right now I am using a l293d chip to control my DC motor which is 6 .pdf
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right now I am using a l293d chip to control my DC motor which is 6 volt. I was wondering if there was a python code that can control the motor based on temperature. I am using tmp102 ic2 temperature sensor also. Solution L293D IC generally comes as a standard 16-pin DIP (dual-in line package). This motor driver IC can simultaneously control two small motors in either direction; forward and reverse with just 4 microcontroller pins (if you do not use enable pins). Some of the features (and drawbacks) of this IC are:.
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which secondary storage devices are appropriate for Saving data durin.pdf
which secondary storage devices are appropriate for Saving data during edit process. Backup
data and applications. Distribute data and applications. Transport data and applications. Archive
data and applications.
Solution
3)
a - magnetic tapes
b- NAS /SAN/tape library
c - external hard disk
d- Flash drives like pendrive,memory card
e - compact disk,magnetic disks,DVD.
What were issues with the Orion Shield ProjectSolutionA good .pdf
What were issues with the Orion Shield Project?
Solution
A good project management plan takes some preparation. In the case study of The Orion Shield
Project that preparation never seemed to occur. From the beginning, the project was plagued with
the subcontractor, Scientific Engineering Corporation (SEC), bidding for a project that they
knew their current product would not be able to do. The case study shows the actions of an
inexperienced professional, his poor planning methods, unethical decision skills and inability to
manage a project to completion. While the project was looked at as a success, ultimately it was
doomed from the start. There was specification issues that initially concerned Gary Allison,
SEC’s project manager, but he was told by the Director of Engineering, Henry Larsen, that this
wasn’t a big issue. By the project manager lying to the primary contractor, Space Technology
Institute (STI), this created an unethical work environment from the start. There are various
issues throughout the project such as contractual, legal, technical and ethical issues. A project
manager should be concerned with time and cost, however, Gary was concerned mainly about
his new career and promotion. With the stakeholders waiting for a complete project and funding
on the line, the project team continuously upset STI until the primary contractor made a decision
to work more closely with SEC to ensure accuracy. With good project management skills, a
project manager uses tools to ensure project accuracy, timeliness and has weekly status meetings
with the primary contractor to keep them aware of any changes that may occur. In this paper you
will see an unorganized project with overrun funding and lack of organization.
Recommendations are made to enhance the performance of the project manager and the overall
improvements that should have been addressed at the start..
What is the MOST likely mode of inheritance of the trait shown in the.pdf
What is the MOST likely mode of inheritance of the trait shown in the pedigree below,
assuming that the trait is rare and fully penetrant? Autosomal recessive X-linked recess X-
linked dominant Y-linked Consider the pedigree shown in Figure 6.3. lf individual [V-7
married a person who was unaffected with Waardenburg syndrome, what is the probability that
their first child would have Waardenburg syndrome? 1.0 3/4 1/2 1/4 0
Solution
1. Autosomal recessive :
Because both male and female affected (autosomal) and there is a skipping generation
(recessive).
2. Answer: 1/2 :
Here waarberg syndrome is a autosomal dominant syndrome. One affected individual is
sufficient to give 50% of their children will get affected..
What is frequency What is wavelength Calculate the frequency of s.pdf
What is frequency? What is wavelength? Calculate the frequency of signals with wavelengths
of 40 m, 5 m, and 8 cm. A commonly used frequency range is 902 to 928 MHz. Compute the
bandwidth of this frequency range? Classify the following applications of electronic
communications as simplex (one-way) or duplex (two-way). [Research Question]
Solution
Answer :
1.What is frequency ?
Answer :
1.The number of cycles per unit of time is called frequency.
2.Frequency is measured in the Hertz units(Hz)
3.Frequency related the number of waves that pass a fixed place in a given amount of time.
..............
2.What is Wavelength ?
Answer :
1.Wavelength is the Distance Between the identical points of two consecutive waves.
2.It represented by Greek letter (lambda)
3.Usually it is measured in meters.
4. =v/f where v=speed & f=frequency
.....................
What are some challenges in adopting HIT to CQISolutionAssump.pdf
What are some challenges in adopting HIT to CQI?
Solution
Assumption
HIT means healthcare in information technology
CQI means Clinical Quality Improvement.
Clinical Quality Improvement (CQI): Ensures a comprehensive approach to integrating clinical
knowledge into health IT. Provides subject matter expertise on policies, standards, and tools that
give providers and consumers the information and tools needed to identify high risk
conditions,assist in decision making, and measure care quality.
All industries are besieged by claims that Internet-based content, services, and applications can
solve a wide range of problems that have plagued prior efforts to implement other information
technologies. In health care, these problems include the difficulty of integrating a diverse,
heterogeneous set of legacy systems; a lack of data standards; the complexity of medical
practice; and complex, politicized health care organizational structures.
It is not clear whether the traditional hurdles to implementation of complex clinical information
systems, such as the electronic medical record, are significantly lowered by the use of the
Internet and related technologies. For example, although the capability to provide access to a set
of services, content, and applications through a wide array of workstations and other devices
might reduce the support costs for an application, it is not clear that the aggregate cost of the
application infrastructure is reduced.
Some of the challenges are
Standards Development and Harmonization- This is important for interoperability of various
systems, testing of features and for intergrity of data. This is yet to be done.
Health Information Exchange- This includes engaging HIE participants and assisting them to
implement HIE services, and providing a focus on the right set of standards, protocols, legal
agreements, specifications, and services needed to manage the exchange of health information.
Certification and Accreditation: Provides vendors and developers with clear criteria for
developing their products by issuing certification criteria for the Certification Program.
Awarness among patients or end user: This will help in engaging consumers directly with IT
system to empower them to meaningfully use their health information and actively participate in
their health care through improved access to health information.
Use the graph to determine the functions domain; the functions .pdf
Use the graph to determine the function\'s domain; the function\'s range; the x-intercepts, if
any; the y-intercept, if any; and e. the missing function values, indicated by question marks,
below. f(-2) = ? f(2) = ? The domain is (-infinity, infinity). (Use interval notation.) The range is
(Use interval notation.)
Solution
Range is possible Values of Y for domain x.
Here, In the graph... Maximum value of Y is 1.
And Minimum value of Y is going to infinity.
Therefore, Range of this function should be ( -infinity , 1]..
Transmembrane proteins _. All of these statements are correct. may pa.pdf
Transmembrane proteins _. All of these statements are correct. may pass entirely through the
lipid bilayer and protrude from both the extracellular and cytoplasmic sides of the membrane.
Includes lipid-anchored proteins. are called peripheral proteins. Using figure 4.13 (below), what
type of interactions do peripheral proteins (shown in light green) use to associate with a
membrane? Peripheral proteins use interactions to associate with a membrane, either on the or
extracellular side. Lipid-anchored proteins (as show in 4.13 below) are bound covalently to a
lipid group within the membrane. Which types of membrane lipids can lipid-anchored proteins
be attached? Select all that apply. a phenyl group (a long-chain hydrocarbon build from five-
carbon units) a fatty acid
Solution
3. A Phosphatidylinositol. These are basically lipid-anchored protein.
Lipid anchored protein are bound covalently to a lipid group. These are proteins which are
located on the surface of the cell membrane and covalently attached to lipid-embedded within the
cell membrane.
The coelom of most deuterostomes develops from outpocketings of the e.pdf
The coelom of most deuterostomes develops from outpocketings of the endoderm of the gut.
Torsion is a unique characteristic of all mollusks. Fishes may rely on their lungs for buoyancy.
Sessile animals are often filter feeders. Fruits are expanded anthers. The spleen filters blood and
removes damaged red blood cells. Coenocytic fungi have many nuclei in one extensive
cytoplasm. Segmentation allowed the development of specialized appendages. Animals that
have nerves always have brains. Mollusks have closed circulatory systems. You could kill a
starfish by cutting off its arms. Reptiles never have gill slits. The heart of an adult amphibian
has three chambers. Phloem transports organic nutrients from leaves to roots. Sieve tube
members are found in xylem. Tracheids are a type of cell in phloem. Xylem transports water
from roots to leaves. Sponges have three tissue layers. Higher plants have a minute diploid
phase. Many fungi depend on plants for mineral absorption. If lichen were sprayed with a
chemical that blocked photosynthesis, it would probably die. The insects include more species
than any other group. Poikilotherms are unable to produce body heat. Leaves are broad and thin
to minimize their absorption of oxygen. Water can hold more oxygen than air can. Book lungs
are found in crustaceans and mollusks, while external feathery gills are found in arachinids.
Saprobes parasitize plants. In coelomates, but not in pseudocoelomates, a peritoneum
completely lines the coelomic cavity Trypanosomes possess a kinetoplast (a mass of
mitochondrial DNA close to the nucleus) Psychrophiles live in extremely hot conditions Above
100 degree C Spirogyra is a plant protist Shigella spp is a bacteria that causes dysentery
Anthrax is a bacteria that can be used in terrorism. Plants have internal clocks.
Solution
17. True
18. true
19. false
20. True
21. false
22. True
23. True
24. True
25.false (platyhelminthes)
26.True
27. false
28. True
29.false (only in croodile, birs and mammals)
30.True
31.false
32. false
33. True
34. false diploblastic
35. false
36 True
37 true
38 True
39 True
40 True
41 True
42 True
43 false
44 True
45 True
46 false
47 True
48 True
49. True
50. True (hormone).
Recommended
which secondary storage devices are appropriate for Saving data durin.pdf
which secondary storage devices are appropriate for Saving data during edit process. Backup
data and applications. Distribute data and applications. Transport data and applications. Archive
data and applications.
Solution
3)
a - magnetic tapes
b- NAS /SAN/tape library
c - external hard disk
d- Flash drives like pendrive,memory card
e - compact disk,magnetic disks,DVD.
What were issues with the Orion Shield ProjectSolutionA good .pdf
What were issues with the Orion Shield Project?
Solution
A good project management plan takes some preparation. In the case study of The Orion Shield
Project that preparation never seemed to occur. From the beginning, the project was plagued with
the subcontractor, Scientific Engineering Corporation (SEC), bidding for a project that they
knew their current product would not be able to do. The case study shows the actions of an
inexperienced professional, his poor planning methods, unethical decision skills and inability to
manage a project to completion. While the project was looked at as a success, ultimately it was
doomed from the start. There was specification issues that initially concerned Gary Allison,
SEC’s project manager, but he was told by the Director of Engineering, Henry Larsen, that this
wasn’t a big issue. By the project manager lying to the primary contractor, Space Technology
Institute (STI), this created an unethical work environment from the start. There are various
issues throughout the project such as contractual, legal, technical and ethical issues. A project
manager should be concerned with time and cost, however, Gary was concerned mainly about
his new career and promotion. With the stakeholders waiting for a complete project and funding
on the line, the project team continuously upset STI until the primary contractor made a decision
to work more closely with SEC to ensure accuracy. With good project management skills, a
project manager uses tools to ensure project accuracy, timeliness and has weekly status meetings
with the primary contractor to keep them aware of any changes that may occur. In this paper you
will see an unorganized project with overrun funding and lack of organization.
Recommendations are made to enhance the performance of the project manager and the overall
improvements that should have been addressed at the start..
What is the MOST likely mode of inheritance of the trait shown in the.pdf
What is the MOST likely mode of inheritance of the trait shown in the pedigree below,
assuming that the trait is rare and fully penetrant? Autosomal recessive X-linked recess X-
linked dominant Y-linked Consider the pedigree shown in Figure 6.3. lf individual [V-7
married a person who was unaffected with Waardenburg syndrome, what is the probability that
their first child would have Waardenburg syndrome? 1.0 3/4 1/2 1/4 0
Solution
1. Autosomal recessive :
Because both male and female affected (autosomal) and there is a skipping generation
(recessive).
2. Answer: 1/2 :
Here waarberg syndrome is a autosomal dominant syndrome. One affected individual is
sufficient to give 50% of their children will get affected..
What is frequency What is wavelength Calculate the frequency of s.pdf
What is frequency? What is wavelength? Calculate the frequency of signals with wavelengths
of 40 m, 5 m, and 8 cm. A commonly used frequency range is 902 to 928 MHz. Compute the
bandwidth of this frequency range? Classify the following applications of electronic
communications as simplex (one-way) or duplex (two-way). [Research Question]
Solution
Answer :
1.What is frequency ?
Answer :
1.The number of cycles per unit of time is called frequency.
2.Frequency is measured in the Hertz units(Hz)
3.Frequency related the number of waves that pass a fixed place in a given amount of time.
..............
2.What is Wavelength ?
Answer :
1.Wavelength is the Distance Between the identical points of two consecutive waves.
2.It represented by Greek letter (lambda)
3.Usually it is measured in meters.
4. =v/f where v=speed & f=frequency
.....................
What are some challenges in adopting HIT to CQISolutionAssump.pdf
What are some challenges in adopting HIT to CQI?
Solution
Assumption
HIT means healthcare in information technology
CQI means Clinical Quality Improvement.
Clinical Quality Improvement (CQI): Ensures a comprehensive approach to integrating clinical
knowledge into health IT. Provides subject matter expertise on policies, standards, and tools that
give providers and consumers the information and tools needed to identify high risk
conditions,assist in decision making, and measure care quality.
All industries are besieged by claims that Internet-based content, services, and applications can
solve a wide range of problems that have plagued prior efforts to implement other information
technologies. In health care, these problems include the difficulty of integrating a diverse,
heterogeneous set of legacy systems; a lack of data standards; the complexity of medical
practice; and complex, politicized health care organizational structures.
It is not clear whether the traditional hurdles to implementation of complex clinical information
systems, such as the electronic medical record, are significantly lowered by the use of the
Internet and related technologies. For example, although the capability to provide access to a set
of services, content, and applications through a wide array of workstations and other devices
might reduce the support costs for an application, it is not clear that the aggregate cost of the
application infrastructure is reduced.
Some of the challenges are
Standards Development and Harmonization- This is important for interoperability of various
systems, testing of features and for intergrity of data. This is yet to be done.
Health Information Exchange- This includes engaging HIE participants and assisting them to
implement HIE services, and providing a focus on the right set of standards, protocols, legal
agreements, specifications, and services needed to manage the exchange of health information.
Certification and Accreditation: Provides vendors and developers with clear criteria for
developing their products by issuing certification criteria for the Certification Program.
Awarness among patients or end user: This will help in engaging consumers directly with IT
system to empower them to meaningfully use their health information and actively participate in
their health care through improved access to health information.
Use the graph to determine the functions domain; the functions .pdf
Use the graph to determine the function\'s domain; the function\'s range; the x-intercepts, if
any; the y-intercept, if any; and e. the missing function values, indicated by question marks,
below. f(-2) = ? f(2) = ? The domain is (-infinity, infinity). (Use interval notation.) The range is
(Use interval notation.)
Solution
Range is possible Values of Y for domain x.
Here, In the graph... Maximum value of Y is 1.
And Minimum value of Y is going to infinity.
Therefore, Range of this function should be ( -infinity , 1]..
Transmembrane proteins _. All of these statements are correct. may pa.pdf
Transmembrane proteins _. All of these statements are correct. may pass entirely through the
lipid bilayer and protrude from both the extracellular and cytoplasmic sides of the membrane.
Includes lipid-anchored proteins. are called peripheral proteins. Using figure 4.13 (below), what
type of interactions do peripheral proteins (shown in light green) use to associate with a
membrane? Peripheral proteins use interactions to associate with a membrane, either on the or
extracellular side. Lipid-anchored proteins (as show in 4.13 below) are bound covalently to a
lipid group within the membrane. Which types of membrane lipids can lipid-anchored proteins
be attached? Select all that apply. a phenyl group (a long-chain hydrocarbon build from five-
carbon units) a fatty acid
Solution
3. A Phosphatidylinositol. These are basically lipid-anchored protein.
Lipid anchored protein are bound covalently to a lipid group. These are proteins which are
located on the surface of the cell membrane and covalently attached to lipid-embedded within the
cell membrane.
The coelom of most deuterostomes develops from outpocketings of the e.pdf
The coelom of most deuterostomes develops from outpocketings of the endoderm of the gut.
Torsion is a unique characteristic of all mollusks. Fishes may rely on their lungs for buoyancy.
Sessile animals are often filter feeders. Fruits are expanded anthers. The spleen filters blood and
removes damaged red blood cells. Coenocytic fungi have many nuclei in one extensive
cytoplasm. Segmentation allowed the development of specialized appendages. Animals that
have nerves always have brains. Mollusks have closed circulatory systems. You could kill a
starfish by cutting off its arms. Reptiles never have gill slits. The heart of an adult amphibian
has three chambers. Phloem transports organic nutrients from leaves to roots. Sieve tube
members are found in xylem. Tracheids are a type of cell in phloem. Xylem transports water
from roots to leaves. Sponges have three tissue layers. Higher plants have a minute diploid
phase. Many fungi depend on plants for mineral absorption. If lichen were sprayed with a
chemical that blocked photosynthesis, it would probably die. The insects include more species
than any other group. Poikilotherms are unable to produce body heat. Leaves are broad and thin
to minimize their absorption of oxygen. Water can hold more oxygen than air can. Book lungs
are found in crustaceans and mollusks, while external feathery gills are found in arachinids.
Saprobes parasitize plants. In coelomates, but not in pseudocoelomates, a peritoneum
completely lines the coelomic cavity Trypanosomes possess a kinetoplast (a mass of
mitochondrial DNA close to the nucleus) Psychrophiles live in extremely hot conditions Above
100 degree C Spirogyra is a plant protist Shigella spp is a bacteria that causes dysentery
Anthrax is a bacteria that can be used in terrorism. Plants have internal clocks.
Solution
17. True
18. true
19. false
20. True
21. false
22. True
23. True
24. True
25.false (platyhelminthes)
26.True
27. false
28. True
29.false (only in croodile, birs and mammals)
30.True
31.false
32. false
33. True
34. false diploblastic
35. false
36 True
37 true
38 True
39 True
40 True
41 True
42 True
43 false
44 True
45 True
46 false
47 True
48 True
49. True
50. True (hormone).
The portion of the fungus that projects above the substrate and conta.pdf
The portion of the fungus that projects above the substrate and contain the spores is known as A)
septate hyphae B) septate mycellum C) aerial mycellum D) vegetative mycelium E) none of the
above = identity Bacteria contamination is suppressed in plain Sabouard\'s agar by the addition
of peptone and dextrose B) brain heart infusion media C) Lactophenol Cotton Blue D) all of the
above E) none of the above = identity
Solution
16. C. Aerial mycelium - the portion of the mycelium that project above the substrate and
produce spores. Because of its capability to produce spores, it is also known as reproductive
mycelium.
17. D. all of the above.
the appearance of symptoms of a genetic disorder at earlier ages in .pdf
the appearance of symptoms of a genetic disorder at earlier ages in successive Generations is
called
Solution
In genetics, anticipation is a phenomenon whereby as a genetic disorder is passed on the next
generation, the symptoms of the genetic disorder become apparent at an earlier age with each
generation..
Stuck on hw problem Assume that the forces acting on a mass are such.pdf
Stuck on hw problem Assume that the forces acting on a mass are such that the potential energy
is the function of x shown in Fig. 24-5: Sketch the solutions in the phase plane. Describe the
different kinds of motion that can occur.
Solution
Every sound we hear, every photon of light that hits our eyes, the movement of grass blown by
the wind and the regular beat of the tides are all examples of waves. They are all around us.
Visible, physical wave*s such as those we see when a rock is thrown into water are what many
people think about when they first began to think about waves. These waves have distinct
properties specific to their type but also exhibit characteristics in common with more abstract
waves such as sound waves and light (electromagnetic) waves.
Below is an interactive illustration that explores particle and energy motion for the three wave
types listed above. Once you have reviewed the illustration and accompanying text, test your
understanding of these concepts with the wave type
At the simplest level, waves are a disturbances that propagate energy through a medium.
Propagation of the energy depend on interactions between the particles that make up the
medium. Particles move as the waves pass through but there is no net motion of particles. This
means, once a wave has passed the particles return to their original position. As a result, energy,
not matter, is propagated by waves.
Different types of waves exhibits specific characteristics. These characteristics are used to
distinguish between wave types. Orientation of particle motion relative to the direction of energy
propagation is one way waves are characterized. There are three categories:
A second way that waves are characterized is by the types of matter they are able to travel
through.Electromagnetic waves (including light) can move through a vacuum. Physical waves
require matter through which to propagate. Physical waves are further distinguished by the
phases of matter through which they can move. Longitudinal waves can pass through liquids and
gasses while transverse waves require a material to be solid to propagate..
Some Giant Sulphur Bacteria are so large as to be visible to the nak.pdf
Some Giant Sulphur Bacteria are so large as to be visible to the naked eye. Nonetheless, they
have been very difficult to cultivate and classify. Why?
Solution
Cultivate
Grow very slowly. Metabolism is based on lithotrophic utilization of sulphide oxidation coupled
to nitrate reduction which is gives poor yields. In natural environments they grow in gradients
where one end of filament is sulphide rich, the other end nitrate rich. Difficult to simulate in the
lab.
Classify
Morphology has been used, but this is not linked to phylogeny. 16S rRNA hard to amplify due to
containing an intron.
Problem 1 Eclipses and occults Saturn has five satellites that rotat.pdf
Problem 1: Eclipses and occults Saturn has five satellites that rotate around the planet. The list
below records events occurring during the month of August, 1995, involving eclipses or occults
of Saturnian recorded satellites. For each event, the percent of light lost by the eclipsed or
occulted satelite is Event Aug 2 Aug 4 Aug 5 Aug 6 Aug 8 Light Loss (%) 61 56 Eclipse Eclipse
Occult Eclipse Eclipse Occult Occult Occult Occult Occult Eclipse Occult Occult Occult Occult
Occult Occult Occult Occult 46 Aug 11 39 Aug 12 Aug 14 Aug 14 Aug 15 Aug 15 AUg 16 Aug
20 Aug 23 Aug 23 AUg 25 Aug 28 1 20 12 Source: Astronomy magazine. Aug. 1995, p.60 via
Statistics (10m ed.] McClave and Sincich, p. 53
Solution
a).
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19
b).
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19
(eclipse are marked red)
c).
Based on above graph, eclipse is more likely lead to greater light loss.
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19.
please show work A hospital reports that two patients have been admi.pdf
please show work A hospital reports that two patients have been admitted who have contracted
Crohn\'s disease. Suppose our experiment consists of observing whether each patient survives or
dies as a result of the disease. The simple events and probabilities of their occurrences arc shown
in the table (where 5 in the first position means that patient 1 survives, D in the first position
means that patient 1 dies, etc.).
Solution.
Person.java is givenSolutionSince I didnt have the implement.pdf
Person.java is given
Solution
Since I didn\'t have the implementation of the Disease class, I made some assumptions and wrote
the code accordingly.
public class Person{
public final String name;
protected int health;
protected Disease[] diseases;
public Person(String name, int health){
this.name = name;
this.health = health;
}
public int getHealth(){
return this.health;
}
public void addDisease(Disease dis){
if (!diseasePresent(dis)){
Disease[] updated_diseases[diseases.length+1] = new Disease[];
for (int i<0; i.
In java languageWhich of the following is falseA) A JPasswordFi.pdf
In java language
Which of the following is false?
A) A JPasswordField shows that characters are being entered, but hides the actual character with
an echo character.
B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField.
C) Both JTextFields and JPasswordFields are single-line areas in which the user can enter text
via the keyboard.
D) JPasswordFields generate events; JTextFields do not.
When the user presses Enter in a JPasswordField, the GUI component generates an ________,
which is processed by an object that implements the interface _______.
A) ActionEvent, ActionListener
B) ActionEvent, ActionEventListener
C) TextEvent, TextListener
D) TextEvent, TextEventListener
JPasswordField contains method ___________to obtain the data entered.
A) getInput
B) getPassword
C) getText
D) getEcho
A JRadioButton is different from a JCheckBox in that _________.
A) a JRadioButton is a subclass of JToggleButton, JCheckBox is not
B) normally several JRadioButtons are grouped together and are mutually exclusive
C) a JRadioButton is a type of button, JCheckBox is not
D) a JRadioButton is a state button, JCheckBox is not
The fact that class Graphics is abstract contributes to Java’s portability because:
A) drawing is performed differently on every platform that supports Java. A subclass of Graphics
must be created that uses the drawing capabilities of the current platform.
B) objects of non-abstract classes can only be instantiated on the Windows platform.
C) drawing should not be performed on non-Linux platforms.
D) Class Graphics is not abstract.
JSliders generate what type of event?
A) ActionEvents
B) ChangeEvents
C) WindowEvents
D) MouseEvents
Menus are attached to windows by calling method ____________.
A) addMenuBar
B) setJMenuBar
C) setMenu
D) addJMenuBar
Solution
B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField.
A) ActionEvent, ActionListener
B) getPassword
B) normally several JRadioButtons are grouped together and are mutually exclusive
A) drawing is performed differently on every platform that supports Java. A subclass of Graphics
must be created that uses the drawing capabilities of the current platform.
A) ActionEvents
B) setJMenuBar.
in a completely randomized (one-way) anova, with other things being .pdf
in a completely randomized (one-way) anova, with other things being equal as the sample means
get closer to each other, the probability of rejecting the null hypothesis decreases.
true
or false
Solution
TRUE
because when other things being equal as the sample means get closer to each other
its true that the probability of rejecting the Ho decreases.
I want you to go the Next Generation Sunshine State standards (http.pdf
I want you to go the Next Generation Sunshine State standards
(http://www.cpalms.org/Public/search/Standard) homepage. Choose a grade level (not K) and
look at all the LIFE SCIENCE standards that are found under each of the Big Idea/Supporting
Idea topics. To see all the standards, click on the bolded letters.
please provide the grade level and copy the standards
Summarize what you, as a teacher, would be responsible for teaching your students.
Discuss how this framework changed your thoughts on what you, as a teacher, would be
responsible for teaching your students for the standards?
Discuss how insect topics might be used to help illustrate each of those LIFE SCIENCE
standards for that grade level. What insect information do you think could fit with that standard?
How could insects be used in related lessons?
Solution
Summary of grade 7 according to my skills are :-
Cell Biology
1. All living organisms are composed of cells, from just one to many trillions, whose details
usually are visible only through a microscope. As a basis for understanding this concept:
a. Students know cells function similarly in all living organisms.
b. Students know the characteristics that distinguish plant cells from animal cells, including
chloroplasts and cell walls.
c. Students know the nucleus is the repository for genetic information in plant and animal cells.
d. Students know that mitochondria liberate energy for the work that cells do and that
chloroplasts capture sunlight energy for photosynthesis.
e. Students know cells divide to increase their numbers through a process of mitosis, which
results in two daughter cells with identical sets of chromosomes.
f. Students know that as multicellular organisms develop, their cells differentiate.
Genetics
2. A typical cell of any organism contains genetic instructions that specify its traits. Those traits
may be modified by environmental influences. As a basis for understanding this concept:
a. Students know the differences between the life cycles and reproduction methods of sexual and
asexual organisms.
b. Students know sexual reproduction produces offspring that inherit half their genes from each
parent.
c. Students know an inherited trait can be determined by one or more genes.
d. Students know plant and animal cells contain many thousands of different genes and typically
have two copies of every gene. The two copies (or alleles) of the gene may or may not be
identical, and one may be dominant in determining the phenotype while the other is recessive.
e. Students know DNA (deoxyribonucleic acid) is the genetic material of living organisms and is
located in the chromosomes of each cell.
Evolution
3. Biological evolution accounts for the diversity of species developed through gradual processes
over many generations. As a basis for understanding this concept:
a. Students know both genetic variation and environmental factors are causes of evolution and
diversity of organisms.
b. Students know the reasoning used by C.
How are macromolecules transported across the nuclear envelope Desc.pdf
How are macromolecules transported across the nuclear envelope? Describe the structure that
regulates nuclear membrane permeability.
Solution
The transportation of molecules in and out of the nucleus is controlled by the nuclear pore
complex(NPC). Small molecules move through the nuclear membrane with ease but for
macromolecules like proteins, RNA etc, they require an association with transportation factors
like importins, karyopherins and exportins.
Proteins that need to be carried into the nucleus have nuclear localization signals (NLS) which
bind to the importins. NLS is a sequence of Amino acids which act as a tag. They are mostly
hydrophilic but hydrophobic sequences have also been documented.
Nuclear Import:
Importins bind to the macromolecule to be transported in the cytoplasm itself. The complex then
interacts with the NPC and passes through the channel. Once inside the nucleus, it interacts with
Ran-GTP and the importin dissociates from the macromolecule it was carrying. Then the
importin-Ran-GTP complex is transported to the cytoplasm where it is separated from importin
by the Ran Binding Protein (RanBP). After being separated from importin, GTPase activating
protein(GAP) binds with Ran-GTP and induces hydrolysis of GTP to GDP. The Ran-GDP
produced binds with the nuclear transport factor(NUTF2) and then returns to the nucleoplasm.
Inside the nucleus, Ran-GDP interacts with guanine nucleotide exchange factor(GEF) which
replaces GDP with GTP and Ran-GTP is born again to repeat the cycle.
Nuclear Export:
In the nucleoplasm, exportin binds the macromolecule to be transported outside the nucleus and
Ran-GTP. The complex diffuses through the pore to the cytoplasm where it dissociates. Ran-
GTP binds to GAP and hydrolysis resulting in Ran-GDP. Ran-GDP is returned to the nucleus
where it exchanges its ligand for GTP.
Nuclear Pore:
The nuclear pore complex is made of proteins called nucleoporins. Around 50% of nucleoporins
contains solenoid protein domains (alpha or beta-propeller fold) and the rest might have both as
separate structural domains. Each NPS has about 456 individual protein molecules with 30
distinct nucleoproteins. These are highly flexible proteins that lack an ordered secondary
structure..
GeneticsBackground info About 30 of the Irish population has red.pdf
Genetics
Background info: About 30% of the Irish population has red hair and an inability to tan (MC1R).
This is due to an unusual variant of the pigment molecule melanin, called red melanin. Another
fairly common inherited condition is restless legs syndrome (RLS). It is characterized by an
irresistible urge to move ones legs and can interfere with sleep, leading to chronic sleep
deprivation. Both red melanin and restless leg syndrome are recessive, and their genes are on
different chromosomes.
Molly and Sean met in San Francisco and were drawn to each other because of their striking red
hair and very pale skin (both had parents who emigrated from Ireland). Neither has a family
history of restless leg syndrome. A few years later they married and had identical twins, Jerry
and Garcia, both are red heads and their legs are fine. Danny and Betty lived in San Francisco
too. They both had brown hair, and Danny suffers from restless leg syndrome. Their daughter,
Cali, has dark brown hair and her legs are fine. Jerry and Cali have a son, Henry. He has red hair,
does not tan and has restless leg syndrome. Complete the family tree by assigning the genotypes
based on the phenotypes described above (Hint: obviously you can get some information from
the phenotype of the individual. Further clues can be obtained from the phenotypes of the
progeny). Use R or r for restless leg syndrome alleles or an underscore if the allele could be
either R or r. For example, R_ would represent either Rr or RR. Use M or m for the red melanin
alleles, M_ would represent MM or Mm.
A) (1.5 point) Which individuals must be heterozygous for restless leg syndrome?
B) (1.5 point) Which individuals might be heterozygous for restless leg syndrome?
C) (1 point) Which individuals must be heterozygous for the red melanin gene?
D) (2 point) Garcia marries a woman who has restless leg syndrome and red hair, what is the
probability that a child of theirs will inherit both traits? Show your logic using probability, or
draw a Punnett square to show the potential genotypes of their progeny. M S D B C H
Solution
A) Cali, Jerry; Since Henry has restless leg syndrome (rr), both of his parents should carry atleast
one copy of r allele. So, Jerry and Cali are heterozygous for the restless leg syndrome.
B) Molly and Sean; Since Both Molly and Sean have fine legs but their son is heterozygous for
the syndrome, either of them is also heterozygous for the restless leg syndrome.
C) Since Henry has red hair and no tan tendency (homozygous for red melanin gene), both of
his parents should have one copy of red melanin gene. Jerry has red hair (mm) but Cali has dark
brown hair, so, Cali is heterozygous for the red melanin gene.
D) Genotype of woman to whom he is married = mmrr
promability of getting mm = mm (the woman) x mm (garcia)= 1
Probabiity of getting rr= Rr x rr = 1/4
Probability of Garcia to be Rr is 1/2.
Total probability of rr = 1/4 + 1/2= 3/4
Total probability of rrmm= 3/4 x 1= 3/4.
Fishes are at the base of the vertebrate Tree of Life and fishes do .pdf
Fishes are at the base of the vertebrate Tree of Life and fishes do not have four limbs! However,
earlier tetrapods are sister groups of fishes. What fish species are sister to or considered an
earlier tetrapod?
Solution
The evolution of Tetrapods occur 0.4 billion years ago. It occur in Devonian period. The
ancestors for tetrapods are Lobe-finned fishes or Sarcoptyrigaens. The living sarcoptygians are
Coelacanths and lung fishes.
Consider two point sources separated by a distance of 10 meters. The .pdf
Consider two point sources separated by a distance of 10 meters. The longitudinal waves
emitted from the two sources have the wavelength of = lm. The velocity of the wave is 2m/sec.
What is the frequency f and the period T?
Solution
Frequency=Velocity/wavelength
=(2m/sec)/1m
=2 Hz
Time period= 1/Frequency
=1/2
=0.5 sec
a)
Amplitude at C =0.1+0.1=0.2 m (Since waves are in phase)
b)
Waveform will be at same amplitude that is 0.2 m
c)
If the distance between the points A and B is changed. The waves interfere distructively.
Therefore, amplitude is zero. That is 0.1-0.1=0
d)
If source A is delayed by 0.5 sec
The path difference is= 2(0.5)=1 m
Therefore, waves interfere constructevely.
The wave amplitude at point C is.,
0.1+0.1=0.2 m.
Consider the following system microprocessor based system Program t.pdf
Consider the following system microprocessor based system: Program the PLD such that
interface memory chips at following starting addresses: What is the capacity of each memory
chip? What is the MPU memory space?
Solution
Implementation of SAP SCM Modules for DP, SNP, EIS
Regions, US, Canada, Mexico, Chile, Peru, Columbia and Venezuela.
Provide technical solution leadership to the Program Director, Project Managers and GSLT
Manage IS and vendor stakeholder relationships relevant to the technical solution
Ensure technical project and product meet governance, security and CARM guidelines
Ensure product is accepted into the support organization
Provide status and support to Project Managers in the above activities.
Develop an algorithm for the 4-sum problem.SolutionThe algorit.pdf
Develop an algorithm for the 4-sum problem.
Solution
The algorithm requires creating another array TWOSUMS which has the sum of all possible pair
of numbers in the array. We also need to store the index of numbers along with the sum to keep
track that the numbers are distinct.
Here’s how each entry in the TWOSUMS array will be stored.
class Entry{
int sum;
int indexa;
int indexb;
}
Entry TWOSUMS[];
Algorithm.
Describe the repair mechanisms used during DNA replication.Soluti.pdf
Describe the repair mechanisms used during DNA replication.
Solution
DNA like any other molecule can undergo a variety of chemical reactions. Because DNA
uniquely serves as a permanent copy of the cell genome, however, changes in its structure are of
much greater consequence than are alterations in other cell components, i.e RNA’s and Proteins.
Mutations can consider the incorporation of incorrect bases during DNA replication. And also,
various chemical changes occur in DNA either spontaneously or as a result of exposure to
chemicals or radiation. Such damage to DNA can block replication or transcription, and can
result in a high frequency of mutations—consequences that are unacceptable from the standpoint
of cell reproduction.
To maintain the integrity of their genomes, cells have therefore had to evolve mechanisms to
repair damaged DNA. DNA repair mechanism can be divided into two general classes: (1) direct
reversal of the chemical reaction responsible for DNA damage, and (2) removal of the damaged
bases followed by their replacement with newly synthesized DNA. The rate of DNA repair is
dependent on many factors, including the cell type, the age of the cell, and the extracellular
environment. A cell that has accumulated a large amount of DNA damage, or one that no longer
effectively repairs damage incurred to its DNA, can enter one of three possible states:
1. an irreversible state of dormancy, known as senescence
2. cell suicide, also known as apoptosis or programmed cell death
3. unregulated cell division, which can lead to the formation of a tumor that is cancerous
The DNA repair ability of a cell is vital to the integrity of its genome and thus to the normal
functionality of that organism.
Direct reversal
Cells are known to eliminate three types of damage to their DNA by chemically reversing it.
These mechanisms do not require a template,the types of damage they counteract can occur in
only one of the four bases. Such direct reversal mechanisms are specific to the type of damage
incurred and do not involve breakage of the phosphodiester backbone. The formation of
pyrimidine dimers upon irradiation with UV light results in an abnormal covalent bond between
adjacent pyrimidine bases. The photo reactivation process directly reverses this damage by the
action of the enzyme photolyase, whose activation is obligately dependent on energy absorbed
from blue/UV light (300–500 nm wavelength) to promote catalysis.
The second type of damage, methylation of guanine bases, is directly reversed by the protein
methyl guanine methyl transferase (MGMT),
The third type of DNA damage reversed by cells is certain methylation of the bases cytosine and
adenine.
Excision Repair mechanisms
Single strand damage:
When only one of the two strands of a double helix has a defect, the other strand can be used as a
template to guide the correction of the damaged strand. In order to repair damage to one of the
two paired molecules of DNA, there exist a number of excis.
Describe the current state of the human beta globin gene family and .pdf
Describe the current state of the human beta globin gene family and the processes that were
involved in generating it from a single ancestral globin gene. How do the members of the gene
family differ and how do these differences relate to their expression and function? How does the
human beta globin gene family differ from the beta globin gene families in other vertebrates?
What do these differences imply about the evolution of this gene family in different species; are
the same processes that shaped the human gene family also at work in the other species and if so
why are the gene families different in the other species? What differences in the beta globin
genes are found among the current human population and what are the genetic consequences of
these differences?
Solution
human -globin locus are composed of five gene, that are located on short chromosome number
11, This locus also contains the delta, gamma-A, gamma-G, and epsilon globin. Expression of all
genes are controlled by a single locus control region, and all beta genes are differentially
expressed during development.
Duplication + divergence of ancestral gene is the major process involve in generation of beta
gene form single ancestral globin gene.
Human beta globulin gene come under gene family because of gene family is set of various
similar gene that form by duplication of single original Gene & almost similar biochemical
function.
Gene family member are different in several way 1. arrangement of genes directly reflect to
differentiation of their the expression during development. For proper Functioning required
another alpha globin for complete function for example- Hemoglobin protein is made of two
alpha and two beta polypeptide are encoded by different gene. Globin genes are similar in
structure Each gene encoded similar but different protein.
2. Expression of gene is regulated in the embryonic erythropoiesis by the many transcription
factors like [KLF1] and that which is associated with upregulation of the adult and another
[KLF2] which are vital to expression of the embryonic hemoglobin.
beta-globin gene are clustered & work as unit of gene regulation, embryo --> express epsilon,
fetal --> express gamma, adult --> express beta (and delta) Gene order-position-location-
important for regulation of gene.
the human beta globin gen family different form beta globin gene family in vertebrate-
1.Polyploidy occurs more often in plant genome than in animal one or at least vertebrate
genomes.
2. loss of intron- first & third intron only two intron found in vertebrate hemoglobin &
myoglobin genes. Thecentral intron was lost before the divergence of annelids & arthropods and
absent in all the vertebrate hemoglobin and myoglobin genes. Intron loss has in different
organisms & there is extreme example of the total intron loss in arthropod Chironomus. 3. major
gene duplications in vertebrate globin’s.
evolution of this gen family in difference species-
Gene families are the groups of the.
Derived characteristics that define the order Primate include a.pdf
Derived characteristics that define the order Primate include
a. highly developed olfaction (sense of smell).
b. increased dependence on complex behavior, learning, and behavioral flexibility.
c. specialized molars, up to six incisors, and a lack of premolars.
d. bipedalism.
Solution
In animal order primate including lemurs,tarsiers,monkeys,apes and human beings. So all these
animals have complex beahvior,learning and beahvioral felxibility which is common feature in
all.Primates have limb structure, one upper limb bone, two lower limb bones, many mammalian
orders have lost various bones, especially fusing of the two lower limb bones.primates have the
internal regulation of body temperature and feeding their young with milk produced by
mammary glands..
Consider the (Mining dataset shown in the following table The las.pdf
Consider the (Mining dataset shown in the following table The last attribute Cheat is the class
attribute. For a three-way -split of these data using attribute Marital Status, compute the
information gain using the entropy function fix measuring impurity. Consider that the three-way
(multi-way) spin a does by using the three possible values Single. Married, and Divorced of the
.Marital Status attribute
Solution
Info(D) = -MI=1pi log2(pi);
Pi= |Ci,D|/|D|
InfoA(D) = vj=1 |Dj|/ |D| *Info(Dj):
Gain(A) = Info(D)-InfoA(D):
Info(D)=- (3/10) log2 (3/10) – (7/10) log2 (7/10)
=0.522+0.361
=0.883
Info Marital status (D) = (4/10)* (-(2/4) log2(2/4) - (2/4) log2(2/4)) + (4/10)*(-(0/4) log2(0/2)-
(4/4) log2(4/4) +(2/10)*(-(1/2) log2(1/2) -(1/2)
log2(1/2))
= 0.4 + 0 + 0.2
= 0.6
Therefore Gain (Marital Status) = 0.883-0.6 = 0.283.
The Challenger.pdf DNHS Official Publication
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Francesca Gottschalk - How can education support child empowerment.pptx
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The portion of the fungus that projects above the substrate and conta.pdf
The portion of the fungus that projects above the substrate and contain the spores is known as A)
septate hyphae B) septate mycellum C) aerial mycellum D) vegetative mycelium E) none of the
above = identity Bacteria contamination is suppressed in plain Sabouard\'s agar by the addition
of peptone and dextrose B) brain heart infusion media C) Lactophenol Cotton Blue D) all of the
above E) none of the above = identity
Solution
16. C. Aerial mycelium - the portion of the mycelium that project above the substrate and
produce spores. Because of its capability to produce spores, it is also known as reproductive
mycelium.
17. D. all of the above.
the appearance of symptoms of a genetic disorder at earlier ages in .pdf
the appearance of symptoms of a genetic disorder at earlier ages in successive Generations is
called
Solution
In genetics, anticipation is a phenomenon whereby as a genetic disorder is passed on the next
generation, the symptoms of the genetic disorder become apparent at an earlier age with each
generation..
Stuck on hw problem Assume that the forces acting on a mass are such.pdf
Stuck on hw problem Assume that the forces acting on a mass are such that the potential energy
is the function of x shown in Fig. 24-5: Sketch the solutions in the phase plane. Describe the
different kinds of motion that can occur.
Solution
Every sound we hear, every photon of light that hits our eyes, the movement of grass blown by
the wind and the regular beat of the tides are all examples of waves. They are all around us.
Visible, physical wave*s such as those we see when a rock is thrown into water are what many
people think about when they first began to think about waves. These waves have distinct
properties specific to their type but also exhibit characteristics in common with more abstract
waves such as sound waves and light (electromagnetic) waves.
Below is an interactive illustration that explores particle and energy motion for the three wave
types listed above. Once you have reviewed the illustration and accompanying text, test your
understanding of these concepts with the wave type
At the simplest level, waves are a disturbances that propagate energy through a medium.
Propagation of the energy depend on interactions between the particles that make up the
medium. Particles move as the waves pass through but there is no net motion of particles. This
means, once a wave has passed the particles return to their original position. As a result, energy,
not matter, is propagated by waves.
Different types of waves exhibits specific characteristics. These characteristics are used to
distinguish between wave types. Orientation of particle motion relative to the direction of energy
propagation is one way waves are characterized. There are three categories:
A second way that waves are characterized is by the types of matter they are able to travel
through.Electromagnetic waves (including light) can move through a vacuum. Physical waves
require matter through which to propagate. Physical waves are further distinguished by the
phases of matter through which they can move. Longitudinal waves can pass through liquids and
gasses while transverse waves require a material to be solid to propagate..
Some Giant Sulphur Bacteria are so large as to be visible to the nak.pdf
Some Giant Sulphur Bacteria are so large as to be visible to the naked eye. Nonetheless, they
have been very difficult to cultivate and classify. Why?
Solution
Cultivate
Grow very slowly. Metabolism is based on lithotrophic utilization of sulphide oxidation coupled
to nitrate reduction which is gives poor yields. In natural environments they grow in gradients
where one end of filament is sulphide rich, the other end nitrate rich. Difficult to simulate in the
lab.
Classify
Morphology has been used, but this is not linked to phylogeny. 16S rRNA hard to amplify due to
containing an intron.
Problem 1 Eclipses and occults Saturn has five satellites that rotat.pdf
Problem 1: Eclipses and occults Saturn has five satellites that rotate around the planet. The list
below records events occurring during the month of August, 1995, involving eclipses or occults
of Saturnian recorded satellites. For each event, the percent of light lost by the eclipsed or
occulted satelite is Event Aug 2 Aug 4 Aug 5 Aug 6 Aug 8 Light Loss (%) 61 56 Eclipse Eclipse
Occult Eclipse Eclipse Occult Occult Occult Occult Occult Eclipse Occult Occult Occult Occult
Occult Occult Occult Occult 46 Aug 11 39 Aug 12 Aug 14 Aug 14 Aug 15 Aug 15 AUg 16 Aug
20 Aug 23 Aug 23 AUg 25 Aug 28 1 20 12 Source: Astronomy magazine. Aug. 1995, p.60 via
Statistics (10m ed.] McClave and Sincich, p. 53
Solution
a).
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19
b).
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19
(eclipse are marked red)
c).
Based on above graph, eclipse is more likely lead to greater light loss.
Stem and Leaf plot for
light loss
stem unit =
10
leaf unit =
1
Frequency
Stem
Leaf
8
0
1 1 2 3 4 5 5 9
3
1
1 2 3
2
2
0 0
1
3
9
1
4
6
1
5
6
2
6
1 5
0
7
0
8
0
9
1
10
0
19.
please show work A hospital reports that two patients have been admi.pdf
please show work A hospital reports that two patients have been admitted who have contracted
Crohn\'s disease. Suppose our experiment consists of observing whether each patient survives or
dies as a result of the disease. The simple events and probabilities of their occurrences arc shown
in the table (where 5 in the first position means that patient 1 survives, D in the first position
means that patient 1 dies, etc.).
Solution.
Person.java is givenSolutionSince I didnt have the implement.pdf
Person.java is given
Solution
Since I didn\'t have the implementation of the Disease class, I made some assumptions and wrote
the code accordingly.
public class Person{
public final String name;
protected int health;
protected Disease[] diseases;
public Person(String name, int health){
this.name = name;
this.health = health;
}
public int getHealth(){
return this.health;
}
public void addDisease(Disease dis){
if (!diseasePresent(dis)){
Disease[] updated_diseases[diseases.length+1] = new Disease[];
for (int i<0; i.
In java languageWhich of the following is falseA) A JPasswordFi.pdf
In java language
Which of the following is false?
A) A JPasswordField shows that characters are being entered, but hides the actual character with
an echo character.
B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField.
C) Both JTextFields and JPasswordFields are single-line areas in which the user can enter text
via the keyboard.
D) JPasswordFields generate events; JTextFields do not.
When the user presses Enter in a JPasswordField, the GUI component generates an ________,
which is processed by an object that implements the interface _______.
A) ActionEvent, ActionListener
B) ActionEvent, ActionEventListener
C) TextEvent, TextListener
D) TextEvent, TextEventListener
JPasswordField contains method ___________to obtain the data entered.
A) getInput
B) getPassword
C) getText
D) getEcho
A JRadioButton is different from a JCheckBox in that _________.
A) a JRadioButton is a subclass of JToggleButton, JCheckBox is not
B) normally several JRadioButtons are grouped together and are mutually exclusive
C) a JRadioButton is a type of button, JCheckBox is not
D) a JRadioButton is a state button, JCheckBox is not
The fact that class Graphics is abstract contributes to Java’s portability because:
A) drawing is performed differently on every platform that supports Java. A subclass of Graphics
must be created that uses the drawing capabilities of the current platform.
B) objects of non-abstract classes can only be instantiated on the Windows platform.
C) drawing should not be performed on non-Linux platforms.
D) Class Graphics is not abstract.
JSliders generate what type of event?
A) ActionEvents
B) ChangeEvents
C) WindowEvents
D) MouseEvents
Menus are attached to windows by calling method ____________.
A) addMenuBar
B) setJMenuBar
C) setMenu
D) addJMenuBar
Solution
B) Class JTextField extends JTextComponent; Class JPasswordField extends JTextField.
A) ActionEvent, ActionListener
B) getPassword
B) normally several JRadioButtons are grouped together and are mutually exclusive
A) drawing is performed differently on every platform that supports Java. A subclass of Graphics
must be created that uses the drawing capabilities of the current platform.
A) ActionEvents
B) setJMenuBar.
in a completely randomized (one-way) anova, with other things being .pdf
in a completely randomized (one-way) anova, with other things being equal as the sample means
get closer to each other, the probability of rejecting the null hypothesis decreases.
true
or false
Solution
TRUE
because when other things being equal as the sample means get closer to each other
its true that the probability of rejecting the Ho decreases.
I want you to go the Next Generation Sunshine State standards (http.pdf
I want you to go the Next Generation Sunshine State standards
(http://www.cpalms.org/Public/search/Standard) homepage. Choose a grade level (not K) and
look at all the LIFE SCIENCE standards that are found under each of the Big Idea/Supporting
Idea topics. To see all the standards, click on the bolded letters.
please provide the grade level and copy the standards
Summarize what you, as a teacher, would be responsible for teaching your students.
Discuss how this framework changed your thoughts on what you, as a teacher, would be
responsible for teaching your students for the standards?
Discuss how insect topics might be used to help illustrate each of those LIFE SCIENCE
standards for that grade level. What insect information do you think could fit with that standard?
How could insects be used in related lessons?
Solution
Summary of grade 7 according to my skills are :-
Cell Biology
1. All living organisms are composed of cells, from just one to many trillions, whose details
usually are visible only through a microscope. As a basis for understanding this concept:
a. Students know cells function similarly in all living organisms.
b. Students know the characteristics that distinguish plant cells from animal cells, including
chloroplasts and cell walls.
c. Students know the nucleus is the repository for genetic information in plant and animal cells.
d. Students know that mitochondria liberate energy for the work that cells do and that
chloroplasts capture sunlight energy for photosynthesis.
e. Students know cells divide to increase their numbers through a process of mitosis, which
results in two daughter cells with identical sets of chromosomes.
f. Students know that as multicellular organisms develop, their cells differentiate.
Genetics
2. A typical cell of any organism contains genetic instructions that specify its traits. Those traits
may be modified by environmental influences. As a basis for understanding this concept:
a. Students know the differences between the life cycles and reproduction methods of sexual and
asexual organisms.
b. Students know sexual reproduction produces offspring that inherit half their genes from each
parent.
c. Students know an inherited trait can be determined by one or more genes.
d. Students know plant and animal cells contain many thousands of different genes and typically
have two copies of every gene. The two copies (or alleles) of the gene may or may not be
identical, and one may be dominant in determining the phenotype while the other is recessive.
e. Students know DNA (deoxyribonucleic acid) is the genetic material of living organisms and is
located in the chromosomes of each cell.
Evolution
3. Biological evolution accounts for the diversity of species developed through gradual processes
over many generations. As a basis for understanding this concept:
a. Students know both genetic variation and environmental factors are causes of evolution and
diversity of organisms.
b. Students know the reasoning used by C.
How are macromolecules transported across the nuclear envelope Desc.pdf
How are macromolecules transported across the nuclear envelope? Describe the structure that
regulates nuclear membrane permeability.
Solution
The transportation of molecules in and out of the nucleus is controlled by the nuclear pore
complex(NPC). Small molecules move through the nuclear membrane with ease but for
macromolecules like proteins, RNA etc, they require an association with transportation factors
like importins, karyopherins and exportins.
Proteins that need to be carried into the nucleus have nuclear localization signals (NLS) which
bind to the importins. NLS is a sequence of Amino acids which act as a tag. They are mostly
hydrophilic but hydrophobic sequences have also been documented.
Nuclear Import:
Importins bind to the macromolecule to be transported in the cytoplasm itself. The complex then
interacts with the NPC and passes through the channel. Once inside the nucleus, it interacts with
Ran-GTP and the importin dissociates from the macromolecule it was carrying. Then the
importin-Ran-GTP complex is transported to the cytoplasm where it is separated from importin
by the Ran Binding Protein (RanBP). After being separated from importin, GTPase activating
protein(GAP) binds with Ran-GTP and induces hydrolysis of GTP to GDP. The Ran-GDP
produced binds with the nuclear transport factor(NUTF2) and then returns to the nucleoplasm.
Inside the nucleus, Ran-GDP interacts with guanine nucleotide exchange factor(GEF) which
replaces GDP with GTP and Ran-GTP is born again to repeat the cycle.
Nuclear Export:
In the nucleoplasm, exportin binds the macromolecule to be transported outside the nucleus and
Ran-GTP. The complex diffuses through the pore to the cytoplasm where it dissociates. Ran-
GTP binds to GAP and hydrolysis resulting in Ran-GDP. Ran-GDP is returned to the nucleus
where it exchanges its ligand for GTP.
Nuclear Pore:
The nuclear pore complex is made of proteins called nucleoporins. Around 50% of nucleoporins
contains solenoid protein domains (alpha or beta-propeller fold) and the rest might have both as
separate structural domains. Each NPS has about 456 individual protein molecules with 30
distinct nucleoproteins. These are highly flexible proteins that lack an ordered secondary
structure..
GeneticsBackground info About 30 of the Irish population has red.pdf
Genetics
Background info: About 30% of the Irish population has red hair and an inability to tan (MC1R).
This is due to an unusual variant of the pigment molecule melanin, called red melanin. Another
fairly common inherited condition is restless legs syndrome (RLS). It is characterized by an
irresistible urge to move ones legs and can interfere with sleep, leading to chronic sleep
deprivation. Both red melanin and restless leg syndrome are recessive, and their genes are on
different chromosomes.
Molly and Sean met in San Francisco and were drawn to each other because of their striking red
hair and very pale skin (both had parents who emigrated from Ireland). Neither has a family
history of restless leg syndrome. A few years later they married and had identical twins, Jerry
and Garcia, both are red heads and their legs are fine. Danny and Betty lived in San Francisco
too. They both had brown hair, and Danny suffers from restless leg syndrome. Their daughter,
Cali, has dark brown hair and her legs are fine. Jerry and Cali have a son, Henry. He has red hair,
does not tan and has restless leg syndrome. Complete the family tree by assigning the genotypes
based on the phenotypes described above (Hint: obviously you can get some information from
the phenotype of the individual. Further clues can be obtained from the phenotypes of the
progeny). Use R or r for restless leg syndrome alleles or an underscore if the allele could be
either R or r. For example, R_ would represent either Rr or RR. Use M or m for the red melanin
alleles, M_ would represent MM or Mm.
A) (1.5 point) Which individuals must be heterozygous for restless leg syndrome?
B) (1.5 point) Which individuals might be heterozygous for restless leg syndrome?
C) (1 point) Which individuals must be heterozygous for the red melanin gene?
D) (2 point) Garcia marries a woman who has restless leg syndrome and red hair, what is the
probability that a child of theirs will inherit both traits? Show your logic using probability, or
draw a Punnett square to show the potential genotypes of their progeny. M S D B C H
Solution
A) Cali, Jerry; Since Henry has restless leg syndrome (rr), both of his parents should carry atleast
one copy of r allele. So, Jerry and Cali are heterozygous for the restless leg syndrome.
B) Molly and Sean; Since Both Molly and Sean have fine legs but their son is heterozygous for
the syndrome, either of them is also heterozygous for the restless leg syndrome.
C) Since Henry has red hair and no tan tendency (homozygous for red melanin gene), both of
his parents should have one copy of red melanin gene. Jerry has red hair (mm) but Cali has dark
brown hair, so, Cali is heterozygous for the red melanin gene.
D) Genotype of woman to whom he is married = mmrr
promability of getting mm = mm (the woman) x mm (garcia)= 1
Probabiity of getting rr= Rr x rr = 1/4
Probability of Garcia to be Rr is 1/2.
Total probability of rr = 1/4 + 1/2= 3/4
Total probability of rrmm= 3/4 x 1= 3/4.
Fishes are at the base of the vertebrate Tree of Life and fishes do .pdf
Fishes are at the base of the vertebrate Tree of Life and fishes do not have four limbs! However,
earlier tetrapods are sister groups of fishes. What fish species are sister to or considered an
earlier tetrapod?
Solution
The evolution of Tetrapods occur 0.4 billion years ago. It occur in Devonian period. The
ancestors for tetrapods are Lobe-finned fishes or Sarcoptyrigaens. The living sarcoptygians are
Coelacanths and lung fishes.
Consider two point sources separated by a distance of 10 meters. The .pdf
Consider two point sources separated by a distance of 10 meters. The longitudinal waves
emitted from the two sources have the wavelength of = lm. The velocity of the wave is 2m/sec.
What is the frequency f and the period T?
Solution
Frequency=Velocity/wavelength
=(2m/sec)/1m
=2 Hz
Time period= 1/Frequency
=1/2
=0.5 sec
a)
Amplitude at C =0.1+0.1=0.2 m (Since waves are in phase)
b)
Waveform will be at same amplitude that is 0.2 m
c)
If the distance between the points A and B is changed. The waves interfere distructively.
Therefore, amplitude is zero. That is 0.1-0.1=0
d)
If source A is delayed by 0.5 sec
The path difference is= 2(0.5)=1 m
Therefore, waves interfere constructevely.
The wave amplitude at point C is.,
0.1+0.1=0.2 m.
Consider the following system microprocessor based system Program t.pdf
Consider the following system microprocessor based system: Program the PLD such that
interface memory chips at following starting addresses: What is the capacity of each memory
chip? What is the MPU memory space?
Solution
Implementation of SAP SCM Modules for DP, SNP, EIS
Regions, US, Canada, Mexico, Chile, Peru, Columbia and Venezuela.
Provide technical solution leadership to the Program Director, Project Managers and GSLT
Manage IS and vendor stakeholder relationships relevant to the technical solution
Ensure technical project and product meet governance, security and CARM guidelines
Ensure product is accepted into the support organization
Provide status and support to Project Managers in the above activities.
Develop an algorithm for the 4-sum problem.SolutionThe algorit.pdf
Develop an algorithm for the 4-sum problem.
Solution
The algorithm requires creating another array TWOSUMS which has the sum of all possible pair
of numbers in the array. We also need to store the index of numbers along with the sum to keep
track that the numbers are distinct.
Here’s how each entry in the TWOSUMS array will be stored.
class Entry{
int sum;
int indexa;
int indexb;
}
Entry TWOSUMS[];
Algorithm.
Describe the repair mechanisms used during DNA replication.Soluti.pdf
Describe the repair mechanisms used during DNA replication.
Solution
DNA like any other molecule can undergo a variety of chemical reactions. Because DNA
uniquely serves as a permanent copy of the cell genome, however, changes in its structure are of
much greater consequence than are alterations in other cell components, i.e RNA’s and Proteins.
Mutations can consider the incorporation of incorrect bases during DNA replication. And also,
various chemical changes occur in DNA either spontaneously or as a result of exposure to
chemicals or radiation. Such damage to DNA can block replication or transcription, and can
result in a high frequency of mutations—consequences that are unacceptable from the standpoint
of cell reproduction.
To maintain the integrity of their genomes, cells have therefore had to evolve mechanisms to
repair damaged DNA. DNA repair mechanism can be divided into two general classes: (1) direct
reversal of the chemical reaction responsible for DNA damage, and (2) removal of the damaged
bases followed by their replacement with newly synthesized DNA. The rate of DNA repair is
dependent on many factors, including the cell type, the age of the cell, and the extracellular
environment. A cell that has accumulated a large amount of DNA damage, or one that no longer
effectively repairs damage incurred to its DNA, can enter one of three possible states:
1. an irreversible state of dormancy, known as senescence
2. cell suicide, also known as apoptosis or programmed cell death
3. unregulated cell division, which can lead to the formation of a tumor that is cancerous
The DNA repair ability of a cell is vital to the integrity of its genome and thus to the normal
functionality of that organism.
Direct reversal
Cells are known to eliminate three types of damage to their DNA by chemically reversing it.
These mechanisms do not require a template,the types of damage they counteract can occur in
only one of the four bases. Such direct reversal mechanisms are specific to the type of damage
incurred and do not involve breakage of the phosphodiester backbone. The formation of
pyrimidine dimers upon irradiation with UV light results in an abnormal covalent bond between
adjacent pyrimidine bases. The photo reactivation process directly reverses this damage by the
action of the enzyme photolyase, whose activation is obligately dependent on energy absorbed
from blue/UV light (300–500 nm wavelength) to promote catalysis.
The second type of damage, methylation of guanine bases, is directly reversed by the protein
methyl guanine methyl transferase (MGMT),
The third type of DNA damage reversed by cells is certain methylation of the bases cytosine and
adenine.
Excision Repair mechanisms
Single strand damage:
When only one of the two strands of a double helix has a defect, the other strand can be used as a
template to guide the correction of the damaged strand. In order to repair damage to one of the
two paired molecules of DNA, there exist a number of excis.
Describe the current state of the human beta globin gene family and .pdf
Describe the current state of the human beta globin gene family and the processes that were
involved in generating it from a single ancestral globin gene. How do the members of the gene
family differ and how do these differences relate to their expression and function? How does the
human beta globin gene family differ from the beta globin gene families in other vertebrates?
What do these differences imply about the evolution of this gene family in different species; are
the same processes that shaped the human gene family also at work in the other species and if so
why are the gene families different in the other species? What differences in the beta globin
genes are found among the current human population and what are the genetic consequences of
these differences?
Solution
human -globin locus are composed of five gene, that are located on short chromosome number
11, This locus also contains the delta, gamma-A, gamma-G, and epsilon globin. Expression of all
genes are controlled by a single locus control region, and all beta genes are differentially
expressed during development.
Duplication + divergence of ancestral gene is the major process involve in generation of beta
gene form single ancestral globin gene.
Human beta globulin gene come under gene family because of gene family is set of various
similar gene that form by duplication of single original Gene & almost similar biochemical
function.
Gene family member are different in several way 1. arrangement of genes directly reflect to
differentiation of their the expression during development. For proper Functioning required
another alpha globin for complete function for example- Hemoglobin protein is made of two
alpha and two beta polypeptide are encoded by different gene. Globin genes are similar in
structure Each gene encoded similar but different protein.
2. Expression of gene is regulated in the embryonic erythropoiesis by the many transcription
factors like [KLF1] and that which is associated with upregulation of the adult and another
[KLF2] which are vital to expression of the embryonic hemoglobin.
beta-globin gene are clustered & work as unit of gene regulation, embryo --> express epsilon,
fetal --> express gamma, adult --> express beta (and delta) Gene order-position-location-
important for regulation of gene.
the human beta globin gen family different form beta globin gene family in vertebrate-
1.Polyploidy occurs more often in plant genome than in animal one or at least vertebrate
genomes.
2. loss of intron- first & third intron only two intron found in vertebrate hemoglobin &
myoglobin genes. Thecentral intron was lost before the divergence of annelids & arthropods and
absent in all the vertebrate hemoglobin and myoglobin genes. Intron loss has in different
organisms & there is extreme example of the total intron loss in arthropod Chironomus. 3. major
gene duplications in vertebrate globin’s.
evolution of this gen family in difference species-
Gene families are the groups of the.
Derived characteristics that define the order Primate include a.pdf
Derived characteristics that define the order Primate include
a. highly developed olfaction (sense of smell).
b. increased dependence on complex behavior, learning, and behavioral flexibility.
c. specialized molars, up to six incisors, and a lack of premolars.
d. bipedalism.
Solution
In animal order primate including lemurs,tarsiers,monkeys,apes and human beings. So all these
animals have complex beahvior,learning and beahvioral felxibility which is common feature in
all.Primates have limb structure, one upper limb bone, two lower limb bones, many mammalian
orders have lost various bones, especially fusing of the two lower limb bones.primates have the
internal regulation of body temperature and feeding their young with milk produced by
mammary glands..
Consider the (Mining dataset shown in the following table The las.pdf
Consider the (Mining dataset shown in the following table The last attribute Cheat is the class
attribute. For a three-way -split of these data using attribute Marital Status, compute the
information gain using the entropy function fix measuring impurity. Consider that the three-way
(multi-way) spin a does by using the three possible values Single. Married, and Divorced of the
.Marital Status attribute
Solution
Info(D) = -MI=1pi log2(pi);
Pi= |Ci,D|/|D|
InfoA(D) = vj=1 |Dj|/ |D| *Info(Dj):
Gain(A) = Info(D)-InfoA(D):
Info(D)=- (3/10) log2 (3/10) – (7/10) log2 (7/10)
=0.522+0.361
=0.883
Info Marital status (D) = (4/10)* (-(2/4) log2(2/4) - (2/4) log2(2/4)) + (4/10)*(-(0/4) log2(0/2)-
(4/4) log2(4/4) +(2/10)*(-(1/2) log2(1/2) -(1/2)
log2(1/2))
= 0.4 + 0 + 0.2
= 0.6
Therefore Gain (Marital Status) = 0.883-0.6 = 0.283.
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