Many of the techniques for fate mapping fade over time and cannot be.pdf
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Many of the techniques for fate mapping fade over time and cannot be used long term for tracing cell lineages. Describe a technique used for fate mapping that permanently marks cells and the advantages of this technique over other techniques (besides the longevity of its use). Solution Fate mapping is a method of understanding the embryonic origin of various tissues in the adult organisms by establishing the correspondence between individuals cells at one stage of development, and their progeny at later stages of development. When carried out at single-cell resolution, this process is termed cell-lineage tracing. Technique: Artificial markings: It is a method for labelling the only blastomere by which their fate can be traced:.
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II. Please provide complete correct detailed solution. Thank you. Co.pdf
II. Please provide complete correct detailed solution. Thank you. Consider the distribution of
individual incomes, yi in a population that is divided into two subgroups, indicated by the
random variable D {0,1}. Assume: So mu 0, v 0 are the mean and variance of yi in group 0, and
mu1, v1 are the mean and variance of y, in group 1. Suppose that P(D,i = 1) = p. Find mu =
E[yi]> the overall mean of yi.
Solution.
In a Windows 8 file system there is a file named direct.awk in a fol.pdf
In a Windows 8 file system there is a file named direct.awk in a folder named test in a folder
named kennesaw. The kennesaw folder is in the root of the C: drive. Give the absolute path to
the file direct.awk
Solution
In c drive you have created a folder named Kennesaw and within that folder we have one more
folder named test .In that folder we have our file named direct.awk
The path is as follows:
C:\\kennesaw\\test\\direct.awk.
If using an ANOVA Single factor and the alpha is set at .05 and the.pdf
If using an ANOVA: Single factor and the alpha is set at .05 and the p-value ends up showing
0.173354...what does this explain about the difference in the means between the groups?
Solution
As the p value (= 0.173354) > Significance level (= 0.05), we fail to reject null hypothesis.
The verbal conclusion is that there is no difference in the means between the groups.
Hope this helped!.
I need help please with these 2 assignments. of ul Uus Oll ook.pdf
I need help please with these 2 assignments. of ul Uus Oll ook value mpact hical Dilemma
What\'s an expense? differences nge over the ATC 6-5 Several years ago, Wilson Blowhard
founded a communications company. The company became suc of its competitors. In fact, most ,
Wi cesful and grew by expanding its customer base and acquiring some f its growth resulted
from acquiring other companies. Mr. Blowhard is adamant about continuing the company\'s
growth and increasing its continue to increase at a rapid pace nal assets net worth. To achieve
these goals, the business\'s net income must If the company\'s net worth continues to rise, Mr.
Blowhard plans to sell the company and retire. -term assets licon micro- ften thought rms 10-K
or ing require- tructions in bsite; Intel\'s He is, therefore, focused on improving the company\'s
profit any way he can In the communications business, companies often use the lines of other
communications compa nies. This line usage is a significant operating expense for Mr.
Blowhard\'s company. Generally ac urred cepted accounting principles require operating costs
like line use to be expensed as they are ino each year. Each dollar of line cost reduces net income
by a dollar. After reviewing the company\'s operations, Mr. Blowhard concluded that the
company did not cur rently need all of the line use it was paying for. It was really paying the
owner of the lines now so that the line use would be available in the future for all of Mr.
Blowhard\'s expected new customers. Mr. Blowhard instructed his accountant to capitalize all of
the line cost charges and depreciate them over 10 years. The accountant reluctantly followed Mr.
Blowhard\'s instructions and the company\'s net income for the cur- rent year showed a
significant increase over the prior year\'s net income. Mr. Blowhard had found a way they are not
plant, to report continued growth in the company\'s net income and increase the value of the
company Required operty, a. How does Mr. Blowhard\'s scheme affect the amount of income
that the company would otherwise eview the AICPA\'s Articles of Professional Conduct (see
Chapter 4) and comment on any of the the fraud triangle discussed in Chapter 4 and comment on
the features of the fraud triangle eer? re port in its financial statements and how does the scheme
affect the company\'s balance sh Explain your answer standards that were violated that are
evident in this case ssets 00%
Solution
ATC 6-5
Answer to Assignment 1 :
a. Operating expenses have to be charged to the Statement of Profit and Loss instead of treating
it as a capital expenditure although the company did not utilise all of the expenses that it was
paying for. The accountants treatment of operating expenses as capital expenditure will overstate
profits particularly the EBIDTA of the enterprise. It will subsequently increase the retained
profits in case the company is profitable in the year. Also, assets will be orreneously increased
since the expenditure .
How is water transported across the membrane What are the two types.pdf
How is water transported across the membrane? What are the two types of transport proteins?
Describe how they function. Controlling what enters and leaves the cell in an important function
of the nucleus vesicle plasma membrane Golgi apparatus. During diffusion, substances move
from an area of _____ concentration to an area of _____ - concentration. Higher, lower lower,
higher higher, equal
Solution
2. During diffusion substances move from high concentration to low concentration. It is passive
process and substances move along the Concentration gradient
1. Plasma membrane controls in and out of the molecules due to its selective nature. So plasma
membrane is selective permiable
5. Plasma membrane contain two types of transport proteins. They are Passive transport proteins,
that include carriers and channels. They transport substances in passive manner, without
expending ATP. Active transporters, that transport substances by using ATP
4. Water transport across the Plasma membrane by a special type of protein called Aquaporins..
Gradebook ignment FULL Question 1 Which of the following types of ent.pdf
Gradebook ignment FULL Question 1 Which of the following types of entities are required to
prepare consolidated financial statements? Subsidiary entities Investee entities O Sibling entities
Parent entities Question Attempts: 0 of 1 us { Privacy Polcy I 2000-2018 John Wiley & Sons
Inc. All Rghts Reserved. A Division of 20hnwley asons to
Solution
Parent entities are required to prepare consolidated financial statements A parent entity is
one which owns more than 50% of the voting power of subsidiary entity or controls the Board of
Directors of a company Option 4is correct.
Explain the DuPont identity. How is it used in finance Locate the f.pdf
Explain the DuPont identity. How is it used in finance? Locate the financial statements for two
firms in the same industry. Calculate all four terms of the DuPont identity, and present the results
but do not analyze them.
PLEASE DO NOT POST PICTURES. WRITE IR DOWN. THANK YOU.
Solution
As per the Du Pont Identity,
ROE= Profit Margin* Total asset turnover * Equity multiplier
= (Net Income/Revenues) * (Revenues/Total Assets) * (Total Assets/ Shareholders\' Equity)
It helps the managers to understand what is driving the ROE and pinpoints the exact determinant
of ROE. It helps the management to understand why the company is weak or strong.
APPLE INC- 30 Sep 2017- Source:NASDAQ
ROE= (48,351,000/229,234,000) * (229,234,000/ 375,319,000) * (375,319,000/134,047,000)
=48,351,000/ 134,047,000
=36.07%
HP Enterprise Company- 31/10-2017-Source-NASDAQ
ROE= (344000 /28,871,000)* (28,871,000/61406000) * (61406000/23466000)
=1.47%.
Files to Use httpwww.cse.msu.edu~cse231ProjectsProject05Red.pdf
Files to Use
http://www.cse.msu.edu/~cse231/Projects/Project05/RedCedarRiver.txt
http://www.cse.msu.edu/~cse231/Projects/Project05/plot.py
Assignment Overview
This assignment focuses on the implementation of Python programs to read files and process
data by using lists and functions.
Assignment Background
U.S. Geological Survey (USGS) provides scientific information to understand Earth, manage
resources and minimize loss from natural disasters. You can download interesting data from the
website https://www.usgs.gov/
We have downloaded the flow rate data for the Red Cedar River in East Lansing starting from
1932 . Your task is to design and implement a Python program that draws two plots from the
data (and displays a table of data for each plot).
Here is the first line of the file. The numbers that we are interested are the last three numbers
which are year, month and flow rate in cubic feet per second (CFS). Notice that the year and
month are ints and the flow rate is float:
Assignment Specifications
1. The program must provide following functions to extract some statistics.
a) open_file()prompts the user to enter a file name. The program will try to open the data
file. Appropriate error message should be shown if the data file cannot be opened. This function
will loop until it receives proper input and successfully opens the file. It returns a file pointer.
b) read_file()calls the open_file() function and uses the returned file pointer to read the data file.
This function returns a list of your choosing containing data you need for other parts of this
project.
c) draw_plot( x, y, plt_title, x_label, y_label) provided by us takes two equal-length lists of
numbers and plots them. You need to pass the label strings and plot title
to the function. Include the range of years in the title (Hint: use string concatenation and check
sample output).
d) annual_average(L) takes a list as an argument and returns a list of tuples in the form (year,
average_flow).
e) month_average(L) takes a list as an argument and returns a list of tuples in the form (year,
month_flow).
f) You may use extra functions, if you wish.
2. The program should read the file only once.
3. The program should plot the average flow for each year. That is, find the average flow for
each year and then generate a plot with years on the x-axis and average flow for that year on the
y-axis. Then generate a table that has the year and average flow on each line. Put a title on the
table and label each column.
4. Next, the program should prompt the user for a number in the range 1-12 and should plot the
flow rates for only that month. For example if user enters 1, the program should extract the flow
rate of Jan 1932, Jan 1933, ... , Jan 2015 and plot them. (Note: do not loop and ask for more.)
5. The program should re-prompt the user if an invalid input (either wrong type or wrong value),
was entered for the month. If something other than an integer is input, your error message should
include.
Recommended
II. Please provide complete correct detailed solution. Thank you. Co.pdf
II. Please provide complete correct detailed solution. Thank you. Consider the distribution of
individual incomes, yi in a population that is divided into two subgroups, indicated by the
random variable D {0,1}. Assume: So mu 0, v 0 are the mean and variance of yi in group 0, and
mu1, v1 are the mean and variance of y, in group 1. Suppose that P(D,i = 1) = p. Find mu =
E[yi]> the overall mean of yi.
Solution.
In a Windows 8 file system there is a file named direct.awk in a fol.pdf
In a Windows 8 file system there is a file named direct.awk in a folder named test in a folder
named kennesaw. The kennesaw folder is in the root of the C: drive. Give the absolute path to
the file direct.awk
Solution
In c drive you have created a folder named Kennesaw and within that folder we have one more
folder named test .In that folder we have our file named direct.awk
The path is as follows:
C:\\kennesaw\\test\\direct.awk.
If using an ANOVA Single factor and the alpha is set at .05 and the.pdf
If using an ANOVA: Single factor and the alpha is set at .05 and the p-value ends up showing
0.173354...what does this explain about the difference in the means between the groups?
Solution
As the p value (= 0.173354) > Significance level (= 0.05), we fail to reject null hypothesis.
The verbal conclusion is that there is no difference in the means between the groups.
Hope this helped!.
I need help please with these 2 assignments. of ul Uus Oll ook.pdf
I need help please with these 2 assignments. of ul Uus Oll ook value mpact hical Dilemma
What\'s an expense? differences nge over the ATC 6-5 Several years ago, Wilson Blowhard
founded a communications company. The company became suc of its competitors. In fact, most ,
Wi cesful and grew by expanding its customer base and acquiring some f its growth resulted
from acquiring other companies. Mr. Blowhard is adamant about continuing the company\'s
growth and increasing its continue to increase at a rapid pace nal assets net worth. To achieve
these goals, the business\'s net income must If the company\'s net worth continues to rise, Mr.
Blowhard plans to sell the company and retire. -term assets licon micro- ften thought rms 10-K
or ing require- tructions in bsite; Intel\'s He is, therefore, focused on improving the company\'s
profit any way he can In the communications business, companies often use the lines of other
communications compa nies. This line usage is a significant operating expense for Mr.
Blowhard\'s company. Generally ac urred cepted accounting principles require operating costs
like line use to be expensed as they are ino each year. Each dollar of line cost reduces net income
by a dollar. After reviewing the company\'s operations, Mr. Blowhard concluded that the
company did not cur rently need all of the line use it was paying for. It was really paying the
owner of the lines now so that the line use would be available in the future for all of Mr.
Blowhard\'s expected new customers. Mr. Blowhard instructed his accountant to capitalize all of
the line cost charges and depreciate them over 10 years. The accountant reluctantly followed Mr.
Blowhard\'s instructions and the company\'s net income for the cur- rent year showed a
significant increase over the prior year\'s net income. Mr. Blowhard had found a way they are not
plant, to report continued growth in the company\'s net income and increase the value of the
company Required operty, a. How does Mr. Blowhard\'s scheme affect the amount of income
that the company would otherwise eview the AICPA\'s Articles of Professional Conduct (see
Chapter 4) and comment on any of the the fraud triangle discussed in Chapter 4 and comment on
the features of the fraud triangle eer? re port in its financial statements and how does the scheme
affect the company\'s balance sh Explain your answer standards that were violated that are
evident in this case ssets 00%
Solution
ATC 6-5
Answer to Assignment 1 :
a. Operating expenses have to be charged to the Statement of Profit and Loss instead of treating
it as a capital expenditure although the company did not utilise all of the expenses that it was
paying for. The accountants treatment of operating expenses as capital expenditure will overstate
profits particularly the EBIDTA of the enterprise. It will subsequently increase the retained
profits in case the company is profitable in the year. Also, assets will be orreneously increased
since the expenditure .
How is water transported across the membrane What are the two types.pdf
How is water transported across the membrane? What are the two types of transport proteins?
Describe how they function. Controlling what enters and leaves the cell in an important function
of the nucleus vesicle plasma membrane Golgi apparatus. During diffusion, substances move
from an area of _____ concentration to an area of _____ - concentration. Higher, lower lower,
higher higher, equal
Solution
2. During diffusion substances move from high concentration to low concentration. It is passive
process and substances move along the Concentration gradient
1. Plasma membrane controls in and out of the molecules due to its selective nature. So plasma
membrane is selective permiable
5. Plasma membrane contain two types of transport proteins. They are Passive transport proteins,
that include carriers and channels. They transport substances in passive manner, without
expending ATP. Active transporters, that transport substances by using ATP
4. Water transport across the Plasma membrane by a special type of protein called Aquaporins..
Gradebook ignment FULL Question 1 Which of the following types of ent.pdf
Gradebook ignment FULL Question 1 Which of the following types of entities are required to
prepare consolidated financial statements? Subsidiary entities Investee entities O Sibling entities
Parent entities Question Attempts: 0 of 1 us { Privacy Polcy I 2000-2018 John Wiley & Sons
Inc. All Rghts Reserved. A Division of 20hnwley asons to
Solution
Parent entities are required to prepare consolidated financial statements A parent entity is
one which owns more than 50% of the voting power of subsidiary entity or controls the Board of
Directors of a company Option 4is correct.
Explain the DuPont identity. How is it used in finance Locate the f.pdf
Explain the DuPont identity. How is it used in finance? Locate the financial statements for two
firms in the same industry. Calculate all four terms of the DuPont identity, and present the results
but do not analyze them.
PLEASE DO NOT POST PICTURES. WRITE IR DOWN. THANK YOU.
Solution
As per the Du Pont Identity,
ROE= Profit Margin* Total asset turnover * Equity multiplier
= (Net Income/Revenues) * (Revenues/Total Assets) * (Total Assets/ Shareholders\' Equity)
It helps the managers to understand what is driving the ROE and pinpoints the exact determinant
of ROE. It helps the management to understand why the company is weak or strong.
APPLE INC- 30 Sep 2017- Source:NASDAQ
ROE= (48,351,000/229,234,000) * (229,234,000/ 375,319,000) * (375,319,000/134,047,000)
=48,351,000/ 134,047,000
=36.07%
HP Enterprise Company- 31/10-2017-Source-NASDAQ
ROE= (344000 /28,871,000)* (28,871,000/61406000) * (61406000/23466000)
=1.47%.
Files to Use httpwww.cse.msu.edu~cse231ProjectsProject05Red.pdf
Files to Use
http://www.cse.msu.edu/~cse231/Projects/Project05/RedCedarRiver.txt
http://www.cse.msu.edu/~cse231/Projects/Project05/plot.py
Assignment Overview
This assignment focuses on the implementation of Python programs to read files and process
data by using lists and functions.
Assignment Background
U.S. Geological Survey (USGS) provides scientific information to understand Earth, manage
resources and minimize loss from natural disasters. You can download interesting data from the
website https://www.usgs.gov/
We have downloaded the flow rate data for the Red Cedar River in East Lansing starting from
1932 . Your task is to design and implement a Python program that draws two plots from the
data (and displays a table of data for each plot).
Here is the first line of the file. The numbers that we are interested are the last three numbers
which are year, month and flow rate in cubic feet per second (CFS). Notice that the year and
month are ints and the flow rate is float:
Assignment Specifications
1. The program must provide following functions to extract some statistics.
a) open_file()prompts the user to enter a file name. The program will try to open the data
file. Appropriate error message should be shown if the data file cannot be opened. This function
will loop until it receives proper input and successfully opens the file. It returns a file pointer.
b) read_file()calls the open_file() function and uses the returned file pointer to read the data file.
This function returns a list of your choosing containing data you need for other parts of this
project.
c) draw_plot( x, y, plt_title, x_label, y_label) provided by us takes two equal-length lists of
numbers and plots them. You need to pass the label strings and plot title
to the function. Include the range of years in the title (Hint: use string concatenation and check
sample output).
d) annual_average(L) takes a list as an argument and returns a list of tuples in the form (year,
average_flow).
e) month_average(L) takes a list as an argument and returns a list of tuples in the form (year,
month_flow).
f) You may use extra functions, if you wish.
2. The program should read the file only once.
3. The program should plot the average flow for each year. That is, find the average flow for
each year and then generate a plot with years on the x-axis and average flow for that year on the
y-axis. Then generate a table that has the year and average flow on each line. Put a title on the
table and label each column.
4. Next, the program should prompt the user for a number in the range 1-12 and should plot the
flow rates for only that month. For example if user enters 1, the program should extract the flow
rate of Jan 1932, Jan 1933, ... , Jan 2015 and plot them. (Note: do not loop and ask for more.)
5. The program should re-prompt the user if an invalid input (either wrong type or wrong value),
was entered for the month. If something other than an integer is input, your error message should
include.
explain why the jovian planet atmospheres are called primitive atmos.pdf
explain why the jovian planet atmospheres are called primitive atmospheres while terrestrial
planets are called evolved atmospheres.
Solution
Jupiter\'s atmosphere is composed mostly of hydrogen and helium, which are the same gases that
dominate the Sun\'s and other stars\' composition. Likewise the early Earth atmosphere was most
likely H2 and He.
But the Earth\'s gravity is too weak to retain these gases for a long time. So contrary to the
Jovian atmosphere the Earth\'s atmosphere changed and evolved. Hydrogen and helium were
quickly replaced by an atmosphere produced by volcanic out-gassing. It contained gases like
H2O, CO2, SO2, CO, S2, Cl2, N2, H2 and NH3 and CH4. But no free O2 as this is not found in
volcanic gases. A lot more water came from outer space brought by comets bombarding the
proto-Earth.
As the Earth cooled, H2O rained out and the oceans formed.
It didn\'t stop there, as oxygen started to buildup as soon as the Archean era, more than 2.5
billion years ago, by means of photochemical dissociation. That is breakup of water and
ammonia molecules by ultraviolet. The O2 level rose to 1-2% of the current level.
At these levels O3 (Ozone) can form to shield the Earth surface from ultraviolet light.
Once the Earth was properly shielded, photosynthesis by bacteria and plants could start: CO2 +
H2O + sunlight = organic compounds + O2, increasing the oxygen level to today\'s levels..
During digestion, in the intestinal lumen certain glycerols, FFA an.pdf
During digestion, in the intestinal lumen certain glycerols, FFA and bite salts from micelles
which can then bond to ___ and ___ and promote their ___. fat soluble vitamins, cholesterol,
absorption. water soluble vitamins, cholesterol, absorption. fat soluble vitamins, minerals,
absorption. fat soluble vitamins, cholesterol, digestion.
Solution
The dietary lipids contains fat, triglycerides , phospholipids, cholesterol and fat soluble vitamins.
They are absorbed back by the processes emulsification and diffusion. So the answer is
Fat soluble vitamins and cholesterol- absorption.
Describe the structures and functions associated with the water-vascu.pdf
Describe the structures and functions associated with the water-vascular system echinoderms.
Solution
Echinodermata represents the phylum of animals in which spiny-skinned organisms are included.
All the organisms of this phylum are aquatic in nature and follow either sedentary or mobile life-
style. They have penta-radial symmetry in morphology and normally contain tube feets. The
peculiar system in these organisms is the presence of a water-vascular canal. The features of
these water vascular canals are enlisted as below:
Various structures associated with water vascular system are discussed as below:
Thus, the above given explanations decipher that the water vascular system of echinoderms is
complex and has high physiological value.
Madreporite: The madreporite represents the central tube along with which all the other tubules
are associated. These water canals open directly on the surface of the organism and generate a
water pore through which water can move to and fro.
Stone canal: The stone canal is a tender tube associated with the madreporite. It has two
openings, one of which opens towards the oral cavity and other towards the axial sinus. It runs
longitudinally across the breadth of the organism\'s body. It is normally curved shaped but can be
\"S\' shaped in some organisms.
Axial sinus: The axial sinus and organ represent some accumulated connective tissues and are of
germinal in nature. These axial sinuses directly communicate with the stone canal on their
posterior end.
Ring canal: The ring canal represents one of the most important organs of the echinoderm. It is a
ring shaped structure located near the madreporite on the centre and have extensions of tubules
which help in circulation of water and fluid. Since echinoderms have pentaradial symmetry, the
ring canal often gives riset to atleast 5 extensions.
Lateral vessels and the tube-feet: These are the organs of locomotion present just below the
madreporite and lateral pentameric extensions of ring canal. Internally, the tube feet and lateral
vessel communicate with the skeleton of water-vascular system..
Considering the atoms bonded and the shape of an H_2O molecule, is th.pdf
Considering the atoms bonded and the shape of an H_2O molecule, is the H_2O molecule polar
or non-polar? polar non-polar This is not the right answer. Do NOT click this answer!
Solution
A water molecule is polar because its O-H bonds are significantly polar and its bent geometry
makes the distribution of those polar bonds asymmetrical.The side of the water molecule
containing the more electronegative oxygen atom is partially negative and the side of the
molecule containing the less electronegative hydrogen atoms is partially positive..
describe the differences and similarities between viruses, worms, an.pdf
describe the differences and similarities between viruses, worms, and Trojan horses. Provide an
example of at least one of these and describe that damage that can be done when a person or
corporation is infected.
Solution
Lets me make it clear actually what these are first and then we move to the diffrences and
similarities
A computer virus connects itself to a program or record empowering it to spread starting with
one PC then onto the next, leaving contamination as it travels. Like a human infection, a PC
virus can extend in seriousness: some may bring about just somewhat irritating impacts while
others can harm your hardware, programming or records. All viruses are joined to an executable
record, which implies the infection may exist on your PC yet it really can\'t contaminate your PC
unless you run or open the pernicious program.
A worm is like an virus by outline and is thought to be a sub-class of an infection. Worms spread
from PC to PC, yet not at all like an virus, it has the capacity to go with no human activity. A
worm exploits document or data transport includes on your framework, which is the thing that
permits it to travel unaided.
A Trojan Horse is loaded with as much cunning as the fanciful Trojan Horse it was named after.
The Trojan Horse, at first look will seem, by all accounts, to be valuable software yet will really
do harm once introduced or keep running on your PC. Those on the less than desirable end of a
Trojan Horse are normally deceived into opening them since they give off an impression of
being getting true blue programming or documents from a honest to goodness source.
Now I will Focus on the differences between Viruses, Worms and Torjan Horses
On the basis of its mode of spread
A virus attach itself to an executable file and then require a human to propogate, a worm does
not require a human action to spread,torjan horse does not replicates itself and require human
input to take into action.
similarity is that they all are malicious programs that can cause damage to your computer.
Examples:-
VIRUSES
WORMS
Torjan.
A window has the rectangle surmounted by a semi circle the total per.pdf
A window has the rectangle surmounted by a semi circle the total perimeter of window is 14 and
window has a maximum area
Find the ratio of the semicircle to the height of the rectangle
Show work
Solution
2*r+2*l+pi*r=14 l=(14-2*r-pi*r)/2 Area=l*2*r+1/2*pi*r^2 =(14-2*r-
pi*r)/2*2*r+1/2*pi*r^2 derivative=14-4*r-2*pi+pi*r=0 r*(4+pi)=14 r=14/(4+pi)=1.96 l=1.96
ratio is 1.
Can you identify how energy flows through an ecosystemCan you i.pdf
Can you identify how energy flows through an ecosystem?
Can you identify how energy flows through an ecosystem?
Solution
Energy flow in ecosystem is basically in the form of heat by food chain. Ecosystem includes
living (plants, animals, microrganismand non living things (sun, soil, climate, atmosphere etc)
and the interactions between them. The food chain in our ecosystem comprises of herbivores,
carnivores and omnivores and with every passage of energy from one organism to another there
is a 70-80% loss of energy which is in the form of heat. So we can take an example where in
green plants absorb sunlight, Co2 and water from surroundings and synthesise carbohydrates.
These green plants are then eaten by herbivores like deer, cow, so the energy is tranferred from
green plants to deer/ cow. Deer will be eaten by lion/ jackle, so again energy will be transferred
from deer to lion/ jackle. Once lion/ jackle die, they will be eaten by eagle/ vulture (scavengers).
Scavengers will then release nutrients in the environment and again the nutrients will be
recycled. So the energy is never exhaused but recycled..
You must use the full name, no abbreviations. Name the dynamic found.pdf
You must use the full name, no abbreviations. Name the dynamic found around the cell. You
will find this on the outside of a bacterium or a plant cell, but you will never find this n
extracellular structure around an animal/mammal cell. The organelle is found in the cell, has a
double membrane and is the location of DNA. This non-membrane organelle translates proteins.
This membranous organelle has #4 associated with it. The membrane organelle receives material
from #5 and modifies the lipids and the proteins. The digestive organelle of the cell is the The
membrane organelle that produces ATP and other products is the: The network of proteins
found through the cell, aiding in the placement and movement of the organelles is the: The
specific proteins that comprise #9: The large \"long\" column of globular protein polymerized
together: The microfilaments are composed of these 2 proteins. Name the membrane bound
structure that moves material out of the cell. #8 is referred to as an endosymbiant. Why? Note
the progression to form a protein.
Solution
1. ANS: Cell membrane.
2. ANS: Cell wall.
3. ANS: Mitochondria (Circular DNA is present in this).
4. ANS: Ribosome\'s (The rRNA which present in the ribosomes can helps to translate mRNA
into proteins).
5. ANS: Rough endoplasmic reticulum
6. ANS: Smooth endoplasmic reticulum
7. ANS: Lysosomes.
8. ANS: Mitochondria.
9. ANS: Cytoskeleton.
10- 12 ANS: G-Actin and F-Actin.
Each microfilament is formed with hundreds of globular subunits G-Actin organized into a
double stranded helix with a 36 nm repeat F-Actin.
13. ANS: Exocytosis (In this plasma membrane surrounding substances inside the cell and
moves it outside the cell).
14. ANS: Mitochondria is a endosymbiont
1. Mitochondria had its own DNA (Which is naked and circular).
2. Mitochondria have ribosomes that are mostly similar to prokaryotes.
3. Mitochondria is only produced by pre-existing Mitochondria.
4. Mitochondria can able to synthesizes their own proteins.
15-18. ANS:
Protein synthesis is a process in which cells build proteins from information in a DNA gene in a
DNA gene in a two major steps.
A. Transcription: In which a molecule of DNA is copied into a complimentary strand of RNA.
B. Translation: In this the message in RNA is made into a protein..
You are burning the latest song you bought on ITunes to a disk. The .pdf
You are burning the latest song you bought on ITunes to a disk. The song is 3 minutes, 23
seconds long and was sampled as a stereo signal (two sides, left and right) at 44.1 kHz using 16-
bit encoding. How much space on the disk does this one track take up? If your CD is
640MBytes, how many songs of this size can you fit on one disk (assume 1 byte is 8-bits in
length)? Remember, you don’t want fractional songs! How many songs could be saved on this
same CD if the sampling rate was 128kHz using 32 bit encoding?
Solution
A)
This stream of audio frames, as a whole, is then subjected to CIRC encoding, which segments
and rearranges the data and expands it with parity bits in a way that allows occasional read errors
to be detected and corrected. CIRC encoding also interleaves the audio frames throughout the
disc over several consecutive frames so that the information will be more resistant to burst errors.
Therefore, a physical frame on the disc will actually contain information from multiple logical
audio frames. This process adds 64 bits of error correction data to each frame. After this, 8 bits
of subcode or subchannel data are added to each of these encoded frames, which is used for
control and addressing when playing the CD.
CIRC encoding plus the subcode byte generate 33-bytes long frames, called \"channel-data\"
frames. These frames are then modulated through eight-to-fourteen modulation (EFM), where
each 8-bit word is replaced with a corresponding 14-bit word designed to reduce the number of
transitions between 0 and 1. This reduces the density of physical pits on the disc and provides an
additional degree of error tolerance. Three \"merging\" bits are added before each 14-bit word for
disambiguation and synchronization. In total there are 33 × (14 + 3) = 561 bits. A 27-bit word (a
24-bit pattern plus 3 merging bits) is added to the beginning of each frame to assist with
synchronization, so the reading device can locate frames easily. With this, a frame ends up
containing 588 bits of \"channel data\" (which are decoded to only 192 bits music).
The frames of channel data are finally written to disc physically in the form of pits and lands,
with each pit or land representing a series of zeroes, and with the transition points—the edge of
each pit—representing 1. A Red Book-compatible CD-R has pit-and-land-shaped spots on a
layer of organic dye instead of actual pits and lands; a laser creates the spots by altering the
reflective properties of the dye.
The audio data stream in an audio CD is continuous, but has three parts. The main portion, which
is further divided into playable audio tracks, is the program area. This section is preceded by a
lead-in track and followed by a lead-out track. The lead-in and lead-out tracks encode only silent
audio, but all three sections contain subcode data streams.
The lead-in\'s subcode contains repeated copies of the disc\'s Table Of Contents (TOC), which
provides an index of the start positions of the track.
Write a scientific research paper that investigates the properties of.pdf
Write a scientific research paper that investigates the properties of Digital (hardware) design
from its inception till today. Be sure to pay special attention to the advent of Boolean algebra.
Logic Gates and how they revolutionized today\'s technology. The paper must include the
following sections: Introduction 1:5 pages Methodology 2 pages Project Analysis 2-3 pages
Conclusion 1 page Annotated Bibliography The introduction should be no more than a page and
a half long and should clearly state the objective(s) of your research paper. It should provide a
brief history of the digital technology and the nature of your project. The methodology section
should explain in great details to the layman, how the digital technology works from hardware
and software point of view. The project analysis section should compare features of the digital
technology with features from other technology (of your choice) from all aspects. Explain the
advantages and disadvantages of both types of technologies. The conclusion should relate to the
objective(s) stated in the introduction and summarizes the highlights of the paper. The annotated
bibliography must conform to APA standards and you must remember to reference all your
sources. Pay attention to the content, grammar and style.
Solution
DIGITAL TECHNOLOGY
INTRODUCTION:
Digital electronics or digital (electronic) circuits are electronics that handle digital signals
– discrete bands of analog levels – rather than by continuous ranges (as used in analogue
electronics). All levels within a band of values represent the same numeric value. Because of this
discretization, relatively small changes to the analog signal levels due to manufacturing
tolerance, signal attenuation or parasitic noise do not leave the discrete envelope, and as a result
are ignored by signal state sensing circuitry.
In most cases, the number of these states is two, and they are represented by two voltage bands:
one near a reference value (typically termed as \"ground\" or zero volts), and the other a value
near the supply voltage. These correspond to the \"false\" (\"0\") and \"true\" (\"1\") values of the
Boolean domain respectively, named after its inventor, George Boole, yielding binary code.
Digital techniques are useful because it is easier to get an electronic device to switch into one of
a number of known states than to accurately reproduce a continuous range of values.
Digital electronic circuits are usually made from large assemblies of logic gates, simple
electronic representations of Boolean logic functions.
METHODOLOGY:
A digital circuit is a circuit where the signal must be one of two discrete levels. Each level is
interpreted as one of two different states (for example, on/off, 0/1, true/false). Digital circuits use
transistors to create logic gates in order to perform Boolean logic. This logic is the foundation of
digital electronics and computer processing. Digital circuits are less susceptible to noise or
degradation in qualit.
Write a function that multiplies each element in the array myA.pdf
// Write a function that multiplies each element in the array \"myArray\" // by the variable
\"multiplyMe\". #include using namespace std; // TODO - Write your function prototype here
int main() { const int SIZE = 10; int myArray [SIZE] = {5, 10, 15, 20, 25, 30, 35, 40, 45,
50}; int multiplyMe = 5; // TODO - Add your function call here // print the array
for(int i=0; i < SIZE; i++) { cout << myArray[i] << \" \"; } cout << endl; return
0; } // TODO - Write your function definition here
Solution.
Which of the following is NOT a common feature of monosaccharidesA..pdf
Which of the following is NOT a common feature of monosaccharides?A. similar or equal
numbers of carbon and oxygen atomsB. a carbonyl (C=O) groupC. numerous hydroxyl (-OH)
groupsD. All of the listed responses are correct.
Solution
The general formula for monosaccharides: (CH2O)n here n=3 or >3
On the basis of number of carbon atoms in the monosaccharides, these are classified as triose,
tetrose, pentose, hexose. If n=3 then monosaccharides are called Trioses or if n=4
monosaccharides are known as Tetroses.
Monosaccharides are also classified on the basis of carbonyl functional groups. If
monosaccharide contains ketone group (>C=O) in the structure then these are classified as
Ketose (example Dihydroxyacetone, Mannoheptulose), if monosaccharide contains aldehyde
group (-CHO) then these are known as Aldoses (example Glucose, Glyceraldehyde).
Monosaccharides contain numerous hydroxyl (-OH) groups. Triose, tetrose, pentose, and hexose
contain 2, 3, 4 and 5 -OH groups respectively in their structure.
Ans D will be correct answer..
Who are the owners of a cor[poration,and how is their ownership repr.pdf
Who are the owners of a cor[poration,and how is their ownership represented?
Solution
Shareholders are the real owners of a cooperation. they hold the share capital of the Company
i.e., they bring the fund for the establishment and working of the Company. The shareholders
own the corporation and are responsible for electing the directors also.
Their ownership represented by the the number, value and class of shares they hold..
Which of these is not a determinant of diseaserandom eventaging.pdf
Which of these is not a determinant of disease?
random event
aging
occupational exposure
socio-political factorsA.
random eventB.
agingC.
occupational exposureD.
socio-political factors
Solution
B- Aging ... Is not a determinat of a disease..... because. aging is man\'s natural process... Which
can not posses any determination about disease..
2A) You are the first to arrive at the scene of a house fire. You he.pdf
2A) You are the first to arrive at the scene of a house fire. You hear somebody in the house
calling for help. Ignore the notion of helping people for the sake of helping people (a noble thing
and highly recommended, but not immediately relevant here). In your own words, explain why
it may be selectively advantageous for you to help, and risk serious injury or death in the
process, if the person calling is a close relative. What if that person is a stranger? I am looking
for a narrative explanation, not just a formula.
2B) Jane is an ant, an organism with a haplo-diploid mating system (females are diploid, males
are haploid). In a parallel of J.B.S. Haldane’s statement, should Jane give up her life to save two
of her sisters? Assume all ants can reproduce and will produce 2 offspring (neither is correct in
reality, but assume it for your calculations). List all of your calculations, and your reasoning on
why you plugged in specific numbers.
Solution
Answers:
2. (A) The concept of self-sacrificial behaviour was provided by William D. Hamilton.
According to his arguements, natural selection favours genetic success, not reproductive success,
and that individuals can pass copies of their genes on to future generations.
Genes are passed from direct parentage and by assisting the reproduction of close relatives a
concept referred to as kin selection.
The conditions under which reproductive altruism evolves:
r × B > C, where r = genetic relatedness of the altruist to the beneficiary, B = Benefit received by
the recipient, C = Cost incurred/suffered by the donor.
The coefficient of relatedness is different between everyone
r =1/2 for parents/offspring
r = 1/2 for siblings
r = 1 for offspring and themselves
r = 1/4 for half sibling
Siblings are related by 1 to themselves but 1/2 to each other, while parents are equally related to
all their offspring.
So, its a fair deal in evolutionary terms if the person to be saved is a close relative like
offsprings, siblings aorparents, but not for a stranger since r = 0.0 in a life and death situation.
(B) Half of its genome comes from the father (haploid) and half from the mother (diploid). That
means she carries all of her fathers genes and half of her mother’s genes. So does her sister,
implying that they share of course the entire genome of their common father (i.e. already 50% of
their genome), plus, on average, a quarter of their mother’s genome, yielding a coefficient of
relatedness of 0.75.
r for one sister = 0.75
So, r for two sisters = 0.75 + 0.75 = 1.5
Thus, the cost suffered by Jane is less than the benefit gained by the two sisters. So, Jane should
give up her life to save her two sisters..
Which of the following best describes link encryptionA. Data is e.pdf
Which of the following best describes link encryption?
A. Data is encrypted at the point of origin and is decrypted at the point of destination.
B. The message is decrypted and re-encrypted as it passes through each successive node using a
key common to the two nodes.
C. The KDC shares a user-unique key with each user.
D. It requires a session key that the KDC shares between the originator and the final destination.
Solution
B. With link encryption, the message is decrypted and re-encrypted as it passes through each
successive node using a key common to the two nodes. Answers A, C, and D are incorrect
because they all describe end-to-end encryption.
What structure surrounding the cytoplasm is composed mostly of a phos.pdf
The document asks a series of questions about cellular structures and processes. It asks about the organelle that produces ATP (mitochondria), the structure that makes proteins outside the cell (rough endoplasmic reticulum), the organelle that produces lipids (smooth endoplasmic reticulum), and the structure that controls most cell functions and houses DNA (nucleus). It also asks about the three parts of the cell cycle (G1, S, and G2 phases), when DNA is duplicated (S phase), and when chromosomes first split (M phase).
What is happening in the starkeythiosulfate broth that results in a.pdf
What is happening in the starkey/thiosulfate broth that results in a color change from purple to
yellow? be specific.
Solution
Growth of sulfur oxidizing bacteria would result in oxidation of sulfur which results in decreased
pH of the media causing change in color from original purple to yellow..
What is the relationship between molecular biology and cell biology.pdf
What is the relationship between molecular biology and cell biology?
Solution
Molecular biology is the study of biology at molecular level. Cell biology deals with the study of
cell properties like the physiological properties, their structure, the organelles they contain. It
also studies the interactions with their environment, their life cycle, division and death.
Molecular and cell biology study the properties and function of a cell at molecular level..
what are some examples of hazaedous waste remediationSolution.pdf
what are some examples of hazaedous waste remediation?
Solution
Hazardous Waste - Hazardous Waste is a Waste which is potentially dangerous or harmful to
human health or the environment. Hazardous wastes can be liquids, solids, or contained gases.
They can be the by-products of manufacturing processes, discarded used materials, or discarded
unused commercial products, such as cleaning fluids (solvents) or pesticides.
Wastes may be hazardous wastes if they exhibit any of the four characteristics of a hazardous
waste - Ignitability, Corrosivity, Reactivity, and Toxicity.
Hazardous Waste Remediation - Hazardous waste Remediation is the protection of human health
and the environment by reducing risk.
Primary approaches which can be used in site remediation to achieve acceptable levels of risk
Hazardous waste site remediation usually addresses soils and groundwater . However, it can also
include wastes, surface water, sediment , sludges, bedrock, buildings, and other man-made items.
The hazardous constituents may be organic, inorganic and, occasionally, radioactive. They may
be elemental ionic, dissolved, sorbed, liquid, gaseous, vaporous, solid, or any combination of
these.
Examples of Hazardous Waste Remediation -
Incineration - Incineration is a waste treatment process that involves the combustion of organic
substances contained in waste materials.Incineration of waste materials converts the waste into
ash, flue gas and heat.Incineration with energy recovery is one of several waste-to-energy (WtE)
technologies such as gasification, pyrolysis and anaerobic digestion.
Bioremediation- Bioremediation is a process used to treat contaminated media, including water,
soil and subsurface material, by altering environmental conditions to stimulate growth of
microorganisms and degrade the target pollutants.
Solidification / Stabilization- Solidification/stabilization (S/S) is a widely used treatment for the
management/disposal of a broad range of contaminated media and wastes; particularly those
contaminated with substances classified as hazardous. The treatment involves mixing a binding
reagent into the contaminated media or waste.
Thermal desorption -It is an environmental remediation technology that utilizes heat to increase
the volatility of contaminants such that they can be removed (separated) from the solid matrix
(typically soil, sludge or filter cake).
Chemical Destructions -Chemical oxidation-reduction, acid-base neutralisation, precipitation,
hydrolysis, ion exchange, thermal treatment methods, wet air oxidation photolysis.are various
examples..
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh - RAYH...
Traditional Musical Instruments of Arunachal Pradesh and Uttar Pradesh
Beyond Degrees - Empowering the Workforce in the Context of Skills-First.pptx
Iván Bornacelly, Policy Analyst at the OECD Centre for Skills, OECD, presents at the webinar 'Tackling job market gaps with a skills-first approach' on 12 June 2024
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explain why the jovian planet atmospheres are called primitive atmos.pdf
explain why the jovian planet atmospheres are called primitive atmospheres while terrestrial
planets are called evolved atmospheres.
Solution
Jupiter\'s atmosphere is composed mostly of hydrogen and helium, which are the same gases that
dominate the Sun\'s and other stars\' composition. Likewise the early Earth atmosphere was most
likely H2 and He.
But the Earth\'s gravity is too weak to retain these gases for a long time. So contrary to the
Jovian atmosphere the Earth\'s atmosphere changed and evolved. Hydrogen and helium were
quickly replaced by an atmosphere produced by volcanic out-gassing. It contained gases like
H2O, CO2, SO2, CO, S2, Cl2, N2, H2 and NH3 and CH4. But no free O2 as this is not found in
volcanic gases. A lot more water came from outer space brought by comets bombarding the
proto-Earth.
As the Earth cooled, H2O rained out and the oceans formed.
It didn\'t stop there, as oxygen started to buildup as soon as the Archean era, more than 2.5
billion years ago, by means of photochemical dissociation. That is breakup of water and
ammonia molecules by ultraviolet. The O2 level rose to 1-2% of the current level.
At these levels O3 (Ozone) can form to shield the Earth surface from ultraviolet light.
Once the Earth was properly shielded, photosynthesis by bacteria and plants could start: CO2 +
H2O + sunlight = organic compounds + O2, increasing the oxygen level to today\'s levels..
During digestion, in the intestinal lumen certain glycerols, FFA an.pdf
During digestion, in the intestinal lumen certain glycerols, FFA and bite salts from micelles
which can then bond to ___ and ___ and promote their ___. fat soluble vitamins, cholesterol,
absorption. water soluble vitamins, cholesterol, absorption. fat soluble vitamins, minerals,
absorption. fat soluble vitamins, cholesterol, digestion.
Solution
The dietary lipids contains fat, triglycerides , phospholipids, cholesterol and fat soluble vitamins.
They are absorbed back by the processes emulsification and diffusion. So the answer is
Fat soluble vitamins and cholesterol- absorption.
Describe the structures and functions associated with the water-vascu.pdf
Describe the structures and functions associated with the water-vascular system echinoderms.
Solution
Echinodermata represents the phylum of animals in which spiny-skinned organisms are included.
All the organisms of this phylum are aquatic in nature and follow either sedentary or mobile life-
style. They have penta-radial symmetry in morphology and normally contain tube feets. The
peculiar system in these organisms is the presence of a water-vascular canal. The features of
these water vascular canals are enlisted as below:
Various structures associated with water vascular system are discussed as below:
Thus, the above given explanations decipher that the water vascular system of echinoderms is
complex and has high physiological value.
Madreporite: The madreporite represents the central tube along with which all the other tubules
are associated. These water canals open directly on the surface of the organism and generate a
water pore through which water can move to and fro.
Stone canal: The stone canal is a tender tube associated with the madreporite. It has two
openings, one of which opens towards the oral cavity and other towards the axial sinus. It runs
longitudinally across the breadth of the organism\'s body. It is normally curved shaped but can be
\"S\' shaped in some organisms.
Axial sinus: The axial sinus and organ represent some accumulated connective tissues and are of
germinal in nature. These axial sinuses directly communicate with the stone canal on their
posterior end.
Ring canal: The ring canal represents one of the most important organs of the echinoderm. It is a
ring shaped structure located near the madreporite on the centre and have extensions of tubules
which help in circulation of water and fluid. Since echinoderms have pentaradial symmetry, the
ring canal often gives riset to atleast 5 extensions.
Lateral vessels and the tube-feet: These are the organs of locomotion present just below the
madreporite and lateral pentameric extensions of ring canal. Internally, the tube feet and lateral
vessel communicate with the skeleton of water-vascular system..
Considering the atoms bonded and the shape of an H_2O molecule, is th.pdf
Considering the atoms bonded and the shape of an H_2O molecule, is the H_2O molecule polar
or non-polar? polar non-polar This is not the right answer. Do NOT click this answer!
Solution
A water molecule is polar because its O-H bonds are significantly polar and its bent geometry
makes the distribution of those polar bonds asymmetrical.The side of the water molecule
containing the more electronegative oxygen atom is partially negative and the side of the
molecule containing the less electronegative hydrogen atoms is partially positive..
describe the differences and similarities between viruses, worms, an.pdf
describe the differences and similarities between viruses, worms, and Trojan horses. Provide an
example of at least one of these and describe that damage that can be done when a person or
corporation is infected.
Solution
Lets me make it clear actually what these are first and then we move to the diffrences and
similarities
A computer virus connects itself to a program or record empowering it to spread starting with
one PC then onto the next, leaving contamination as it travels. Like a human infection, a PC
virus can extend in seriousness: some may bring about just somewhat irritating impacts while
others can harm your hardware, programming or records. All viruses are joined to an executable
record, which implies the infection may exist on your PC yet it really can\'t contaminate your PC
unless you run or open the pernicious program.
A worm is like an virus by outline and is thought to be a sub-class of an infection. Worms spread
from PC to PC, yet not at all like an virus, it has the capacity to go with no human activity. A
worm exploits document or data transport includes on your framework, which is the thing that
permits it to travel unaided.
A Trojan Horse is loaded with as much cunning as the fanciful Trojan Horse it was named after.
The Trojan Horse, at first look will seem, by all accounts, to be valuable software yet will really
do harm once introduced or keep running on your PC. Those on the less than desirable end of a
Trojan Horse are normally deceived into opening them since they give off an impression of
being getting true blue programming or documents from a honest to goodness source.
Now I will Focus on the differences between Viruses, Worms and Torjan Horses
On the basis of its mode of spread
A virus attach itself to an executable file and then require a human to propogate, a worm does
not require a human action to spread,torjan horse does not replicates itself and require human
input to take into action.
similarity is that they all are malicious programs that can cause damage to your computer.
Examples:-
VIRUSES
WORMS
Torjan.
A window has the rectangle surmounted by a semi circle the total per.pdf
A window has the rectangle surmounted by a semi circle the total perimeter of window is 14 and
window has a maximum area
Find the ratio of the semicircle to the height of the rectangle
Show work
Solution
2*r+2*l+pi*r=14 l=(14-2*r-pi*r)/2 Area=l*2*r+1/2*pi*r^2 =(14-2*r-
pi*r)/2*2*r+1/2*pi*r^2 derivative=14-4*r-2*pi+pi*r=0 r*(4+pi)=14 r=14/(4+pi)=1.96 l=1.96
ratio is 1.
Can you identify how energy flows through an ecosystemCan you i.pdf
Can you identify how energy flows through an ecosystem?
Can you identify how energy flows through an ecosystem?
Solution
Energy flow in ecosystem is basically in the form of heat by food chain. Ecosystem includes
living (plants, animals, microrganismand non living things (sun, soil, climate, atmosphere etc)
and the interactions between them. The food chain in our ecosystem comprises of herbivores,
carnivores and omnivores and with every passage of energy from one organism to another there
is a 70-80% loss of energy which is in the form of heat. So we can take an example where in
green plants absorb sunlight, Co2 and water from surroundings and synthesise carbohydrates.
These green plants are then eaten by herbivores like deer, cow, so the energy is tranferred from
green plants to deer/ cow. Deer will be eaten by lion/ jackle, so again energy will be transferred
from deer to lion/ jackle. Once lion/ jackle die, they will be eaten by eagle/ vulture (scavengers).
Scavengers will then release nutrients in the environment and again the nutrients will be
recycled. So the energy is never exhaused but recycled..
You must use the full name, no abbreviations. Name the dynamic found.pdf
You must use the full name, no abbreviations. Name the dynamic found around the cell. You
will find this on the outside of a bacterium or a plant cell, but you will never find this n
extracellular structure around an animal/mammal cell. The organelle is found in the cell, has a
double membrane and is the location of DNA. This non-membrane organelle translates proteins.
This membranous organelle has #4 associated with it. The membrane organelle receives material
from #5 and modifies the lipids and the proteins. The digestive organelle of the cell is the The
membrane organelle that produces ATP and other products is the: The network of proteins
found through the cell, aiding in the placement and movement of the organelles is the: The
specific proteins that comprise #9: The large \"long\" column of globular protein polymerized
together: The microfilaments are composed of these 2 proteins. Name the membrane bound
structure that moves material out of the cell. #8 is referred to as an endosymbiant. Why? Note
the progression to form a protein.
Solution
1. ANS: Cell membrane.
2. ANS: Cell wall.
3. ANS: Mitochondria (Circular DNA is present in this).
4. ANS: Ribosome\'s (The rRNA which present in the ribosomes can helps to translate mRNA
into proteins).
5. ANS: Rough endoplasmic reticulum
6. ANS: Smooth endoplasmic reticulum
7. ANS: Lysosomes.
8. ANS: Mitochondria.
9. ANS: Cytoskeleton.
10- 12 ANS: G-Actin and F-Actin.
Each microfilament is formed with hundreds of globular subunits G-Actin organized into a
double stranded helix with a 36 nm repeat F-Actin.
13. ANS: Exocytosis (In this plasma membrane surrounding substances inside the cell and
moves it outside the cell).
14. ANS: Mitochondria is a endosymbiont
1. Mitochondria had its own DNA (Which is naked and circular).
2. Mitochondria have ribosomes that are mostly similar to prokaryotes.
3. Mitochondria is only produced by pre-existing Mitochondria.
4. Mitochondria can able to synthesizes their own proteins.
15-18. ANS:
Protein synthesis is a process in which cells build proteins from information in a DNA gene in a
DNA gene in a two major steps.
A. Transcription: In which a molecule of DNA is copied into a complimentary strand of RNA.
B. Translation: In this the message in RNA is made into a protein..
You are burning the latest song you bought on ITunes to a disk. The .pdf
You are burning the latest song you bought on ITunes to a disk. The song is 3 minutes, 23
seconds long and was sampled as a stereo signal (two sides, left and right) at 44.1 kHz using 16-
bit encoding. How much space on the disk does this one track take up? If your CD is
640MBytes, how many songs of this size can you fit on one disk (assume 1 byte is 8-bits in
length)? Remember, you don’t want fractional songs! How many songs could be saved on this
same CD if the sampling rate was 128kHz using 32 bit encoding?
Solution
A)
This stream of audio frames, as a whole, is then subjected to CIRC encoding, which segments
and rearranges the data and expands it with parity bits in a way that allows occasional read errors
to be detected and corrected. CIRC encoding also interleaves the audio frames throughout the
disc over several consecutive frames so that the information will be more resistant to burst errors.
Therefore, a physical frame on the disc will actually contain information from multiple logical
audio frames. This process adds 64 bits of error correction data to each frame. After this, 8 bits
of subcode or subchannel data are added to each of these encoded frames, which is used for
control and addressing when playing the CD.
CIRC encoding plus the subcode byte generate 33-bytes long frames, called \"channel-data\"
frames. These frames are then modulated through eight-to-fourteen modulation (EFM), where
each 8-bit word is replaced with a corresponding 14-bit word designed to reduce the number of
transitions between 0 and 1. This reduces the density of physical pits on the disc and provides an
additional degree of error tolerance. Three \"merging\" bits are added before each 14-bit word for
disambiguation and synchronization. In total there are 33 × (14 + 3) = 561 bits. A 27-bit word (a
24-bit pattern plus 3 merging bits) is added to the beginning of each frame to assist with
synchronization, so the reading device can locate frames easily. With this, a frame ends up
containing 588 bits of \"channel data\" (which are decoded to only 192 bits music).
The frames of channel data are finally written to disc physically in the form of pits and lands,
with each pit or land representing a series of zeroes, and with the transition points—the edge of
each pit—representing 1. A Red Book-compatible CD-R has pit-and-land-shaped spots on a
layer of organic dye instead of actual pits and lands; a laser creates the spots by altering the
reflective properties of the dye.
The audio data stream in an audio CD is continuous, but has three parts. The main portion, which
is further divided into playable audio tracks, is the program area. This section is preceded by a
lead-in track and followed by a lead-out track. The lead-in and lead-out tracks encode only silent
audio, but all three sections contain subcode data streams.
The lead-in\'s subcode contains repeated copies of the disc\'s Table Of Contents (TOC), which
provides an index of the start positions of the track.
Write a scientific research paper that investigates the properties of.pdf
Write a scientific research paper that investigates the properties of Digital (hardware) design
from its inception till today. Be sure to pay special attention to the advent of Boolean algebra.
Logic Gates and how they revolutionized today\'s technology. The paper must include the
following sections: Introduction 1:5 pages Methodology 2 pages Project Analysis 2-3 pages
Conclusion 1 page Annotated Bibliography The introduction should be no more than a page and
a half long and should clearly state the objective(s) of your research paper. It should provide a
brief history of the digital technology and the nature of your project. The methodology section
should explain in great details to the layman, how the digital technology works from hardware
and software point of view. The project analysis section should compare features of the digital
technology with features from other technology (of your choice) from all aspects. Explain the
advantages and disadvantages of both types of technologies. The conclusion should relate to the
objective(s) stated in the introduction and summarizes the highlights of the paper. The annotated
bibliography must conform to APA standards and you must remember to reference all your
sources. Pay attention to the content, grammar and style.
Solution
DIGITAL TECHNOLOGY
INTRODUCTION:
Digital electronics or digital (electronic) circuits are electronics that handle digital signals
– discrete bands of analog levels – rather than by continuous ranges (as used in analogue
electronics). All levels within a band of values represent the same numeric value. Because of this
discretization, relatively small changes to the analog signal levels due to manufacturing
tolerance, signal attenuation or parasitic noise do not leave the discrete envelope, and as a result
are ignored by signal state sensing circuitry.
In most cases, the number of these states is two, and they are represented by two voltage bands:
one near a reference value (typically termed as \"ground\" or zero volts), and the other a value
near the supply voltage. These correspond to the \"false\" (\"0\") and \"true\" (\"1\") values of the
Boolean domain respectively, named after its inventor, George Boole, yielding binary code.
Digital techniques are useful because it is easier to get an electronic device to switch into one of
a number of known states than to accurately reproduce a continuous range of values.
Digital electronic circuits are usually made from large assemblies of logic gates, simple
electronic representations of Boolean logic functions.
METHODOLOGY:
A digital circuit is a circuit where the signal must be one of two discrete levels. Each level is
interpreted as one of two different states (for example, on/off, 0/1, true/false). Digital circuits use
transistors to create logic gates in order to perform Boolean logic. This logic is the foundation of
digital electronics and computer processing. Digital circuits are less susceptible to noise or
degradation in qualit.
Write a function that multiplies each element in the array myA.pdf
// Write a function that multiplies each element in the array \"myArray\" // by the variable
\"multiplyMe\". #include using namespace std; // TODO - Write your function prototype here
int main() { const int SIZE = 10; int myArray [SIZE] = {5, 10, 15, 20, 25, 30, 35, 40, 45,
50}; int multiplyMe = 5; // TODO - Add your function call here // print the array
for(int i=0; i < SIZE; i++) { cout << myArray[i] << \" \"; } cout << endl; return
0; } // TODO - Write your function definition here
Solution.
Which of the following is NOT a common feature of monosaccharidesA..pdf
Which of the following is NOT a common feature of monosaccharides?A. similar or equal
numbers of carbon and oxygen atomsB. a carbonyl (C=O) groupC. numerous hydroxyl (-OH)
groupsD. All of the listed responses are correct.
Solution
The general formula for monosaccharides: (CH2O)n here n=3 or >3
On the basis of number of carbon atoms in the monosaccharides, these are classified as triose,
tetrose, pentose, hexose. If n=3 then monosaccharides are called Trioses or if n=4
monosaccharides are known as Tetroses.
Monosaccharides are also classified on the basis of carbonyl functional groups. If
monosaccharide contains ketone group (>C=O) in the structure then these are classified as
Ketose (example Dihydroxyacetone, Mannoheptulose), if monosaccharide contains aldehyde
group (-CHO) then these are known as Aldoses (example Glucose, Glyceraldehyde).
Monosaccharides contain numerous hydroxyl (-OH) groups. Triose, tetrose, pentose, and hexose
contain 2, 3, 4 and 5 -OH groups respectively in their structure.
Ans D will be correct answer..
Who are the owners of a cor[poration,and how is their ownership repr.pdf
Who are the owners of a cor[poration,and how is their ownership represented?
Solution
Shareholders are the real owners of a cooperation. they hold the share capital of the Company
i.e., they bring the fund for the establishment and working of the Company. The shareholders
own the corporation and are responsible for electing the directors also.
Their ownership represented by the the number, value and class of shares they hold..
Which of these is not a determinant of diseaserandom eventaging.pdf
Which of these is not a determinant of disease?
random event
aging
occupational exposure
socio-political factorsA.
random eventB.
agingC.
occupational exposureD.
socio-political factors
Solution
B- Aging ... Is not a determinat of a disease..... because. aging is man\'s natural process... Which
can not posses any determination about disease..
2A) You are the first to arrive at the scene of a house fire. You he.pdf
2A) You are the first to arrive at the scene of a house fire. You hear somebody in the house
calling for help. Ignore the notion of helping people for the sake of helping people (a noble thing
and highly recommended, but not immediately relevant here). In your own words, explain why
it may be selectively advantageous for you to help, and risk serious injury or death in the
process, if the person calling is a close relative. What if that person is a stranger? I am looking
for a narrative explanation, not just a formula.
2B) Jane is an ant, an organism with a haplo-diploid mating system (females are diploid, males
are haploid). In a parallel of J.B.S. Haldane’s statement, should Jane give up her life to save two
of her sisters? Assume all ants can reproduce and will produce 2 offspring (neither is correct in
reality, but assume it for your calculations). List all of your calculations, and your reasoning on
why you plugged in specific numbers.
Solution
Answers:
2. (A) The concept of self-sacrificial behaviour was provided by William D. Hamilton.
According to his arguements, natural selection favours genetic success, not reproductive success,
and that individuals can pass copies of their genes on to future generations.
Genes are passed from direct parentage and by assisting the reproduction of close relatives a
concept referred to as kin selection.
The conditions under which reproductive altruism evolves:
r × B > C, where r = genetic relatedness of the altruist to the beneficiary, B = Benefit received by
the recipient, C = Cost incurred/suffered by the donor.
The coefficient of relatedness is different between everyone
r =1/2 for parents/offspring
r = 1/2 for siblings
r = 1 for offspring and themselves
r = 1/4 for half sibling
Siblings are related by 1 to themselves but 1/2 to each other, while parents are equally related to
all their offspring.
So, its a fair deal in evolutionary terms if the person to be saved is a close relative like
offsprings, siblings aorparents, but not for a stranger since r = 0.0 in a life and death situation.
(B) Half of its genome comes from the father (haploid) and half from the mother (diploid). That
means she carries all of her fathers genes and half of her mother’s genes. So does her sister,
implying that they share of course the entire genome of their common father (i.e. already 50% of
their genome), plus, on average, a quarter of their mother’s genome, yielding a coefficient of
relatedness of 0.75.
r for one sister = 0.75
So, r for two sisters = 0.75 + 0.75 = 1.5
Thus, the cost suffered by Jane is less than the benefit gained by the two sisters. So, Jane should
give up her life to save her two sisters..
Which of the following best describes link encryptionA. Data is e.pdf
Which of the following best describes link encryption?
A. Data is encrypted at the point of origin and is decrypted at the point of destination.
B. The message is decrypted and re-encrypted as it passes through each successive node using a
key common to the two nodes.
C. The KDC shares a user-unique key with each user.
D. It requires a session key that the KDC shares between the originator and the final destination.
Solution
B. With link encryption, the message is decrypted and re-encrypted as it passes through each
successive node using a key common to the two nodes. Answers A, C, and D are incorrect
because they all describe end-to-end encryption.
What structure surrounding the cytoplasm is composed mostly of a phos.pdf
The document asks a series of questions about cellular structures and processes. It asks about the organelle that produces ATP (mitochondria), the structure that makes proteins outside the cell (rough endoplasmic reticulum), the organelle that produces lipids (smooth endoplasmic reticulum), and the structure that controls most cell functions and houses DNA (nucleus). It also asks about the three parts of the cell cycle (G1, S, and G2 phases), when DNA is duplicated (S phase), and when chromosomes first split (M phase).
What is happening in the starkeythiosulfate broth that results in a.pdf
What is happening in the starkey/thiosulfate broth that results in a color change from purple to
yellow? be specific.
Solution
Growth of sulfur oxidizing bacteria would result in oxidation of sulfur which results in decreased
pH of the media causing change in color from original purple to yellow..
What is the relationship between molecular biology and cell biology.pdf
What is the relationship between molecular biology and cell biology?
Solution
Molecular biology is the study of biology at molecular level. Cell biology deals with the study of
cell properties like the physiological properties, their structure, the organelles they contain. It
also studies the interactions with their environment, their life cycle, division and death.
Molecular and cell biology study the properties and function of a cell at molecular level..
what are some examples of hazaedous waste remediationSolution.pdf
what are some examples of hazaedous waste remediation?
Solution
Hazardous Waste - Hazardous Waste is a Waste which is potentially dangerous or harmful to
human health or the environment. Hazardous wastes can be liquids, solids, or contained gases.
They can be the by-products of manufacturing processes, discarded used materials, or discarded
unused commercial products, such as cleaning fluids (solvents) or pesticides.
Wastes may be hazardous wastes if they exhibit any of the four characteristics of a hazardous
waste - Ignitability, Corrosivity, Reactivity, and Toxicity.
Hazardous Waste Remediation - Hazardous waste Remediation is the protection of human health
and the environment by reducing risk.
Primary approaches which can be used in site remediation to achieve acceptable levels of risk
Hazardous waste site remediation usually addresses soils and groundwater . However, it can also
include wastes, surface water, sediment , sludges, bedrock, buildings, and other man-made items.
The hazardous constituents may be organic, inorganic and, occasionally, radioactive. They may
be elemental ionic, dissolved, sorbed, liquid, gaseous, vaporous, solid, or any combination of
these.
Examples of Hazardous Waste Remediation -
Incineration - Incineration is a waste treatment process that involves the combustion of organic
substances contained in waste materials.Incineration of waste materials converts the waste into
ash, flue gas and heat.Incineration with energy recovery is one of several waste-to-energy (WtE)
technologies such as gasification, pyrolysis and anaerobic digestion.
Bioremediation- Bioremediation is a process used to treat contaminated media, including water,
soil and subsurface material, by altering environmental conditions to stimulate growth of
microorganisms and degrade the target pollutants.
Solidification / Stabilization- Solidification/stabilization (S/S) is a widely used treatment for the
management/disposal of a broad range of contaminated media and wastes; particularly those
contaminated with substances classified as hazardous. The treatment involves mixing a binding
reagent into the contaminated media or waste.
Thermal desorption -It is an environmental remediation technology that utilizes heat to increase
the volatility of contaminants such that they can be removed (separated) from the solid matrix
(typically soil, sludge or filter cake).
Chemical Destructions -Chemical oxidation-reduction, acid-base neutralisation, precipitation,
hydrolysis, ion exchange, thermal treatment methods, wet air oxidation photolysis.are various
examples..
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