art F You decide to cross the reciprocal translocation strain to a pu.pdf
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art F You decide to cross the reciprocal translocation strain to a pure (black, sepia) line to generate female F1 fies that are both translocaton heterozygotes and Bbss dhybrids. You then backcross these F, females to males from the pure (black sepia)Ine. This diagram shows synaptis in the F females. (In the diagram, normal chromosome ll: TI translocated chromosome with the chromosome l centromere NIll normal chromosome Ill TIl translocated chromosome with the chromosome Ill centromere.) You wish to recover several categories of F2fies. wtat event must occur to produce each category ofF2fies stedin the table below? Complete the table by dragging the labels to the correct locations. Labels can be used once, more than once, or not at all. (Note: CO crossover NCO no crossover, b allele on TIll s allele on TIII Region b Region a Gamete from Dihybrid Required Event at Required Event at b allele on TIII (CO2 or NCO?) s allele on Till B allele on TTII s allele on TI allele on NII s allele on N b allele on NII sepia, fully fertile s allele on N previous l of 4 retum assign Solution Reciprocal translocation is usually the exchange of genetic material between non-homologous chromosomes. As In the below diagram: Crossing Over is the exchange of genetic material between homologous chromosomes resulting in recombinant chromosome during sexual reproduction. In the given question the F1 female flies (BbSs dihybrids) translocation heterozygotes are produced when reciprocal translocation strain is crossed to pure black and sepia line (BBSS). These females (BbSs) are backcrossed to pure black and sepia males (BBSS) to produce F2 progeny flies of following categories: F2 progeny black and semi-sterileRegion a:- Crossing Over- Label B allele on NII and s allele on NIII Region b: - Non Crossing Over- Label B allele on TIII and s allele on TIII sepia, fully fertile Region a:- Non Crossing Over- Label b allele on TIII and S allele on TIII Region b: - Non Crossing Over-Label b allele on NII and S allele on NIII (black and sepia), semi-sterile Region a:- Crossing Over- Label b allele on NII and s allele on NIII Region b: - Crossing Over- Label b allele on TIII and s allele on TIII..
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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..
Bacteriophage T2 normally infects the different E. coli strains. B an.pdf
Bacteriophage T2 normally infects the different E. coli strains. B and C. However, certain
mutations, designated r, block growth on strain B. Ten r mutations were mapped by two-point
recombination frequencies to a small region of the T2 chromosome, shown below. When the ten
r mutants were tested for complementation by mixed-infection of strain B, certain pairs could kill
the B host (+), while other pairs failed to grow completely (-), Assuming the mutations a, b, and
c are known to lie in the genes A, B. and C. use the complementation data in the table below to
mark gene boundaries on the genetic map. Do the data reveal the presence of a fourth gene in
this region? If so, please show it\'s location on the map.
Solution
Crosses between strains which produce wild type progenies, suggest that these genes are present
in different gene. On the basis of this we would determine the following boundaries:
agdj are present in one gene because crosses between these gene are producing mutant
progenies; representing cis effect. The other genes are bh, efi and kc. Therefore, there are four
different genes.
Agdj, bh, efi and kc..
Give an expression for the pattern inventory of 2-colorings of the ed.pdf
Give an expression for the pattern inventory of 2-colorings of the edges of the unoriented figure
on the right (rotations and Hips allowed).
Solution
It is basically one-thrd of a rectagular cubiod. The volume of a rectagular cubiod is
2*(lb+bh+lh). As such the expression of the figure can be written as
f(x,y,z) = (2/3)*(xy+yz+zx).
For the 4 arguments below, proceed as followsBREAK DOWN THE ARGU.pdf
For the 4 arguments below, proceed as follows:
:BREAK DOWN THE ARGUMENT- Explicitly identify the component statements (p,q,r,etc)
that make up the argument and show the symbolic form of the argument.
:ANALYZE THE ARGUMENT- Decide whether the argument is valid or invalid by
manipulating the premises, if necessary, and eventually recognizing the logical form of the
argument (e.g Modus Ponens, Modus Tollens, Fallacy of the Inverse, etc)
1. “If we evolved a race of pure logicians, that would not be progress; but we have not evolved a
race of pure logicians; so that’s progress.”-
2. “Homer is a visionary or he’s a fool; but he’s certainly a fool; so he’s not a visionary.”
3. “If the fossil is an ammonite, then it is from the Crestaceous period. If the fossil is not from
the Mesozoic era, then it is not from the Crestaceous period. If the fossil is from the Mesozoic
era, then it is at least 65 million years old. Therefore, if the fossil is an ammonite, it is at least
65million years old.”
4. “Babies are illogical. Nobody is despised who can manage a crocodile. Illogical persons are
despised. Hence, babies cannot manage crocodiles.”
Solution.
External respiration includes all of these processes EXCEPT _____. r.pdf
External respiration includes all of these processes EXCEPT _____. release of carbon dioxide
by tissue cells gas transport gas diffusion across respiratory membrane breathing
Solution
External respiration includes all of these processes except release of carbon dioxide by tissue
cells.
External respiration occurs in the lungs where oxygen diffuses into the blood and carbon dioxide
diffuses into the alveolar air.
Internal respiration occurs in the tissues where the oxygen diffuses out of the blood and carbon
dioxide diffuses out of the cells.
Gas exchange takes place by the diffusion process where the movement of the substance from
higher concentration to lower concentration of that substance occurs.
Gas diffusion occurs across respiratory membrane..
Ecosystem ecologyWhy are rain forests wet and deserts dry Compare.pdf
Ecosystem ecology
Why are rain forests wet and deserts dry? Compare and contrast biotic aspects of these very
different terrestrial ecosystems.
Solution
The major difference between each type of ecosystem is the climate.Tropical rainforests cover
the area between the Tropic of Capricorn and the Tropic of Cancer and tend to be warm and
wet.Deserts, on the other hand,are hot and dry and have very little annual rainfall.
Tropical rainforests are found only near the Earth\'s equator,which is also one of the reasons
why they are humid.Being near the Equator,tropical rainforests receive more radiation from the
sun than any other ecosystem on earth.This is one of the main reasons for their extreme
biodiversity.Deserts occur at all latitudes.
While most tropical rainforests typically receive from 60 to 160 inches of rain annually, the
warmest tropical forests can receive as much as 400 inches of rain.In comparison,deserts are
defined largely by their lack of precipitation,although colder deserts do experience snowfall.On
average, deserts can receive half an inch to 15 inches of annual rainfall.
Tropical rainforests have the greatest diversity of animals of any biome,with about half of all
animal species in the world living in tropical forests.This incredible species richness is due
largely to the combination of a tropical climate and high,year-round rainfall,which provide
optimal growing conditions for the largest number of plant species.Deserts are less than ideal
environments for animals;with scarce sources of water and food,as well as very high
temperatures.
The desert biome receives twice as much radiation, but also lose twice as much at night in
comparison with more humid areas such as a Rainforest.Deserts tend to be warm because of
vegetation, which causes water to evaporate.Absorbed sunlight by the plants goes straight into
the ground and causes it to warm..
Recommended
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..
Bacteriophage T2 normally infects the different E. coli strains. B an.pdf
Bacteriophage T2 normally infects the different E. coli strains. B and C. However, certain
mutations, designated r, block growth on strain B. Ten r mutations were mapped by two-point
recombination frequencies to a small region of the T2 chromosome, shown below. When the ten
r mutants were tested for complementation by mixed-infection of strain B, certain pairs could kill
the B host (+), while other pairs failed to grow completely (-), Assuming the mutations a, b, and
c are known to lie in the genes A, B. and C. use the complementation data in the table below to
mark gene boundaries on the genetic map. Do the data reveal the presence of a fourth gene in
this region? If so, please show it\'s location on the map.
Solution
Crosses between strains which produce wild type progenies, suggest that these genes are present
in different gene. On the basis of this we would determine the following boundaries:
agdj are present in one gene because crosses between these gene are producing mutant
progenies; representing cis effect. The other genes are bh, efi and kc. Therefore, there are four
different genes.
Agdj, bh, efi and kc..
Give an expression for the pattern inventory of 2-colorings of the ed.pdf
Give an expression for the pattern inventory of 2-colorings of the edges of the unoriented figure
on the right (rotations and Hips allowed).
Solution
It is basically one-thrd of a rectagular cubiod. The volume of a rectagular cubiod is
2*(lb+bh+lh). As such the expression of the figure can be written as
f(x,y,z) = (2/3)*(xy+yz+zx).
For the 4 arguments below, proceed as followsBREAK DOWN THE ARGU.pdf
For the 4 arguments below, proceed as follows:
:BREAK DOWN THE ARGUMENT- Explicitly identify the component statements (p,q,r,etc)
that make up the argument and show the symbolic form of the argument.
:ANALYZE THE ARGUMENT- Decide whether the argument is valid or invalid by
manipulating the premises, if necessary, and eventually recognizing the logical form of the
argument (e.g Modus Ponens, Modus Tollens, Fallacy of the Inverse, etc)
1. “If we evolved a race of pure logicians, that would not be progress; but we have not evolved a
race of pure logicians; so that’s progress.”-
2. “Homer is a visionary or he’s a fool; but he’s certainly a fool; so he’s not a visionary.”
3. “If the fossil is an ammonite, then it is from the Crestaceous period. If the fossil is not from
the Mesozoic era, then it is not from the Crestaceous period. If the fossil is from the Mesozoic
era, then it is at least 65 million years old. Therefore, if the fossil is an ammonite, it is at least
65million years old.”
4. “Babies are illogical. Nobody is despised who can manage a crocodile. Illogical persons are
despised. Hence, babies cannot manage crocodiles.”
Solution.
External respiration includes all of these processes EXCEPT _____. r.pdf
External respiration includes all of these processes EXCEPT _____. release of carbon dioxide
by tissue cells gas transport gas diffusion across respiratory membrane breathing
Solution
External respiration includes all of these processes except release of carbon dioxide by tissue
cells.
External respiration occurs in the lungs where oxygen diffuses into the blood and carbon dioxide
diffuses into the alveolar air.
Internal respiration occurs in the tissues where the oxygen diffuses out of the blood and carbon
dioxide diffuses out of the cells.
Gas exchange takes place by the diffusion process where the movement of the substance from
higher concentration to lower concentration of that substance occurs.
Gas diffusion occurs across respiratory membrane..
Ecosystem ecologyWhy are rain forests wet and deserts dry Compare.pdf
Ecosystem ecology
Why are rain forests wet and deserts dry? Compare and contrast biotic aspects of these very
different terrestrial ecosystems.
Solution
The major difference between each type of ecosystem is the climate.Tropical rainforests cover
the area between the Tropic of Capricorn and the Tropic of Cancer and tend to be warm and
wet.Deserts, on the other hand,are hot and dry and have very little annual rainfall.
Tropical rainforests are found only near the Earth\'s equator,which is also one of the reasons
why they are humid.Being near the Equator,tropical rainforests receive more radiation from the
sun than any other ecosystem on earth.This is one of the main reasons for their extreme
biodiversity.Deserts occur at all latitudes.
While most tropical rainforests typically receive from 60 to 160 inches of rain annually, the
warmest tropical forests can receive as much as 400 inches of rain.In comparison,deserts are
defined largely by their lack of precipitation,although colder deserts do experience snowfall.On
average, deserts can receive half an inch to 15 inches of annual rainfall.
Tropical rainforests have the greatest diversity of animals of any biome,with about half of all
animal species in the world living in tropical forests.This incredible species richness is due
largely to the combination of a tropical climate and high,year-round rainfall,which provide
optimal growing conditions for the largest number of plant species.Deserts are less than ideal
environments for animals;with scarce sources of water and food,as well as very high
temperatures.
The desert biome receives twice as much radiation, but also lose twice as much at night in
comparison with more humid areas such as a Rainforest.Deserts tend to be warm because of
vegetation, which causes water to evaporate.Absorbed sunlight by the plants goes straight into
the ground and causes it to warm..
b) Analyze your network IP packets headers and contents.Solu.pdf
b) Analyze your network \"IP packets\" headers and contents.
Solution
ip packets is like communication network from one device to another..ip packets uses the packets
to send the data packets from one ntework to another
and its like transmission of networks
IP HEADER : ip header contains the informetion about which ip vesrion we are using and what
is the source and destinetion of ip header
IP CONTENT : content is nothing but the all informetion which is covered by everything is one
set like photos videos.
A type A man is the son of a type O father and type A mother. If he .pdf
A type A man is the son of a type O father and type A mother. If he has children with a type O
woman, what are the following probabilities?
A. A type AB child?
B. A type O child?
C. The first three children with type A blood?
Solution
The child can have only two blood types possible type A and type O.
a) Probability for type AB =0
b) Probability for type 0 =0.5
c) Probablity for The first three children with type A blood = 3/4.
A One-Way Analysis of Variance is a way to test the equality of thre.pdf
A One-Way Analysis of Variance is a way to test the equality of three or more means at one time
by using variances. Which of the following are necessary assumptions? The variances of the
populations must be homogeneous.
Solution
a. The samples must be independent.
d. The populations from which the samples were obtained must be normally or approximately
normally distributed..
5 000-0 SolutionEquity multiplier = Total Assets Equityor, .pdf
5 000-0
Solution
Equity multiplier = Total Assets / Equity
or, Equity Multiplier = (Total Assets / Equity) x (Net Profit / Net Profit) [ Muliply Numerator
and denominator by Net Profit]
or, Equity Multiplier = (Total Assets / Net profit) x (Equity / Net profit)
or, Equity Multiplier = (1 / ROA) x (ROE) = ROE / ROA = 9% / 1% = 9 (first option).
1sequences and sampling. Suppose we went to sample the x-axis from X.pdf
1sequences and sampling. Suppose we went to sample the x-axis from Xmin to Xmax using a
step size of step
A)Draw a picture of what is going on.
B) Write a expression for n the total number of samples involved (in terms of Xmin, Xmax and
step)
C) Write out the sequence of x-samples
D) Write a direct and general expression for xi that captures the sequence
E) Write a recursive expression for the sequence
F) Write a program to compute and store the x-samples over the range -5x5 using a step size of
0.1 do everything in main ()
2 . We talked about the following string functions that are available in C (as long as you include
string.h):
int strlen(char str[])
void strcpy(char str1[], char str2[])
void strcat(char str1[], str2[])
Write your own versions of these functions; for example: int paul_strlen(int char str[]). Hint: for
your version of the strlen function, start at the first character in the array and keep counting until
you find the ‘\\0’ character (use a while loop for this). Note: Use your version of the strlen
function in the strcpy and strcat functions.
9. We want to insert a number into an array.
(a) Formulate the problem mathematically with two sequences: x and y. (b) Write a function of
the form:
insertNumIntoArray(int n, int array[], int num, int index)
The function inserts num into the array at the specified index. The rest of the array then follows.
For example, if num = 9 and index = 3 and array = [7 2 8 8 3 1 2] then the function will produce:
array = [7 2 8 9 8 3 1 2]
Note: assume that array is properly dimensioned to have at least 1 extra space for storage.
10. Repeat #2 by for the delete operation; that is, we want to delete a single element (at a
specified index) from an array; for example, suppose index = 3 and array = [50 70 10 90 60 20],
then the result will be
array: [50 70 10 60 20]
11. Repeat #2 by for an insert operation where we are inserting several values into the array. The
function should be of the form:
int insertArrayIntoArray(int n, int inArray[],
int nInsert, int insertArray[], int outArray[], int index)
The dimension of outArray is returned (explicitly). For example:
inArrayarray: [7 2 8 6 3 9]
insertArray: [50 60 70]
index: 2
outArray: [7 2 50 60 70 8 6 3 9]
Assume that outArray is large enough to hold all n + nInsert values.
Solution
#include
//Simulates strlen() library function
int paul_strlen(char str[])
{
int l;
for(l = 0; str[l] != \'\\0\'; l++) ;
return l;
}
//Simulates strcpy() library function
void paul_strcpy(char str1[], char str2[])
{
int c;
for(c = 0; str1[c] != \'\\0\'; c++)
str2[c] = str1[c];
str2[c] = \'\\0\';
printf(\"\ Original String: %s\", str1);
printf(\"\ Copied String: %s\", str2);
}
//Simulates strcat() library function
void paul_strcat(char str1[], char str2[])
{
int i, j;
for(i = 0; str1[i] != \'\\0\'; i++) ;
for (j = 0; str2[j] != \'\\0\'; i++, j++)
{
str1[i] = str2[j];
}
str1[i] = \'\\0\';
printf(\"\ Concatenated String: %s\", str1);
}
int main()
{
char data1[20], data2[20];
pri.
4 (4 points). How would you describe the difference in cell structur.pdf
4 (4 points). How would you describe the difference in cell structures between a thermophile and
a psychrophile? Include mention of cell membrane construction and enzyme/protein structure at
least.
Solution
Thermophilic bacteria have more long chain fatty acids or lipids which maintain the membrane
fluidity. They have more tetraether lipids that increase the membrane stability. Hence they
survive in high temperatures. Thermophiles also have a high level of K+ salts which protect the
DNA and proteins by forming hydrophilic shell. Thermophiles have have acidic aminoacids.
psychrophiles grow at very low temperatures. They have high polyunsaturated and methyl-
branched fatty acids, and/or a shorter acyl-chain length which are generally present in cis
configuration . This increases membrane fluidity and reduces the interactions between the
membrane preventing freezing at low temperature. They have cryoprotectants and antifreeze
agents..
Distinguish between cell fate and cell commitment. How can one assay.pdf
Distinguish between cell fate and cell commitment. How can one assay these aspects of cellular
phenotype by experimental embryology?
Solution
Cell fate--It means the development of the cell into an embryo and the events which govern the
development of the egg cell throught the process of embryogenesis.The factors which determine
the cell fate include cell signaling pathways like notch pathway etc, regulation and control by
pathways like mRNA,siRNA inside the cell,development of different cavity inside the embryo,
development of ectoderm,mesoderm and endoderm, and the regulation of cell fate determination
by release of various controlling hormones.
cell commitment--it involves the orientation and decision by the cell to undergo differentiation
and follow a particular developmental pathway.It involves the release of various growth
promoting factors by stem cells which lead to the differentiation of stem cells into mature cells
by triggering and activating a particular organ development pathway example--blood cells.
There are several aspects of determination of the cellular phenotype by experimental
embryology----1)sex determination pathways in insects and dosage compensation 2)regulation of
different genes in different climates,temperature, and background environment determines the
activation and inhibition of development genes in caterpillar larvae of insects3) the division of
fertilized egg into morula, blastula, gastrula ,and development of different layers by cell
regulatory protein factors 4)role of cell adhesion protein molecules during cell differentiation
pathway 5)development of cell symmetry and asymmetry of the insects 6)development of
embryo cavity and germline layers of development after fertilization 6)development of cells to
tissues to organs with the help of secreting chemicals, protein factors and hormones which
activate the differentiation pathway..
Differentiate the processes of oogenesis and spermatogenesis. Are th.pdf
Differentiate the processes of oogenesis and spermatogenesis. Are there similarities between the
two processes? Differences? Consider the cells, their structures and functions.
Solution
Comparison between Spermatogenesis and Oogenesis
Characteristics
Spermatogenesis
Oogenesis
Definition
The process of formation of sperm cells is called spermatogenesis
Formation of ovum is called oogenesis
Primary organ
Testis (in males)
Ovary (in females)
The source cell/ structure that undergoes gametogenesis
Spermatogonium
Primordial follicle
Source cell is preserved
Yes, Spermatogonium is a stem cell.
No
Type of cell division
Meiosis
Meiosis
Onset of gametogenesis
At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty.
During embryonic development.
Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive
cycles.
Continuous or cyclic
Continuous. Millions of sperms are produced each day.
Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days).
Occurrence of gametogenesis in life
Begins at puberty
Continues throughout life
Begins at menarche
Ends at menopause
No. of functional gametes in one cycle
4
1
Polar cells (that degenerate later) produced
No
Yes, one polar cell in each of meiosis I and II.
In spite of major difference of these two processes, they both share a common path and fate- they
meiotic divisions that produce haploid gametes.
Characteristics
Spermatogenesis
Oogenesis
Definition
The process of formation of sperm cells is called spermatogenesis
Formation of ovum is called oogenesis
Primary organ
Testis (in males)
Ovary (in females)
The source cell/ structure that undergoes gametogenesis
Spermatogonium
Primordial follicle
Source cell is preserved
Yes, Spermatogonium is a stem cell.
No
Type of cell division
Meiosis
Meiosis
Onset of gametogenesis
At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty.
During embryonic development.
Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive
cycles.
Continuous or cyclic
Continuous. Millions of sperms are produced each day.
Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days).
Occurrence of gametogenesis in life
Begins at puberty
Continues throughout life
Begins at menarche
Ends at menopause
No. of functional gametes in one cycle
4
1
Polar cells (that degenerate later) produced
No
Yes, one polar cell in each of meiosis I and II..
Define induction and give an example of deductive reasoningSolu.pdf
Define induction and give an example of deductive reasoning?
Solution
There are two methods of reasoning. One is Deductive approach and another is Inductive
approach.
Deductive reasoning:-This kind of reasoning works from the general to more specific. It is also
referred to as \"Top-down\" approach.
Examples:-
1) If X = Y and Y = Z, then by deductive reasoning X = Z.
2) Snakes are reptiles and reptiles are cold-blooded, by deductive reasoning it can be said that
snakes are cold-blooded.
Inductive reasoning:-
Inductive reasoning as opposed to deductive reasoning work from specific observations to
broader generalizations and theories. It is also referred to as \"bottom up\" approach in informal
sense.
Example:-
1) A just moved here from Hongkong. A has red hair, therefore by inductive reasoning, it can be
said that People from Hongkong have red hair..
A report says that the between-subjects factor of participants sal.pdf
A report says that the between-subjects factor of participants\' salary produced significant
differences in self-esteem. (a) What does this tell you about the design? (b) What does it tell you
about the results?
Solution
It means the design was effective enough to study the relation between the self esteem and
salary.
Secondly the variance ratio for the salary was significant.
City GUIWrite a Java GUI program which reads data about US cities..pdf
City GUI
Write a Java GUI program which reads data about US cities. The data file is different, but is in
the same format as in Lab #01. The GUI provides a combo box with the cities in the data file. As
cities are selected in the combo box, information about the city is displayed.
Requirements
1. The program must have a graphical user interface similar to the one on the bottom of the
Requirements section. It must have a text field, a button, a combo box, and labels.
2. One or more layout managers must be used. This example was created using a border layout.
The east and west panels each used a grid layout in addition.
3. A message must be displayed in the GUI after the data has been successfully read. A message
should also be displayed if the file is not found. The program should not exit in that case, but
allow the user to enter a different file name.
4. The combo box must be populated by the names of the cities in the data file in alphabetical
order. Note that this means the GUI starts with an empty combo box. Items are added to it as or
after the data file is read.
5. The program must provide an ActionListener for the Read button. It must also provide either
an ActionListener or an ItemListener for the combo box, which displays the state, zip code, and
time zone for that city.
6. The information for each city must be stored in its own object. Another object must be used to
manage the collection of cities. This is similar to Lab #01. For this assignment you may assume
that there will be no more than 100 cities.
7. The time zone should be displayed as the correct abbreviation for that city’s time zone in July.
Here are the abbreviations. If the city observes daylight savings time, the S should be changed to
a D. Note that some cities in the data file do not observe daylight savings time.
Hours relative to GMT -4 -5 -6 -7 -8 -9 -10
Standard Time Abr. AST EST CST MST PST AKST HST
I have attached an executable .jar file to the assignment that you can run to see how this program
should work. Copy it to your computer. Copy the data file into the same directory as the .jar file.
Suggestions
1. Create class diagrams first. This will help you think through and organize your program. There
are many ways to do this. Start with the class diagrams from Lab #01. Both classes will require
some additions.
2. Create the GUI and handle the IO in a separate class which extends JFrame. The main method
can be included in this class.
3. Don’t worry about making your implementation efficient or using wonderful abstract data
types (ADTs). You will be required to do that soon enough.
4. This program will take you several days to complete. Break it down into smaller tasks. This is
the order I used, yours may be different.
*Modify the class diagrams from Lab #01 to include new requirements that you identify now.
You can add additional functionality later as you discover the need.
*Implement those changes in the classes.
* Create a GUI class. Implement a non-functiona.
What is the role of sulfur chemoautotrophs in the sulfur cycle Deco.pdf
What is the role of sulfur chemoautotrophs in the sulfur cycle? Decomposition reduction
oxidation assimilation Reduction of CO2 to generate carbohydrates can occur when coupled to
the oxidation of reduced sulfur compounds (for examples, H2S). This latter process is carried out
by what type of sulfate oxidizers sulfate reducers
Solution
10
Oxidation
Sulfur chemotrophs oxidize sufur to derive energy for the synthesis of ATP.
What is the PDU at Layer 4 calledA. DataB. SegmentC. Packet.pdf
What is the PDU at Layer 4 called?
A. Data
B. Segment
C. Packet
D. Frame
E. Bit
Solution
B is correct. The Layer 4 PDU is called a segment.
A, C, D, and E are incorrect. A is incorrect because the data is the PDU for the top three layers.
C is incorrect because the packet is the PDU at the Network layer. D is incorrect because the
frame is the PDU for the Data Link layer. E is incorrect because the bit is the PDU at the
Physical layer..
Briefly describe the contributions to the quality movement made by e.pdf
Briefly describe the contributions to the quality movement made by each of the following
pioneers.
Shewhart
Dodge and Romig
Sarasohn
Deming
Juran
Solution
Walter Shewhart - The Grandfather of Total Quality Management.
His Ideas
The original notions of Total Quality Management and continuous improvement trace back to a
former Bell Telephone employee named Walter Shewhart. One of W. Edwards Deming\'s
teachers, he preached the importance of adapting management processes to create profitable
situations for both businesses and consumers, promoting the utilization of his own creation -- the
SPC control chart.
Dr. Shewhart believed that lack of information greatly hampered the efforts of control and
management processes in a production environment. In order to aid a manager in making
scientific, efficient, economical decisions, he developed Statistical Process Control methods.
Many of the modern ideas regarding quality owe their inspirtation to Dr. Shewhart.
He also developed the Shewhart Cycle Learning and Improvement cycle, combining both
creative management thinking with statistical analysis. This cycle contains four continuous steps:
Plan, Do, Study and Act. These steps (commonly refered to as the PDSA cycle), Shewhart
believed, ultimately lead to total quality improvement. The cycle draws its structure from the
notion that constant evaluation of management practices -- as well as the willingness of
management to adopt and disregard unsupported ideas --are keys to the evolution of a successful
enterprise.
Harold French Dodge (January 23, 1893 in Lowell, Massachusetts – December 10, 1976) was
one of the principal architects of the science of statistical quality control. He is universally
known for his work in originating acceptance sampling plans for putting inspection operations on
a scientific basis in terms of controllable risks.
From 1917 to 1958 worked at quality assurance department at Bell Laboratories with Walter
Shewhart, George Edwards, Harry Romig, R. L. Jones, Paul Olmstead, E.G.D. Paterson, and
Mary N. Torrey. At that time the basic concepts of acceptance sampling was developed, such as :
· Consumer\'s Risk,
· Producer\'s Risk,
· double sampling,
· lot tolerance percent defective (LTPD), and
· average outgoing quality limit (AOQL).
Also he originated several types of:
· acceptance sampling schemes,
· CSP type continuous sampling plans,
· chain sampling plans, and
· skip-lot sampling plans.
During World War II, Dodge served as a consultant to the Secretary of War, and was chairman
of the American Standards Association (now the American National Standards Institute) War
Committee Z1, which prepared the Z1.1, Z1.2, and Zl.3 quality control standards.
After he retired from Bell Labs in 1958, Dodge became a professor of applied mathematical
statistics at Rutgers. He also served as a consultant to NASA and the Sandia Corporation.
The American Society for Testing and Materials honors Harold Dodge\'s memory with the
Harold F. Dodge Award.
Ho.
Background Angiosperms (flowering plants) are the largest Phylum in .pdf
Background Angiosperms (flowering plants) are the largest Phylum in the plant kingdom
Plantae, (Note that the Phylum A may also called Anthophyta or Magnoliophyta in different
classification systems.) These plants have true roots, stems, leaves, and flowers. The roots grow
into the soil to anchor the plant in place and take up water and nutrients. The leaves are above
ground and act as the main organs for photosynthesis. Stems provide above ground support for
the plant and operate as conduits to move nutrients and water throughout the plant. Flowers
contain the male and female reproductive organs of the plant. Assignment Details Part 1 Plants
develop specialized roots, stems, leaves, and flowers to make them better adapted to their
environment. People exploit many modifications as vegetables. An example of this is the taproot
of carrots. The carrot plant stores sugars in the taproot to supply energy for the formation of
flowers in later development. People dig up and eat the taproots while the sugar content is still
high. Complete the following: Describe the anatomy (structure) and physiology (function) of a
plant. Is the structure of your plant a modification from the standard root, stem, leaf, or flower?
Explain how this structure differs from the standard root, stem, leaf, or flower. Compare the
plant that you have chosen, its structure, and its function with 1 or more of your classmates\'
choices. Part 2 Animals are important for humans in ways other than providing food.
Medicines, clothing, beauty, fuel, air quality, building materials, and industrial products are a
few of the ways in which humans exploit animals. Complete the following: Choose l product or
benefit that people get from animals (besides food), describe it, and explain how people benefit.
Which do you think is the most important benefit of animals? Select 1 benefit and explain. List
3 adaptations that animals have that plants do not have. Provide references in APA format. This
includes a reference list and in-text citations for references used throughout the assignment
Solution
1. the normal function of roots is fixation of the plant in the soil absorption of water and
Minerals from the soil and conduction of the materials to the stem but in some plants roots
perform special functions like storage of food materials(carrot,raddish) respiration(rhizopora)
vegetative reproduction (sweet potato)support(betel vine) photosynthesis(orchids).
Betel vine:
Weak Climbing Plants produce some advantitious roots which cling to the supports and help the
plants in climbing they sometimes they secrete sticky substance or a disc like structures for this
purpose.
Betel Vine is a plant having climbing roots these adventitious roots are produced from their
nodes which help them to climb over a support another example is money plant.
These roots are aerial roots which twist and clasp the support to help plant to vlimb.
Normal rootsd show positive geotropism but these climbing roots show negativ.
Write a Pep8 Assembly program that reads in and stores two integers .pdf
Write a Pep/8 Assembly program that reads in and stores two integers from the user and prints
out their sum in the form: Total = Submit: Pep/8 formatted program in a text file
Solution
Answer:
Assembly Language Program :
.zero 1
.LC0:
.string \"Enter two integers: \"
.LC1:
.string \"Total\"
.LC2:
.string \" = \"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov esi, OFFSET FLAT:.LC0
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
lea rax, [rbp-8]
mov rsi, rax
mov edi, OFFSET FLAT:std::cin
call std::basic_istream >::operator>>(int&)
mov rdx, rax
lea rax, [rbp-12]
mov rsi, rax
mov rdi, rdx
call std::basic_istream >::operator>>(int&)
mov edx, DWORD PTR [rbp-8]
mov eax, DWORD PTR [rbp-12]
add eax, edx
mov DWORD PTR [rbp-4], eax
mov esi, OFFSET FLAT:.LC1
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov esi, OFFSET FLAT:.LC2
mov rdi, rax
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov rdx, rax
mov eax, DWORD PTR [rbp-4]
mov esi, eax
mov rdi, rdx
call std::basic_ostream >::operator<<(int)
mov eax, 0
leave
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L5
cmp DWORD PTR [rbp-8], 65535
jne .L5
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L5:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret.
Why is it important that the duration of cell-cycle phases be contro.pdf
Why is it important that the duration of cell-cycle phases be controllable in adult multicellular
organisms? b. What is MFP? Describe how this complex affects the cell cycle.
Solution
SOME BRIEF DESCRIPTION WHICH HELP YOU TO UNDERSTAND QUESTION AND IT
S ANSWER FOR MY KNOWLEDGE IT HELPS YOU
NOTE ;------ (The cell grows . The duration of these cell cycle phases varies considerably in
different kinds of cells. For a typical rapidly proliferating human cell with a total cycle time of
24 hours, the G1 phase might last about 11 hours, S phase about 8 hours, G2 about 4 hours, and
M about 1 hour.)
The cell cycle or cell-division cycle is the series of
events that take place in a cell leading to its division and duplication of its DNA (DNA
replication) to produce two daughter cells. In bacteria, which lack a cell nucleus, the cell cycle is
divided into the B, C, and D periods. The B period extends from the end of cell division to the
beginning of DNA replication. DNA replication occurs during the C period. The D period refers
to the stage between the end of DNA replication and the splitting of the bacterial cell into two
daughter cells. In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into three
periods: interphase, the mitotic (M) phase, and cytokinesis. During interphase, the cell grows,
accumulating nutrients needed for mitosis, preparing it for cell division and duplicating its DNA.
During the mitotic phase, the chromosomes separate. During the final stage, cytokinesis, the
chromosomes and cytoplasm separate into two new daughter cells. To ensure the proper division
of the cell, there are control mechanisms known as cell cycle checkpoints.
The cell-division cycle is a vital process by which a
single-celled fertilized egg develops into a mature organism, as well as the process by which
hair, skin, blood cells, and some internal organs are renewed. After cell division, each of the
daughter cells begin the interphase of a new cycle. Although the various stages of interphase are
not usually morphologically distinguishable, each phase of the cell cycle has a distinct set of
specialized biochemical processes that prepare the cell for initiation of cell divisions.
Cell cycle checkpoints are control mechanisms in eukaryotic cells which
ensure proper division of the cell. Each checkpoint serves as a potential point along the cell
cycle, during which the conditions of the cell are assessed, with progression through the various
phases of the cell cycle occurring when favorable conditions are met. Currently, there are three
known checkpoints: the G1 checkpoint, also known as the restriction or start checkpoint or
(Major Checkpoint); the G2/M checkpoint; and the metaphase checkpoint, also known as the
spindle checkpoint.
Effects of MULTIFOLIATE-PINNA, AFILA, TENDRIL-LESS and UNIFOLIATA genes on
leafblade architecture in Pisum sativum
In order to dissect the genetic regulation of leafblade
morphogenesis, 16 genotypes of pea, constructed.
How does the Rob proteins phosphorylation stage affect the cell cy.pdf
How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression?
How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression?
Solution
The RB(Retinonlastoma) is a tumor-suppressor protein and it has capability to inhibit
progressing the cell cycle meticulously at the G1 checkpoint, Hyperphosphorylation of RB
trigger release of “E2F” factor. G1 cyclin-cyclin dependent kinases (CDK) complex is mainly
involved in the phosphorylation of RB to activate E2F result in initiation of transcription rate to
promote DNA synthesis. These events are leading higher DNA synthesis to promote cell division
in cell cycle result in tumor cells formation.
Would you use physical segmentation or VLANs. Why or why not and whe.pdf
Would you use physical segmentation or VLANs. Why or why not and where would you use
each?
In the Diagram below where would you put them. HU HUB Roud er Router Switeh Sreh Routen
pe Wineless access point DMZ Zone auess point out ey acess poinf rele 70adba Broadbond
wrreless access auuess point server access point access
Solution
Vlans and physical segmentation are both different things which both are used when we want to
divided network. Basically physical segmentation means to have many different network on the
basis of physical connectivity and VLANs may be defined as the division of network into logical
parts and to have different networks with different nodes. Suppose if you have switch then it is
better to have vlans rather than physical segmentation since as we know that switch is
broadcasting frams and each interface of switch is present into different collision domain. So in
order to have different broadcast domian we have to configure vlans into the switch. Which can
certainly provide us security on the basis of different network.
In above diagram we can have vlans on the every switch where each node connected to the
switch would be placed into different network thus saves us from the collisions and broadcasting
to the other nodes unnecessarily which may cause unnecessary overheads and less security in the
network..
Which plants produce leaves (Select all that apply.)a.Gametophyte.pdf
Which plants produce leaves? (Select all that apply.)
a.Gametophytes of bryophytes
b.Gametophytes of seedless vascular plants
c.Sporophytes of bryophytes
d.Sporophytes of seedless vascular plants
e.None of the above
a.Gametophytes of bryophytes
b.Gametophytes of seedless vascular plants
c.Sporophytes of bryophytes
d.Sporophytes of seedless vascular plants
e.None of the above
Solution
c. Sporophytes of bryophytes
d. Sporophytes of seedless vascular plants
The sporophyte is associated with the green parts of the plant roots, stems, leaves and flowers.
The gametophytes are the haploid structures, which are present in the multicellular stage of
plants, which give rise haploid gametes (both male and female gametes)..
Chapter 3 - Islamic Banking Products and Services.pptx
Chapter 3 - Islamic Banking Products and Services.pptxMohd Adib Abd Muin, Senior Lecturer at Universiti Utara Malaysia
This slide is prepared for master's students (MIFB & MIBS) UUM. May it be useful to all.Sectors of the Indian Economy - Class 10 Study Notes pdf
The Indian economy is classified into different sectors to simplify the analysis and understanding of economic activities. For Class 10, it's essential to grasp the sectors of the Indian economy, understand their characteristics, and recognize their importance. This guide will provide detailed notes on the Sectors of the Indian Economy Class 10, using specific long-tail keywords to enhance comprehension.
For more information, visit-www.vavaclasses.com
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b) Analyze your network IP packets headers and contents.Solu.pdf
b) Analyze your network \"IP packets\" headers and contents.
Solution
ip packets is like communication network from one device to another..ip packets uses the packets
to send the data packets from one ntework to another
and its like transmission of networks
IP HEADER : ip header contains the informetion about which ip vesrion we are using and what
is the source and destinetion of ip header
IP CONTENT : content is nothing but the all informetion which is covered by everything is one
set like photos videos.
A type A man is the son of a type O father and type A mother. If he .pdf
A type A man is the son of a type O father and type A mother. If he has children with a type O
woman, what are the following probabilities?
A. A type AB child?
B. A type O child?
C. The first three children with type A blood?
Solution
The child can have only two blood types possible type A and type O.
a) Probability for type AB =0
b) Probability for type 0 =0.5
c) Probablity for The first three children with type A blood = 3/4.
A One-Way Analysis of Variance is a way to test the equality of thre.pdf
A One-Way Analysis of Variance is a way to test the equality of three or more means at one time
by using variances. Which of the following are necessary assumptions? The variances of the
populations must be homogeneous.
Solution
a. The samples must be independent.
d. The populations from which the samples were obtained must be normally or approximately
normally distributed..
5 000-0 SolutionEquity multiplier = Total Assets Equityor, .pdf
5 000-0
Solution
Equity multiplier = Total Assets / Equity
or, Equity Multiplier = (Total Assets / Equity) x (Net Profit / Net Profit) [ Muliply Numerator
and denominator by Net Profit]
or, Equity Multiplier = (Total Assets / Net profit) x (Equity / Net profit)
or, Equity Multiplier = (1 / ROA) x (ROE) = ROE / ROA = 9% / 1% = 9 (first option).
1sequences and sampling. Suppose we went to sample the x-axis from X.pdf
1sequences and sampling. Suppose we went to sample the x-axis from Xmin to Xmax using a
step size of step
A)Draw a picture of what is going on.
B) Write a expression for n the total number of samples involved (in terms of Xmin, Xmax and
step)
C) Write out the sequence of x-samples
D) Write a direct and general expression for xi that captures the sequence
E) Write a recursive expression for the sequence
F) Write a program to compute and store the x-samples over the range -5x5 using a step size of
0.1 do everything in main ()
2 . We talked about the following string functions that are available in C (as long as you include
string.h):
int strlen(char str[])
void strcpy(char str1[], char str2[])
void strcat(char str1[], str2[])
Write your own versions of these functions; for example: int paul_strlen(int char str[]). Hint: for
your version of the strlen function, start at the first character in the array and keep counting until
you find the ‘\\0’ character (use a while loop for this). Note: Use your version of the strlen
function in the strcpy and strcat functions.
9. We want to insert a number into an array.
(a) Formulate the problem mathematically with two sequences: x and y. (b) Write a function of
the form:
insertNumIntoArray(int n, int array[], int num, int index)
The function inserts num into the array at the specified index. The rest of the array then follows.
For example, if num = 9 and index = 3 and array = [7 2 8 8 3 1 2] then the function will produce:
array = [7 2 8 9 8 3 1 2]
Note: assume that array is properly dimensioned to have at least 1 extra space for storage.
10. Repeat #2 by for the delete operation; that is, we want to delete a single element (at a
specified index) from an array; for example, suppose index = 3 and array = [50 70 10 90 60 20],
then the result will be
array: [50 70 10 60 20]
11. Repeat #2 by for an insert operation where we are inserting several values into the array. The
function should be of the form:
int insertArrayIntoArray(int n, int inArray[],
int nInsert, int insertArray[], int outArray[], int index)
The dimension of outArray is returned (explicitly). For example:
inArrayarray: [7 2 8 6 3 9]
insertArray: [50 60 70]
index: 2
outArray: [7 2 50 60 70 8 6 3 9]
Assume that outArray is large enough to hold all n + nInsert values.
Solution
#include
//Simulates strlen() library function
int paul_strlen(char str[])
{
int l;
for(l = 0; str[l] != \'\\0\'; l++) ;
return l;
}
//Simulates strcpy() library function
void paul_strcpy(char str1[], char str2[])
{
int c;
for(c = 0; str1[c] != \'\\0\'; c++)
str2[c] = str1[c];
str2[c] = \'\\0\';
printf(\"\ Original String: %s\", str1);
printf(\"\ Copied String: %s\", str2);
}
//Simulates strcat() library function
void paul_strcat(char str1[], char str2[])
{
int i, j;
for(i = 0; str1[i] != \'\\0\'; i++) ;
for (j = 0; str2[j] != \'\\0\'; i++, j++)
{
str1[i] = str2[j];
}
str1[i] = \'\\0\';
printf(\"\ Concatenated String: %s\", str1);
}
int main()
{
char data1[20], data2[20];
pri.
4 (4 points). How would you describe the difference in cell structur.pdf
4 (4 points). How would you describe the difference in cell structures between a thermophile and
a psychrophile? Include mention of cell membrane construction and enzyme/protein structure at
least.
Solution
Thermophilic bacteria have more long chain fatty acids or lipids which maintain the membrane
fluidity. They have more tetraether lipids that increase the membrane stability. Hence they
survive in high temperatures. Thermophiles also have a high level of K+ salts which protect the
DNA and proteins by forming hydrophilic shell. Thermophiles have have acidic aminoacids.
psychrophiles grow at very low temperatures. They have high polyunsaturated and methyl-
branched fatty acids, and/or a shorter acyl-chain length which are generally present in cis
configuration . This increases membrane fluidity and reduces the interactions between the
membrane preventing freezing at low temperature. They have cryoprotectants and antifreeze
agents..
Distinguish between cell fate and cell commitment. How can one assay.pdf
Distinguish between cell fate and cell commitment. How can one assay these aspects of cellular
phenotype by experimental embryology?
Solution
Cell fate--It means the development of the cell into an embryo and the events which govern the
development of the egg cell throught the process of embryogenesis.The factors which determine
the cell fate include cell signaling pathways like notch pathway etc, regulation and control by
pathways like mRNA,siRNA inside the cell,development of different cavity inside the embryo,
development of ectoderm,mesoderm and endoderm, and the regulation of cell fate determination
by release of various controlling hormones.
cell commitment--it involves the orientation and decision by the cell to undergo differentiation
and follow a particular developmental pathway.It involves the release of various growth
promoting factors by stem cells which lead to the differentiation of stem cells into mature cells
by triggering and activating a particular organ development pathway example--blood cells.
There are several aspects of determination of the cellular phenotype by experimental
embryology----1)sex determination pathways in insects and dosage compensation 2)regulation of
different genes in different climates,temperature, and background environment determines the
activation and inhibition of development genes in caterpillar larvae of insects3) the division of
fertilized egg into morula, blastula, gastrula ,and development of different layers by cell
regulatory protein factors 4)role of cell adhesion protein molecules during cell differentiation
pathway 5)development of cell symmetry and asymmetry of the insects 6)development of
embryo cavity and germline layers of development after fertilization 6)development of cells to
tissues to organs with the help of secreting chemicals, protein factors and hormones which
activate the differentiation pathway..
Differentiate the processes of oogenesis and spermatogenesis. Are th.pdf
Differentiate the processes of oogenesis and spermatogenesis. Are there similarities between the
two processes? Differences? Consider the cells, their structures and functions.
Solution
Comparison between Spermatogenesis and Oogenesis
Characteristics
Spermatogenesis
Oogenesis
Definition
The process of formation of sperm cells is called spermatogenesis
Formation of ovum is called oogenesis
Primary organ
Testis (in males)
Ovary (in females)
The source cell/ structure that undergoes gametogenesis
Spermatogonium
Primordial follicle
Source cell is preserved
Yes, Spermatogonium is a stem cell.
No
Type of cell division
Meiosis
Meiosis
Onset of gametogenesis
At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty.
During embryonic development.
Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive
cycles.
Continuous or cyclic
Continuous. Millions of sperms are produced each day.
Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days).
Occurrence of gametogenesis in life
Begins at puberty
Continues throughout life
Begins at menarche
Ends at menopause
No. of functional gametes in one cycle
4
1
Polar cells (that degenerate later) produced
No
Yes, one polar cell in each of meiosis I and II.
In spite of major difference of these two processes, they both share a common path and fate- they
meiotic divisions that produce haploid gametes.
Characteristics
Spermatogenesis
Oogenesis
Definition
The process of formation of sperm cells is called spermatogenesis
Formation of ovum is called oogenesis
Primary organ
Testis (in males)
Ovary (in females)
The source cell/ structure that undergoes gametogenesis
Spermatogonium
Primordial follicle
Source cell is preserved
Yes, Spermatogonium is a stem cell.
No
Type of cell division
Meiosis
Meiosis
Onset of gametogenesis
At puberty. Both meiosis I and II in spermatogonium begin at onset of puberty.
During embryonic development.
Meiosis I begins in fetal ovary. Meiosis II completes after onset of puberty in reproductive
cycles.
Continuous or cyclic
Continuous. Millions of sperms are produced each day.
Cyclic. Only one mature ovum is produced per reproductive cycle (of 28 days).
Occurrence of gametogenesis in life
Begins at puberty
Continues throughout life
Begins at menarche
Ends at menopause
No. of functional gametes in one cycle
4
1
Polar cells (that degenerate later) produced
No
Yes, one polar cell in each of meiosis I and II..
Define induction and give an example of deductive reasoningSolu.pdf
Define induction and give an example of deductive reasoning?
Solution
There are two methods of reasoning. One is Deductive approach and another is Inductive
approach.
Deductive reasoning:-This kind of reasoning works from the general to more specific. It is also
referred to as \"Top-down\" approach.
Examples:-
1) If X = Y and Y = Z, then by deductive reasoning X = Z.
2) Snakes are reptiles and reptiles are cold-blooded, by deductive reasoning it can be said that
snakes are cold-blooded.
Inductive reasoning:-
Inductive reasoning as opposed to deductive reasoning work from specific observations to
broader generalizations and theories. It is also referred to as \"bottom up\" approach in informal
sense.
Example:-
1) A just moved here from Hongkong. A has red hair, therefore by inductive reasoning, it can be
said that People from Hongkong have red hair..
A report says that the between-subjects factor of participants sal.pdf
A report says that the between-subjects factor of participants\' salary produced significant
differences in self-esteem. (a) What does this tell you about the design? (b) What does it tell you
about the results?
Solution
It means the design was effective enough to study the relation between the self esteem and
salary.
Secondly the variance ratio for the salary was significant.
City GUIWrite a Java GUI program which reads data about US cities..pdf
City GUI
Write a Java GUI program which reads data about US cities. The data file is different, but is in
the same format as in Lab #01. The GUI provides a combo box with the cities in the data file. As
cities are selected in the combo box, information about the city is displayed.
Requirements
1. The program must have a graphical user interface similar to the one on the bottom of the
Requirements section. It must have a text field, a button, a combo box, and labels.
2. One or more layout managers must be used. This example was created using a border layout.
The east and west panels each used a grid layout in addition.
3. A message must be displayed in the GUI after the data has been successfully read. A message
should also be displayed if the file is not found. The program should not exit in that case, but
allow the user to enter a different file name.
4. The combo box must be populated by the names of the cities in the data file in alphabetical
order. Note that this means the GUI starts with an empty combo box. Items are added to it as or
after the data file is read.
5. The program must provide an ActionListener for the Read button. It must also provide either
an ActionListener or an ItemListener for the combo box, which displays the state, zip code, and
time zone for that city.
6. The information for each city must be stored in its own object. Another object must be used to
manage the collection of cities. This is similar to Lab #01. For this assignment you may assume
that there will be no more than 100 cities.
7. The time zone should be displayed as the correct abbreviation for that city’s time zone in July.
Here are the abbreviations. If the city observes daylight savings time, the S should be changed to
a D. Note that some cities in the data file do not observe daylight savings time.
Hours relative to GMT -4 -5 -6 -7 -8 -9 -10
Standard Time Abr. AST EST CST MST PST AKST HST
I have attached an executable .jar file to the assignment that you can run to see how this program
should work. Copy it to your computer. Copy the data file into the same directory as the .jar file.
Suggestions
1. Create class diagrams first. This will help you think through and organize your program. There
are many ways to do this. Start with the class diagrams from Lab #01. Both classes will require
some additions.
2. Create the GUI and handle the IO in a separate class which extends JFrame. The main method
can be included in this class.
3. Don’t worry about making your implementation efficient or using wonderful abstract data
types (ADTs). You will be required to do that soon enough.
4. This program will take you several days to complete. Break it down into smaller tasks. This is
the order I used, yours may be different.
*Modify the class diagrams from Lab #01 to include new requirements that you identify now.
You can add additional functionality later as you discover the need.
*Implement those changes in the classes.
* Create a GUI class. Implement a non-functiona.
What is the role of sulfur chemoautotrophs in the sulfur cycle Deco.pdf
What is the role of sulfur chemoautotrophs in the sulfur cycle? Decomposition reduction
oxidation assimilation Reduction of CO2 to generate carbohydrates can occur when coupled to
the oxidation of reduced sulfur compounds (for examples, H2S). This latter process is carried out
by what type of sulfate oxidizers sulfate reducers
Solution
10
Oxidation
Sulfur chemotrophs oxidize sufur to derive energy for the synthesis of ATP.
What is the PDU at Layer 4 calledA. DataB. SegmentC. Packet.pdf
What is the PDU at Layer 4 called?
A. Data
B. Segment
C. Packet
D. Frame
E. Bit
Solution
B is correct. The Layer 4 PDU is called a segment.
A, C, D, and E are incorrect. A is incorrect because the data is the PDU for the top three layers.
C is incorrect because the packet is the PDU at the Network layer. D is incorrect because the
frame is the PDU for the Data Link layer. E is incorrect because the bit is the PDU at the
Physical layer..
Briefly describe the contributions to the quality movement made by e.pdf
Briefly describe the contributions to the quality movement made by each of the following
pioneers.
Shewhart
Dodge and Romig
Sarasohn
Deming
Juran
Solution
Walter Shewhart - The Grandfather of Total Quality Management.
His Ideas
The original notions of Total Quality Management and continuous improvement trace back to a
former Bell Telephone employee named Walter Shewhart. One of W. Edwards Deming\'s
teachers, he preached the importance of adapting management processes to create profitable
situations for both businesses and consumers, promoting the utilization of his own creation -- the
SPC control chart.
Dr. Shewhart believed that lack of information greatly hampered the efforts of control and
management processes in a production environment. In order to aid a manager in making
scientific, efficient, economical decisions, he developed Statistical Process Control methods.
Many of the modern ideas regarding quality owe their inspirtation to Dr. Shewhart.
He also developed the Shewhart Cycle Learning and Improvement cycle, combining both
creative management thinking with statistical analysis. This cycle contains four continuous steps:
Plan, Do, Study and Act. These steps (commonly refered to as the PDSA cycle), Shewhart
believed, ultimately lead to total quality improvement. The cycle draws its structure from the
notion that constant evaluation of management practices -- as well as the willingness of
management to adopt and disregard unsupported ideas --are keys to the evolution of a successful
enterprise.
Harold French Dodge (January 23, 1893 in Lowell, Massachusetts – December 10, 1976) was
one of the principal architects of the science of statistical quality control. He is universally
known for his work in originating acceptance sampling plans for putting inspection operations on
a scientific basis in terms of controllable risks.
From 1917 to 1958 worked at quality assurance department at Bell Laboratories with Walter
Shewhart, George Edwards, Harry Romig, R. L. Jones, Paul Olmstead, E.G.D. Paterson, and
Mary N. Torrey. At that time the basic concepts of acceptance sampling was developed, such as :
· Consumer\'s Risk,
· Producer\'s Risk,
· double sampling,
· lot tolerance percent defective (LTPD), and
· average outgoing quality limit (AOQL).
Also he originated several types of:
· acceptance sampling schemes,
· CSP type continuous sampling plans,
· chain sampling plans, and
· skip-lot sampling plans.
During World War II, Dodge served as a consultant to the Secretary of War, and was chairman
of the American Standards Association (now the American National Standards Institute) War
Committee Z1, which prepared the Z1.1, Z1.2, and Zl.3 quality control standards.
After he retired from Bell Labs in 1958, Dodge became a professor of applied mathematical
statistics at Rutgers. He also served as a consultant to NASA and the Sandia Corporation.
The American Society for Testing and Materials honors Harold Dodge\'s memory with the
Harold F. Dodge Award.
Ho.
Background Angiosperms (flowering plants) are the largest Phylum in .pdf
Background Angiosperms (flowering plants) are the largest Phylum in the plant kingdom
Plantae, (Note that the Phylum A may also called Anthophyta or Magnoliophyta in different
classification systems.) These plants have true roots, stems, leaves, and flowers. The roots grow
into the soil to anchor the plant in place and take up water and nutrients. The leaves are above
ground and act as the main organs for photosynthesis. Stems provide above ground support for
the plant and operate as conduits to move nutrients and water throughout the plant. Flowers
contain the male and female reproductive organs of the plant. Assignment Details Part 1 Plants
develop specialized roots, stems, leaves, and flowers to make them better adapted to their
environment. People exploit many modifications as vegetables. An example of this is the taproot
of carrots. The carrot plant stores sugars in the taproot to supply energy for the formation of
flowers in later development. People dig up and eat the taproots while the sugar content is still
high. Complete the following: Describe the anatomy (structure) and physiology (function) of a
plant. Is the structure of your plant a modification from the standard root, stem, leaf, or flower?
Explain how this structure differs from the standard root, stem, leaf, or flower. Compare the
plant that you have chosen, its structure, and its function with 1 or more of your classmates\'
choices. Part 2 Animals are important for humans in ways other than providing food.
Medicines, clothing, beauty, fuel, air quality, building materials, and industrial products are a
few of the ways in which humans exploit animals. Complete the following: Choose l product or
benefit that people get from animals (besides food), describe it, and explain how people benefit.
Which do you think is the most important benefit of animals? Select 1 benefit and explain. List
3 adaptations that animals have that plants do not have. Provide references in APA format. This
includes a reference list and in-text citations for references used throughout the assignment
Solution
1. the normal function of roots is fixation of the plant in the soil absorption of water and
Minerals from the soil and conduction of the materials to the stem but in some plants roots
perform special functions like storage of food materials(carrot,raddish) respiration(rhizopora)
vegetative reproduction (sweet potato)support(betel vine) photosynthesis(orchids).
Betel vine:
Weak Climbing Plants produce some advantitious roots which cling to the supports and help the
plants in climbing they sometimes they secrete sticky substance or a disc like structures for this
purpose.
Betel Vine is a plant having climbing roots these adventitious roots are produced from their
nodes which help them to climb over a support another example is money plant.
These roots are aerial roots which twist and clasp the support to help plant to vlimb.
Normal rootsd show positive geotropism but these climbing roots show negativ.
Write a Pep8 Assembly program that reads in and stores two integers .pdf
Write a Pep/8 Assembly program that reads in and stores two integers from the user and prints
out their sum in the form: Total = Submit: Pep/8 formatted program in a text file
Solution
Answer:
Assembly Language Program :
.zero 1
.LC0:
.string \"Enter two integers: \"
.LC1:
.string \"Total\"
.LC2:
.string \" = \"
main:
push rbp
mov rbp, rsp
sub rsp, 16
mov esi, OFFSET FLAT:.LC0
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
lea rax, [rbp-8]
mov rsi, rax
mov edi, OFFSET FLAT:std::cin
call std::basic_istream >::operator>>(int&)
mov rdx, rax
lea rax, [rbp-12]
mov rsi, rax
mov rdi, rdx
call std::basic_istream >::operator>>(int&)
mov edx, DWORD PTR [rbp-8]
mov eax, DWORD PTR [rbp-12]
add eax, edx
mov DWORD PTR [rbp-4], eax
mov esi, OFFSET FLAT:.LC1
mov edi, OFFSET FLAT:std::cout
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov esi, OFFSET FLAT:.LC2
mov rdi, rax
call std::basic_ostream >& std::operator<< >(std::basic_ostream >&, char const*)
mov rdx, rax
mov eax, DWORD PTR [rbp-4]
mov esi, eax
mov rdi, rdx
call std::basic_ostream >::operator<<(int)
mov eax, 0
leave
ret
__static_initialization_and_destruction_0(int, int):
push rbp
mov rbp, rsp
sub rsp, 16
mov DWORD PTR [rbp-4], edi
mov DWORD PTR [rbp-8], esi
cmp DWORD PTR [rbp-4], 1
jne .L5
cmp DWORD PTR [rbp-8], 65535
jne .L5
mov edi, OFFSET FLAT:std::__ioinit
call std::ios_base::Init::Init()
mov edx, OFFSET FLAT:__dso_handle
mov esi, OFFSET FLAT:std::__ioinit
mov edi, OFFSET FLAT:std::ios_base::Init::~Init()
call __cxa_atexit
.L5:
nop
leave
ret
push rbp
mov rbp, rsp
mov esi, 65535
mov edi, 1
call __static_initialization_and_destruction_0(int, int)
pop rbp
ret.
Why is it important that the duration of cell-cycle phases be contro.pdf
Why is it important that the duration of cell-cycle phases be controllable in adult multicellular
organisms? b. What is MFP? Describe how this complex affects the cell cycle.
Solution
SOME BRIEF DESCRIPTION WHICH HELP YOU TO UNDERSTAND QUESTION AND IT
S ANSWER FOR MY KNOWLEDGE IT HELPS YOU
NOTE ;------ (The cell grows . The duration of these cell cycle phases varies considerably in
different kinds of cells. For a typical rapidly proliferating human cell with a total cycle time of
24 hours, the G1 phase might last about 11 hours, S phase about 8 hours, G2 about 4 hours, and
M about 1 hour.)
The cell cycle or cell-division cycle is the series of
events that take place in a cell leading to its division and duplication of its DNA (DNA
replication) to produce two daughter cells. In bacteria, which lack a cell nucleus, the cell cycle is
divided into the B, C, and D periods. The B period extends from the end of cell division to the
beginning of DNA replication. DNA replication occurs during the C period. The D period refers
to the stage between the end of DNA replication and the splitting of the bacterial cell into two
daughter cells. In cells with a nucleus, as in eukaryotes, the cell cycle is also divided into three
periods: interphase, the mitotic (M) phase, and cytokinesis. During interphase, the cell grows,
accumulating nutrients needed for mitosis, preparing it for cell division and duplicating its DNA.
During the mitotic phase, the chromosomes separate. During the final stage, cytokinesis, the
chromosomes and cytoplasm separate into two new daughter cells. To ensure the proper division
of the cell, there are control mechanisms known as cell cycle checkpoints.
The cell-division cycle is a vital process by which a
single-celled fertilized egg develops into a mature organism, as well as the process by which
hair, skin, blood cells, and some internal organs are renewed. After cell division, each of the
daughter cells begin the interphase of a new cycle. Although the various stages of interphase are
not usually morphologically distinguishable, each phase of the cell cycle has a distinct set of
specialized biochemical processes that prepare the cell for initiation of cell divisions.
Cell cycle checkpoints are control mechanisms in eukaryotic cells which
ensure proper division of the cell. Each checkpoint serves as a potential point along the cell
cycle, during which the conditions of the cell are assessed, with progression through the various
phases of the cell cycle occurring when favorable conditions are met. Currently, there are three
known checkpoints: the G1 checkpoint, also known as the restriction or start checkpoint or
(Major Checkpoint); the G2/M checkpoint; and the metaphase checkpoint, also known as the
spindle checkpoint.
Effects of MULTIFOLIATE-PINNA, AFILA, TENDRIL-LESS and UNIFOLIATA genes on
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In order to dissect the genetic regulation of leafblade
morphogenesis, 16 genotypes of pea, constructed.
How does the Rob proteins phosphorylation stage affect the cell cy.pdf
How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression?
How does the Rob proteins phosphorylation stage affect the cell cycle and cancer progression?
Solution
The RB(Retinonlastoma) is a tumor-suppressor protein and it has capability to inhibit
progressing the cell cycle meticulously at the G1 checkpoint, Hyperphosphorylation of RB
trigger release of “E2F” factor. G1 cyclin-cyclin dependent kinases (CDK) complex is mainly
involved in the phosphorylation of RB to activate E2F result in initiation of transcription rate to
promote DNA synthesis. These events are leading higher DNA synthesis to promote cell division
in cell cycle result in tumor cells formation.
Would you use physical segmentation or VLANs. Why or why not and whe.pdf
Would you use physical segmentation or VLANs. Why or why not and where would you use
each?
In the Diagram below where would you put them. HU HUB Roud er Router Switeh Sreh Routen
pe Wineless access point DMZ Zone auess point out ey acess poinf rele 70adba Broadbond
wrreless access auuess point server access point access
Solution
Vlans and physical segmentation are both different things which both are used when we want to
divided network. Basically physical segmentation means to have many different network on the
basis of physical connectivity and VLANs may be defined as the division of network into logical
parts and to have different networks with different nodes. Suppose if you have switch then it is
better to have vlans rather than physical segmentation since as we know that switch is
broadcasting frams and each interface of switch is present into different collision domain. So in
order to have different broadcast domian we have to configure vlans into the switch. Which can
certainly provide us security on the basis of different network.
In above diagram we can have vlans on the every switch where each node connected to the
switch would be placed into different network thus saves us from the collisions and broadcasting
to the other nodes unnecessarily which may cause unnecessary overheads and less security in the
network..
Which plants produce leaves (Select all that apply.)a.Gametophyte.pdf
Which plants produce leaves? (Select all that apply.)
a.Gametophytes of bryophytes
b.Gametophytes of seedless vascular plants
c.Sporophytes of bryophytes
d.Sporophytes of seedless vascular plants
e.None of the above
a.Gametophytes of bryophytes
b.Gametophytes of seedless vascular plants
c.Sporophytes of bryophytes
d.Sporophytes of seedless vascular plants
e.None of the above
Solution
c. Sporophytes of bryophytes
d. Sporophytes of seedless vascular plants
The sporophyte is associated with the green parts of the plant roots, stems, leaves and flowers.
The gametophytes are the haploid structures, which are present in the multicellular stage of
plants, which give rise haploid gametes (both male and female gametes)..
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