Let P[x] denote the vector space of all polynomials. Let E be the set.pdf
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Let P[x] denote the vector space of all polynomials. Let E be the set of polynomials which only contain terms of even degree of even degree. Is E a subspace of P[x]? Explain. Solution Let E be the set as given in the problem E = { f(x)| f(x) contains only terms of even degree} = { g(x2)| g any polynomial in F[x]} If g and h are any polynomials in F[x], clearly g+h and cg are polynomials in F[x] and (g+h)(x2) and cg(x2) belong to E (for any scalar c in F). So E is closed under vector addition and scalar multiplication. So E is a subspace of F.
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![Let P[x] denote the vector space of all polynomials. Let E be the set of polynomials which only
contain terms of even degree of even degree. Is E a subspace of P[x]? Explain.
Solution
Let E be the set as given in the problem
E = { f(x)| f(x) contains only terms of even degree}
= { g(x2)| g any polynomial in F[x]}
If g and h are any polynomials in F[x], clearly g+h and cg are polynomials in F[x] and
(g+h)(x2) and cg(x2) belong to E (for any scalar c in F).
So E is closed under vector addition and scalar multiplication. So E is a subspace of F](https://image.slidesharecdn.com/letpxdenotethevectorspaceofallpolynomials-230704061200-96f1b701/85/Let-P-x-denote-the-vector-space-of-all-polynomials-Let-E-be-the-set-pdf-1-320.jpg)
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On April 18, 1906, an earthquake killed over 3,000 people when it hi.pdf
On April 18, 1906, an earthquake killed over 3,000 people when it hit San Francisco, with a
magnitude of 8.3 on the Richter Scale. On March 11, 2011 the “Tohoku” earthquake hit central
Japan with a magnitude of 9.0 on the Richter Scale, and killed 15, 894 people. How much greater
was the 2011 earthquake than the 1906 earthquake?
Solution
The formula of magnitude of earthquake is
M=log(I/S)
I is the intensity of the earthquake And S is the intensity of the standard earthquake
8.3 = log(I1/S) and 9= log(I2/S)
9-8.3= log(I2/S) -log(I1/S)
.7= log(I2/I1)
I2/I1=10.7
I2=5 I1
Therefore the earhquake occurs in 2011 is 5 times more intense as compared to earthquake
occurs in 1906.
On the planet of Caracas, in the Stellar Solorais Ka Chunk Galaxy, th.pdf
On the planet of Caracas, in the Stellar Solorais Ka Chunk Galaxy, there is a population of
Nugget Arians. Two phenotypes of interest are: Peanuts; and Carmelaoid. Each is a single gene
trait, and for each gene there are two alleles. A cross between a female with Peanuts and a male
with Carmelaoid (each of which is homozygous for the wild-type (non-trait) allele at the other
gene) produces the following F1 counts: The mode of inheritance for each phenotype is: Peanuts
= Autosomal Recessive. Carmelaoid = Autosomal Dominant. Peanuts = Autosomal Recessive.
Carmelaoid = Autosomal Recessive. Peanuts = Autosomal Recessive. Carmelaoid = X-linked
Recessive. Peanuts = Autosomal Dominant. Carmelaoid = Autosomal Recessive. Peanuts =
Autosomal Dominant. Carmelaoid = Autosomal Dominant.
Solution
The answer is C. Peanuts=Autosomal Recessive
Carmeloid=X-linked recessive.
As
Autosomal characters of male are not carried to progeny so autosomal inheritance is othrwise
called maternal inheritance. As carmelaoid phenotype from male is expressed in F1 so it has to
be in cromosome and from the probability of expression it is evident that it is present in X
cromosome. The male counterpart like sperm carry very little of cytoplasmic material but half of
chromosomal meterial. In human type sex determination i.e.XY male and XX female X linked
recessive genes only express in homozygous i.e. XX condition..
Muv is homozygous You have a C. elegans worm with the following geno.pdf
Muv is homozygous You have a C. elegans worm with the following genotype: + dpy/lov +;
muv Where, dpy, lov and muv are alleles for these recessive phenotypes: Dpy = dumpy (short
and fat) Lov= mating behavior defective Muv = multivulva After taking a minute to feel sorry
for this poor worm, answer the following questions: What gene(s) is this animal heterozygous
for? What gene(s) is this animal homozygous for? What genes(s) are located on the same
chromosome? What is the phenotype of this animal? Assuming lov and dpy loci are 20cM
apart, what are the phenotypes, genotypes and expected frequencies of this hermaphrodite\'s self-
progeny? Percentage Phenotype
Solution
5.1: The animal is heterozygous for dpy and lov genes.
5.2: The animal is homozygous for muv gene.
5.3: The dpy and muv genes are located on the same chromosome.
5.4: Phenotype of animal= long and thin; mating behaviour, multivulva..
Material Science (Solidification)What is the solute atoms concent.pdf
Material Science: (Solidification)
What is the solute atoms concentration with position for a directional solidification of alloy?
Please don\'t answer if you don\'t know the right answer. Thanks.
Solution
During directional solidification, the solute atom concentration increases with increase in
distance from the outside surface. For example during solidification of a spherical casting, the
solute concentration will be maximum at the core and minimum on the surface..
Investment in Note ReceivableSuppliesPrepaid RentBuildingsAc.pdf
Investment in Note Receivable
Supplies
Prepaid Rent
Buildings
Accumulated Depreciation—Buildings
Unearned Service RevenueDebitCredit
Investment in Note Receivable$24,000
Supplies23,500
Prepaid Rent3,200
Buildings280,000
Accumulated Depreciation—Buildings$140,000
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Solution
1) Interest receivble (24000*12%/12) 240 Interest income 240 2) Supplies expense
(23500-15200) 8300 Supplies 8300 3) Rent expense (3200/4) 800 Prepaid rent 800
4) Salaries and wages expense 2700 Salaries and wages payable 2700 5) Depreciation
expense-Buildings (6720/12) 560 Accumualated depreciation-Buildings 560 6)
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expense 2300 Accounts payable 2300.
In a tube containing 1 ml of a 15 dilution of A, and 4 ml of a 110.pdf
In a tube containing 1 ml of a 1:5 dilution of A, and 4 ml of a 1:100 dilution of B:
What is the final dilution of substance A?
What is the final dilution of substance B?
Solution
What is the final dilution of substance A?
When 1mL substance A is mixed with 4 mL of Substance B, four dilution of substance A
happens.
original=1/5
Further =1/4
Final = 1/5 x1/4 = 1/20 fold dilution
What is the final dilution of substance B?
When 4mL substance B is mixed with 1 mL of Substance A, = times
original=1/100
Further =4/5
Final = 1/100 x 4/5 = 4/500 fold dilution.
Implement a function called getElements() that takes in two lists. T.pdf
Implement a function called getElements() that takes in two lists. The first list L consists of any
type of items and the second list P consists of a list of integers. The implemented function will
return a list consisting of the element in L that are in positions indicated by P. For example, if
P=1, 7, 3, then the element in position 1, 7 and 3 in the list L must be returned. The function
must be implemented using the C++ list SLT API. Hint: The implementation for this function
will require some thought. Here are some hints for one approach to solve the problem. Use a for-
loop to walk through list P in a similar way as done in the printList() function. Consider creating
an index variable to track the item number in list L. Also, set the value of the iterator equal to a
variable in this loop. For example: int index = 0; // Track the position int position = (*itr) ; // Get
the item in P Create a second loop embedded within the first loop to walk through the elements
in list L. If the index equals position-1 then you know you have found the element. Thus, call the
push_back() function on the resultList to add the element. At the end of this loop increment the
index variable by 1. Note that your iterator variable for the second loop must be named
differently than the first loop. For example you can all it itr2. Output: The output from the
program once the function is implement will be:
/**
* List.cpp - This program implements and tests the getIntersection
* and getUnion functions.
*
* TODO: Include your name and course number here.
*/
#include
#include
#include
using namespace std;
template
void printList(list list1);
template
list getElements(list L, list P);
int main(int argc, char **argv)
{
// Declare list variables
list list1;
list list2;
// Add data to list 1
list1.push_back(\"the\");
list1.push_back(\"quick\");
list1.push_back(\"brown\");
list1.push_back(\"fox\");
list1.push_back(\"jumped\");
list1.push_back(\"over\");
list1.push_back(\"the\");
list1.push_back(\"lazy\");
list1.push_back(\"dog\");
// Add data to list 2
list2.push_back(4);
list2.push_back(3);
list2.push_back(9);
list2.push_back(6);
list2.push_back(7);
// Print out the lists
cout << \"List 1: \" << endl;
printList(list1);
cout << \"List 2: \" << endl;
printList(list2);
cout << endl;
cout << \"Results of getElements():\" << endl;
printList(getElements(list1, list2));
cout << \"\ ** Press any key to continue **\ \";
getchar();
return 0;
}
template
void printList(list list1)
{
cout << \" \";
// Use an iterator to walk the list. Print the value of
// for each node.
typename list::iterator itr;
for (itr = list1.begin(); itr != list1.end(); itr++)
{
// Get the contents of each node
Object value = (*itr);
// Print the value of the node
cout << \" \" << value;
}
cout << endl;
return;
}
template
list getElements(list L, list P)
{
//A result list to return to the caller
list resultList;
// TODO: Implement the details for the function.
return resultList;
}
Solution
/**
* List.cpp - Thi.
gnmenttakeAssignmentMaindoinvoker assignmentstakeAssignmernt Dr. L. .pdf
gnment/takeAssignmentMaindoinvoker assignmentstakeAssignmernt Dr. L. M. Paton is an
ophthalmologist As of December 31, Dr. Patton owned the following property that elated tos
prfessional practice, Pattion Eye Clinic Cash, 2,720 Professional Equipment, $62,400 Office
Equipment, $11,220 On the same date, he owed the following business creditors Munez Supply
Compay $5,320 Martin Equlpment Sales, $6,180 Compute the following amounts in the
accounting equation Assets Equity
Solution
Assets 76330 =2720+62400+11210 Liabilities 11500 =5320+6180 Owner\'s Equity 64830
=76330-11500.
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On April 18, 1906, an earthquake killed over 3,000 people when it hi.pdf
On April 18, 1906, an earthquake killed over 3,000 people when it hit San Francisco, with a
magnitude of 8.3 on the Richter Scale. On March 11, 2011 the “Tohoku” earthquake hit central
Japan with a magnitude of 9.0 on the Richter Scale, and killed 15, 894 people. How much greater
was the 2011 earthquake than the 1906 earthquake?
Solution
The formula of magnitude of earthquake is
M=log(I/S)
I is the intensity of the earthquake And S is the intensity of the standard earthquake
8.3 = log(I1/S) and 9= log(I2/S)
9-8.3= log(I2/S) -log(I1/S)
.7= log(I2/I1)
I2/I1=10.7
I2=5 I1
Therefore the earhquake occurs in 2011 is 5 times more intense as compared to earthquake
occurs in 1906.
On the planet of Caracas, in the Stellar Solorais Ka Chunk Galaxy, th.pdf
On the planet of Caracas, in the Stellar Solorais Ka Chunk Galaxy, there is a population of
Nugget Arians. Two phenotypes of interest are: Peanuts; and Carmelaoid. Each is a single gene
trait, and for each gene there are two alleles. A cross between a female with Peanuts and a male
with Carmelaoid (each of which is homozygous for the wild-type (non-trait) allele at the other
gene) produces the following F1 counts: The mode of inheritance for each phenotype is: Peanuts
= Autosomal Recessive. Carmelaoid = Autosomal Dominant. Peanuts = Autosomal Recessive.
Carmelaoid = Autosomal Recessive. Peanuts = Autosomal Recessive. Carmelaoid = X-linked
Recessive. Peanuts = Autosomal Dominant. Carmelaoid = Autosomal Recessive. Peanuts =
Autosomal Dominant. Carmelaoid = Autosomal Dominant.
Solution
The answer is C. Peanuts=Autosomal Recessive
Carmeloid=X-linked recessive.
As
Autosomal characters of male are not carried to progeny so autosomal inheritance is othrwise
called maternal inheritance. As carmelaoid phenotype from male is expressed in F1 so it has to
be in cromosome and from the probability of expression it is evident that it is present in X
cromosome. The male counterpart like sperm carry very little of cytoplasmic material but half of
chromosomal meterial. In human type sex determination i.e.XY male and XX female X linked
recessive genes only express in homozygous i.e. XX condition..
Muv is homozygous You have a C. elegans worm with the following geno.pdf
Muv is homozygous You have a C. elegans worm with the following genotype: + dpy/lov +;
muv Where, dpy, lov and muv are alleles for these recessive phenotypes: Dpy = dumpy (short
and fat) Lov= mating behavior defective Muv = multivulva After taking a minute to feel sorry
for this poor worm, answer the following questions: What gene(s) is this animal heterozygous
for? What gene(s) is this animal homozygous for? What genes(s) are located on the same
chromosome? What is the phenotype of this animal? Assuming lov and dpy loci are 20cM
apart, what are the phenotypes, genotypes and expected frequencies of this hermaphrodite\'s self-
progeny? Percentage Phenotype
Solution
5.1: The animal is heterozygous for dpy and lov genes.
5.2: The animal is homozygous for muv gene.
5.3: The dpy and muv genes are located on the same chromosome.
5.4: Phenotype of animal= long and thin; mating behaviour, multivulva..
Material Science (Solidification)What is the solute atoms concent.pdf
Material Science: (Solidification)
What is the solute atoms concentration with position for a directional solidification of alloy?
Please don\'t answer if you don\'t know the right answer. Thanks.
Solution
During directional solidification, the solute atom concentration increases with increase in
distance from the outside surface. For example during solidification of a spherical casting, the
solute concentration will be maximum at the core and minimum on the surface..
Investment in Note ReceivableSuppliesPrepaid RentBuildingsAc.pdf
Investment in Note Receivable
Supplies
Prepaid Rent
Buildings
Accumulated Depreciation—Buildings
Unearned Service RevenueDebitCredit
Investment in Note Receivable$24,000
Supplies23,500
Prepaid Rent3,200
Buildings280,000
Accumulated Depreciation—Buildings$140,000
Unearned Service Revenue11,900
Solution
1) Interest receivble (24000*12%/12) 240 Interest income 240 2) Supplies expense
(23500-15200) 8300 Supplies 8300 3) Rent expense (3200/4) 800 Prepaid rent 800
4) Salaries and wages expense 2700 Salaries and wages payable 2700 5) Depreciation
expense-Buildings (6720/12) 560 Accumualated depreciation-Buildings 560 6)
Unearned service revenue 4800 Service revenue 4800 7) Repairs and maintenane
expense 2300 Accounts payable 2300.
In a tube containing 1 ml of a 15 dilution of A, and 4 ml of a 110.pdf
In a tube containing 1 ml of a 1:5 dilution of A, and 4 ml of a 1:100 dilution of B:
What is the final dilution of substance A?
What is the final dilution of substance B?
Solution
What is the final dilution of substance A?
When 1mL substance A is mixed with 4 mL of Substance B, four dilution of substance A
happens.
original=1/5
Further =1/4
Final = 1/5 x1/4 = 1/20 fold dilution
What is the final dilution of substance B?
When 4mL substance B is mixed with 1 mL of Substance A, = times
original=1/100
Further =4/5
Final = 1/100 x 4/5 = 4/500 fold dilution.
Implement a function called getElements() that takes in two lists. T.pdf
Implement a function called getElements() that takes in two lists. The first list L consists of any
type of items and the second list P consists of a list of integers. The implemented function will
return a list consisting of the element in L that are in positions indicated by P. For example, if
P=1, 7, 3, then the element in position 1, 7 and 3 in the list L must be returned. The function
must be implemented using the C++ list SLT API. Hint: The implementation for this function
will require some thought. Here are some hints for one approach to solve the problem. Use a for-
loop to walk through list P in a similar way as done in the printList() function. Consider creating
an index variable to track the item number in list L. Also, set the value of the iterator equal to a
variable in this loop. For example: int index = 0; // Track the position int position = (*itr) ; // Get
the item in P Create a second loop embedded within the first loop to walk through the elements
in list L. If the index equals position-1 then you know you have found the element. Thus, call the
push_back() function on the resultList to add the element. At the end of this loop increment the
index variable by 1. Note that your iterator variable for the second loop must be named
differently than the first loop. For example you can all it itr2. Output: The output from the
program once the function is implement will be:
/**
* List.cpp - This program implements and tests the getIntersection
* and getUnion functions.
*
* TODO: Include your name and course number here.
*/
#include
#include
#include
using namespace std;
template
void printList(list list1);
template
list getElements(list L, list P);
int main(int argc, char **argv)
{
// Declare list variables
list list1;
list list2;
// Add data to list 1
list1.push_back(\"the\");
list1.push_back(\"quick\");
list1.push_back(\"brown\");
list1.push_back(\"fox\");
list1.push_back(\"jumped\");
list1.push_back(\"over\");
list1.push_back(\"the\");
list1.push_back(\"lazy\");
list1.push_back(\"dog\");
// Add data to list 2
list2.push_back(4);
list2.push_back(3);
list2.push_back(9);
list2.push_back(6);
list2.push_back(7);
// Print out the lists
cout << \"List 1: \" << endl;
printList(list1);
cout << \"List 2: \" << endl;
printList(list2);
cout << endl;
cout << \"Results of getElements():\" << endl;
printList(getElements(list1, list2));
cout << \"\ ** Press any key to continue **\ \";
getchar();
return 0;
}
template
void printList(list list1)
{
cout << \" \";
// Use an iterator to walk the list. Print the value of
// for each node.
typename list::iterator itr;
for (itr = list1.begin(); itr != list1.end(); itr++)
{
// Get the contents of each node
Object value = (*itr);
// Print the value of the node
cout << \" \" << value;
}
cout << endl;
return;
}
template
list getElements(list L, list P)
{
//A result list to return to the caller
list resultList;
// TODO: Implement the details for the function.
return resultList;
}
Solution
/**
* List.cpp - Thi.
gnmenttakeAssignmentMaindoinvoker assignmentstakeAssignmernt Dr. L. .pdf
gnment/takeAssignmentMaindoinvoker assignmentstakeAssignmernt Dr. L. M. Paton is an
ophthalmologist As of December 31, Dr. Patton owned the following property that elated tos
prfessional practice, Pattion Eye Clinic Cash, 2,720 Professional Equipment, $62,400 Office
Equipment, $11,220 On the same date, he owed the following business creditors Munez Supply
Compay $5,320 Martin Equlpment Sales, $6,180 Compute the following amounts in the
accounting equation Assets Equity
Solution
Assets 76330 =2720+62400+11210 Liabilities 11500 =5320+6180 Owner\'s Equity 64830
=76330-11500.
Every cell in our bodies expresses the same genes. True or False.pdf
Every cell in our bodies expresses the same genes. True or False?
Solution
Answer : The given statement is FALSE. Although all the cells in our body contains same
amount of DNA/number of genes, different cells expresses different genes which gives their
identity. This phenamenan is called differential gene expression..
Describe the partition of Z resulting from the equivalence relati.pdf
Describe the partition of Z resulting from the equivalence relation = (mod 4).
Solution
a=b mod 4
implies and b give the same remainder modulo 4
There are only 4 least residues modulo 4: 0,1,2,3
So total 4 equivalence classes
The partitions are
[0]= Multiples of 4
[1]=INtegers of the form 4k+1
[2]=Integers of the form 4k+2
[3]=Integers of the form 4k+3.
describe what is meant by Rh factor when typing blood.Solution.pdf
describe what is meant by Rh factor when typing blood.
Solution
Ans). Rh derives from rhesus and the terms rhesus blood group system. Rh factor is an inherited
protein on the surface of red blood cells. It consists of many blood-group antigens, among which
the five antigens D, C, c, E, and e are the most important..
Describe the causes, symptoms, and treatment of conjunctivitisSo.pdf
Describe the causes, symptoms, and treatment of conjunctivitis
Solution
Conjuctivits is a condition that is associated with the redness and inflammation of the conjuctiva.
Conjuctiva is a thin transparent layer of tissue lining the inner surface of the eyelid covering the
white part of the eye. It is a common disease and is usually referred toas the \"pink eye\".
CAUSES OF CONJUCTIVITIS
Three types of conjuctivitis include:
Allergic Conjunctivitis
Infectious Conjuctivitis
Chemical Conjunctivitis
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TREATMENT OF CONJUCTIVITIS
The appropriate treatment for conjunctivitis depends on its cause:.
Discuss the relationship between gap rule gene mutations, pair-rule .pdf
Discuss the relationship between gap rule gene mutations, pair-rule gene mutations, and segment
polarity gene mutations in Drosophila melanogaster.
Solution
Segmentation in Drosophila is controlled by gene expression cascade that occur in a sequential
order. The maternal effect genes form the anterior and posterior poles in the embryo. The spatial
pattern of segments is controlled by transcription regulation of gap genes. The gap genes express
the overlapping segments of anterior and posterior poles. The gap genes regulate next set of
genes called pair rule genes. These genes are expressed in 7 stripes corresponding to each
segment. Pair rule genes express the transcription factors regulating the segment polarity genes.
These genes are expressed as 14 segmentally repeated stripes. The segment polarity genes
express regulatory proteins and transcription factors that maintain interaction between cells and
finally, homeotic genes control the character of each segment.
The missing segments producing gaps are known by Krüppel mutants resulting in missing of T1-
A5 segments. Hunchback mutants are missing head segments and knirps mutants lack posterior
segments. Mutations in pair-rule genes- result in half the number of parasegments.The fushi-
tarazu mutants lack odd-numbered parasegments and even-skipped mutants lack even-numbered
parasegments. Mutations in segment polarity genes- result in duplication of the anterior or
posterior half within each segment. The gooseberry and engrailed mutants have the posterior part
of each segment replaced by the anterior part of the adjacent segment..
Create a directory your virtual machine. Within that directory creat.pdf
Create a directory your virtual machine. Within that directory create the html file index.html.
This should be an HTML5 formatted file, with the following elements and text defined
Links to an external JavaScript file q2.js in the same directory as index.html.
Contains a button with the text \"ClickMe\" (no quotes) that will invoke the callback function
\"go()\" when the button is clicked.
Contains a paragraph with id \"output\" (no quotes) which contains the text \"Change Me\". (no
quotes)
When the ClickMe button is pushed, the text in the HTML element with id \"output\" should be
updated as follows using JavaScript code in q2.js. (Note: the number given below should reset
when the page is reloaded and all effects below are cumulative.)
The first time the button is pushed, the colour of the text in output to be set to red.
The second time the button is pushed, the font of the text in output should be set to \"sans-serif\"
The third time the button is pressed, the text should be changed to five copies of whatever text is
currently found as the innerHTML of the output element. Do not add extra spaces or line breaks
to this value.
If the button is pushed four or more times, the text in output should be set to \"QUESTION
DONE\" (no quotes).
Solution
index.html
Title
Change MeClick Me
q2.js
var num_clicks = 0;
function clickme(){
num_clicks = num_clicks + 1;
if (num_clicks==1){
document.getElementById(\"output\").style.color=\"red\"; //change color to red on 1 click
}
else if (num_clicks==2){
document.getElementById(\"output\").style.fontFamily=\"sans-serif\"; //change font family
on 2 clicks
}
else if (num_clicks==3){
text = document.getElementById(\"output\").innerHTML;
document.getElementById(\"output\").innerHTML = text + text + text + text + text;
//change text on 3 clicks
}
else if (num_clicks>=4){
document.getElementById(\"output\").innerHTML = \"QUESTION DONE\"; //question
done text on 4 clicks
}
}.
Compare between the two light microscope SolutionFeature1. .pdf
Compare between the two light microscope:
Solution
Feature
1. Similar functional parts : source of light is same - visible light is used by focusing the light
either through lenses or prism on the specimen.
2. Dissimilar functional parts : Compound microscope- has multiple objective lenses with one
eyepiece;
Dissecting Microscope- has single objective lens with two eyepiece lenses
3. Purpose of use : Dissecting microscope: used to perform small dissections under the
microscope and to view live specimen , compound microscope- used to visualise things that are
too small to be seen via the naked eye.
7. Maximum magnification: Compound microscope- up to 1000x ; Dissecting microscope- up to
70x
4. Image dimension : -
5. Image orientation : -
6. Image development: no image development or orientation or dimensions- these are all
properties of electron microscopes
3. Purpose of use : Dissecting microscope: used to perform small dissections under the
microscope and to view live specimen , compound microscope- used to visualise things that are
too small to be seen via the naked eye.
7. Maximum magnification: Compound microscope- up to 1000x ; Dissecting microscope- up to
70x
4. Image dimension : -.
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Chordin, Tolloid and Sizzled all play important roles in BMP signal regulation. Discuss Name
the three embryonic germ layers and briefly discuss how the BMP-Chordin axis determines the
fate of one of these germ layers
Solution
Ques-1: In Xenopus embryos, the cell differentiation of ectodermal and mesodermal layers are
mainly through the dorso- ventral axis and this differentiation is determined by the bone
morphogenetic proteins (BMPs) where cells are mainly exposed in the excellylar region.
Therefore, chordin activity (chondrin and Noggin are released by Spearmann organizer) has
mainly increased and tolloid family members can control these chondrin activity to promote
BMPs in generating organ systems such as blood or neural tubes. Sizzeled and chonrin activities
are mainly according to BMP gradient during gastrulation of embryonic development in xenopus
Answer: Ques-2:
Three embryonic layers: ectoderm, mesoderm and endoderm
The BMP-chondrin axis is going to determine the fate of formation of ectoderm or mesoderm
organ formation for example, inhibition of BMP is leading to formation of neural tube from
mesoderm instead of blood. Therefore, the factors such as Noggin, chondrin are going to be
regulated by these BMP factors during embryonic development.
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Chills are used in casting where the part is likely to form a cavity.They are used in thicker
regions to increase the solidification rate.
A. False
B. True
Solution
TRUE
To maximize the heat tranfer rate a metalic object is provided called CHILL. Chill is inserted
into the cavity created due to removal of pattern and inserted in moulding sand.
By changing the direction of the chill the direction of solidification also changes.
Chill is a metalic object which is conductive so heat transfer rate is high compare to moulding
sand..
A drug that disrupts tubulin polymerzation is applied to some cell.pdf
A drug that disrupts tubulin polymerzation is applied to some cells. What specific part of the cell
cycle will be disrupted?
A drug that disrupts tubulin polymerzation is applied to some cells. What specific part of the cell
cycle will be disrupted?
Solution
Microtubule assembly .... anaphase ... as chromosomes pass along microtubules towards poles of
new dividing cells ....and if micotubule disassembly occurs they cannot move also cargo cannot
move in the cells from one location to other location hence affecting the cells overall.
At the second metaphase of meiosis Chromosomes have single chromatid.pdf
At the second metaphase of meiosis Chromosomes have single chromatids and align in
homologous pairs. Chromosomes have single chromatids and do not align in homologous pairs.
Chromosomes have two chromatids and align in homologous pairs. Chromosomes have two
chromatids and do not align in homologous pairs. Chromosomes have two chromatids and are
separating to opposite poles of the cell. Incomplete dominance at a single locus always results
from the mutant protein having partial function results in a self-crossed population with identical
phenotypic and genotypic ratios cannot be distinguished from epistasis can generate a true
breeding intermediate phenotype gives the same genetic ratios as 2 unlinked genes each showing
complete dominance
Solution
1. The correct option is e. In the second metaphase of meiosis, the chromosomes have two
chromatids and are separating to opposite poles of the cell which is very similar to mitosis.
2. The correct option is d. This is because in incomplete dominance both the alleles are unable to
express themselves completely. Thus resulting in an F1 hybrid with an intermediate phenotype of
the two parental phenotype..
BEFO Ca.nin view shows longi tudimal vertical view or alde viu is a.pdf
BEFO Ca.nin view shows longi tudimal vertical view or \"alde viu is a relatively short vertical
view of the ground drawn aE EghEangles or perpendicular toaroute centeriine of 43. A 1ine that
identiries and separates adjoining tracts of privately ofr 44. The division ot a large parcel of
ground into smailer parcels for sale 45 Title to land may be transferred or conveyed from one
owner to another highway or sewe s publiely owned land in called a as separate building lots is
called a primarily by a legal document called a 46. Streets and roads are conveyed to the public
for right-ot way tor publie use by a developer or landowner by a recorded legal instrument
known as a 47. are neighboring property owners identitied in a metes and bounds boundary
description 48. Taking possession of someone else\'s property by open and continuous
encroachment for a statutory period of time is 49. The privilege or right, which the owner of one
parcel has to use the land of another for a certain defined purpose, is called an 50. The right of a
government to use private property for a specitic purpose for the pubiic benefit and welfare is
called 51. The owner of property adjacent to a body of water has certain privileges with respect
to the use of the water called_ 52. In many states a Surveyor\'s liability for damages due to
mistakes and inaccurate work is covered under the statute of limitations, more specifically under
the that liability begins from the time that the error vas discovered, not from when the work was
done. , which states 53. A 1inear measuring devise having a nominal length of 66\', and the
distances described in very old deeds which are referred to as perches or rods, and whose length
has a direct relationship to the acre.
Solution
41-Profile view
42-cross section view
43-Property line
44-Subdivision
45-Deed
46-Easement
47-Adjoiners
48-Adverse possession
49-Deed of dedication
50-Condemnation (Eminent domain)
51-Riparian right
52-Accidental errors
53-Scale.
A LuriaDelbruck fluctuation test was performed to assess the frequen.pdf
A Luria/Delbruck fluctuation test was performed to assess the frequency that T1-resistant E. coli
mutants arise in a culture. Four different E. coil cultures were allowed to grow for 6 hours, and
then T1 page was added. Cells were then plated and the number of phage resistant bacteria was
tabulated. Explain how the results of this experiment support the hypothesis that the environment
selects from pre-existing spontaneously occurring mutations.
Solution
According Luria Delbrück experiment, principle suggests that if spontaneous mutation occurs,
bacteria grown on different plates would have different numbers. On the vice versa of that, if
bacterial population occurs equal in numbers in different plates or culture then the mutation are
induced mutation. Therefore, the result suggests that survivals from spontaneous mutations get
select only..
8 Which of the following is not one of the tissue layers of the respi.pdf
8 Which of the following is not one of the tissue layers of the respiratory tract? a. Muscularis b.
Mucosa c. Adventitia d. Submucosa 9 What three structures constitute the respiratory
membrane? 10 hich of the following is not a trend we find within the respiratory tract? a. The
epithelium gets progressively shorter in the lower passages b. The amount of hyaline cartilage
gradually increases as we move to smaller passages c. The amount of smooth muscle increases in
the smaller pa d. The amount of elastic fibers increases in the smaller passages
Solution
8 a. Muscularis is not one of the tissue layer of respiratory tract.
The tracheal wall is made up of three distinct tissue layers, the mucosa, submucosa, and
adventitia (internal to external).Mucosa: composed of goblet cell which contains pseudostratified
ciliated epithelium.Submucosa: a connective tissue that contains seromucous glands,produces
mucous within the trachea. Adventitia:outermost connective tissue which is strengthened
internally by 16-20 C- shaped rings of hyaline cartilage.
9. The three structures that constitute the respiratory membrane are
a squamous epithelial lining of alveolus
b endothelial cells lining of adjacent capillary
c fused basal laminae between alveolar and edothelial cells
The respiratory membrane is made up of the alveolar and capillary walls and their fused basal
laminae, which acts as a air-blood barrier. It is also called the alveolar-capillary membrane. In
the lungs gase exchange occurs across the respiratory membrane by simple diffusion, as it has
gas on one side and blood flow on the other side of
10. Within the respiratory tract we do not find
b. The amount of hyaline cartilage gradually increases as we move to the smaller passages.
Rather the smaller passages or the bronchioles do not have the hyaline cartilage and they have
club cells in their epithelial lining.
5. Goblet cells are more numerous in the upper respiratory passage than in the lower respiratory
passages because more dust and debris are present in the upper respiratory passage and they are
needed to be removed and since the goblet cells secrete mucus, which traps these and the ciliated
cells help in removing it out of the respiratory tract . The goblet cells are present in a ratio of 1:6
ciliated cells. In the lower respiratory passages these dust and debris decreases and hence the
number of goblet cell also decreases..
4. Sickle cell anemia is an autosomal recessive disorder. Using S.pdf
4. Sickle cell anemia is an autosomal recessive disorder. Using \"S\" and \"s\" to represent the
alleles, what is the genotype of a person afflicted with sickle cell anemia?
5. Two parents are apparently normal, but their daughter was born with cystic fibrosis, an
inherited disorder. If cystic fibrosis is like most other human hereditary disorders, the probability
of their next child being born with cystic fibrosis is _____.
Solution
4.To express sicle cell anemia in a person,genotype must be SS.But the disease remain silent in
persons with Ss trait.
5.1/4th probability of having the next child with cystic fibrosis are there.it is caused by defective
allele that encodes a defective chloride-channel membrane transport protein..
112.Twenty-one executives in a large corporation were randomly select.pdf
112.Twenty-one executives in a large corporation were randomly selected to study the effect of
several factors on annual salary (expressed in $000s) The factors selected were age, seniority,
years of college, number of company divisions they had been exposed to, and the level of their
responsibility. The results of the regression analysis follow: Which independent variable has
the most significant effect on annual salary?
Solution
the level has most significant effect on annual salary because of regression coefficient of level is
higher comparison to the other.
1.Which of the following features is found in the common ancestor of.pdf
1.Which of the following features is found in the common ancestor of the deuterostomes?Radial
symmetryNotochordPharyngeal slits2.Which of the deuterostome groups does not display
bilateral symmetry at some time during their
development?EchinodermsHemichordatesChordatesNone of the above3.Which of the following
deuterostome groups include species that live on
land?EchinodermsHemichordatesChordates1.Which of the following features is found in the
common ancestor of the deuterostomes?Radial symmetryNotochordPharyngeal slits
Solution
1. Both the hemichordata and chordata have gill slits and primitive fossil echinodermeta also
show signs of gill slit . Based on above fact deuterostome evolved from common ancestor that
had pharyngeal gill.
2 All deuterostome including echinoderms, hemichordates and chordates show bilateral
symmetry during development.
3 chordates among deuterostome group live on land..
Write an essay on E-Businness and Compare the four categories of ebu.pdf
Write an essay on E-Businness and Compare the four categories of ebusiness models.
Solution
The fast and dramatic changes in information technology specially in last one decade has given
new concept of marketing in which buyer and seller do not see each other face to face nor see the
goods physically; the whole transaction is carried out with the help of ‘on line’ communication.
The entire deal is carried out with the help of computer – telecommunication and net working
with associated hardware.
In the e – commerce internet provides information about goods and services “It is” a way of
conducting imaging and executing business transactions and services through elec tronic media
and net working in computers and communication net work, websites, e-mail are resorted.
Customers know about goods and services sitting at home. The manufacturers, distributors,
suppliers and services providers let the consumers know about their products quality, price, size,
color etc. through multi-colored catalogues on website. The consumers can ‘surface various web
sites and compare their relative prices, quality characteristic, features etc.
These details can be obtained from suppliers around the globe. The websites are available beside
for goods for direct selling, context selling, financial and other services such as hospitals,
education, training, advertise ments, property, entertainment, product demonstrations, bill
payment, exchange and all other ser vices which one can think of.
Advantage to Consumers:
The consumer has number of advantages and convenience and therefore the system is becoming
popular.
1. Consumer has wider choice not from his town or country but also round the globe unless there
are import restrictions.
2. Customized or personalized product and service. For instance if some lady wants a bra of
exact size, her size can be measured through internet and stored and she will be supplied bra of
her requirement.
3. In case of purchase, one is not required to go from store to store to see the products to collect
their details, prices etc. Sitting at home he gets all the required information and that too very fast
without spending much time.
4. There is absolute flexibility of time, place and distance is no hurdle; one can open the site any
time day or night to get details, there is no problem of shops/stores opening/closing hours.
Website can be opened any time. In physical sales place and distance is also a problem which is
no problem in e-commerce because one can see sites all over the world without moving out of
the house.
5. Goods are available at cheaper price because there are lot of economies of space, rent, interest
to the seller Further, he manages with much lesser number of outlets and cost of marketing is
reduced. Part of these savings is passed on to consumer and therefore, he gets the products
cheaper than from conventional shops/departmental stores, grocers etc.
6. It helps to globalize retail trading. One can buy things without geograp.
Which Pirl ofiherenrie cinardad an acthm Feinia DI 3rd 4 Ideatry .pdf
Which Pirl ofiherenrie cinardad an acthm Feinia DI 3rd 4 Ideatry the streatuselabeled 5.
lrympkleinirah E) dendrthK SS res dendrime spines ng Fgure ti-2 The Nene Action Paureal Use
neare 12-2 ss auner eke ilacoresteac 6) Which af the graihehows ween pa ve dpularizarkn b
wing? which area ofthe Eraph occurs when fee is aruddex irauhei sdhmlan? which ama ofthe
Fapi, shows when membrane piterkialapproaches potanium quilhnus po metal? A) D)4 9) Which
point of the Faph sheen when potassimianouinou earedssdimininnem? 014
Solution
Answers: (Figure 12.2)
(6) (A) 1
In the graph above. area 1 shows when passive depolarization is occuring.
(7) (B) 2
Area 2 in the graph shows when there is a sudden rush of soudium ions.
(8) (D) 4
Area 4 of the graph shows membrane potential approaches the potassium equilibrium potential.
(9) (C) 3
Area 3 of the graph shows when potassium ion outflow exceeds sodium ion inflow..
Which of the following observations is supporting evidence that all p.pdf
Which of the following observations is supporting evidence that all populations are linked by
descent with modification? living animals all conduct transcription the same way fossils of
early horses show that they stood on three toes different geological strata contain species not
alive today today, pest species are constantly changing genetically all of these observations are
evidence that evolution occurs White eyes is caused by an X-linked recessive allele in fruit flies.
Suppose that we found the following offspring that resulted from one mating between two flies,
called Fred and Ethel: 25% while eye females, 25% red eye females, 25% white eye mates, 25%
red eye mates. What can be concluded about Fred and Ethel? Fred must have had red eyes Ethel
must have been heterozygous Ethel might have had white eyes Fred must have had alleles for
both red and white eyes none of the above is true
Solution
5) Correct answer is e) all of these observations are evidences that evolution occurs.
6) Correct answer is c) Ethel must have been heterozygous..
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...
BÀI TẬP BỔ TRỢ TIẾNG ANH GLOBAL SUCCESS LỚP 3 - CẢ NĂM (CÓ FILE NGHE VÀ ĐÁP Á...Nguyen Thanh Tu Collection
https://app.box.com/s/hqnndn05v4q5a4k4jd597rkdbda0fniiPalestine last event orientationfvgnh .pptx
An EFL lesson about the current events in Palestine. It is intended to be for intermediate students who wish to increase their listening skills through a short lesson in power point.
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Every cell in our bodies expresses the same genes. True or False.pdf
Every cell in our bodies expresses the same genes. True or False?
Solution
Answer : The given statement is FALSE. Although all the cells in our body contains same
amount of DNA/number of genes, different cells expresses different genes which gives their
identity. This phenamenan is called differential gene expression..
Describe the partition of Z resulting from the equivalence relati.pdf
Describe the partition of Z resulting from the equivalence relation = (mod 4).
Solution
a=b mod 4
implies and b give the same remainder modulo 4
There are only 4 least residues modulo 4: 0,1,2,3
So total 4 equivalence classes
The partitions are
[0]= Multiples of 4
[1]=INtegers of the form 4k+1
[2]=Integers of the form 4k+2
[3]=Integers of the form 4k+3.
describe what is meant by Rh factor when typing blood.Solution.pdf
describe what is meant by Rh factor when typing blood.
Solution
Ans). Rh derives from rhesus and the terms rhesus blood group system. Rh factor is an inherited
protein on the surface of red blood cells. It consists of many blood-group antigens, among which
the five antigens D, C, c, E, and e are the most important..
Describe the causes, symptoms, and treatment of conjunctivitisSo.pdf
Describe the causes, symptoms, and treatment of conjunctivitis
Solution
Conjuctivits is a condition that is associated with the redness and inflammation of the conjuctiva.
Conjuctiva is a thin transparent layer of tissue lining the inner surface of the eyelid covering the
white part of the eye. It is a common disease and is usually referred toas the \"pink eye\".
CAUSES OF CONJUCTIVITIS
Three types of conjuctivitis include:
Allergic Conjunctivitis
Infectious Conjuctivitis
Chemical Conjunctivitis
SYMPTOMS OF CONJUCTIVITIS
TREATMENT OF CONJUCTIVITIS
The appropriate treatment for conjunctivitis depends on its cause:.
Discuss the relationship between gap rule gene mutations, pair-rule .pdf
Discuss the relationship between gap rule gene mutations, pair-rule gene mutations, and segment
polarity gene mutations in Drosophila melanogaster.
Solution
Segmentation in Drosophila is controlled by gene expression cascade that occur in a sequential
order. The maternal effect genes form the anterior and posterior poles in the embryo. The spatial
pattern of segments is controlled by transcription regulation of gap genes. The gap genes express
the overlapping segments of anterior and posterior poles. The gap genes regulate next set of
genes called pair rule genes. These genes are expressed in 7 stripes corresponding to each
segment. Pair rule genes express the transcription factors regulating the segment polarity genes.
These genes are expressed as 14 segmentally repeated stripes. The segment polarity genes
express regulatory proteins and transcription factors that maintain interaction between cells and
finally, homeotic genes control the character of each segment.
The missing segments producing gaps are known by Krüppel mutants resulting in missing of T1-
A5 segments. Hunchback mutants are missing head segments and knirps mutants lack posterior
segments. Mutations in pair-rule genes- result in half the number of parasegments.The fushi-
tarazu mutants lack odd-numbered parasegments and even-skipped mutants lack even-numbered
parasegments. Mutations in segment polarity genes- result in duplication of the anterior or
posterior half within each segment. The gooseberry and engrailed mutants have the posterior part
of each segment replaced by the anterior part of the adjacent segment..
Create a directory your virtual machine. Within that directory creat.pdf
Create a directory your virtual machine. Within that directory create the html file index.html.
This should be an HTML5 formatted file, with the following elements and text defined
Links to an external JavaScript file q2.js in the same directory as index.html.
Contains a button with the text \"ClickMe\" (no quotes) that will invoke the callback function
\"go()\" when the button is clicked.
Contains a paragraph with id \"output\" (no quotes) which contains the text \"Change Me\". (no
quotes)
When the ClickMe button is pushed, the text in the HTML element with id \"output\" should be
updated as follows using JavaScript code in q2.js. (Note: the number given below should reset
when the page is reloaded and all effects below are cumulative.)
The first time the button is pushed, the colour of the text in output to be set to red.
The second time the button is pushed, the font of the text in output should be set to \"sans-serif\"
The third time the button is pressed, the text should be changed to five copies of whatever text is
currently found as the innerHTML of the output element. Do not add extra spaces or line breaks
to this value.
If the button is pushed four or more times, the text in output should be set to \"QUESTION
DONE\" (no quotes).
Solution
index.html
Title
Change MeClick Me
q2.js
var num_clicks = 0;
function clickme(){
num_clicks = num_clicks + 1;
if (num_clicks==1){
document.getElementById(\"output\").style.color=\"red\"; //change color to red on 1 click
}
else if (num_clicks==2){
document.getElementById(\"output\").style.fontFamily=\"sans-serif\"; //change font family
on 2 clicks
}
else if (num_clicks==3){
text = document.getElementById(\"output\").innerHTML;
document.getElementById(\"output\").innerHTML = text + text + text + text + text;
//change text on 3 clicks
}
else if (num_clicks>=4){
document.getElementById(\"output\").innerHTML = \"QUESTION DONE\"; //question
done text on 4 clicks
}
}.
Compare between the two light microscope SolutionFeature1. .pdf
Compare between the two light microscope:
Solution
Feature
1. Similar functional parts : source of light is same - visible light is used by focusing the light
either through lenses or prism on the specimen.
2. Dissimilar functional parts : Compound microscope- has multiple objective lenses with one
eyepiece;
Dissecting Microscope- has single objective lens with two eyepiece lenses
3. Purpose of use : Dissecting microscope: used to perform small dissections under the
microscope and to view live specimen , compound microscope- used to visualise things that are
too small to be seen via the naked eye.
7. Maximum magnification: Compound microscope- up to 1000x ; Dissecting microscope- up to
70x
4. Image dimension : -
5. Image orientation : -
6. Image development: no image development or orientation or dimensions- these are all
properties of electron microscopes
3. Purpose of use : Dissecting microscope: used to perform small dissections under the
microscope and to view live specimen , compound microscope- used to visualise things that are
too small to be seen via the naked eye.
7. Maximum magnification: Compound microscope- up to 1000x ; Dissecting microscope- up to
70x
4. Image dimension : -.
Chordin, Tolloid and Sizzled all play important roles in BMP signal r.pdf
Chordin, Tolloid and Sizzled all play important roles in BMP signal regulation. Discuss Name
the three embryonic germ layers and briefly discuss how the BMP-Chordin axis determines the
fate of one of these germ layers
Solution
Ques-1: In Xenopus embryos, the cell differentiation of ectodermal and mesodermal layers are
mainly through the dorso- ventral axis and this differentiation is determined by the bone
morphogenetic proteins (BMPs) where cells are mainly exposed in the excellylar region.
Therefore, chordin activity (chondrin and Noggin are released by Spearmann organizer) has
mainly increased and tolloid family members can control these chondrin activity to promote
BMPs in generating organ systems such as blood or neural tubes. Sizzeled and chonrin activities
are mainly according to BMP gradient during gastrulation of embryonic development in xenopus
Answer: Ques-2:
Three embryonic layers: ectoderm, mesoderm and endoderm
The BMP-chondrin axis is going to determine the fate of formation of ectoderm or mesoderm
organ formation for example, inhibition of BMP is leading to formation of neural tube from
mesoderm instead of blood. Therefore, the factors such as Noggin, chondrin are going to be
regulated by these BMP factors during embryonic development.
Chills are used in casting where the part is likely to form a cavity.pdf
Chills are used in casting where the part is likely to form a cavity.They are used in thicker
regions to increase the solidification rate.
A. False
B. True
Solution
TRUE
To maximize the heat tranfer rate a metalic object is provided called CHILL. Chill is inserted
into the cavity created due to removal of pattern and inserted in moulding sand.
By changing the direction of the chill the direction of solidification also changes.
Chill is a metalic object which is conductive so heat transfer rate is high compare to moulding
sand..
A drug that disrupts tubulin polymerzation is applied to some cell.pdf
A drug that disrupts tubulin polymerzation is applied to some cells. What specific part of the cell
cycle will be disrupted?
A drug that disrupts tubulin polymerzation is applied to some cells. What specific part of the cell
cycle will be disrupted?
Solution
Microtubule assembly .... anaphase ... as chromosomes pass along microtubules towards poles of
new dividing cells ....and if micotubule disassembly occurs they cannot move also cargo cannot
move in the cells from one location to other location hence affecting the cells overall.
At the second metaphase of meiosis Chromosomes have single chromatid.pdf
At the second metaphase of meiosis Chromosomes have single chromatids and align in
homologous pairs. Chromosomes have single chromatids and do not align in homologous pairs.
Chromosomes have two chromatids and align in homologous pairs. Chromosomes have two
chromatids and do not align in homologous pairs. Chromosomes have two chromatids and are
separating to opposite poles of the cell. Incomplete dominance at a single locus always results
from the mutant protein having partial function results in a self-crossed population with identical
phenotypic and genotypic ratios cannot be distinguished from epistasis can generate a true
breeding intermediate phenotype gives the same genetic ratios as 2 unlinked genes each showing
complete dominance
Solution
1. The correct option is e. In the second metaphase of meiosis, the chromosomes have two
chromatids and are separating to opposite poles of the cell which is very similar to mitosis.
2. The correct option is d. This is because in incomplete dominance both the alleles are unable to
express themselves completely. Thus resulting in an F1 hybrid with an intermediate phenotype of
the two parental phenotype..
BEFO Ca.nin view shows longi tudimal vertical view or alde viu is a.pdf
BEFO Ca.nin view shows longi tudimal vertical view or \"alde viu is a relatively short vertical
view of the ground drawn aE EghEangles or perpendicular toaroute centeriine of 43. A 1ine that
identiries and separates adjoining tracts of privately ofr 44. The division ot a large parcel of
ground into smailer parcels for sale 45 Title to land may be transferred or conveyed from one
owner to another highway or sewe s publiely owned land in called a as separate building lots is
called a primarily by a legal document called a 46. Streets and roads are conveyed to the public
for right-ot way tor publie use by a developer or landowner by a recorded legal instrument
known as a 47. are neighboring property owners identitied in a metes and bounds boundary
description 48. Taking possession of someone else\'s property by open and continuous
encroachment for a statutory period of time is 49. The privilege or right, which the owner of one
parcel has to use the land of another for a certain defined purpose, is called an 50. The right of a
government to use private property for a specitic purpose for the pubiic benefit and welfare is
called 51. The owner of property adjacent to a body of water has certain privileges with respect
to the use of the water called_ 52. In many states a Surveyor\'s liability for damages due to
mistakes and inaccurate work is covered under the statute of limitations, more specifically under
the that liability begins from the time that the error vas discovered, not from when the work was
done. , which states 53. A 1inear measuring devise having a nominal length of 66\', and the
distances described in very old deeds which are referred to as perches or rods, and whose length
has a direct relationship to the acre.
Solution
41-Profile view
42-cross section view
43-Property line
44-Subdivision
45-Deed
46-Easement
47-Adjoiners
48-Adverse possession
49-Deed of dedication
50-Condemnation (Eminent domain)
51-Riparian right
52-Accidental errors
53-Scale.
A LuriaDelbruck fluctuation test was performed to assess the frequen.pdf
A Luria/Delbruck fluctuation test was performed to assess the frequency that T1-resistant E. coli
mutants arise in a culture. Four different E. coil cultures were allowed to grow for 6 hours, and
then T1 page was added. Cells were then plated and the number of phage resistant bacteria was
tabulated. Explain how the results of this experiment support the hypothesis that the environment
selects from pre-existing spontaneously occurring mutations.
Solution
According Luria Delbrück experiment, principle suggests that if spontaneous mutation occurs,
bacteria grown on different plates would have different numbers. On the vice versa of that, if
bacterial population occurs equal in numbers in different plates or culture then the mutation are
induced mutation. Therefore, the result suggests that survivals from spontaneous mutations get
select only..
8 Which of the following is not one of the tissue layers of the respi.pdf
8 Which of the following is not one of the tissue layers of the respiratory tract? a. Muscularis b.
Mucosa c. Adventitia d. Submucosa 9 What three structures constitute the respiratory
membrane? 10 hich of the following is not a trend we find within the respiratory tract? a. The
epithelium gets progressively shorter in the lower passages b. The amount of hyaline cartilage
gradually increases as we move to smaller passages c. The amount of smooth muscle increases in
the smaller pa d. The amount of elastic fibers increases in the smaller passages
Solution
8 a. Muscularis is not one of the tissue layer of respiratory tract.
The tracheal wall is made up of three distinct tissue layers, the mucosa, submucosa, and
adventitia (internal to external).Mucosa: composed of goblet cell which contains pseudostratified
ciliated epithelium.Submucosa: a connective tissue that contains seromucous glands,produces
mucous within the trachea. Adventitia:outermost connective tissue which is strengthened
internally by 16-20 C- shaped rings of hyaline cartilage.
9. The three structures that constitute the respiratory membrane are
a squamous epithelial lining of alveolus
b endothelial cells lining of adjacent capillary
c fused basal laminae between alveolar and edothelial cells
The respiratory membrane is made up of the alveolar and capillary walls and their fused basal
laminae, which acts as a air-blood barrier. It is also called the alveolar-capillary membrane. In
the lungs gase exchange occurs across the respiratory membrane by simple diffusion, as it has
gas on one side and blood flow on the other side of
10. Within the respiratory tract we do not find
b. The amount of hyaline cartilage gradually increases as we move to the smaller passages.
Rather the smaller passages or the bronchioles do not have the hyaline cartilage and they have
club cells in their epithelial lining.
5. Goblet cells are more numerous in the upper respiratory passage than in the lower respiratory
passages because more dust and debris are present in the upper respiratory passage and they are
needed to be removed and since the goblet cells secrete mucus, which traps these and the ciliated
cells help in removing it out of the respiratory tract . The goblet cells are present in a ratio of 1:6
ciliated cells. In the lower respiratory passages these dust and debris decreases and hence the
number of goblet cell also decreases..
4. Sickle cell anemia is an autosomal recessive disorder. Using S.pdf
4. Sickle cell anemia is an autosomal recessive disorder. Using \"S\" and \"s\" to represent the
alleles, what is the genotype of a person afflicted with sickle cell anemia?
5. Two parents are apparently normal, but their daughter was born with cystic fibrosis, an
inherited disorder. If cystic fibrosis is like most other human hereditary disorders, the probability
of their next child being born with cystic fibrosis is _____.
Solution
4.To express sicle cell anemia in a person,genotype must be SS.But the disease remain silent in
persons with Ss trait.
5.1/4th probability of having the next child with cystic fibrosis are there.it is caused by defective
allele that encodes a defective chloride-channel membrane transport protein..
112.Twenty-one executives in a large corporation were randomly select.pdf
112.Twenty-one executives in a large corporation were randomly selected to study the effect of
several factors on annual salary (expressed in $000s) The factors selected were age, seniority,
years of college, number of company divisions they had been exposed to, and the level of their
responsibility. The results of the regression analysis follow: Which independent variable has
the most significant effect on annual salary?
Solution
the level has most significant effect on annual salary because of regression coefficient of level is
higher comparison to the other.
1.Which of the following features is found in the common ancestor of.pdf
1.Which of the following features is found in the common ancestor of the deuterostomes?Radial
symmetryNotochordPharyngeal slits2.Which of the deuterostome groups does not display
bilateral symmetry at some time during their
development?EchinodermsHemichordatesChordatesNone of the above3.Which of the following
deuterostome groups include species that live on
land?EchinodermsHemichordatesChordates1.Which of the following features is found in the
common ancestor of the deuterostomes?Radial symmetryNotochordPharyngeal slits
Solution
1. Both the hemichordata and chordata have gill slits and primitive fossil echinodermeta also
show signs of gill slit . Based on above fact deuterostome evolved from common ancestor that
had pharyngeal gill.
2 All deuterostome including echinoderms, hemichordates and chordates show bilateral
symmetry during development.
3 chordates among deuterostome group live on land..
Write an essay on E-Businness and Compare the four categories of ebu.pdf
Write an essay on E-Businness and Compare the four categories of ebusiness models.
Solution
The fast and dramatic changes in information technology specially in last one decade has given
new concept of marketing in which buyer and seller do not see each other face to face nor see the
goods physically; the whole transaction is carried out with the help of ‘on line’ communication.
The entire deal is carried out with the help of computer – telecommunication and net working
with associated hardware.
In the e – commerce internet provides information about goods and services “It is” a way of
conducting imaging and executing business transactions and services through elec tronic media
and net working in computers and communication net work, websites, e-mail are resorted.
Customers know about goods and services sitting at home. The manufacturers, distributors,
suppliers and services providers let the consumers know about their products quality, price, size,
color etc. through multi-colored catalogues on website. The consumers can ‘surface various web
sites and compare their relative prices, quality characteristic, features etc.
These details can be obtained from suppliers around the globe. The websites are available beside
for goods for direct selling, context selling, financial and other services such as hospitals,
education, training, advertise ments, property, entertainment, product demonstrations, bill
payment, exchange and all other ser vices which one can think of.
Advantage to Consumers:
The consumer has number of advantages and convenience and therefore the system is becoming
popular.
1. Consumer has wider choice not from his town or country but also round the globe unless there
are import restrictions.
2. Customized or personalized product and service. For instance if some lady wants a bra of
exact size, her size can be measured through internet and stored and she will be supplied bra of
her requirement.
3. In case of purchase, one is not required to go from store to store to see the products to collect
their details, prices etc. Sitting at home he gets all the required information and that too very fast
without spending much time.
4. There is absolute flexibility of time, place and distance is no hurdle; one can open the site any
time day or night to get details, there is no problem of shops/stores opening/closing hours.
Website can be opened any time. In physical sales place and distance is also a problem which is
no problem in e-commerce because one can see sites all over the world without moving out of
the house.
5. Goods are available at cheaper price because there are lot of economies of space, rent, interest
to the seller Further, he manages with much lesser number of outlets and cost of marketing is
reduced. Part of these savings is passed on to consumer and therefore, he gets the products
cheaper than from conventional shops/departmental stores, grocers etc.
6. It helps to globalize retail trading. One can buy things without geograp.
Which Pirl ofiherenrie cinardad an acthm Feinia DI 3rd 4 Ideatry .pdf
Which Pirl ofiherenrie cinardad an acthm Feinia DI 3rd 4 Ideatry the streatuselabeled 5.
lrympkleinirah E) dendrthK SS res dendrime spines ng Fgure ti-2 The Nene Action Paureal Use
neare 12-2 ss auner eke ilacoresteac 6) Which af the graihehows ween pa ve dpularizarkn b
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Fapi, shows when membrane piterkialapproaches potanium quilhnus po metal? A) D)4 9) Which
point of the Faph sheen when potassimianouinou earedssdimininnem? 014
Solution
Answers: (Figure 12.2)
(6) (A) 1
In the graph above. area 1 shows when passive depolarization is occuring.
(7) (B) 2
Area 2 in the graph shows when there is a sudden rush of soudium ions.
(8) (D) 4
Area 4 of the graph shows membrane potential approaches the potassium equilibrium potential.
(9) (C) 3
Area 3 of the graph shows when potassium ion outflow exceeds sodium ion inflow..
Which of the following observations is supporting evidence that all p.pdf
Which of the following observations is supporting evidence that all populations are linked by
descent with modification? living animals all conduct transcription the same way fossils of
early horses show that they stood on three toes different geological strata contain species not
alive today today, pest species are constantly changing genetically all of these observations are
evidence that evolution occurs White eyes is caused by an X-linked recessive allele in fruit flies.
Suppose that we found the following offspring that resulted from one mating between two flies,
called Fred and Ethel: 25% while eye females, 25% red eye females, 25% white eye mates, 25%
red eye mates. What can be concluded about Fred and Ethel? Fred must have had red eyes Ethel
must have been heterozygous Ethel might have had white eyes Fred must have had alleles for
both red and white eyes none of the above is true
Solution
5) Correct answer is e) all of these observations are evidences that evolution occurs.
6) Correct answer is c) Ethel must have been heterozygous..
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