Which of the following observations is supporting evidence that all p.pdf
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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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Clouded sulfur butterflies are common in our area and can be either .pdf
Clouded sulfur butterflies are common in our area and can be either yellow or white. While most
individuals in the total population are yellow, it is known that a single autosomal gene controls
color and that the white allele is completely dominant over the yellow allele. However, when
analyzing crosses we find the following results:
Yellow male x yellow female = 100% yellow males:100% yellow females
Yellow male x yellow female = 100% yellow males:50% white females:50% yellow females
Yellow male x white female = 100% yellow males:100% white females
Yellow male x yellow female = 100% yellow males:100% white females
Yellow male x white female = 100% yellow males:75% white females:25% yellow females
What do these results suggest about the genetic inheritance patterns of this trait in clouded sulfur
butterflies?
Solution
Colias philodice, the common yellow butterfly, also sometimes called the clouded sulphur
butterfly.
The different forms of a gene that are found at a locus along a given chromosome are known as
alleles. Diploid organisms have two alleles for each autosomal gene; one inherited from the
mother, one inherited from the father.
If the phenotype associated with a gene is observed, in an individual having only one copy, the
allele is said to be autosomal dominant. The phenotype will be observed whether the individual
has one copy of the allele (heterozygous) or has two copies of the allele (homozygous).
Females carrying a dominant allele w+ at a single autosomal locus are white instead of the
species-specific red, orange, or yellow displayed by their homozygous recessive ww sisters and
by all males.
The cross is outlined below: We know that females are XX homogametic males are XY
heterogametic.
P generation:
Female w+w+ (white) × Male w (yellow)
F1 generation:
Female w+w (white) × Male w+ (white)
The fact that all the F1 progeny had white color wings, suggests that the wild-type (w+) allele is
dominant over the yellow allele..
Linked inheritance
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
Recommended
Clouded sulfur butterflies are common in our area and can be either .pdf
Clouded sulfur butterflies are common in our area and can be either yellow or white. While most
individuals in the total population are yellow, it is known that a single autosomal gene controls
color and that the white allele is completely dominant over the yellow allele. However, when
analyzing crosses we find the following results:
Yellow male x yellow female = 100% yellow males:100% yellow females
Yellow male x yellow female = 100% yellow males:50% white females:50% yellow females
Yellow male x white female = 100% yellow males:100% white females
Yellow male x yellow female = 100% yellow males:100% white females
Yellow male x white female = 100% yellow males:75% white females:25% yellow females
What do these results suggest about the genetic inheritance patterns of this trait in clouded sulfur
butterflies?
Solution
Colias philodice, the common yellow butterfly, also sometimes called the clouded sulphur
butterfly.
The different forms of a gene that are found at a locus along a given chromosome are known as
alleles. Diploid organisms have two alleles for each autosomal gene; one inherited from the
mother, one inherited from the father.
If the phenotype associated with a gene is observed, in an individual having only one copy, the
allele is said to be autosomal dominant. The phenotype will be observed whether the individual
has one copy of the allele (heterozygous) or has two copies of the allele (homozygous).
Females carrying a dominant allele w+ at a single autosomal locus are white instead of the
species-specific red, orange, or yellow displayed by their homozygous recessive ww sisters and
by all males.
The cross is outlined below: We know that females are XX homogametic males are XY
heterogametic.
P generation:
Female w+w+ (white) × Male w (yellow)
F1 generation:
Female w+w (white) × Male w+ (white)
The fact that all the F1 progeny had white color wings, suggests that the wild-type (w+) allele is
dominant over the yellow allele..
Linked inheritance
Breaking down Biology into simpler bits is the most effective way to learn hence this presentation aims to simplify the concept of 'Linked Inheritance' which makes understanding Inheritance better.
In 1905, the French biologist Lucien Cuénot discovered an interestin.pdf
In 1905, the French biologist Lucien Cuénot discovered an interesting mutant mouse displaying a
yellow coat color. He bred the mouse and its progeny for several generations but could never
obtain a pure breeding stock of yellow mice! Normal mice have a coat color called Agouti which
appears gray. When he crossed pure breeding gray Agouti mice with the yellow mouse, half the
progeny had the gray Agouti coat and half had the light yellow coat. When he crossed the F1
yellow mice with each other, he found that 1/3 of the F2 progeny had the gray Agouti coat and
2/3 had the yellow coat. When he crossed the F2 yellow progeny with pure breeding Agouti
mice, he found again that they produced 50% gray Agouti mice and 50% yellow mice.
Based on these results, you would conclude that relative to the Agouti coat color, the Yellow
coat color is:
A. Completely Dominant
B. Incompletely Dominant
C. Co-dominant
D. Incompletely Penetrant
E. Recessive
Cuénot was never able to produce a pure breeding strain of yellow mice even after trying several
crosses for several years. Provide a one-sentence explanation for why Cuénot was unable to
obtain yellow mice but not a pure breeding stock of yellow mice.
Solution
(A) It is complete dominant. As per definition, dominant allele shows phenotype in homozygous.
Genes are located on chromosomes and consist of DNA. They are passed from parents to their
offspring through reproduction. Genes also contain information about a specific characteristic or
trait and can either be dominant or recessive. Each gene has a designated place on every
chromosome, called a locus. All copies of a gene are not identical and alternative forms of a gene
is called alleles that lead to the alternative or different form of one trait. Alleles are helpful in
identification of the two members of a gene pair, which produce opposite contrasting
phenotypes, e.g., b is an allele of B and vice versa. When the two alleles of a gene are identical,
the individual is homozygous for that trail, and on the other hand, if there are two different
alleles, the individual is heterozygous. A homozygous pair can be either dominant (AA, BB) or
recessive (aa, bb). Heterozygous pairs are made up of one dominant and one recessive allele (Aa,
Bb). In heterozygous individuals, only one allele, the dominant one, is able to express itself,
while the other allele, the recessive, is hidden but still present. The dominant genes are denoted
by upper case letters and recessive genes are denoted by lower case letters. The word genotype
was created to identify genes of an individual and phenotype for the external appearance of the
trait and genes.Phenotype and genotype are terms used to describe the difference between the
visible expressions of the trait versus the actual gene makeup. An individual, which expresses a
dominant trait may carry a recessive allele, but the recessive expression is hidden by its dominant
partner. Mendel\'s observations from these experiments can be summar.
Sex linked inheritance aand interaction of genes
This was group presentation... with lots of good & memorable time....
chromosomal basis of inheritance
in this presentation how chromosomes are basis of inheritance is explained....i have combined 3 or 4 presentations..i hope so it will be helpfull
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers.docx
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers
15.1 Mendelian Inheritance and chromosome theory______________________________
1. Thomas Hunt Morgan identified the first solid evidence associating a specific___gene_____ on a specific chromosome.
2. Why did Morgan choose Drosophila as his experimental organism? (List 3 reasons)
They reproduce quickly, a new generation of adults forms every two weeks!
Prolific (single matting = hundreds of offspring)
Only 4 chromosome pairs = simple genetics
3A. The normal phenotype for a character (phenotype most common in nature) is called___wild type_____.
3B. Provide at least two examples this phenotype in Drosophila.
Red eyes
gray body
4A. An alternative phenotype for a character (phenotype not common in nature) is called____mutant____.
4B. Provide at least two examples of this phenotype in Drosophila.
White eyes
Black body
5. Morgan and his students invented a notation to symbolize alleles in Drosophila that differed from the notation Mendel used to represent alleles. Describe how Morgan’s Drosophila characters are named and the symbol used for the allele type.
You can think of Morgan’s wild-type allele as equivalent to the dominant allele in Mendel’s naming system. And you can think of Morgan’s mutant allele as equivalent to Mendel’s recessive allele. So whenever you have one wild-type allele and one mutant allele, that organism will have the wild-type phenotype.
Mendel
Used the first letter of the dominant character name to represent the dominant allele. This letter was capitalized for the dominant allele, lower case for the recessive allele
Ex: purple flower allele dominant to white flower allele: P= dominant allele, p=recessive allele
Morgan
-Used the first letter of the mutant character name to represent the wild-type allele. The letter for the wild-type and mutant alleles are both lower case. The wild-type allele gets a + sign and the mutant does not.
Ex: red eye allele wild-type to white eye mutant: w+ = wild-type allele, w= recessive allele.
6. Morgan and his students invented a notation to symbolize alleles in Drosophila. Which of the following genotypes would produce a fly that is wild-type for eye color (red vs. white eyes)?
w+ w+
w+ w
w w
w+w+ and w+w
7. Morgan performed an experiment that yielded a 3:1 ratio of offspring in the F2 generation; however, only the males of this F2 generation had white eyes.
White- eye females DO EXIST, so why were only white-eye males observed in this cross?
Complete the following chart and describe why Morgan observed these results and how it allowed him to conclusively determine that the gene for eye color was located on the X chromosome.
Morgan’s experiment (see Figure 15.4) Use notation developed by Morgan and his students
If the gene for eye color is located on the X chromosome….
P generation
Red eye Female genotype for eye color = w+w+
White eye Male genotype for eye color= w
F1 generatio.
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..
More Related Content
Similar to Which of the following observations is supporting evidence that all p.pdf
In 1905, the French biologist Lucien Cuénot discovered an interestin.pdf
In 1905, the French biologist Lucien Cuénot discovered an interesting mutant mouse displaying a
yellow coat color. He bred the mouse and its progeny for several generations but could never
obtain a pure breeding stock of yellow mice! Normal mice have a coat color called Agouti which
appears gray. When he crossed pure breeding gray Agouti mice with the yellow mouse, half the
progeny had the gray Agouti coat and half had the light yellow coat. When he crossed the F1
yellow mice with each other, he found that 1/3 of the F2 progeny had the gray Agouti coat and
2/3 had the yellow coat. When he crossed the F2 yellow progeny with pure breeding Agouti
mice, he found again that they produced 50% gray Agouti mice and 50% yellow mice.
Based on these results, you would conclude that relative to the Agouti coat color, the Yellow
coat color is:
A. Completely Dominant
B. Incompletely Dominant
C. Co-dominant
D. Incompletely Penetrant
E. Recessive
Cuénot was never able to produce a pure breeding strain of yellow mice even after trying several
crosses for several years. Provide a one-sentence explanation for why Cuénot was unable to
obtain yellow mice but not a pure breeding stock of yellow mice.
Solution
(A) It is complete dominant. As per definition, dominant allele shows phenotype in homozygous.
Genes are located on chromosomes and consist of DNA. They are passed from parents to their
offspring through reproduction. Genes also contain information about a specific characteristic or
trait and can either be dominant or recessive. Each gene has a designated place on every
chromosome, called a locus. All copies of a gene are not identical and alternative forms of a gene
is called alleles that lead to the alternative or different form of one trait. Alleles are helpful in
identification of the two members of a gene pair, which produce opposite contrasting
phenotypes, e.g., b is an allele of B and vice versa. When the two alleles of a gene are identical,
the individual is homozygous for that trail, and on the other hand, if there are two different
alleles, the individual is heterozygous. A homozygous pair can be either dominant (AA, BB) or
recessive (aa, bb). Heterozygous pairs are made up of one dominant and one recessive allele (Aa,
Bb). In heterozygous individuals, only one allele, the dominant one, is able to express itself,
while the other allele, the recessive, is hidden but still present. The dominant genes are denoted
by upper case letters and recessive genes are denoted by lower case letters. The word genotype
was created to identify genes of an individual and phenotype for the external appearance of the
trait and genes.Phenotype and genotype are terms used to describe the difference between the
visible expressions of the trait versus the actual gene makeup. An individual, which expresses a
dominant trait may carry a recessive allele, but the recessive expression is hidden by its dominant
partner. Mendel\'s observations from these experiments can be summar.
Sex linked inheritance aand interaction of genes
This was group presentation... with lots of good & memorable time....
chromosomal basis of inheritance
in this presentation how chromosomes are basis of inheritance is explained....i have combined 3 or 4 presentations..i hope so it will be helpfull
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers.docx
Study Guide Chapter 15 -Chromosomal Basis of Inheritance-Answers
15.1 Mendelian Inheritance and chromosome theory______________________________
1. Thomas Hunt Morgan identified the first solid evidence associating a specific___gene_____ on a specific chromosome.
2. Why did Morgan choose Drosophila as his experimental organism? (List 3 reasons)
They reproduce quickly, a new generation of adults forms every two weeks!
Prolific (single matting = hundreds of offspring)
Only 4 chromosome pairs = simple genetics
3A. The normal phenotype for a character (phenotype most common in nature) is called___wild type_____.
3B. Provide at least two examples this phenotype in Drosophila.
Red eyes
gray body
4A. An alternative phenotype for a character (phenotype not common in nature) is called____mutant____.
4B. Provide at least two examples of this phenotype in Drosophila.
White eyes
Black body
5. Morgan and his students invented a notation to symbolize alleles in Drosophila that differed from the notation Mendel used to represent alleles. Describe how Morgan’s Drosophila characters are named and the symbol used for the allele type.
You can think of Morgan’s wild-type allele as equivalent to the dominant allele in Mendel’s naming system. And you can think of Morgan’s mutant allele as equivalent to Mendel’s recessive allele. So whenever you have one wild-type allele and one mutant allele, that organism will have the wild-type phenotype.
Mendel
Used the first letter of the dominant character name to represent the dominant allele. This letter was capitalized for the dominant allele, lower case for the recessive allele
Ex: purple flower allele dominant to white flower allele: P= dominant allele, p=recessive allele
Morgan
-Used the first letter of the mutant character name to represent the wild-type allele. The letter for the wild-type and mutant alleles are both lower case. The wild-type allele gets a + sign and the mutant does not.
Ex: red eye allele wild-type to white eye mutant: w+ = wild-type allele, w= recessive allele.
6. Morgan and his students invented a notation to symbolize alleles in Drosophila. Which of the following genotypes would produce a fly that is wild-type for eye color (red vs. white eyes)?
w+ w+
w+ w
w w
w+w+ and w+w
7. Morgan performed an experiment that yielded a 3:1 ratio of offspring in the F2 generation; however, only the males of this F2 generation had white eyes.
White- eye females DO EXIST, so why were only white-eye males observed in this cross?
Complete the following chart and describe why Morgan observed these results and how it allowed him to conclusively determine that the gene for eye color was located on the X chromosome.
Morgan’s experiment (see Figure 15.4) Use notation developed by Morgan and his students
If the gene for eye color is located on the X chromosome….
P generation
Red eye Female genotype for eye color = w+w+
White eye Male genotype for eye color= w
F1 generatio.
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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
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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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