new_state[index] new_state[index] = temp
return new_state else: return None def move_right( state ): new_state =
state[:] index = new_state.index( 0 ) if index not in [6, 7, 8]: temp =
new_state[index + 3] new_state[index + 3] = new_state[index]
new_state[index] = temp return new_state else: return None def
create_node( state, parent, operator, depth, cost ): return Node( state, parent, operator,
depth, cost ) def expand_node( node, nodes ): expanded_nodes = []
expanded_nodes.append( create_node( move_up( node.state ), node, \"u\", node.depth + 1, 0 ) )
expanded_nodes.append( create_node( move_down( node.state ), node, \"d\", node.depth + 1,
0 ) ) expanded_nodes.append( create_node( move_left( node.state ), node, \"l\", node.depth
+ 1, 0 ) ) expanded_nodes.append( create_node( move_right( node.state), node, \"r\",
node.depth + 1, 0 ) ) # Filter the list and remove the nodes that are impossible (move
function returned None) expanded_nodes = [node for node in expanded_nodes if node.state
!= None] #list comprehension! expanded_nodes = [node for node in expanded_nodes if
node not in nodes] #list comprehension! #expanded_nodes = [node for node in
expanded_nodes if node.state!= ((node.parent).parent).state] #list comprehension!
expanded_nodes1 = [] for node in expanded_nodes: temp=node
temp=temp.parent temp=temp.parent if temp==None:
expanded_nodes1.extend(expanded_nodes) break if node.state
!=temp.state: expanded_nodes1.append(node) return
expanded_nodes1 def bfs( start, goal ): nodes = [] count=0 # Create the queue
with the root node in it. nodes.append( create_node( start, None, None, 0, 0 ) ) while
True: # We\'ve run out of states, no solution. if len( nodes ) == 0:
return None # take the node from the front of the queue node = nodes.pop(0)
count=count+1 #display_board(node.state) # Append the move we made to
moves # if this node is the goal, return the moves it took to get here. if node.state
== goal: #display_board(node.state) moves = []
temp = node print cou.
7 steps are required as 9 is the 7th element and the search is line.pdf
7 steps are required as 9 is the 7th element and the search is linear
Solution
7 steps are required as 9 is the 7th element and the search is linear.
(D) the number of moles of hydroxide ion added and the number of mol.pdf
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present
Solution
(D) the number of moles of hydroxide ion added and the number of moles of monoprotic acid
initially present.
Look for changes in oxidation numbers. These occ.pdf
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions.
Solution
Look for changes in oxidation numbers. These occur in reactions (c) and (e) only.
Thus, reactions (c) and (e) are REDOX reactions..
Use Daltons Law of partial pressures. P(Total).pdf
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm
Solution
Use Dalton\'s Law of partial pressures. P(Total)=P(H2)+P(H20) p(H2)=P(Total)-
P(H20) Look up the partial pressure of water at the given temperature. It comes out to be 0.0351
atm. P(H2)=1-0.0351 = 0.9649 atm.
The Br was originally neutral, but picks up an ex.pdf
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear.
Solution
The Br was originally neutral, but picks up an extra electron from the N and
becomes Br-. The Br- leaves the molecule The O was originally negative, but gives up an
electron to form a double bond with C1 and becomes neutral. Leaving O=C=N-C as a neutral
molecule. Keeping the labeling of the two carbons it would be O=C1=N-C2. With the double
bonds on both sides of C1, the O, C1, N backbone is linear..
Step1 firstly balance the Phophorous atoms by mul.pdf
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4
Solution
Step1 firstly balance the Phophorous atoms by multiplying Right hand side by 4
Step 2 now balance oxygen atoms by multiplying H2O by 6 Step3 P4O10 +6 H2O------->
4H3PO4.
Year 2000 is the period when the productivity growth rate was at min.pdf
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved.
Solution
Year 2000 is the period when the productivity growth rate was at minimum ie., 0.7. Degrowth
refers to those years where productivity growth rate was in negative. Minimum growth rate
refers to the minimum positive growth rate achieved..
You have tab characters in your text file which is making console to.pdf
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it.
Solution
You have tab characters in your text file which is making console to display following both lines
aligned at middle.
dress
across the porch as the
There are two ways of doing it..
the frequency of the allele I^A in this population-0.22the frequen.pdf
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08
Solution
the frequency of the allele I^A in this population-0.22
the frequency of the allele I^B in this population-0.08.
The reaction isSr(s) + 2 H2O(l) = Sr(OH)2(s) + H2(g)Solution.pdf
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g)
Solution
The reaction is:
Sr(s) + 2 H2O(l) => Sr(OH)2(s) + H2(g).
Solution!DOCTYPE html html head style type=textcss.pdf
Solution
IDENameCostOpen SourceDestinationNETBEANSFreeyescross platform IDEINTELLIJ
IDEA$199noit has great moderen UIECLIPSEfreeyesit features huge library of
pluginsJDEVELOPERfreeyesit has fully loades tool set for creating javaDR. JAVAFreeyesit
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TCPIP is mostly uaed network protocol in Linux and Mac os.But app.pdf
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os.
Solution
TCP/IP is mostly uaed network protocol in Linux and Mac os.
But appletalk also used in both the os..
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