What is the medium in bottom picture Is selective ... Why What doe.pdf
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What is the medium in bottom picture? Is selective ... Why? What does it grow? Is differential ...Why? What does the differential property reveal about the bacteria growing there? Solution The medium in bottom pictures a gel made up of agar. It is a selective & differential media. It is a selective because the agar plate in this in the diagram shows that only specific bacteria can be grown and resist the growth of all other bacterial species. EXAMPLE: Mannitol sat agar, Mac conkey agar, Eosin mythylene blue agar etc..... It grows Staphylococcus aureus. USES:-Differential media is used to check water pollution, diary industries and in medical science . The differential property reveal that the bacteria grow there is mainly Escherichia coli. It can be used to know the amount of bacterial colonies present in water, air etc....,.
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Answer 12:
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Solution
Answers are :
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contaminated so.
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To avoid contamination to grow unwanted cultures and spoiling the medium components.
To know the perfect study of the unknown characteristics, contamination shouldn’t be seen .
To avoid infecting the person who handling the sample, some may cause severe infections and
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To obtain the pure culture techniques to study the characteristics of the cultures.
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Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of
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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
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gene) produces the following F1 counts: The mode of inheritance for each phenotype is: Peanuts
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Recessive. Peanuts = Autosomal Dominant. Carmelaoid = Autosomal Recessive. Peanuts =
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Solution
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As
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Solution
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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.
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During directional solidification, the solute atom concentration increases with increase in
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leaves and lemon grass oil extracts were evaluated. Oil extracts of neem leaves and
lemon grass were obtained by solvent extraction method using hexane and ethanol.
Antimicrobial activity screening of plants’ oil extracts were conducted using agar well
diffusion method and the oil extracts were tested against three gram negative bacteria
(Pseudomonas aeruginosa, Klebsiella specie, Escherichia coli), one gram positive
bacteria (Staphylococcus aureus) and two fungi (Candida albicans, Rhizopus specie).
Phytochemical components of the ethanolic oil extracts were anthocyanin and
betacyanin; quinones; terpenoids and acid for lemon grass. In addition to other
phytochemicals present in lemon grass ethanolic oil extract, neem ethanolic oil extracts
tested positive to flavonoids. Lemon grass oil extract shows high activity against
Escherichia coli, Staphylococcus aureus and Candida albicans which are
representative of the three categories of microorganisms considered. Neem leaves oil
extracts have relatively low activity against most of the selected microorganisms.
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Microorganisms can be studied through cultivation, isolation, purification, and characterization techniques. Samples are collected from various sources and streaked onto agar plates using aseptic technique to prevent contamination. Single colonies are picked and restreaked to obtain a pure culture of a single microbial species. The isolated microbes are then characterized through morphological examination, staining techniques, and biochemical tests to identify the microorganism.
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Microorganisms can be studied through cultivation, isolation, purification, and characterization techniques. Samples are collected from various sources and streaked onto agar plates using aseptic technique to prevent contamination. Single colonies are then purified by streaking and incubating to obtain a pure culture of a single microbial species. The isolated microbe can be characterized through morphological analysis under a microscope, biochemical testing of metabolic activities, and other identification methods. This allows scientists to determine the type of microorganism being studied.
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2. Agar is commonly added to culture media to solidify it for growing bacteria on solid surfaces like Petri dishes and slants. Agar solidifies the media without degrading.
3. Pure cultures of microbes are necessary for most bacteriological work and are obtained using methods like streak plating that isolate single colonies from mixed cultures.
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Excoecaria agallocha L. is an important medicinal plant inhabited in mangrove regions. Early researches focused on antimicrobial activity of leaves of concerned plant with various solvents among which ethanol, chloroform and methanol were Used. Ella. Sai Kumar "Evaluation of Antimicrobial Activity of Excoecaria Agallocha L" Published in International Journal of Trend in Scientific Research and Development (ijtsrd), ISSN: 2456-6470, Volume-4 | Issue-3 , April 2020, URL: https://www.ijtsrd.com/papers/ijtsrd30276.pdf Paper Url :https://www.ijtsrd.com/biological-science/microbiology/30276/evaluation-of-antimicrobial-activity-of-excoecaria-agallocha-l/ella-sai-kumar
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Considering the cultures used to inoculate each medium in this laboratory, how many different
microbial types should you expect to see in/on each medium? Which medium was most difficult
for you to transfer from? which medium was most difficult for you to inoculate? Explain your
difficulties. What is the purpose of flaming in the aseptic techniques? Provide 3 (three) reasons
why the use of aseptic techniques is essential when handling microbial cultures. Methoos of
isolation What is a microbial colony? Can a pure culture by prepared from a mixd broth or
mixed agar slant? Explain.
Solution
Answers are :
When the cultures are inoculated onto the surface of the medium, different colonies are
obtained, based on the mother culture we have chosen colonies are obtained. Until unless it is not
contaminated so.
Preparation of agar slants is difficult to transfer, and inoculate, because it needs to place
correct angle while solidifying, less surface area. Though among the medium we can say some
difficult but not so rather difficult. Nutrient broth is also difficult to transfer because more prone
of contamination and determination of the bacterial growth is also difficult.
Purpose of flaming during aseptic transfer :- Flaming causes direct burning of the cultures
samples which stick onto the inoculating loop to avoid from contamination, which technique is
called Incineration i.e., Flame sterilization.
Aseptic techniques are safe transfer of culture onto the fresh medium this is very essential
because :-
To avoid contamination to grow unwanted cultures and spoiling the medium components.
To know the perfect study of the unknown characteristics, contamination shouldn’t be seen .
To avoid infecting the person who handling the sample, some may cause severe infections and
death of the investigator.
To obtain the pure culture techniques to study the characteristics of the cultures.
Microbial colony means, group of the similar cells which shows same characteristic features ,
which obtained on the surface of the solid agar medium. These are varied in shapes and surface
but shows same morphological and physiological characteristics.
Yes, we can prepare pure culture from mixed sample either broth or solid agar, in case of
broth prepare serial dilution of the sample to obtain the pure cultures. In case of the solid slant
prepare quadrant streaking on the fresh agar plate..
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1. The document discusses the importance of microbiologists developing strong bench skills through dedicated practice in order to properly identify pathogens and separate bacterial mixtures.
2. It emphasizes how early microbiologists had limited resources but were still able to isolate and differentiate bacteria through techniques like streak plating and using differential and selective media.
3. The author stresses that modern microbiologists should not neglect hands-on work and rely only on technicians, as skill in laboratory techniques is essential to distinguish bacterial species and identify true pathogens.
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The document discusses meningococcal meningitis, a disease caused by the bacterium Neisseria meningitidis. It enters the body through contact and symptoms take 3-7 days to present, including fever, rash, vomiting, and altered mental status. The disease is most prevalent in children under two. Treatments are available to combat the microbe and its impact on human health.
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Here is a summary of the key points about blood hemolysis testing from the passage:
- Blood hemolysis testing is used to evaluate potential probiotic properties of bacterial isolates. It involves streaking overnight cultures of isolates on blood agar plates.
- The plates are then incubated at 37°C for 24 hours and examined for signs of hemolysis. Hemolysis refers to the breakdown of red blood cells.
- There are three possible outcomes when bacteria are tested on blood agar: alpha hemolysis causes greening of the agar around colonies, beta hemolysis causes clearing of the agar, and gamma hemolysis shows no lysis of red blood cells.
- This allows researchers to determine whether bacterial isolates
Wagner College Forum for Undergraduate Research, Vol. 14 No. 2
The Spring 2016 issue contains papers by Joseph V. Agro, Kendra Best, Katie Murphy, Jessica Catanzaro, Nicole Bianco, Sandra G. Minchala, Karina Cusumano, Avika Sagwal, Alyssa Thompson and Juliana R. Ohanian.
Clinical Micro.pptx
The document provides an overview of the process for identifying bacteria in a clinical microbiology laboratory. Samples are first examined microscopically and using gram staining to look for bacteria and determine if they are gram-positive or gram-negative. If bacteria are present, the sample is cultured to grow colonies. Technicians then isolate individual colonies and use their morphology, biochemical tests, and antibiotic susceptibility testing to fully identify the pathogenic bacteria causing infection and determine the most effective antibiotic treatment. The entire process from receiving a sample to identifying bacteria typically takes around 48 hours.
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Considering the cultures used to inoculate each medium in this labora.pdf
Considering the cultures used to inoculate each medium in this labora.pdf
Wagner College Forum for Undergraduate Research, Vol. 14 No. 2
Wagner College Forum for Undergraduate Research, Vol. 14 No. 2
More from arracollection
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..
Let P[x] denote the vector space of all polynomials. Let E be the set.pdf
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.
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
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..
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Discuss the relationship between gap rule gene mutations, pair-rule .pdf
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You will hear from Liz Willett, the Head of Nonprofits, and hear about what Walmart is doing to help nonprofits, including Walmart Business and Spark Good. Walmart Business+ is a new offer for nonprofits that offers discounts and also streamlines nonprofits order and expense tracking, saving time and money.
The webinar may also give some examples on how nonprofits can best leverage Walmart Business+.
The event will cover the following::
Walmart Business + (https://business.walmart.com/plus) is a new shopping experience for nonprofits, schools, and local business customers that connects an exclusive online shopping experience to stores. Benefits include free delivery and shipping, a 'Spend Analytics” feature, special discounts, deals and tax-exempt shopping.
Special TechSoup offer for a free 180 days membership, and up to $150 in discounts on eligible orders.
Spark Good (walmart.com/sparkgood) is a charitable platform that enables nonprofits to receive donations directly from customers and associates.
Answers about how you can do more with Walmart!"
Wound healing PPT
This document provides an overview of wound healing, its functions, stages, mechanisms, factors affecting it, and complications.
A wound is a break in the integrity of the skin or tissues, which may be associated with disruption of the structure and function.
Healing is the body’s response to injury in an attempt to restore normal structure and functions.
Healing can occur in two ways: Regeneration and Repair
There are 4 phases of wound healing: hemostasis, inflammation, proliferation, and remodeling. This document also describes the mechanism of wound healing. Factors that affect healing include infection, uncontrolled diabetes, poor nutrition, age, anemia, the presence of foreign bodies, etc.
Complications of wound healing like infection, hyperpigmentation of scar, contractures, and keloid formation.
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Presented at the Stoke Newington Literary Festival on 9th June 2024
www.StokeNewingtonHistory.com
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Chapter wise All Notes of First year Basic Civil Engineering.pptx
Chapter wise All Notes of First year Basic Civil Engineering
Syllabus
Chapter-1
Introduction to objective, scope and outcome the subject
Chapter 2
Introduction: Scope and Specialization of Civil Engineering, Role of civil Engineer in Society, Impact of infrastructural development on economy of country.
Chapter 3
Surveying: Object Principles & Types of Surveying; Site Plans, Plans & Maps; Scales & Unit of different Measurements.
Linear Measurements: Instruments used. Linear Measurement by Tape, Ranging out Survey Lines and overcoming Obstructions; Measurements on sloping ground; Tape corrections, conventional symbols. Angular Measurements: Instruments used; Introduction to Compass Surveying, Bearings and Longitude & Latitude of a Line, Introduction to total station.
Levelling: Instrument used Object of levelling, Methods of levelling in brief, and Contour maps.
Chapter 4
Buildings: Selection of site for Buildings, Layout of Building Plan, Types of buildings, Plinth area, carpet area, floor space index, Introduction to building byelaws, concept of sun light & ventilation. Components of Buildings & their functions, Basic concept of R.C.C., Introduction to types of foundation
Chapter 5
Transportation: Introduction to Transportation Engineering; Traffic and Road Safety: Types and Characteristics of Various Modes of Transportation; Various Road Traffic Signs, Causes of Accidents and Road Safety Measures.
Chapter 6
Environmental Engineering: Environmental Pollution, Environmental Acts and Regulations, Functional Concepts of Ecology, Basics of Species, Biodiversity, Ecosystem, Hydrological Cycle; Chemical Cycles: Carbon, Nitrogen & Phosphorus; Energy Flow in Ecosystems.
Water Pollution: Water Quality standards, Introduction to Treatment & Disposal of Waste Water. Reuse and Saving of Water, Rain Water Harvesting. Solid Waste Management: Classification of Solid Waste, Collection, Transportation and Disposal of Solid. Recycling of Solid Waste: Energy Recovery, Sanitary Landfill, On-Site Sanitation. Air & Noise Pollution: Primary and Secondary air pollutants, Harmful effects of Air Pollution, Control of Air Pollution. . Noise Pollution Harmful Effects of noise pollution, control of noise pollution, Global warming & Climate Change, Ozone depletion, Greenhouse effect
Text Books:
1. Palancharmy, Basic Civil Engineering, McGraw Hill publishers.
2. Satheesh Gopi, Basic Civil Engineering, Pearson Publishers.
3. Ketki Rangwala Dalal, Essentials of Civil Engineering, Charotar Publishing House.
4. BCP, Surveying volume 1
LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
How to Setup Warehouse & Location in Odoo 17 Inventory
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