Help if you can Genetics question. thanks beforehand! You continue y.pdf
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Help if you can Genetics question. thanks beforehand! You continue your experiments with the black mutation on chromosome II. You cross the reciprocal translocation strain to the black pure line to generate F_1 flies that are both translocation heterozygotes and Bb heterozygotes. As before, you testcross F_1 males and females in two separate experiments. What can you conclude from these results? Select all statements that are true. The F_2 progeny in both experiments contain recombinants. The F_2 progeny contain recombinants only when F?_1 females are testcrossed. None of the F_2 progeny classes require crossing over. Some of the F_2 progeny classes require crossing over, others do not. Chromosome II is one of the chromosomes involved in the reciprocal balanced translocation. Solution Since the recombinants can be formed only when there is cross over in atleast one the parent; the F2 progeny from cross between F1 males and pure black females do not have recombinants. Only the F2 progeny from the cross between F1 females and pure black males have the recombinants as cross over occurs in F1 female. The F2 progeny has both parental types and recombinants; formation of parental types does not require the crossover. Therefore, some of the F2 progeny class require cross over while others do not. Since the gene is present on the chromosome II, it is involved in the reciprocal translocation. Correct statements are 1: The F2 progeny has recombinants only when F1 females are testcrossed 2: Some of the F2 progeny classes require crossing over; others donot. 3: Chromosome II is one of the chromosome for reciprocal translocation..
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AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf
Answer:
Answer:
Part B).
The F2 progeny contain recombinants only when F1 females are test crossed;
Some of the F2 progeny classes require crossing over; others do not.
Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio.
Explanation:
To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing
over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing
over.
A balanced, or chromosomal translocation is a condition in which part of a chromosome has
broken off and reattached in another location. In other words, it means that sections of two
chromosomes have switched places.
Solution
Answer:
Answer:
Part B).
The F2 progeny contain recombinants only when F1 females are test crossed;
Some of the F2 progeny classes require crossing over; others do not.
Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio.
Explanation:
To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing
over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing
over.
A balanced, or chromosomal translocation is a condition in which part of a chromosome has
broken off and reattached in another location. In other words, it means that sections of two
chromosomes have switched places..
Figare 11-1- When these F1 offspring self-pollinated- the next generat.pdf
Figare 11.1. When these F 1 offspring self-pollinated, the next generation ( F 2 ) showed both
traits in a 3.1 ratio. The trait shown in the F 1 was always the more commen trait in the F 2 , with
3 out of every 4 offspring expressing it. See Figure 11.2 . Figure 11.2. Essentially, Mendel's
explanation for what he saw was: 1. Traits are controlled by what he called "factors" and which
we now call genes. 2. An organism has a pair of genes for each trait. We now know that cach
gene is located on one of two similar (homologous) chromosomes. 3. An organism mandomly
inherits one gene from each of its parents' pair of genes. Thus, its own pair consists of one gene
that is puternal and one gene that is maternal. 4. The pairs of genes can be identical, or they can
be different. If they are the same, the organism is said to be homoxygous; if different, the
organism is said to be heterozygous. Different versions of genes are called alleles. Thus, an
organism with two different alleles is heterozygous and one with identical alleles is homozygous.
5. Contrasting alleles can vary in their effect. Since his F 1 generation resembled only one of the
parents, while both of the traits present in the parents reappeared in the F 2 generation, he
concluded that the factons do not blend or disappear. He defined a rule of dominance to explain
this: The allele that determines the appearance of a heterozygote is called the dominant allele.
The one that is hidden is the recessive allele. The visible outcome of the genes is the phenotype,
while the actual genetic outcome is the genotype. The 3 : 1 ratio in the F 2 generation refers
specifically to 3 dominant traits being expressed for every one recessive. Today you will
examine several traits that illustrate Mendel's basic explanations for patterns of inheritance. You
will also see some ways in which these basic explanations can be used to study human genetics
and to understand how to do pedigree analysis. EXERCISE 1. REVIEW OF BASIC GENETICS
TERMS PROCEDURE Using your lecture materials and textbook as a reference, review the
following genetic situation and answer the questions that follow. When a true-breeding brown
mink is crossed with a true-breeding silver-blue mink they produce offspring all with the same
phenotype: brown. When these F 1 mink were crossed amongst themselves, they produced 47
brown animals and 15 silver-blue animals ( F 2 generation). 1. What does the gene being
discussed code for? 2. What allelles for this gene are being discussed? 3. Which one of these
alleles is dominant and how do you know it is dominant? 4. Which one of the alleles is recessive
and how do you know it is recessive? 5. What does F 1 stand for? Explain what the F 1
generation is. F = childien 1 = Grotaeneration 6. What is the meaning of the word phenotype? In
this particular case, what was the phenotype of the F 1 ? 7. If you were studying genotypes, what
would you be looking at? In general, what different genotypes can you find i.
5- Note that questions 5 and 6 are -linked-- Question 5 is the first s.pdf
5. Note that questions 5 and 6 are "linked". Question 5 is the first step, crossing two "true
breeding parents (P) in order to create the F 1 offspring that are double heterozygotes. Question 6
is the second step, crossing two F 1 double heterozygotes to create the F 2 offspring. In Mendel's
experiments, the pod color gene and the seed shape gene were located on different chromosomes
(in other words, they were not linked. They assort independently). Green pods (G) are dominant
over yellow pods ( g ) and tall stems ( T ) are dominant over dwarf stems ( t ) . You are given
two true-breeding plants (homozygous at both genes). In the parental (P) generation, one parent
has green pods and dwarf stems and the second parent has yellow pods and tall stems. Your key:
G is the dominant allele, it stands for what trait? g is the recessive allele, it stands for what trait?
T is the dominant allele, it stands for what trait? t is the recessive allele, it stands for what trait?
Genotype of true breeding Parent 1: Genotype(s) of Parent 1's gametes (also write these above
the columns): Genotype of true breeding Parent 2: Genotype(s) of Parent 2 's gametes (also write
these to the left of the rows): Proportion and phenotype(s) of offspring: 6. Note that questions 5
and 6 are "linked". Question 5 is the first step, crossing two "true breeding" parents ( P ) in order
to create the F 1 offspring that are double heterozygotes. Question 6 is the second step, crossing
two F 1 double heterozygotes to create the F 2 offspring. As in question 5 , the pod color gene
and the seed shape gene were located on different chromosomes (in other words, they were not
linked. They assort independently). Green pods (G) are dominant over yellow pods ( g ) and tall
stems ( T ) are dominant over dwarf stems (f). If you cross an F 1 from question 5 (above) with
another F 1 from question 5 , this is a dihybrid cross. Remember that you are dealing with two
genes. Your key: G is the dominant allele, it stands for what trait? g is the recessive allele, it
stands for what trait? T is the dominant allele, it stands for what trait? t is the recessive allele, it
stands for what trait? Genotype of Parent 1 (who is also the F 1 offspring from question 5):
Genotype(s) of Parent 1's gametes (also write these above the columns): Genotype of Parent 2
(who is also the F 1 offspring from question 5): Genotype(s) of Parent 2 's gametes (also write
these to the left of the rows): Four different phenotypes of offspring in this dihybrid cross: Ratio
of phenotypes of offspring that we see in a dihybrid cross:.
Recommended
AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf
Answer:
Answer:
Part B).
The F2 progeny contain recombinants only when F1 females are test crossed;
Some of the F2 progeny classes require crossing over; others do not.
Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio.
Explanation:
To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing
over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing
over.
A balanced, or chromosomal translocation is a condition in which part of a chromosome has
broken off and reattached in another location. In other words, it means that sections of two
chromosomes have switched places.
Solution
Answer:
Answer:
Part B).
The F2 progeny contain recombinants only when F1 females are test crossed;
Some of the F2 progeny classes require crossing over; others do not.
Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio.
Explanation:
To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing
over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing
over.
A balanced, or chromosomal translocation is a condition in which part of a chromosome has
broken off and reattached in another location. In other words, it means that sections of two
chromosomes have switched places..
Figare 11-1- When these F1 offspring self-pollinated- the next generat.pdf
Figare 11.1. When these F 1 offspring self-pollinated, the next generation ( F 2 ) showed both
traits in a 3.1 ratio. The trait shown in the F 1 was always the more commen trait in the F 2 , with
3 out of every 4 offspring expressing it. See Figure 11.2 . Figure 11.2. Essentially, Mendel's
explanation for what he saw was: 1. Traits are controlled by what he called "factors" and which
we now call genes. 2. An organism has a pair of genes for each trait. We now know that cach
gene is located on one of two similar (homologous) chromosomes. 3. An organism mandomly
inherits one gene from each of its parents' pair of genes. Thus, its own pair consists of one gene
that is puternal and one gene that is maternal. 4. The pairs of genes can be identical, or they can
be different. If they are the same, the organism is said to be homoxygous; if different, the
organism is said to be heterozygous. Different versions of genes are called alleles. Thus, an
organism with two different alleles is heterozygous and one with identical alleles is homozygous.
5. Contrasting alleles can vary in their effect. Since his F 1 generation resembled only one of the
parents, while both of the traits present in the parents reappeared in the F 2 generation, he
concluded that the factons do not blend or disappear. He defined a rule of dominance to explain
this: The allele that determines the appearance of a heterozygote is called the dominant allele.
The one that is hidden is the recessive allele. The visible outcome of the genes is the phenotype,
while the actual genetic outcome is the genotype. The 3 : 1 ratio in the F 2 generation refers
specifically to 3 dominant traits being expressed for every one recessive. Today you will
examine several traits that illustrate Mendel's basic explanations for patterns of inheritance. You
will also see some ways in which these basic explanations can be used to study human genetics
and to understand how to do pedigree analysis. EXERCISE 1. REVIEW OF BASIC GENETICS
TERMS PROCEDURE Using your lecture materials and textbook as a reference, review the
following genetic situation and answer the questions that follow. When a true-breeding brown
mink is crossed with a true-breeding silver-blue mink they produce offspring all with the same
phenotype: brown. When these F 1 mink were crossed amongst themselves, they produced 47
brown animals and 15 silver-blue animals ( F 2 generation). 1. What does the gene being
discussed code for? 2. What allelles for this gene are being discussed? 3. Which one of these
alleles is dominant and how do you know it is dominant? 4. Which one of the alleles is recessive
and how do you know it is recessive? 5. What does F 1 stand for? Explain what the F 1
generation is. F = childien 1 = Grotaeneration 6. What is the meaning of the word phenotype? In
this particular case, what was the phenotype of the F 1 ? 7. If you were studying genotypes, what
would you be looking at? In general, what different genotypes can you find i.
5- Note that questions 5 and 6 are -linked-- Question 5 is the first s.pdf
5. Note that questions 5 and 6 are "linked". Question 5 is the first step, crossing two "true
breeding parents (P) in order to create the F 1 offspring that are double heterozygotes. Question 6
is the second step, crossing two F 1 double heterozygotes to create the F 2 offspring. In Mendel's
experiments, the pod color gene and the seed shape gene were located on different chromosomes
(in other words, they were not linked. They assort independently). Green pods (G) are dominant
over yellow pods ( g ) and tall stems ( T ) are dominant over dwarf stems ( t ) . You are given
two true-breeding plants (homozygous at both genes). In the parental (P) generation, one parent
has green pods and dwarf stems and the second parent has yellow pods and tall stems. Your key:
G is the dominant allele, it stands for what trait? g is the recessive allele, it stands for what trait?
T is the dominant allele, it stands for what trait? t is the recessive allele, it stands for what trait?
Genotype of true breeding Parent 1: Genotype(s) of Parent 1's gametes (also write these above
the columns): Genotype of true breeding Parent 2: Genotype(s) of Parent 2 's gametes (also write
these to the left of the rows): Proportion and phenotype(s) of offspring: 6. Note that questions 5
and 6 are "linked". Question 5 is the first step, crossing two "true breeding" parents ( P ) in order
to create the F 1 offspring that are double heterozygotes. Question 6 is the second step, crossing
two F 1 double heterozygotes to create the F 2 offspring. As in question 5 , the pod color gene
and the seed shape gene were located on different chromosomes (in other words, they were not
linked. They assort independently). Green pods (G) are dominant over yellow pods ( g ) and tall
stems ( T ) are dominant over dwarf stems (f). If you cross an F 1 from question 5 (above) with
another F 1 from question 5 , this is a dihybrid cross. Remember that you are dealing with two
genes. Your key: G is the dominant allele, it stands for what trait? g is the recessive allele, it
stands for what trait? T is the dominant allele, it stands for what trait? t is the recessive allele, it
stands for what trait? Genotype of Parent 1 (who is also the F 1 offspring from question 5):
Genotype(s) of Parent 1's gametes (also write these above the columns): Genotype of Parent 2
(who is also the F 1 offspring from question 5): Genotype(s) of Parent 2 's gametes (also write
these to the left of the rows): Four different phenotypes of offspring in this dihybrid cross: Ratio
of phenotypes of offspring that we see in a dihybrid cross:.
12- Analyzing Coat Color Mutants - Wild type gerbils from Arizona- Mon.docx
12. Analyzing Coat Color Mutants - Wild type gerbils from Arizona, Montana, indiana and Califomia are black. - You have obtained pure-breeding mutant varieties of gerbils that are white or pink from these different places. - You know that each gerbil strain has a mutation in a single autosomal gene. - The table below shows E1 proseny obtained when crossing purebreeding WT and mutants to each other. - Bolded boxes at the top and side indicate the purebreeding parent strains you crossed to each other. - The intersection of these boxes represents the phenotype of the F 1 offspring of a particular cross. For each cross, you can assume 100% of the F1 progeny are of the color indicated. a) What is the purpose of crossing the mutant to wild type (black) and observing F1 progeny phenotypes? b) What is the purpose of crossing the mutants to each other and observing F1 progeny phenotypes? c) Which of the strains cannot be used for the analysis you described in (b)? Why? d) When you cross pure-breeding white from Arizona and indiana to each other, you get black in the F 1 . What does this indicate about the mutations that give rise to white in these 2 strains? e) To further explain your answer in (d), assign genotypes for the cross between the 2 purebreeding strains from Indiana and Arizona and their resulting F 1 . Use the letter A if only alleles of one gene are segregating in the cross, letters A and B if 2 genes are involved, etc. Denote dominant alleles with a capital letter and recessive alleles with a small letter.
.
For a single Mendelian trait, a homozygous dominant individual is cr.pdf
For a single Mendelian trait, a homozygous dominant individual is crossed with a homozygous
recessive individual. Which of the following is true of the individual offspring of the F1
generation from this cross?
Two individuals of the F1 generation from the cross in the previous question are crossed together
(F1 crossed with F1). Which of the following is true about individual offspring from the F2
generation of this cross? Select all that apply.
only the dominant phenotype will be expressed
both dominant and recessive phenotypes will be expressed
None of the other answers fit.
there will be two different genotypes
there will be three different genotypes
only the recessive phenotype will be expresseda.
only the dominant phenotype will be expressedb.
both dominant and recessive phenotypes will be expressedc.
None of the other answers fit.d.
there will be two different genotypese.
there will be three different genotypesf.
only the recessive phenotype will be expressed.
Pls solve this with the Punnett squared table with the given informati.docx
Pls solve this with the Punnett squared table with the given information and why is the 5th answer correct?
A mutant rabbit with pointy teeth has been found in a cave somewhere in Canada. To characterize the mutant at the genetic level, it was crossed to a wild-type rabbit. The progeny (F1) were found to have all normal shaped teeth like wild-type rabbits. Intercrossing the F1 progeny gave a F 2 population with the following ratio: 3/4 with normal shaped teeth 3/16 with intermediately pointy teeth 1/16 with pointy teeth. Which one of the following is the best description of the pointy-toothed phenotype? caused by recessive mutation of a single gene. caused by dominant mutation of a single gene caused by recessive mutation of one gene and dominant mutation of a second gene caused by recessive mutation of one genes and suppressor mutation of a second gene (no phenotype on its own) caused by recessive mutation of one genes and recessive modifier mutation of a second gene.
.
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
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.
1. How is the recipient cell different from before the two cells mat.pdf
1. How is the recipient cell different from before the two cells mated? Phenotypically as well as
genetically.
2. What do F+ cells do?
3. What are all properties of a recipient cell (which receives the F plasmid)?
Solution
1. Recipient cell (F- cell) is different after it mates with the F+ cell. Conjugation/mating is
equivalent to sex (however no new individuals are produced in this case) in that it combines
genetic material from two different individuals into a single genome.It occurs through F pili
which originate from cell membrane and project outward beyond the cell wall. Thus, like sexual
reproduction conjugation produces new genetic combinations. The pilus is essential for
recognition of recipient cell. The sex plasmid passes from the donor F+ to the recipient F– cell
through the conjugation tube, and seperates after the event is over and now called as
exconjugates. F- cell after matinh with F+ cell, now becomes F+ and can act as donor during
next conjugation event.
After conjugation event, recipient cell now contains the F factor. Genetic changes takes place in
the recipient cell. Once inside the F cell, the linear single-stranded DNA molecule acts as a
polymerization template and is converted into a DNA double helix. This linear donor fragment is
termed as the exogenote, and the resident F chromosome is termed as the endogenote. Crossover
events take place which integrate a part of the donor genome into the recipient. It is in this way
that recombination takes place.
2. F+ cells are the bacterial cells which possess F factor as a plasmid independent of the bacterial
genome. The F cell plasmid contains F factor as the only DNA and no DNA from the bacterial
genome. Mating or conjugation of an F+ cell with an F cell, result in two F+ cells, both of which
are capable of transmitting the plasmid to other F cells by conjugation. Thus F+ bacterial cells
act as donor. The F+ cells also contain surface exclusion proteins; TraS and TraT on the bacterial
surface. These proteins have the capability to prevent secondary mating events involving
plasmids belonging to the same incompatibility group. Thus, each F+ bacterium can host only a
single plasmid type of any given incompatibility group.
3. Properties of Recipient cells (receiving the F plasmid)
Recipients are basically F bacteria that do not contain F factor (F plasmid), and are considered as
female. The F-plasmid is a class of conjugative plasmids that control sexual functions of bacteria
with a fertility inhibition (Fin) system. In this system, FinO, which is a trans-acting factor and
Fin P, an antisense RNAs, combine to repress the expression of the activator gene TraJ. TraJ is a
transcription factor that upregulates the tra operon, which contains the genes required for
conjugation and plasmid transfer. The F- plasmid can exist in two alternative states, viz. it can
either remain free in the cell or it can be integrated into the E. coli chromosome.called as
episomes. The F-plasmid also carr.
Suppose that a single gene controls fruit colour in mongo. Yellow fru.pdf
Suppose that a single gene controls fruit colour in mongo. Yellow fruit (Y-) is dominant to red
fruit(yy). Suppose a true breeding yellow mongo plant was crossed with a red-fruited plant and
the resulting F_1 was selfed. The F_2 segregation pattern was as expected. If one of the yellow-
fruited plant was randomly selected and selfed, what is the probability that the progeny would
segregate for fruit colour (i.e. red- and yellow-fruited progeny)? 3/4 2/3 1/4 1/2 none would
segregate
Solution
Answer: b). 2/3
Explanation:
Parents are true breeding means that they are homozygous so the F1 gen would be Yy
do a punnet and see that a self cross will produce YY, Yy. yY and yy out of the 4 possible there
are 3 that are yellow. One in 3 would be true breeding (all yellow offspring) and 2/3 would be
heterozygous and produce the typical 3:1 yellow to red offspring.
The F2 genotypes are YY (1/4), Yy(1/2) and yy (1/4). Thus, the yellow-fruited plant that was
randomly picked could be either YY or Yy. There is a 1/3 chance that it was YY and 2/3 chance
that it was Yy. If the YY plant was selected and selfed, the progeny would not segregate for fruit
color. If the Yy plant was selected, the progeny would segregate for fruit color..
BI 211 Graded problem set 9 The drosophila alleles for black.pdf
BI 211 Graded problem set 9 The drosophila alleles for black body (instead of the normal tan
color, which is the wild type) and curved Wings (instead of the wild-type straight wings) are both
recessive and on chromosome 2. Gene "C" designates color, with alleles C representing the
dominant tan allele and c representing the recessive black allele, while gene " W " designates the
wing shape gene with W being the dominant straight allele and w the recessive curved allele. You
want to discover how close together these two genes are on chromosome 2. You place one male
from a pure breeding stock and one female from a pure breeding stock (these are the 'parental'
flies) into vial 1. The offspring of this cross (cross 1) hatch in vial 1 . These F1 offspring are
heterozygous for both genes. 1. What is (are) the genotype(s) of the offspring of cross 1 (the F1
flies)? b. To perform the mapping experiment, you select a single male with the genotype coww
and place it into a vial you label 'vial 2' along with a female F1; this is cross 2 . You get 500
offspring in vial 2 (these are the F2 fies). Among these F2 flies, 65 have tan bodies and curved
wings, 70 have black bodies and straight wings, 190 have black bodies and curved wings, and 175
have tan bodies and straight wings. What are the genotypes of offspring with parental phenotypes
(hint there are twol). Show your work. c. What are the genotypes of the recombinant offspring
(hint: there are also twol). d. Two true breeding parents (parents of cross 1) gave rise to the F 1
flies in vial 1. What is the genotype of each parent? parent 1: parent 2: e. What is the percent of
recombinants for these two genes? Show your work..
AP Biology Inheritance and chromosomal mutations
multiple modes of inheritance and chromosomal mutations
Write the following using javaGiven a class ‘Node’ and ‘NodeList’,.pdf
Write the following using java
Given a class ‘Node’ and ‘NodeList’, that contains the below information diagram.
-id: int
-name: String
-next: Node
+Node(id: int, name: String)
+setId(id: int): void
+getId(): int
+setName(name: String) : void
+getName(): String
+setNext(node: Node): void
+getNext(): Node
NodeList
-size: int
-root: Node
+add(node: Node): void
+size(): int
+findNode(node: Node): boolean
Implement add(Node), findNode(Node) and size methods in the NodeList class which is
provided to you. The methods should work as:
Using the following main method:
public static void main(String[] args)
{
NodeList list = new NodeList();
Node node = new Node(1, \"Book\");
Node node2 = new Node(2, \"Lappy\");
list.add(node);
list.add(node2);
System.out.println(\"Length : \"+list.size());
Node node3 = new Node(3, \"Glass\");
Node node4 = new Node(4, \"Pen\");
list.add(node3);
System.out.println(\"Length : \"+list.size());
if(list.findNode(node3)) System.out.println(\"Node found: \"+ node3.getName());
else
System.out.println(\"Node not found: \"+ node3.getName());
if(list.findNode(node4)) System.out.println(\"Node found: \"+ node4.getName());
else
System.out.println(\"Node not found: \"+ node4.getName());
}
Then it should return the following output:
Length : 2
Length : 3
Node found: Glass
Node not found: Pen
The given Node class is:
public class Node
{
private int id = 0;
private String name = \"\";
private Node next;
public Node(int id, String name)
{
this.id = id;
this.name = name;
this.next = null;
}
public Node getNext()
{
return next;
}
public void setNext(Node node)
{
this.next = node
}
public int getId()
{
return id;
{
public void setId(int id)
{
this.id = id;
}
public String getName()
{
return name;
}
public void setName(String name)
{
this.name = name;
}
public String toString()
{
return \"ID : \"+this.id+\" Name : \"+this.name;
}
}
The given NodeList class is:
public class NodeList
{
private int size = 0;
private Node root = null;
/*
* It has to take a new Node and add that to the next address of previous Node.
* If the list is empty, assign it as the \"root\"
* @Param - Node
*/
public void add(Node node)
{
// Implement this method!!!
}
/*
* It has to return the size of the NodeList
*
* @return size
*/
public int size()
{
// Implement this method!!!
}
/*
* It has to take a Node and checks if the node is in the list.
* If it finds the node, it returns true, otherwise false
*
* @param - Node
* @return boolean true/false
*/
public boolean findNode(Node node)
{
// Implement this method!!!
}
}Node
-id: int
-name: String
-next: Node
+Node(id: int, name: String)
+setId(id: int): void
+getId(): int
+setName(name: String) : void
+getName(): String
+setNext(node: Node): void
+getNext(): Node
Solution
Hi, Please find my implementation.
Please let me know in case of any issue.
############ Node.java #############
public class Node
{
private int id = 0;
private String name = \"\";
private Node next;
public Node(int id, String name)
{
this.id = id;
this..
Write a program in C++ that declares a structure to store the code n.pdf
Write a program in C++ that declares a structure to store the code number, salary and grade of an
employee. The program defines two structures variables, inputs records of two employees and
then displays the record of the employees.
Solution
#include
#include
using namespace std;
struct employee
{
int codenumber[7];
char grade[14];
float salary;
};
int main(void)
{
struct employee ;
cout<<\'\ .......................................: \" ;
cout<<\"\ enter the employee code number: \" ;
gets(employee.codenumber);
cout<<\"\ enter the employee grade: \" ;
gets(employee.grade);
cout<<\"\ enter the employee salary: \" ;
gets(employee.salary);
return (employee);
}
int list_function(employee list)
{
cout<<\"\ employee codenumber: \" <.
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12- Analyzing Coat Color Mutants - Wild type gerbils from Arizona- Mon.docx
12. Analyzing Coat Color Mutants - Wild type gerbils from Arizona, Montana, indiana and Califomia are black. - You have obtained pure-breeding mutant varieties of gerbils that are white or pink from these different places. - You know that each gerbil strain has a mutation in a single autosomal gene. - The table below shows E1 proseny obtained when crossing purebreeding WT and mutants to each other. - Bolded boxes at the top and side indicate the purebreeding parent strains you crossed to each other. - The intersection of these boxes represents the phenotype of the F 1 offspring of a particular cross. For each cross, you can assume 100% of the F1 progeny are of the color indicated. a) What is the purpose of crossing the mutant to wild type (black) and observing F1 progeny phenotypes? b) What is the purpose of crossing the mutants to each other and observing F1 progeny phenotypes? c) Which of the strains cannot be used for the analysis you described in (b)? Why? d) When you cross pure-breeding white from Arizona and indiana to each other, you get black in the F 1 . What does this indicate about the mutations that give rise to white in these 2 strains? e) To further explain your answer in (d), assign genotypes for the cross between the 2 purebreeding strains from Indiana and Arizona and their resulting F 1 . Use the letter A if only alleles of one gene are segregating in the cross, letters A and B if 2 genes are involved, etc. Denote dominant alleles with a capital letter and recessive alleles with a small letter.
.
For a single Mendelian trait, a homozygous dominant individual is cr.pdf
For a single Mendelian trait, a homozygous dominant individual is crossed with a homozygous
recessive individual. Which of the following is true of the individual offspring of the F1
generation from this cross?
Two individuals of the F1 generation from the cross in the previous question are crossed together
(F1 crossed with F1). Which of the following is true about individual offspring from the F2
generation of this cross? Select all that apply.
only the dominant phenotype will be expressed
both dominant and recessive phenotypes will be expressed
None of the other answers fit.
there will be two different genotypes
there will be three different genotypes
only the recessive phenotype will be expresseda.
only the dominant phenotype will be expressedb.
both dominant and recessive phenotypes will be expressedc.
None of the other answers fit.d.
there will be two different genotypese.
there will be three different genotypesf.
only the recessive phenotype will be expressed.
Pls solve this with the Punnett squared table with the given informati.docx
Pls solve this with the Punnett squared table with the given information and why is the 5th answer correct?
A mutant rabbit with pointy teeth has been found in a cave somewhere in Canada. To characterize the mutant at the genetic level, it was crossed to a wild-type rabbit. The progeny (F1) were found to have all normal shaped teeth like wild-type rabbits. Intercrossing the F1 progeny gave a F 2 population with the following ratio: 3/4 with normal shaped teeth 3/16 with intermediately pointy teeth 1/16 with pointy teeth. Which one of the following is the best description of the pointy-toothed phenotype? caused by recessive mutation of a single gene. caused by dominant mutation of a single gene caused by recessive mutation of one gene and dominant mutation of a second gene caused by recessive mutation of one genes and suppressor mutation of a second gene (no phenotype on its own) caused by recessive mutation of one genes and recessive modifier mutation of a second gene.
.
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
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.
1. How is the recipient cell different from before the two cells mat.pdf
1. How is the recipient cell different from before the two cells mated? Phenotypically as well as
genetically.
2. What do F+ cells do?
3. What are all properties of a recipient cell (which receives the F plasmid)?
Solution
1. Recipient cell (F- cell) is different after it mates with the F+ cell. Conjugation/mating is
equivalent to sex (however no new individuals are produced in this case) in that it combines
genetic material from two different individuals into a single genome.It occurs through F pili
which originate from cell membrane and project outward beyond the cell wall. Thus, like sexual
reproduction conjugation produces new genetic combinations. The pilus is essential for
recognition of recipient cell. The sex plasmid passes from the donor F+ to the recipient F– cell
through the conjugation tube, and seperates after the event is over and now called as
exconjugates. F- cell after matinh with F+ cell, now becomes F+ and can act as donor during
next conjugation event.
After conjugation event, recipient cell now contains the F factor. Genetic changes takes place in
the recipient cell. Once inside the F cell, the linear single-stranded DNA molecule acts as a
polymerization template and is converted into a DNA double helix. This linear donor fragment is
termed as the exogenote, and the resident F chromosome is termed as the endogenote. Crossover
events take place which integrate a part of the donor genome into the recipient. It is in this way
that recombination takes place.
2. F+ cells are the bacterial cells which possess F factor as a plasmid independent of the bacterial
genome. The F cell plasmid contains F factor as the only DNA and no DNA from the bacterial
genome. Mating or conjugation of an F+ cell with an F cell, result in two F+ cells, both of which
are capable of transmitting the plasmid to other F cells by conjugation. Thus F+ bacterial cells
act as donor. The F+ cells also contain surface exclusion proteins; TraS and TraT on the bacterial
surface. These proteins have the capability to prevent secondary mating events involving
plasmids belonging to the same incompatibility group. Thus, each F+ bacterium can host only a
single plasmid type of any given incompatibility group.
3. Properties of Recipient cells (receiving the F plasmid)
Recipients are basically F bacteria that do not contain F factor (F plasmid), and are considered as
female. The F-plasmid is a class of conjugative plasmids that control sexual functions of bacteria
with a fertility inhibition (Fin) system. In this system, FinO, which is a trans-acting factor and
Fin P, an antisense RNAs, combine to repress the expression of the activator gene TraJ. TraJ is a
transcription factor that upregulates the tra operon, which contains the genes required for
conjugation and plasmid transfer. The F- plasmid can exist in two alternative states, viz. it can
either remain free in the cell or it can be integrated into the E. coli chromosome.called as
episomes. The F-plasmid also carr.
Suppose that a single gene controls fruit colour in mongo. Yellow fru.pdf
Suppose that a single gene controls fruit colour in mongo. Yellow fruit (Y-) is dominant to red
fruit(yy). Suppose a true breeding yellow mongo plant was crossed with a red-fruited plant and
the resulting F_1 was selfed. The F_2 segregation pattern was as expected. If one of the yellow-
fruited plant was randomly selected and selfed, what is the probability that the progeny would
segregate for fruit colour (i.e. red- and yellow-fruited progeny)? 3/4 2/3 1/4 1/2 none would
segregate
Solution
Answer: b). 2/3
Explanation:
Parents are true breeding means that they are homozygous so the F1 gen would be Yy
do a punnet and see that a self cross will produce YY, Yy. yY and yy out of the 4 possible there
are 3 that are yellow. One in 3 would be true breeding (all yellow offspring) and 2/3 would be
heterozygous and produce the typical 3:1 yellow to red offspring.
The F2 genotypes are YY (1/4), Yy(1/2) and yy (1/4). Thus, the yellow-fruited plant that was
randomly picked could be either YY or Yy. There is a 1/3 chance that it was YY and 2/3 chance
that it was Yy. If the YY plant was selected and selfed, the progeny would not segregate for fruit
color. If the Yy plant was selected, the progeny would segregate for fruit color..
BI 211 Graded problem set 9 The drosophila alleles for black.pdf
BI 211 Graded problem set 9 The drosophila alleles for black body (instead of the normal tan
color, which is the wild type) and curved Wings (instead of the wild-type straight wings) are both
recessive and on chromosome 2. Gene "C" designates color, with alleles C representing the
dominant tan allele and c representing the recessive black allele, while gene " W " designates the
wing shape gene with W being the dominant straight allele and w the recessive curved allele. You
want to discover how close together these two genes are on chromosome 2. You place one male
from a pure breeding stock and one female from a pure breeding stock (these are the 'parental'
flies) into vial 1. The offspring of this cross (cross 1) hatch in vial 1 . These F1 offspring are
heterozygous for both genes. 1. What is (are) the genotype(s) of the offspring of cross 1 (the F1
flies)? b. To perform the mapping experiment, you select a single male with the genotype coww
and place it into a vial you label 'vial 2' along with a female F1; this is cross 2 . You get 500
offspring in vial 2 (these are the F2 fies). Among these F2 flies, 65 have tan bodies and curved
wings, 70 have black bodies and straight wings, 190 have black bodies and curved wings, and 175
have tan bodies and straight wings. What are the genotypes of offspring with parental phenotypes
(hint there are twol). Show your work. c. What are the genotypes of the recombinant offspring
(hint: there are also twol). d. Two true breeding parents (parents of cross 1) gave rise to the F 1
flies in vial 1. What is the genotype of each parent? parent 1: parent 2: e. What is the percent of
recombinants for these two genes? Show your work..
AP Biology Inheritance and chromosomal mutations
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Write the following using javaGiven a class ‘Node’ and ‘NodeList’,.pdf
Write the following using java
Given a class ‘Node’ and ‘NodeList’, that contains the below information diagram.
-id: int
-name: String
-next: Node
+Node(id: int, name: String)
+setId(id: int): void
+getId(): int
+setName(name: String) : void
+getName(): String
+setNext(node: Node): void
+getNext(): Node
NodeList
-size: int
-root: Node
+add(node: Node): void
+size(): int
+findNode(node: Node): boolean
Implement add(Node), findNode(Node) and size methods in the NodeList class which is
provided to you. The methods should work as:
Using the following main method:
public static void main(String[] args)
{
NodeList list = new NodeList();
Node node = new Node(1, \"Book\");
Node node2 = new Node(2, \"Lappy\");
list.add(node);
list.add(node2);
System.out.println(\"Length : \"+list.size());
Node node3 = new Node(3, \"Glass\");
Node node4 = new Node(4, \"Pen\");
list.add(node3);
System.out.println(\"Length : \"+list.size());
if(list.findNode(node3)) System.out.println(\"Node found: \"+ node3.getName());
else
System.out.println(\"Node not found: \"+ node3.getName());
if(list.findNode(node4)) System.out.println(\"Node found: \"+ node4.getName());
else
System.out.println(\"Node not found: \"+ node4.getName());
}
Then it should return the following output:
Length : 2
Length : 3
Node found: Glass
Node not found: Pen
The given Node class is:
public class Node
{
private int id = 0;
private String name = \"\";
private Node next;
public Node(int id, String name)
{
this.id = id;
this.name = name;
this.next = null;
}
public Node getNext()
{
return next;
}
public void setNext(Node node)
{
this.next = node
}
public int getId()
{
return id;
{
public void setId(int id)
{
this.id = id;
}
public String getName()
{
return name;
}
public void setName(String name)
{
this.name = name;
}
public String toString()
{
return \"ID : \"+this.id+\" Name : \"+this.name;
}
}
The given NodeList class is:
public class NodeList
{
private int size = 0;
private Node root = null;
/*
* It has to take a new Node and add that to the next address of previous Node.
* If the list is empty, assign it as the \"root\"
* @Param - Node
*/
public void add(Node node)
{
// Implement this method!!!
}
/*
* It has to return the size of the NodeList
*
* @return size
*/
public int size()
{
// Implement this method!!!
}
/*
* It has to take a Node and checks if the node is in the list.
* If it finds the node, it returns true, otherwise false
*
* @param - Node
* @return boolean true/false
*/
public boolean findNode(Node node)
{
// Implement this method!!!
}
}Node
-id: int
-name: String
-next: Node
+Node(id: int, name: String)
+setId(id: int): void
+getId(): int
+setName(name: String) : void
+getName(): String
+setNext(node: Node): void
+getNext(): Node
Solution
Hi, Please find my implementation.
Please let me know in case of any issue.
############ Node.java #############
public class Node
{
private int id = 0;
private String name = \"\";
private Node next;
public Node(int id, String name)
{
this.id = id;
this..
Write a program in C++ that declares a structure to store the code n.pdf
Write a program in C++ that declares a structure to store the code number, salary and grade of an
employee. The program defines two structures variables, inputs records of two employees and
then displays the record of the employees.
Solution
#include
#include
using namespace std;
struct employee
{
int codenumber[7];
char grade[14];
float salary;
};
int main(void)
{
struct employee ;
cout<<\'\ .......................................: \" ;
cout<<\"\ enter the employee code number: \" ;
gets(employee.codenumber);
cout<<\"\ enter the employee grade: \" ;
gets(employee.grade);
cout<<\"\ enter the employee salary: \" ;
gets(employee.salary);
return (employee);
}
int list_function(employee list)
{
cout<<\"\ employee codenumber: \" <.
Write a generic VBA Sub procedure to compute the value of the follow.pdf
Write a generic VBA Sub procedure to compute the value of the following Integral (I) using the
Trapezoidal Rule. dac
Solution
// Trapazoidal rule calculates area under the curve
// n is number of interval
Function Integration(a As Integer, b as Integer, n As Integer) As Double
Dim sum as Double
sum = (F(a) + F(b))/2
For i = 1 To n - 1
sum = sum + F(a + ((b-a)/n)*i)
Next i
Integration = sum * (b-a)/n
End Function.
Why are standards needed in data communication and networking What .pdf
Why are standards needed in data communication and networking? What are the advantages and
disadvantages of standards? How do standards fit in with regulations at the federal,
manufacturing, and organizational levels? Give an example of a standard that fits in at your
workplace.
Solution
he Importance of Standards
Standards are necessary in almost every business and public service entity. For example, before
1904, fire hose couplings in the United States were not standard, which meant a fire department
in one community could not help in another community. The transmission of electric current was
not standardized until the end of the nineteenth century, so customers had to choose between
Thomas Edison’s direct current (DC) and George Westinghouse’s alternating current (AC).
The primary reason for standards is to ensure that hardware and software produced by different
vendors can work together. Without networking standards, it would be difficult—if not
impossible—to develop networks that easily share information. Standards also mean that
customers are not locked into one vendor. They can buy hardware and software from any vendor
whose equipment meets the standard. In this way, standards help to promote more competition
and hold down prices.
The use of standards makes it much easier to develop software and hardware that link different
networks because software and hardware can be developed one layer at a time.
The Standards-Making Process
There are two types of standards: formal and de facto. A formal standard is developed by an
official industry or government body. For example, there are formal standards for applications
such as Web browsers (e.g., HTTP, HTML), for network layer software (e.g., IP), data link layer
software (e.g., Ethernet IEEE 802.3), and for physical hardware (e.g., V.90 modems). Formal
standards typically take several years to develop, during which time technology changes, making
them less useful.
De facto standards are those that emerge in the marketplace and are supported by several vendors
but have no official standing. For example, Microsoft Windows is a product of one company and
has not been formally recognized by any standards organization, yet it is a de facto standard. In
the communications industry, de facto standards often become formal standards once they have
been widely accepted.
The formal standardization process has three stages: specification, identification of choices, and
acceptance. The specification stage consists of developing a nomenclature and identifying the
problems to be addressed. In the identification of choices stage, those working on the standard
identify the various solutions and choose the optimum solution from among the alternatives.
Acceptance, which is the most difficult stage, consists of defining the solution and getting
recognized industry leaders to agree on a single, uniform solution. As with many other
organizational processes that have the potential to influence the sales of hardwar.
Which of the following are mismatched A. Giordia - transmitted by f.pdf
Which of the following are mismatched? A. Giordia - transmitted by faces in drinking water B.
Histoplasma - causes Ohio Valley fever C. Trichomonas - sexually transmitted D. Plasmodium
- causes chagas disease E. Saccharomyces - yeast in making bread and beer All of the following
are helminths except A. pinworm. B. flukes. C. ring worm. D. roundworm. E. tapeworm.
Solution
Answer 24]
The mismatched is
D} plasmodium = causes Chagas disease
Plasmodium falciparum is a protozoan parasite a special species of plasmodium cause malaria in
humans
It is transmitted by female anopheles mosquito
it does not causes chagos disease
Incorrect options are=
A] Giardia = transmitted by feces in drinking water
it is the most common cause of waterborne disease in united state
B ] Histoplasma =causes ohio valley fever
is disease caused by fungus Histoplasma capsulatum
C]Tricomonas = Sexually transmitted
it is sexually transmitted
e] Saccharomyces = yeast in making bread & cake
Answer 25=
All of the following are Helminths except
C] Ring worm
Explaination = It is a misnomer it leads to the assumption that this organism is helminths
Actually it is refers to the very common infection caused by some types of fungus
microsporum,Epidermophyton & trycophyton.
Which of the following statements about human evolution is correct.pdf
Which of the following statements about human evolution is correct?
A) The upright posture and enlarged brain of humans evolved simultaneously.
B) Fossil evidence indicates that early anthropoids were arboreal, were cat-sized, and lived about
35-45 million years ago.
C) Human ancestors were virtually identical to chimpanzees. D) Human evolution
occurred by anagenic change within an unbranched lineage.
E) Modern humans are the only human species to have evolved on Earth.
Please explain reasoning. Thank you.
Solution
B.Fossil evidence indicates that early anthropoids were arboreal, were cat-sized, and lived about
35-45 million years ago.
Reason:-The fossil record of Family Felidae extends to the Late Eocene, 34 million years
ago.The first known feline was Aelurogale. Its descendent, Proailurus, gave rise to the major
branches of the cat family we see today.Miacoid species gave rise to Proailurus which appeared
about 30 million years ago, and is generally considered the first true cat.There were other cat-like
species before Proailurus.Archeological evidence shows a relationship between man and cat for
the past 8,000 years..
What are the six major pollutions in the National Ambient Air Qualit.pdf
What are the six major pollutions in the National Ambient Air Quality Standards (NAAQS)?
Briefly explain dry adiabatic lapse rate? What is the value of the dry adiabatic lapse rate? How
does the dry adiabatic lapse rate affect atmospheric stabillity?, Explain briefly In general, how
many stability categories can you name? What are they?
Solution
The six major pollutants per NAAQS are sulphur dioxide, particulate matters (PM10 and
PM2.5), carbon monoxide, ozone, nitrogen dioxide and lead
the lapse rate observed in air when the air has little or no water is called dry adiabatiuc lapse rate
value of dry adiabatic lapse rate is 9.8oC/km.
UPS Worldpart DiscussionWhat do you think are the operational str.pdf
UPS Worldpart Discussion:
What do you think are the operational strengths of UPS?
How have they achieved these strengths?
How do their strengths match their marketing messages, some of which can be seen below.
Solution
1.
Operational strengths of UPS are as following:
2.
They have achieved these strengths through:
3.
It seems to be an incomplete question as marketing messages are not given as mentioned in the
question..
True or False Justify your answer.Using multilevel signaling, it .pdf
True or False? Justify your answer.
Using multilevel signaling, it is possible to arbitrarily increase the bit rateat the physical layer by
increasing the number of levels that each pulsecarries.
Solution
False.
This is because as the number of levels increases, it becomes harder to detect the levels
accurately and this also leads to higher noise levels..
This is question about excel cuers wish to answer shortly how to use.pdf
This is question about excel cuers wish to answer shortly how to use.
How would you use Financial Functions, Data tables and Hyperlinks on a day-to-day functions?
Solution
Excel provides an enormous number of functions to analyze, audit and calculate data.
Some of them are utilized by many Excel users on a daily basis while others are
so very specific that only a small group of financial, statistical or engineering specialists
understand and need them.
financial functions
*******************
Excel has a wide assortment of financial functions that are used by thousands of financial
analysts every day.
Microsoft Excel provides a host of functions to simplify the work of accounting managers,
financial analysts and banking specialists.
Here are some commonly used financial functions in Excel they are
i)Present Value (PV) = the value of the load or savings today.
ii)Interest Rate per period (RATE) = the interest rate, usually monthly (if given an annual rate,
be sure to divide by 12!)
iii)Number of payment periods (NPER) = Number of payment periods (number of months
usually).
iv)Periodic payment (PMT) = The payment per period (monthly payment usually). This value is
the same for every payment period.
Data tables
***********
A data table is a range of cells in which you can change values in some in some of the cells and
come up with different answers to a problem.
You can create one-variable or two-variable data tables, depending on the number of variables
and formulas that you want to test.
You can find some cool uses for tables in your data to day work. They are intuitive, easy to use
and provide great power without added complexity.
Hyperlinks
**********
A Hyperlink is nothing but a reference to a document, a location or an action that the reader can
directly follow by selecting the link.
Excel provides a convenient way for linking up cells data with whatever you want.
To create hyperlinks in the sheet. Select the cell in which you want to create hyperlink and right-
click to select Hyperlink.
When you click the cell that contains the HYPERLINK function, Microsoft Excel opens the file
that is stored at link_location.
The use of Hyperlinks in Excel has been extended to a number of areas and this includes:
Opening Files (of any type)
Opening Web Pages (Internet or Intranet).
There is a requirement to design a system to sense the presence of gl.pdf
There is a requirement to design a system to sense the presence of glass bottles leaving an
automatic washer on a production line conveyor belt. The production line is for food products
and the environment is clean and maintained at a temperature of 15 degree C. The system must
meet the following criteria: The switching distance will lie between 5 and 20 mm. The output
from the sensor is to be of binary form, i.e. logical 1 when the part is present and logical 0 when
no part is present. It is necessary to ensure that a change of sensor state occurs closer to the
sensor when a bottle approaches than when it moves away. Bottles will pass the sensor at the
rate of 10 every second. There is no requirement for protection from contact with objects but it
must be proof against water drips. Recommend a suitable sensor and justify your
recommendation. Sketch a diagram that shows the functional blocks within the sensor and
explain each function.
Solution
lock Diagram of the system
Basic Elements Used
MICRO CONTROLLER
RESISTORS
CAPACITOR
INDUCTOR
POWER DIODE
WATER PUMP
IR SENSORS
DC MOTOR
CONVEYOR BELT
Resistore
A part from resister and inductor, a capacitor is the other basic component used in electronic
circuits. It is a device which
1. Has the ability to store charge which neither a resistor nor an inductor can do;
2. opposes any change of voltage in the circuit in which it is connected;
3. Blocks the passage of direct current through it.
CAPACITOR
The electronic component (mostly a coil) which opposes the change of current in circuit is
called an inductor. The ability of coil or inductor due to which it opposes the change of current
flowing through it is known as its inductance. The inductance of an inductor may also be defined
as the ability of an inductor to produce induced voltage when the current varies through it. The
unit of inductance is Henry which is named after Joseph Henry.
INDUCTOR
Diodes allow electricity to flow in only one direction. The arrow of the circuit symbol shows the
direction in which the current can flow. Diodes are the electrical version of a valve and early
diodes were actually called valves. The construction and the fundamental working of the power
diode and signal diode are same but the power handling capacity of the power diode is higher
than the signal diodes
POWER DIODE
Sensors
IR Sensors work by using a specific light sensor to detect a select light wavelength in the Infra-
Red (IR) spectrum. By using an LED which produces light at the same wavelength as what the
sensor is looking for, you can look at the intensity of the received light. When an object is close
to the sensor, the light from the LED bounces off the object and into the light sensor. This results
in a large jump in the intensity, which can be detected using a threshold
DRIVER CIRCUIT FOR DC MOTOR
We have used the dc motors to drive conveyer belt and to. The driver circuit for dc motors is
shown below.
COMPARATOR CIRCUIT
Comparator circuit is used in thi.
The investments of Charger Inc. include a single investment 11,010 .pdf
The investments of Charger Inc. include a single investment: 11,010 shares of Raiders Inc.
common stock purchased on February 24, Year 1, for $39 per share including brokerage
commission. These shares were classified as trading securities. As of the December 31, Year 1,
balance sheet date, the share price had increased to $42 per share.
Required:A.Journalize the entries to acquire the investment on February 24, and record the
adjustment to fair value on December 31, Year 1. Refer to the Chart of Accounts for exact
wording of account titles.B.How is the unrealized gain or loss for trading investments reported
on the financial statements?
Solution
a) Journal entry :
b) Unrealized gain or loss for trading investment reported on the financial statement that is
recognized in net income.Dateaccount & explanationdebitcreditYear 1 Feb 24Investment in
trading securities (11010*39)429390Cash429390(To record investment)Year 1 Dec
31Investment in trading securities (11010*3)33030Unrealized Gain-Trading securities33030(To
record unrealized gain).
RNA polymerasebinds to DNA after the double strands have been unwoun.pdf
RNA polymerasebinds to DNA after the double strands have been unwound by helicase.
Solution
Answer: binds to the template strand in the promoter region.
Explanation:
RNA polymerases are enzymes that transcribe DNA into RNA. Using a DNA template, RNA
polymerase builds a new RNA molecule through base pairing. It always builds a new RNA
strand in the 5’ to 3’ direction..
Research and explain the deviant actions of the Los Angeles Police D.pdf
Research and explain the deviant actions of the Los Angeles Police Department’s Rampart
CRASH unit during the late 1990s. Did socialization and culture of the unit have a role within
the officer’s deviant behaviors? As a member of the CRASH team, what would have been your
course of action after observing the illegal conduct? submit your answer in 3-5 pages.
Solution
Traditional definitions of police misconduct, especially Apolice corruption,@ are imprecise. In
the past, police scholars have classified acts of police misconduct as Apolice corruption,@
Apolice brutality,@ and Adrug-related misconduct.@ We found, however, that these
classifications are not mutually exclusive, and that determining whether profit-motivated
criminality by police officers involved jobrelated police corruption frequently is impossible. -
Involuntary separations are rare. Separations for cause from the New York City Police
Department (through dismissals, terminations, and forced resignations and retirements) are
infrequent events. Only two percent of the officers employed by the NYPD during the 22 years
(1975-96) we studied were involuntarily separated from the department. - Pre-employment
history matters. Officers whose life histories include records of arrest, traffic violations, and
failure in other jobs are more likely than other officers to be involuntarily separated from the
NYPD. - Education and training matter. Officers who hold associate or higher degrees are less
likely than those who do not to be involuntarily separated. Those who do well in the Police
Academy=s recruit training program are less likely than marginal recruits to be separated as
unsatisfactory probationers. Subsequently, they also are less likely to be involuntarily separated
for cause after successful completion of their probationary periods. - Diversity matters. As the
NYPD has become more diverse, it has become better behaved. We found a very strong inverse
correlation (r=-.71; r2 .50) between the percentage of white male NYPD officers and the
department=s annual rate of involuntary separations. - Race still matters, but apparently only for
black officers: As the representation of Hispanic and Asian officers in the NYPD has increased,
their involuntary separation rates have decreased and become virtually indistinguishable from
those of white officers. Black officers= representation in the NYPD remained relatively flat
during the years studied. Black officers= involuntary separation rates have also decreased, but
remain higher than those for other racial groups. This document is a research report submitted to
the U.S. Department of Justice. This report has not been published by the Department. Opinions
or points of view expressed are those of the author(s) and do not necessarily reflect the official
position or policies of the U.S. Department of Justice. Fyfe and Kane -- xiii The study reaches
these conclusions after examining the life and career histories of all 1,543 officers who were
dis.
Question 1 10 points Save A TALIA Contribute NINA Co Contribute TALIA.pdf
Question 1 10 points Save A TALIA Contribute NINA Co Contribute TALIA EMILY TALIA
Contribute 9 Contribute TALIA Dont information sets and she has possible Consider the
extensive form representation of the game between Emily, Nina and Talia. Emily moves at
information sets and has possible strategies. Talia moves at strategies. (Please fill in the blanks
using numerical values like 0, 1, 75, etc)
Solution
Emily moves at 01 information sets with 02 possible strategies. Talia will have 04 information
sets and 08 strategies.
Q1. Print all the odd numbers from 1 to a user specifiable upper lim.pdf
Q1. Print all the odd numbers from 1 to a user specifiable upper limit (inclusive) using both
while and for loops
Example,
Please enter a number:
9
Odd numbers from 1 to 9 are:
1
3
5
7
9
HINT: The number 9 should be included!!!
2. Write a program called CheckerBoard that displays the following n×n (n=7) checkerboard
pattern using two nested for-loops.
Solution
Q)1)
Using While Loop:
DisplayOddNumbers.java
import java.util.Scanner;
public class DisplayOddNumbers {
public static void main(String[] args) {
//Declaring variables
int number;
int i=1;
//Scanner object is used to get the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number entered by the user
System.out.println(\"Please enter a number:\");
number=sc.nextInt();
//Displaying the odd numbers using while loop
System.out.println(\"Odd numbers from 1 to 9 are:\");
while(number>0)
{
//Checking the number is even or odd
if(i%2!=0)
{
System.out.println(i);
}
i++;
number--;
}
}
}
_____________________
Output:
Please enter a number:
9
Odd numbers from 1 to 9 are:
1
3
5
7
9
_______________
using for loop:
DisplayOddNosUsingForLoop.java
import java.util.Scanner;
public class DisplayOddNosUsingForLoop {
public static void main(String[] args) {
//Declaring variables
int number;
//Scanner object is used to get the inputs entered by the user
Scanner sc=new Scanner(System.in);
//getting the number entered by the user
System.out.println(\"Please enter a number:\");
number=sc.nextInt();
//Displaying the odd numbers using while loop
System.out.println(\"Odd numbers from 1 to 9 are:\");
for(int i=1;i<=number;i++)
{
//Checking the number is even or odd
if(i%2!=0)
{
System.out.println(i);
}
}
}
}
____________________
Output:
Please enter a number:
9
Odd numbers from 1 to 9 are:
1
3
5
7
9
_____________________
2)Q)
package org.students;
import java.util.Scanner;
public class CheckerBoard {
public static void main(String[] args) {
//Displaying the checker board using nested for loop
for(int i=1;i<=7;i++)
{
for(int j=1;j<=7;j++)
{
if(i%2==0)
System.out.print(\" #\");
else
System.out.print(\"# \");
}
System.out.println(\" \");
}
}
}
__________________
output:
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
# # # # # # #
____________Thank You.
Please revise the answer bellow.Q1. What historically have been Ap.pdf
Please revise the answer bellow.
Q1. What historically have been Apple’s competitive advantages?
There have been many competitive advantages that have allowed Apple to be a consistent leader
in the PC market. One competitive advantage has been their innovative spirit. This was shown
with the initial release of the Apple 2. Steve Job’s mission of Apple was to bring an easy-to-use
computer to the market, and this is what he accomplished with the release of the Apple 2. The
Apple 2 was a revolutionary product that changed the PC world forever. The Apple 2 was
launched in 1978 and it lead the PC industry to $1 billion in annual sales in less than three years.
Another competitive advantage for Apple is that they have total control over all of the software,
hardware, and operations related to Apple products. The apple software is horizontally and
vertically integrated, meaning it relies on its own designs and does not license software to third
parties. Apple released the Mac in 1984, which featured an interface that was easy-to-use, a great
industrial design, and technical proficiency. The superior software allowed Apple to enter the
Mac into new markets, desktop publishing and education. In education, Apple grabbed almost
half of the market and by 1990, Apple had $1 billion in cash and was the most profitable PC
Company in the world. Furthermore, Apple hired Tim Cook to streamline the Apple supply
chain, and also Apple launched a website to set up direct sales. This has allowed Apple to be
successful by creating a comprehensive, unified culture with all aspects of the company. Their
most important competitive advantage was Steve Jobs. Being the co-founder, Steve Jobs had the
long-term vision of Apple and worked every day to live out the vision. Steve Jobs was forced out
of 1985, but returned as the interim CEO in 1997 when the company was near bankruptcy. Jobs
refocused Apple to only having 4 product categories, rather than 15 it had previously. Through
Steve Job’s leadership, Apple created the iMac in 1998, which allowed Apple’s sales to outpace
the industry average for the first time in many years. Steve Jobs made strategic decisions to make
Apple products the most competitive by bringing a new culture to Apple, and positively
changing the image and brand of Apple.
Q2. Analyze the personal computer industry. Why did Apple struggle historically in PCs?
It is true that Apple was the company who pioneered the personal computing devices, but IBM
was the company the mainstream PC’s. This was the biggest competitor to the early success of
Apple. The IBM PC operated on an “open” system that allowed producers to clone it. Another
reason why Apple struggled with PCs was the price. Macintosh did have a loyal customer base
which allowed them to sell it at a premium price. IBM- compatible prices started to drop, and the
high Macintosh’s looked unfeasible and too expensive. Another reason for the struggle of Apple
PCs was the rise of “Wintel”. “Wintel” was t.
Negligence Curtis R. Wilhelm owned beehives and kept the hives on pr.pdf
Negligence Curtis R. Wilhelm owned beehives and kept the hives on property he owned. John
Black, who operated a honeybee business, contracted to purchase some beehives from Wilhelm.
Black employed Santos Flores, Sr. to help him pick up the beehives from 101102Wilhelm. Black
provided Flores with a protective suit to wear while picking up the beehives. Neither Wilhelm
nor Black informed Flores of the danger of working with bees. After picking up beehives from
Wilhelm’s home, Black and Flores drove to remote property owned by Wilhelm to pick up other
beehives. Flores opened the veil on his protective suit. After loading one beehive onto the truck,
Flores started staggering and yelling for help. Flores sustained several bee stings, suffered
anaphylactic shock reaction, and died before an ambulance could reach him. Flores’s wife and
children sued Wilhelm and Black for negligence for failing to warn Flores of the dangers of
working with beehives and the possibility of dying of anaphylactic shock if stung by a bee. Did
Wilhelm act negligently by failing to warn Flores of the dangers of working with beehives?
Wilhelm v. Flores, 133 S.W.3d 726, Web 2003 Tex. App. Lexis 9335 (Court of Appeals of
Texas)
Solution
Answer:
Yes. The Court of Appeals held that the owner of the bees owed a duty to warn Flores of
thedangers of working with bees and bee stings, affirming the award of damages..
List and explain at least three popular sixteenth century dance type.pdf
List and explain at least three popular sixteenth century dance types, including the style, common
instrumentation, and use in court, society, or both.
Solution
Sixteemth century dance types:
(1) Galliard:
(i) Renaissance dance
(ii) places-- England, France, Spain, Germany, and Italy
(iii) dance steps :--- five stepsright, left, right, left, cadence
(iv) instruments-- crumhorns, lute
(b)
Courante
(i)Renaissance dance
(ii)places:-- France and Italy
(iii)dance steps:-- fast running and jumping steps
(iv)instruments:- cello, harpsichord, lute and violin
(c)
Canary dance
(i)Renaissance dance
(ii)places:-- France ,Italy, spain and England
(iii)dance steps:-- jumps, stamping of the feet and violent movement
(iv)instruments:- drum and wind and stringed instruments.
Introduction to Database Management SystemConsider the following .pdf
Introduction to Database Management System:
Consider the following relation of students: R(A, B, C, D, E, F, G) (A is underlined)
D is a candidate key. Suppose the following dependencies exist:
FD1: A-> B, C, D, E, F, G
FD2: C, D -> A, B, E, F, G
FD3: D-> G
FD4: F -> C
Apply normalization until you cannot decompose the relations further. State the reasons behind
each decomposition.
Solution
FD1: A-> B, C, D, E, F, G: This Functional dependency uniquely idendifies the table row so, all
other functional dependecies are not need. we simply ignore all other.
>>> for example it any of the FD voilates any of the NF then we will decompose it and create a
new table for that FD. Here FD2 voilates 3NF but if we want to decompose it what could be the
foreign key in R2 since all other attributes are present in relation R1:(A, B, C, D, E, F, G) .
>>>Therefore FD1 is the final one for the Relation..
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