What are the two principles of Protocol Layering (use your own word.pdf
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What additional data could be used to help distinguish between certain classes when you use objected oriented segmentation as a method to classify an image Solution THE OBJECT ORIENTED SEGMENTATION APPROACH IS A PHILOSOPHY OF IMPROVED IMAGE UNDERSTANDING.HUMAN VISION IS VERY CAPABLE OF DETECTING OBJECTS AND OBJECT CLASSES WITH AN IMAGE. TO PURSUE THIS TYPE OF ANALYSIS, 1. IT IS IMPORTANT TO STAY CLOSE TO THE INTUITIVE IMAGE UNDERSTANDING. OBJECT BASED CLASSIFICATION STARTS WITH THE CRUCIAL INITIAL STEP OF GROUPING NEIGHBOURING PIXELS INTO MEANINGFUL AREASESPECIALLY FOR THE END USER. 2. THE SEGMENTATION AND OBJECT GENERATION MUST BE SET ACCORDING TO THE RESOLUTION AND THE SCALE OF THE EXPECTED OBJECTS.THE SPATIAL CONTEXT PLAY A MAJOR ROLE IN PIXEL BASED ANALYSIS.FILTER OPERATIONS WHICH ARE AN IMPORTANT PART OF THE PIXEL BASED SPATIAL ANALYSIS, HAVE THE LIMITATION OF THEIR WINDOW SIZE. 3. IN OBJECT ANALYSIS, THIS LIMITATION DOES NOT EXIST.THE SPATIAL CONTEXT CAN BE DESCRIBEDIN TERMS OF TOPOLOGIC RELATIONSOF NEIGHBOURHOOD OBJECTS. 4. TO DEFINE THE IMAGE OBJECTS THE THUMB RULE FOR A SEGMENTATION OR PRE SEGMENTATION OF IMAGE PRIMITIVES WHICH CAN BUILD UP THE CORRESPONDING OBJECTS, THERE ARE SEVERAL APPROACHES IN IMAGE ANALYSIS AND PATTERN RECOGNITIONARE EXPLOITED TO GENERATE HYPOTHESIS FOR THE SEGMENTATION RULES AS AN ALTERNATIVE TO KNOWLEDGE BASED SEGMENTATION. 5. THE EARTH OBSERVATION DATA ARE NOT SUFFICIENT TO CHARACTERIZE COMPLETELY THE NATURAL ENVIRONMENT. THEY NEED TO BE ASSOSIATED WITH OTHER DATA. 6. WE HAVE TO FURTHER INVESTIGATE SYNERGY EFFECTS BETWEEN SATELLITE IMAGES AND GIS DERIVED VECTOR DATA, SUCH AS DIGITAL TOPOGRAPHIC DATABASE. THE USE OF SUCH TOPOGRAPHIC DATABASES, WHICH ARE BUILT UP IN MOST COUNTRIES CAN SUPPOT THE SEGMENTATION. 7. THIS DIGITAL DATABASE OFFERS A GEOMETRIC AS WELL AS A SEMANTIC PREDICTIONFOR OBJETCS IN THE IMAGE.DEVELOP A METHODOLOGICAL FRAMEWORK OF OBJECT BASED ANALYSISWHICH OFFERS AN OPTION TO IMPROVE AUTOMATIC INFORMATION DERIVATION FROM A RAPIDLY INCREASED AMOUNT OF DATA FROM DIFFERENT SENSORS. 8. ACHIEVE MEANINGFUL REPRESENTATIONS OF SPATIAL PHYSICAL PARAMETERS DERIVED FROM ORIGINAL REFLECTANCE VALUES. 9. DEVELOP NEW METHODS FOR ACCURACY ASSESSMENT BASED ON OBJECTS RATHER THAN PIXELS. 10. DEVELOP DATA ANALYISIS TOOLS FOR ENVIRONMENTAL MANAGEMENT BASED ON MULTI SCALE OBJECT REPRESENTATION IN ORDER TO SUPPORT DECISION MAKING. THANK YOU..
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Question Completion Status QUESTION In the absence of oxygen, animal.pdf
Question Completion Status: QUESTION In the absence of oxygen, animal cells, through a
series of reactions, convert pyruvate to O lactic O CO2 and ethyl alcohol O carbon dioxide O
citric acid QUESTION 2 The prepatory reaction converts the pyruvate into for use in the citric
acid cycle) and (which leaves the cell O NADH: NAD O ATP: ADP O glucose; CO2 O acetyl-
CoA, carbon dioxide QUESTION 3 Energy released as electrons are transported down the
electron transport chain in cellular respiration is directly used to O oxygen O produce co2 O
produce NADH and FADH2 O establish a proton (H gradient OS
Solution
1.pyruvate is converted to CO2 and ethyl alcohol.
2.preparatory reaction convertpyruvate to acetyl coA and CO2
3.to produce oxygen.
4.glucose is ultimately converted to CO2.
5.cellular respiration requires glucose and CO2.
6.TCA, ETS take place in mitochondria.
7.purpose of dellular respiration to produce ATP.
8.ATP SYNTHASE PHOSPHORYLATES ADP TP ATP.
9.all organisms respire.
10.oxygen is not a product of respiration.
11.oxygen is the final electron acceptor.
12.glycolysis break glucose in 2 molecule pyruvate.
13.CO2 is produced in the preparatory reaction of citric acid cycle.
14.FADH2 and NADH carry electron .
15.in CYTRIC ACID CYCLE 34 molecule ATP produced from one glucose..
Short Answer 1. What are the 5 mechanisms of evolutionary ch.pdf
Short Answer: 1. What are the 5 mechanisms of evolutionary change that make up Modern
synthesis?
Solution
Answer = 1]
Five mechanisms of evolutionary change that make up modern synthesis are
1] Natural selection = is the process by which traits that enhance survival & reproduction
becomes more common in succesive generations of population
2] Biased mutation =When there are different probabilities occure at molecular level for different
mutations to occur known as biased mutation
3] Genetic drift = is the change in allele frequency from one generation to next that occure
because alleles are subjected to sampling error
drift acts faster & has more drastic effects in smaller populations
genetic drift contribute to speciation
4] Genetic hitchhiking =When one allele in particular haplotype is strongly beneficial natural
selection can drive a selective sweep that will also cause the other allales in haplotype to become
more common in population is called as genetic hitchhiking
5] Gene flow = Involves the exchange of genes between populations & species
gene flow between population can introduce traits & alleles.
Review Preparation for Statistics A statistics instructor wanted to s.pdf
Review Preparation for Statistics A statistics instructor wanted to see if student participation in
review preparation methods led to higher examination scores. Five students were randomly
selected and placed in each test group for a three-week unit on statistical inference. Everyone
took the same examination at the end of the unit, and the resulting scores are shown below Is
there sufficient evidence at 0.05 to conclude an interaction between the two factors? Is there
sufficient evidence to conclude a difference in mean scores based on formula delivery system? Is
there sufficient evidence to conclude a difference in mean scores based on the review
organization technique? (12-3)
Solution.
Problem 3-8BRIDGEPORT ADVERTISING TRIAL BALANCE DECEMBER 31, 2.pdf
Problem 3-8
BRIDGEPORT ADVERTISING
TRIAL BALANCE
DECEMBER 31, 2017
Unadjusted
Adjusted
Dr.
Cr.
Dr.
Cr.
$148,820
$148,820
$158,950
$158,950
No.
Date
Account Titles and Explanation
Debit
Credit
(To record accrued service revenue)
(To record supplies used)
(To expired insurance)
(To record depreciation on equiment)
(To record interest accrued on the note)
(To record service revenue earned)
(To record accrued wages)
Assets
Liabilities and Stockholders’ Equity
%
Open Show Work
SAVE FOR LATER
SUBMIT ANSWER
Problem 3-8 Bridgeport Advertising was founded by Murali Vedula in January 2015. Presented
below are both the adjusted and unadjusted trial balances as of December 31, 2017.
BRIDGEPORT ADVERTISING
TRIAL BALANCE
DECEMBER 31, 2017
Unadjusted
Adjusted
Dr.
Cr.
Dr.
Cr.Cash$16,740$16,740Accounts Receivable19,17023,012Supplies9,7606,845Prepaid
Insurance4,2202,625Equipment64,70064,700Accumulated Depreciation-
Equipment$24,120$29,050Notes Payable8,6008,600Accounts Payable1,8901,890Interest
Payable0602Unearned Service Revenue5,3503,405Salaries and Wages Payable0756Common
Stock19,22019,220Retained Earnings27,24727,247Dividends19,30019,300Service
Revenue62,39368,180Salaries and Wages Expense9,40010,156Insurance Expense1,595Interest
Expense602Depreciation Expense4,930Supplies Expense2,915Rent Expense5,5305,530
$148,820
$148,820
$158,950
$158,950Journalize the annual adjusting entries that were made. (Credit account titles are
automatically indented when amount is entered. Do not indent manually. If no entry is required,
select \"No entry\" for the account titles and enter 0 for the amounts.)
No.
Date
Account Titles and Explanation
Debit
Credit1.Dec. 31
(To record accrued service revenue)2.Dec. 31
(To record supplies used)3.Dec. 31
(To expired insurance)4.Dec. 31
(To record depreciation on equiment)5.Dec. 31
(To record interest accrued on the note)6.Dec. 31
(To record service revenue earned)7.Dec. 31
(To record accrued wages)Prepare an income statement for the year ended December 31. (Enter
loss using either a negative sign preceding the number e.g. -45 or parentheses e.g. (45).)
BRIDGEPORTADVERTISING
Income Statement
December 31, 2017For the Year Ended December 31, 2017For the Quarter Ended December
31, 2017
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues
$
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues
$
Dividends Expenses Net Income / (Loss) Retained Earnings, Janaury 1 Retained
Earnings, December 31 Revenues Total Expenses Total Revenues
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues$
Prepare a retained earnings statement for the year ended December 31. (List items that increase
retained earnings first.)
BRIDGEPORTADVERTISING
Retained Earnings Statement
December 31, 2017For the Year Ended December 31, 2017For .
Q1 Define the three types of data flow. Q2 What are the t.pdf
Q1: Define the three types of data flow.
Q2: What are the two principles of Protocol Layering?
Q3: A sine wave is offset 1/4 cycle with respect to time 0. What is its phase in degrees and
radians?
Q4: List three properties for Infrared waves.
Solution
: 4) Infrared (IR) waves are :
- electromagnetic in nature with longer wavelengths than visible light
- wavelength is from 0.74 micrometres (µm) to 300 µm.
- frequency range is from 1 to 400 THz
IR is thermal radiation emitted by objects near room temperature.
3) phase in degree :
1 complete cycle = 360 °
Then 1/4th cycle =1/4*360= 90°
Phase in radians :
1° =2/360 radians
Then 90° = 90*2/360 = /2 rad = 1.57 rad
1)
Simplex:All the data flow is unidirectional from transmitter to receiver.
Duplex communication is two way in that we have two types:
Half-duplex: communication may be likened to two way Either can may be used to transmit or
receive, but not at the same time
Fullduplex:Full-duplex communication is more like a true telephone, where two people can talk
at the same time and hear one another simultaneously.
1)
When communication is simple we may need only one protocol. When the communication is
Complex we need a protocol at each layer which is called protocol layering
We need to understand two simple principles for that
a) The first principle dictates that if we want
bidirectional communication, we need to make each
layer so that it is able to perform two opposite tasks,
one in each direction.
b) The second principle that we need to follow in
protocol layering is that the two objects under each
layer at both sides should be identical..
Multiple choice A young girl is staring at the raindrops running do.pdf
Multiple choice: A young girl is staring at the raindrops running down her window. She notices
that the raindrops remain more or less intact, even as they cascade down the windowpane. This is
a result of:
ionic bonds between water molecules.
oxygen bonds between water molecules.
covalent bonds between water molecules.
hydrogen bonds between water molecules.
polar covalent bonds between water molecules.
Solution
d). hydrogen bonds between water molecules.
Because of the hydrogen bonding, water exhibits high surface tension. So, water molecules on
the surface are more strongly attracted to each other than to the air above. So that the raindrops
remain more or less intact, even as they cascade down the windowpane..
mitosis is a special set of events that occur in what region of .pdf
mitosis is a special set of events that occur in what region of the cell?
mitosis is a special set of events that occur in what region of the cell?
mitosis is a special set of events that occur in what region of the cell?
Solution
Mitosis is found to occur in the cell nuclei of eukaryotic cells.
In the process of mitosis, one cell divides into two cells and these two cells will be genetically
identical to the original cell.
Our body will grow and repair itself due to the process of mitosis.
Mitosis includes five phases: prophase, prometaphase, metaphase, anaphase and telophase..
If you were looking at an mRNA and saw the codon AUG, what would you .pdf
If you were looking at an mRNA and saw the codon AUG, what would you conclude about it?
What does it mean to that the genetic code is redundant, but not ambiguous? The genetic code is
nearly universal, meaning the same RNA codon that designates tryptophan in humans, designates
tryptophan in bacteria. a. What has this knowledge allowed us to conclude about the code? b.
What has this knowledge allowed us to do with genes Transcription is the DNA-directed
synthesis of RNA: a closer look Describe the general model of transcription. Include in your
answer the steps of transcription and the key elements of each step. Compare and contrast
transcription between prokaryotes and eukaryotes What makes RNA polymerase start
transcribing in a gene at the right place on the DNA of a prokaryotic cell? What makes RNA
polymerase start transcribing in a gene at the right place on the DNA of a eukaryotic cell?
Solution
A)Transcription is the first step in gene expression. It involves copying a gene\'s DNA sequence
to make an RNA molecule.Transcription can be divided into four distinct stages:
Template recognition
Initiation
Elongation
Termination
Initiation:RNA polymerase binds to a sequence of DNA called the promoter,found near the
beginning of a gene.Each gene has its own promoter.Once bound,RNA polymerase separates the
DNA strands,providing the single-stranded template needed for transcription.
Elongation:One strand of DNA,the template strand,acts as a template for RNA polymerase.As it
reads this template one base at a time,the polymerase builds an RNA molecule out of
complementary nucleotides,making a chain that grows from 5\' to 3\'.The RNA transcript carries
the same information as the non-template (coding) strand of DNA,but it contains the base uracil
instead of thymine .
Termination:Sequences called terminators signal that the RNA transcript is complete.Once they
are transcribed,they cause the transcript to be released from the RNA polymerase.
B)Prokaryotes do not have an organized nucleus,so the nuclear materials or DNA is in the
cytoplasm.Therefore,the transcription occurs in the cytoplasm and all the precursors needed for
the transcription are found in the cytoplasm.Prokaryotic transcription requires the RNA
polymerase enzyme in order for the transcription to be successfully completed.The enzyme binds
to the sigma factor and the promoter region,and then initiate the transcription by completing the
holoenzyme.In prokaryotes,DNA is not bound to histones.Thus,the transcription initiates
directly.This could be advantageous when prokaryotes have overlapping genes.Transcription
starts at the promoter region and elongate through the coding region and ends when the RNA
polymerase reads the termination signal.There are two types of termination signals,Rho-
dependent and independant.Transcribed mRNA will be completely translated during the
transcription,and no post-transcription processing will be undergoing most of the
time.Transcriptionl unit has one or more .
Recommended
Question Completion Status QUESTION In the absence of oxygen, animal.pdf
Question Completion Status: QUESTION In the absence of oxygen, animal cells, through a
series of reactions, convert pyruvate to O lactic O CO2 and ethyl alcohol O carbon dioxide O
citric acid QUESTION 2 The prepatory reaction converts the pyruvate into for use in the citric
acid cycle) and (which leaves the cell O NADH: NAD O ATP: ADP O glucose; CO2 O acetyl-
CoA, carbon dioxide QUESTION 3 Energy released as electrons are transported down the
electron transport chain in cellular respiration is directly used to O oxygen O produce co2 O
produce NADH and FADH2 O establish a proton (H gradient OS
Solution
1.pyruvate is converted to CO2 and ethyl alcohol.
2.preparatory reaction convertpyruvate to acetyl coA and CO2
3.to produce oxygen.
4.glucose is ultimately converted to CO2.
5.cellular respiration requires glucose and CO2.
6.TCA, ETS take place in mitochondria.
7.purpose of dellular respiration to produce ATP.
8.ATP SYNTHASE PHOSPHORYLATES ADP TP ATP.
9.all organisms respire.
10.oxygen is not a product of respiration.
11.oxygen is the final electron acceptor.
12.glycolysis break glucose in 2 molecule pyruvate.
13.CO2 is produced in the preparatory reaction of citric acid cycle.
14.FADH2 and NADH carry electron .
15.in CYTRIC ACID CYCLE 34 molecule ATP produced from one glucose..
Short Answer 1. What are the 5 mechanisms of evolutionary ch.pdf
Short Answer: 1. What are the 5 mechanisms of evolutionary change that make up Modern
synthesis?
Solution
Answer = 1]
Five mechanisms of evolutionary change that make up modern synthesis are
1] Natural selection = is the process by which traits that enhance survival & reproduction
becomes more common in succesive generations of population
2] Biased mutation =When there are different probabilities occure at molecular level for different
mutations to occur known as biased mutation
3] Genetic drift = is the change in allele frequency from one generation to next that occure
because alleles are subjected to sampling error
drift acts faster & has more drastic effects in smaller populations
genetic drift contribute to speciation
4] Genetic hitchhiking =When one allele in particular haplotype is strongly beneficial natural
selection can drive a selective sweep that will also cause the other allales in haplotype to become
more common in population is called as genetic hitchhiking
5] Gene flow = Involves the exchange of genes between populations & species
gene flow between population can introduce traits & alleles.
Review Preparation for Statistics A statistics instructor wanted to s.pdf
Review Preparation for Statistics A statistics instructor wanted to see if student participation in
review preparation methods led to higher examination scores. Five students were randomly
selected and placed in each test group for a three-week unit on statistical inference. Everyone
took the same examination at the end of the unit, and the resulting scores are shown below Is
there sufficient evidence at 0.05 to conclude an interaction between the two factors? Is there
sufficient evidence to conclude a difference in mean scores based on formula delivery system? Is
there sufficient evidence to conclude a difference in mean scores based on the review
organization technique? (12-3)
Solution.
Problem 3-8BRIDGEPORT ADVERTISING TRIAL BALANCE DECEMBER 31, 2.pdf
Problem 3-8
BRIDGEPORT ADVERTISING
TRIAL BALANCE
DECEMBER 31, 2017
Unadjusted
Adjusted
Dr.
Cr.
Dr.
Cr.
$148,820
$148,820
$158,950
$158,950
No.
Date
Account Titles and Explanation
Debit
Credit
(To record accrued service revenue)
(To record supplies used)
(To expired insurance)
(To record depreciation on equiment)
(To record interest accrued on the note)
(To record service revenue earned)
(To record accrued wages)
Assets
Liabilities and Stockholders’ Equity
%
Open Show Work
SAVE FOR LATER
SUBMIT ANSWER
Problem 3-8 Bridgeport Advertising was founded by Murali Vedula in January 2015. Presented
below are both the adjusted and unadjusted trial balances as of December 31, 2017.
BRIDGEPORT ADVERTISING
TRIAL BALANCE
DECEMBER 31, 2017
Unadjusted
Adjusted
Dr.
Cr.
Dr.
Cr.Cash$16,740$16,740Accounts Receivable19,17023,012Supplies9,7606,845Prepaid
Insurance4,2202,625Equipment64,70064,700Accumulated Depreciation-
Equipment$24,120$29,050Notes Payable8,6008,600Accounts Payable1,8901,890Interest
Payable0602Unearned Service Revenue5,3503,405Salaries and Wages Payable0756Common
Stock19,22019,220Retained Earnings27,24727,247Dividends19,30019,300Service
Revenue62,39368,180Salaries and Wages Expense9,40010,156Insurance Expense1,595Interest
Expense602Depreciation Expense4,930Supplies Expense2,915Rent Expense5,5305,530
$148,820
$148,820
$158,950
$158,950Journalize the annual adjusting entries that were made. (Credit account titles are
automatically indented when amount is entered. Do not indent manually. If no entry is required,
select \"No entry\" for the account titles and enter 0 for the amounts.)
No.
Date
Account Titles and Explanation
Debit
Credit1.Dec. 31
(To record accrued service revenue)2.Dec. 31
(To record supplies used)3.Dec. 31
(To expired insurance)4.Dec. 31
(To record depreciation on equiment)5.Dec. 31
(To record interest accrued on the note)6.Dec. 31
(To record service revenue earned)7.Dec. 31
(To record accrued wages)Prepare an income statement for the year ended December 31. (Enter
loss using either a negative sign preceding the number e.g. -45 or parentheses e.g. (45).)
BRIDGEPORTADVERTISING
Income Statement
December 31, 2017For the Year Ended December 31, 2017For the Quarter Ended December
31, 2017
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues
$
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues
$
Dividends Expenses Net Income / (Loss) Retained Earnings, Janaury 1 Retained
Earnings, December 31 Revenues Total Expenses Total Revenues
DividendsExpensesNet Income / (Loss)Retained Earnings, Janaury 1Retained Earnings,
December 31RevenuesTotal ExpensesTotal Revenues$
Prepare a retained earnings statement for the year ended December 31. (List items that increase
retained earnings first.)
BRIDGEPORTADVERTISING
Retained Earnings Statement
December 31, 2017For the Year Ended December 31, 2017For .
Q1 Define the three types of data flow. Q2 What are the t.pdf
Q1: Define the three types of data flow.
Q2: What are the two principles of Protocol Layering?
Q3: A sine wave is offset 1/4 cycle with respect to time 0. What is its phase in degrees and
radians?
Q4: List three properties for Infrared waves.
Solution
: 4) Infrared (IR) waves are :
- electromagnetic in nature with longer wavelengths than visible light
- wavelength is from 0.74 micrometres (µm) to 300 µm.
- frequency range is from 1 to 400 THz
IR is thermal radiation emitted by objects near room temperature.
3) phase in degree :
1 complete cycle = 360 °
Then 1/4th cycle =1/4*360= 90°
Phase in radians :
1° =2/360 radians
Then 90° = 90*2/360 = /2 rad = 1.57 rad
1)
Simplex:All the data flow is unidirectional from transmitter to receiver.
Duplex communication is two way in that we have two types:
Half-duplex: communication may be likened to two way Either can may be used to transmit or
receive, but not at the same time
Fullduplex:Full-duplex communication is more like a true telephone, where two people can talk
at the same time and hear one another simultaneously.
1)
When communication is simple we may need only one protocol. When the communication is
Complex we need a protocol at each layer which is called protocol layering
We need to understand two simple principles for that
a) The first principle dictates that if we want
bidirectional communication, we need to make each
layer so that it is able to perform two opposite tasks,
one in each direction.
b) The second principle that we need to follow in
protocol layering is that the two objects under each
layer at both sides should be identical..
Multiple choice A young girl is staring at the raindrops running do.pdf
Multiple choice: A young girl is staring at the raindrops running down her window. She notices
that the raindrops remain more or less intact, even as they cascade down the windowpane. This is
a result of:
ionic bonds between water molecules.
oxygen bonds between water molecules.
covalent bonds between water molecules.
hydrogen bonds between water molecules.
polar covalent bonds between water molecules.
Solution
d). hydrogen bonds between water molecules.
Because of the hydrogen bonding, water exhibits high surface tension. So, water molecules on
the surface are more strongly attracted to each other than to the air above. So that the raindrops
remain more or less intact, even as they cascade down the windowpane..
mitosis is a special set of events that occur in what region of .pdf
mitosis is a special set of events that occur in what region of the cell?
mitosis is a special set of events that occur in what region of the cell?
mitosis is a special set of events that occur in what region of the cell?
Solution
Mitosis is found to occur in the cell nuclei of eukaryotic cells.
In the process of mitosis, one cell divides into two cells and these two cells will be genetically
identical to the original cell.
Our body will grow and repair itself due to the process of mitosis.
Mitosis includes five phases: prophase, prometaphase, metaphase, anaphase and telophase..
If you were looking at an mRNA and saw the codon AUG, what would you .pdf
If you were looking at an mRNA and saw the codon AUG, what would you conclude about it?
What does it mean to that the genetic code is redundant, but not ambiguous? The genetic code is
nearly universal, meaning the same RNA codon that designates tryptophan in humans, designates
tryptophan in bacteria. a. What has this knowledge allowed us to conclude about the code? b.
What has this knowledge allowed us to do with genes Transcription is the DNA-directed
synthesis of RNA: a closer look Describe the general model of transcription. Include in your
answer the steps of transcription and the key elements of each step. Compare and contrast
transcription between prokaryotes and eukaryotes What makes RNA polymerase start
transcribing in a gene at the right place on the DNA of a prokaryotic cell? What makes RNA
polymerase start transcribing in a gene at the right place on the DNA of a eukaryotic cell?
Solution
A)Transcription is the first step in gene expression. It involves copying a gene\'s DNA sequence
to make an RNA molecule.Transcription can be divided into four distinct stages:
Template recognition
Initiation
Elongation
Termination
Initiation:RNA polymerase binds to a sequence of DNA called the promoter,found near the
beginning of a gene.Each gene has its own promoter.Once bound,RNA polymerase separates the
DNA strands,providing the single-stranded template needed for transcription.
Elongation:One strand of DNA,the template strand,acts as a template for RNA polymerase.As it
reads this template one base at a time,the polymerase builds an RNA molecule out of
complementary nucleotides,making a chain that grows from 5\' to 3\'.The RNA transcript carries
the same information as the non-template (coding) strand of DNA,but it contains the base uracil
instead of thymine .
Termination:Sequences called terminators signal that the RNA transcript is complete.Once they
are transcribed,they cause the transcript to be released from the RNA polymerase.
B)Prokaryotes do not have an organized nucleus,so the nuclear materials or DNA is in the
cytoplasm.Therefore,the transcription occurs in the cytoplasm and all the precursors needed for
the transcription are found in the cytoplasm.Prokaryotic transcription requires the RNA
polymerase enzyme in order for the transcription to be successfully completed.The enzyme binds
to the sigma factor and the promoter region,and then initiate the transcription by completing the
holoenzyme.In prokaryotes,DNA is not bound to histones.Thus,the transcription initiates
directly.This could be advantageous when prokaryotes have overlapping genes.Transcription
starts at the promoter region and elongate through the coding region and ends when the RNA
polymerase reads the termination signal.There are two types of termination signals,Rho-
dependent and independant.Transcribed mRNA will be completely translated during the
transcription,and no post-transcription processing will be undergoing most of the
time.Transcriptionl unit has one or more .
If the proportion of an African population that is susceptible to Ma.pdf
If the proportion of an African population that is susceptible to Malaria is 0.49, what proportion
of the population is affected with Sickle Cell Anemia? A. 0.68. B. 0.09. C. 0.45. D. 0.34.
Solution
B 0.09
In literature it was mentioned that 40% of popullation in africa suffering from malaria are
sucessiptible to sickel cell anemia..
Identify nursing diagnoses goal and outcome criteria and interven.pdf
Identify nursing diagnoses goal and outcome criteria and interventions for the postoperative
patient
Identify nursing diagnoses goal and outcome criteria and interventions for the postoperative
patient
Solution
ASSESSMENT: - After the surgery, when the operation procedure is completed, the patient is
usually shifted to intensive care unit. When patient is admitted to PACU determine the medical
diagnosis, the surgery procedure, pain, consciousness, vital signs, medical history, how patient
tolerated the procedure? what was the medication given, and other assessment data listed earlier.
Check and set up equipment like oxygen, suction devices, urinary drainage etc. A more detailed
assessment should be performed when a patient returns to the nursing unit. It is actually
collecting, organizing, validating, and documenting patient data and its purpose is to establish a
database about the patient’s response to health concerns or illness and the ability to manage
health care needs.
DIAGNOSIS: - Medical diagnoses are important factors influencing patient outcomes during
hospitalization. For example, ICU patients with infectious diseases at admission had higher
mortality than the patients with gastrointestinal diseases. After surgery patients are at an
increased risk of severe complication. Nursing diagnosis is the nurse’s clinical judgment about
the patient’s response to potential health conditions and comorbid medical conditions. Several
studies show that these diagnosis influence 31 different types of patient outcomes (e.g. hospital
mortality, the length of stay, and ICU readmission) and interventions. A standard nursing
diagnosis includes three structural components: the problem, the etiology and supporting data. Its
purpose is to identify patient’s strengths and health problems that can be prevented or resolved
by collaborative and independent nursing interventions. and to develop a list of nursing
diagnoses and collaborative problems.
GOAL AND OUTCOME: - The goal is aimed at the medical diagnosis. The outcomes are aimed
at meeting the goals. A patient goal represents a predicted resolution of a diagnosis, evidence of
progress toward resolution, progress toward improved health, or continued maintenance of good
health. Each goal is limited to a time so the health care team has a common time limit for solving
the problem. For example, the goal of “patient will achieve pain relief. A short-term goal is an
objective behavior or response that you expect a patient to achieve in a short time, usually in a
week. A long-term goal is an objective behavior or response that you expect a patient to achieve
over a longer period, usually over several weeks, or even several months (for example “Patient
will leave nicotine products within two months”). Determine how to prevent, reduce, or resolve
the identified patient problems; how to support patient strengths; and how to implement nursing
interventions in a well organized, individualized, and goal-directed way, .
How would you arrange these concepts in terms of concreteness (from .pdf
How would you arrange these concepts in terms of concreteness (from most to least concrete)?
Justify your arrangement.
Asset
Common stock
Equity
Investment
Security
Stock
Solution
Below is the arrangement of concepts in terms of concreteness (most to least):
1. Stock 2. Common Stock 3. Security 4. Investment 5. Asset 6. Equity
Justification : As per my understanding, I have arranged all concepts based on how easily can we
do the valuation and the scope of a particular concept.
Stock is a current asset i.e. the valuation of stock can be done on an yearly basis. Since, it is short
term in nature and can be valued at cost using methods like FIFO, LIFO, etc. Cost is easily
available by looking at the invoices, purchase orders, etc.
Common Stock is a long term liability also called as equity shares. They can be valued using
different techniques like Dividend Discount model. Stock prices are easily available in the
exchange markets.
Securities comes next as they are wider than Common stock. They are the tradeable financial
instruments. They are divided into three parts - equity securities, debt securities and derivatives.
We can value securities from the market which is determined by the demand and supply forces.
Investments includes short term and long term investments. They are valued at the lesser of cost
or fair market value. The market value of investemnts can not be seen on stock exchanges and
hence lesser concrete than securities.
Assets includes both short term and long term assets. Fixed assets like Long term investments,
building, factory, machinery and current assets like Stock, short term investments, account
receivables, cash are included in the broad defination of assets. Hence, investments are a subset
of assets and hence more concrete than assets. Again, there is no trade market for assets to do the
asset valuation.
Equity is difference between assets and liabilities. Equity is a wider term which includes
common stock, preffered stock, retained earnings (calculated from Income statement), security
premium, revaluation reserves (calculated as the difference between the fair market value and
book value of assets). Hence, they can be regarded as least concrete of the given list..
For 12 pages randomly selected from War and Peace by Leo Tolstoy, th.pdf
For 12 pages randomly selected from War and Peace by Leo Tolstoy, the data set lists the mean
number of words per sentence, the mean number of characters per word.
Words/sentence
Characters/word
20.6
4.3
28.0
4.5
12.0
4.5
11.5
4.5
17.4
4.5
19.7
4.8
20.3
4.3
17.8
4.2
22.1
4.7
31.4
4.3
18.3
4.4
11.7
4.5
1. ( 15 points) Prepare a frequency distribution of words/sentence with a class width of 3 words
and another with class width of 4 words.
Words/sentence
Characters/word
20.6
4.3
28.0
4.5
12.0
4.5
11.5
4.5
17.4
4.5
19.7
4.8
20.3
4.3
17.8
4.2
22.1
4.7
31.4
4.3
18.3
4.4
11.7
4.5
Solution
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-14.4
13.0
3.0
3
25.0
3
25.0
14.5-17.4
16.0
3.0
1
8.3
4
33.3
17.5-20.4
19.0
3.0
4
33.3
8
66.7
20.5-23.4
22.0
3.0
2
16.7
10
83.3
23.5-26.4
25.0
3.0
0
0.0
10
83.3
26.5-29.4
28.0
3.0
1
8.3
11
91.7
29.5-32.4
31.0
3.0
1
8.3
12
100.0
12
100.0
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-15.4
13.5
4.0
3
25.0
3
25.0
15.5-19.4
17.5
4.0
3
25.0
6
50.0
19.5-23.4
21.5
4.0
4
33.3
10
83.3
23.5-27.4
25.5
4.0
0
0.0
10
83.3
27.5-31.4
29.5
4.0
2
16.7
12
100.0
12
100.0
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-14.4
13.0
3.0
3
25.0
3
25.0
14.5-17.4
16.0
3.0
1
8.3
4
33.3
17.5-20.4
19.0
3.0
4
33.3
8
66.7
20.5-23.4
22.0
3.0
2
16.7
10
83.3
23.5-26.4
25.0
3.0
0
0.0
10
83.3
26.5-29.4
28.0
3.0
1
8.3
11
91.7
29.5-32.4
31.0
3.0
1
8.3
12
100.0
12
100.0.
Explain what an enumerated type isSolutionEnum is also called.pdf
Explain what an enumerated type is?
Solution
Enum is also called enumerated Data type
Enum is a user defined data type
Syntax: of enumerated data type
Enum is a data type which contains fixed set of constants
identifier is a user defined datatype variable name
val1,val2,val3.....valn is a set of enum values
enum is just used for generation index values.
index starts from 0 to n
example
solution.cpp
#include//header for input output function
using namespace std;//it tells the compiler to link std namespace
int main()
{//main function
int i;
enum season {winter,summer,spring,fall};//enum data type
for(i=winter;i<=fall;i++)//for loop
cout<.
Explain how the three germ layers participate in the formation of ext.pdf
Explain how the three germ layers participate in the formation of extraembryonic membranes,
and discuss the importance of the placenta as an endocrine organ.
Solution
The three germ layers are not only responsible for the differentiation of specialized tissue rather
are also responsible for the formation of the extraembryonic membrane, that protects and
nourishes the embryo. The three extraembryonic membranes are formed from the primitive germ
layers:
The yolk sac is the membrane that forms from the endoderm and splanchnic mesoderm that
grows from the area opaca to form the sac. the yolk sac surrounds the yolk and draws the. the
endodermal gets more specialized and digest the yolk and pass its nutrients to the vitelline
circulation and thus to the membrane.
The amnion is formed from the ectoderm and somatic mesoderm that surrounds the embryo,
forming a protective cover for the embryo. the inner layer of the main secretes amniotic fluid in
which the embryo floats and prevents it from drying.
The chorion-allantoic membrane in which the chorion is made from the ectoderm and somatic
mesoderm while the allantoic membrane is formed from the endoderm and splanchnic
mesoderm. the allantoic membrane works with the chorion to exchange gasses between the
embryo and the outside environment. The allantois also stores the wastes of the embryo.
The placenta is considered as an endocrine gland present temporarily during pregnancy. the
placenta is responsible for the production of hormones that are responsible for the healthy
pregnancy. It secretes placental lactogen, placental growth relaxin as these hormones are
important for the healthy propagation of the pregnancy..
Ethical issue What would you doTony, a data analyst for a major c.pdf
Ethical issue: What would you do
Tony, a data analyst for a major casino, is working after normal business hours to finish an
important project. He realizes that he is missing data that had been sent to his coworker Robert.
Tony had inadvertently observed Robert typing his password several days ago and decides to log
into Robert\'s computer and resend the data to himself. Upon doing so, Tony sees an open email
regarding gambling bets Robert placed over the last several days with a local sports book. All
employees of the casino are forbidden to engage in gambling activities to avoid any hint of
conflict of interest.
Tony knows he should report this but would have to admit to violating the company\'s
information technology regulations by logging into Robert\'s computer. If he warns Robert to
stop his betting, he would also have to reveal the source of his information. What does Tony do
in this situation?
Solution
In this situation, Tony should warn Robert to stop his betting. He should clearly specify Robert
that he has accidently seen his password few days back. Due to the urgency of data, he logged
onto his computer and seen the mails. He should apologize for his mistake for logging into his
account without his permission. However, at the same time he should warn Robert that if he
found his further involvement in gambling activities through any source, then he would report it
to the management..
Compare and contrast the structure of a nephron with that of an alve.pdf
Compare and contrast the structure of a nephron with that of an alveolus in the lung. Identify
TWO (2) similarities and TWO (2) differences. (2 marks)
Solution
DIFFERENCES:1=Each alveolas is a very small bag_like structure bearing a component known
as Surfactant while as each nephron is a curled tube_like arrangement with a cup_shaped
structure called Bowman\'s capsule.2=Alveoli are present inside the lungs while as nephrons are
lying in kidneys. SIMILARITIES:1=Both Alveoli and nephrons are present in larger quantity in
their respective organs.2=Both the structures act as functional units..
Classless addressing was introduced to extend the life of the IPv ad.pdf
Classless addressing was introduced to extend the life of the IPv address scheme:
a. True b. False
Solution
True.
Explanation:
IP addresses are described as consisting of two groups of bits in the address: the most significant
bits are the network address (or network prefix or network block), which identifies a whole
network or subnet, and the least significant set forms the host identifier, which specifies a
particular interface of a host on that network. This division is used as the basis of traffic routing
between IP networks and for address allocation policies. Classful network design for IPv4 sized
the network address as one or more 8-bit groups, resulting in the blocks of Class A, B, or C
addresses. Classless Inter-Domain Routing allocates address space to Internet service providers
and end users on any address bit boundary, instead of on 8-bit segments. In IPv6, however, the
interface identifier has a fixed size of 64 bits by convention, and smaller subnets are never
allocated to end users..
C++ Complexity analysis Represent the time complexity of the follow.pdf
C++ Complexity analysis: Represent the time complexity of the following recursive algorithm,
T(n), as a recurrence equation:
int pow_7( int n ){
if ( n==1)
return 7;
if ( n > 1)
return ( 7 * pow_7( n-1 ) );
}
Solution
T(n) = T(n-1) + 1
T(n) = T(n-2)+2
suppose after k iteration
T(n) = T(n-k)+k
then T(n) = T(n-k-1)+k+1....T(n) = T(n-(k+1))+(k+1)
=> true for k+1 also.
T(n) = O(n).
Can someone provide a solution for this assignmentPurpose of This.pdf
Can someone provide a solution for this assignment?
Purpose of This Assignment
The purpose of this assignment is to develop your abilities to write a larger application involving
arrays, structures and linked lists. It also introduces switchable tracing.
Graphs and Terminology
This assignment uses weighted graphs, as described in assignment 4. Be sure that you are
familiar with weighted graphs.
Two vertices are said to be adjacent if there is an edge that connects them directly to one another.
A given edge is incident on each of the vertices that it connects.
Think of the vertices of a weighted graph as towns and the edges as roads. The weight of an edge
is the length of the road. One thing that you might like to know is how to get from one town to
another by the shortest possible route. For example, in the following weighted graph, the shortest
route from vertex 1 to vertex 5 is to go from 1 to 3 and then from 3 to 5, and the length of that
route is 27. You add the weights of the edges that you use.
For this assignment, the weights are real numbers (type double). All of the weights are required
to be nonnegative. Edges of weight 0 are allowed.
Assignment Requirements
Functional Requirements
Write a program that reads information about a weighted graph from the standard input. The
input format is described in detail below. After the description of the graph, the input has two
vertex numbers, s and t.
Your program should print a description of the graph, followed by the shortest path from s to t
and the distance from s to t via that path, on the standard output.
For example, the input might look like this.
That says that there are five vertices. There is an edge from vertex 1 to vertex 2 with weight 9.0,
an edge from vertex 1 to vertex 3 with weight 12.0, etc. The start vertex s is 1, and the end vertex
t is 5. The output for this input would be as follows.
Nonfunctional Requirements
Create a directory to hold assignment 6 and call your program dijkstra.cpp. Start with the
standard template.
Below is a description of an algorithm, based on Dijkstra\'s algorithm, for solving this problem.
You are required to use that algorithm, and to follow the guidelines for its implementation. It is
not acceptable to rely on a different approach to the problem. Follow the design.
Make variable and function names sensible. Use sensible terminology. If something is a graph,
do not call it an edge. If something is an edge, do not call it a vertex. Keep functions short and
simple.
As always, you must follow the coding standards for this course.
Representing Graphs: Types Vertex, Edge and Graph
This assignment uses a different graph representation from assignment 4. Here, we use the
adjacency list representation.
1. Type Vertex
Create and document type Vertex. Each vertex v has the following pieces of information.
A pointer to a linked list of edges listing all edges that are incident on v. This list is called an
adjacency list.
A real number indicating v\'s shor.
Before first law of thermodynamics, there is zeroth law of thermodyn.pdf
Before first law of thermodynamics, there is zeroth law of thermodynamics. The thermometer
works on the principle of zeroth law of thermodynamics. Can you explain what is a zeroth law of
thermodynamics?
Solution
Zeroth law states that Objects into contact will reach a thermal equilibrium (and heat transfers
from hotter to colder), this means one can use a device that can measure this difference of this
quantity as follows:
We have a device (called a thermometer) which has an indication. When the \"thermometer
device\" comes into contact with another object, 0th law states that they wil reach thermal
equilibrium . Then the indication of the device will change to reflect the difference of this
quantity that changed or transfered and refered to as temperature .
One such device is the thermometer based on mercury (which is used as the indicator due to its
special heat expansion factor)..
An enzyme that has lost its shape a. is denatured and no longer func.pdf
An enzyme that has lost its shape a. is denatured and no longer functional. b. is denatured but
still functional. c. is not different from a regular enzyme as shape is not important. d. can be
restored by extreme temperatures or pH swings. Predict whether each of the following has ionic,
polar covalent, or not O_2
Solution
7) Each enzyme has specific active sites. When an enzyme loses its shape then its active sites are
no longer functional. As a result, the properties of the enzyme are destroyed. This process is
called denaturation.
Therefore, an enzyme that has lost its shape is denatured and no longer functional.
(Answer) (a)
8) In oxygen molecule (O2), two oxygen atoms are joined by double covalent bond. The bonds
are formed by sharing of electron pairs. The electron pairs are equally shared between two
oxygen atoms. For this reason, oxygen molecule has non-polar covalent bonds..
A valley fog(ground fog) is an example of a _______ fog type.Fill .pdf
A valley fog(ground fog) is an example of a _______ fog type.Fill in the correct answer in the
blank.
A valley fog(ground fog) is an example of a _______ fog type.Fill in the correct answer in the
blank.
Solution
Valley fog forms in mountain valleys, often during winter. It is essentially a radiation fog
confined by local topography,
So, Radiation fog.
A simple pendulum makes 115 complete oscillations in 2.40 min at a lo.pdf
A simple pendulum makes 115 complete oscillations in 2.40 min at a location where g = 9.80
m/s^2. Find the period of the pendulum. Find the length of the pendulum.
Solution
time period = time taken in one oscillation
=> T = 2.4*60)/(115 = 1.25 sec
b)
T = 2pi*sqrt(L/g)
=> L = (1.25/(2*3.14))^2*9.8 = 0.388 m.
A large group of people is to be checked for two common symptoms of .pdf
A large group of people is to be checked for two common symptoms of a certain disease. It is
thought that
20% possess symptom A alone;
30% possess symptom B alone
10% possess both
the remainder have neither symptoms
If one person chosen at random from this group, find the conditional probability P(A | B).
Solution
P(A and B) = 0.1
P(B) = 0.3
so P(A/B) = P(A and B) / P(b) = 0.1/0.3 = 1/3.
1.) What does “exonuclease” activity mean Which enzyme important fo.pdf
1.) What does “exonuclease” activity mean? Which enzyme important for DNA polymerization
has both 5’ – 3’ AND 3’ – 5’ exonuclease activity? Which is important for proofreading
function?
2.) How are the site(s) where replication begin(s) is/are characterized as in both prokaryotes and
eukaryotes? What is beneficial about this? Are the sites the same in prokaryotes and eukaryotes?
3.) Arrange the following proteins in the proper order in which they participate in DNA
replication.
1 = Primase
2 = Helicase
3 = Single-strand binding proteins
4 = DNA polymerase I
5 =DNA polymerase III
4.) When does DNA replicate in the cell cycle of eukaryotes and prokaryotes?
5.) Why are Okazaki fragments only found on the lagging strand, but not on the leading strand of
DNA?
6.) What protein transports the histones of Nucleosomes into the nucleus?
7.) Describe the differences between Topoisomerase I and Topoisomerase II.
8.) During DNA replication, ___________ adds an RNA primer to naked template strands of
DNA. Why is this necessary during replication? What fills the gap after removal of this RNA
primer?
9.) What is the problem with the ends of linear eukaryotic chromosomes? How does telomerase
help the issue?
Solution
1. Exonuclease activity means that enzymes work by cleaving nucleotides one at a time from the
end (exo) of a polynucleotide chain. A hydrolyzing reaction breaks phosphodiester bonds at
either the 3’ or the 5’ ends.
DNA polymerase I is the enzyme important for DNA polymerization and has both 5’ – 3’ and 3’
– 5’ exonuclease activity, which is used in editing and proofreading DNA for errors. The 3\' to 5\'
can remove only one mononucleotide at a time, whereas the 5\' to 3\' activity can remove up to
10 nucleotides at a time.
2. Sites from which the process of replication starts are different in prokaryotes and Eukaryotes.
Prokaryotes have only one origin of replication per DNA molecule, whereas in ekaryotes, there
are multiple sites for origin of replication.
To initiate DNA replication in Eukaryotes, multiple replicative proteins assemble on and
dissociate from these replicative origins. DNA double helix is unwound and an initial priming
event by DNA polymerase occurs on the leading strand.To initiate the process of DNA
replication, multiple replicative proteins assemble on and dissociate from these replicative
origins. The individual factors work together to direct the formation of the pre-replication
complex (pre-RC), a key intermediate in the replication initiation process.
In case of prokaryotes, Helicase separates the DNA to form a replication fork at the origin of
replication where DNA replication begins. Replication forks extend bi-directionally as
replication continues. The initiation of DNA replication is mediated by DnaA, a protein that
binds to a region of the origin known as the DnaA box. E. coli, has 4 DnaA boxes, each of which
contains a highly conserved 9 bp consensus sequence 5\' - TTATCCACA - 3\'.
3. Enzymes that participate in DNa.
1. The case of albinism among the Hopi Indians of Black M.pdf
1. The case of albinism among the Hopi Indians of Black Mesa is different from other
cases of albinism because of all of the following EXCEPT:
A. the incidence is much higher among the Hopis.
B. the albinism seems to have provided a reproductive advantage.
C. the albinism seems to have conferred social advantages.
D. the albinism did not involve vision defects.
E. Actually, all of the above are accurate.
2. Genetics has NOT played a key role in which of the following areas?
A. agriculture
B. medicine
C. pharmaceutical production
D. urban design and layout
E. Actually, it has played a key role in all of these areas.
3. Genetic variation in populations:
A. demonstrates descent from a common ancestor.
B. provides the basis of adaptation.
C. explains the use of similar genetic systems in all organisms.
D. is a consequence of origin of species.
E. All of the above.
4. Genetics as a discipline can be subdivided into all of the following EXCEPT:
A. transmission genetics.
B. acquisition genetics.
C. molecular genetics.
D. population genetics.
E. Actually, all of these are found in the discipline.
5. While modern genetics starts with the experiments of Gregor Mendel in 1866, it wasn\'t
until the start of the 20th century that those studies became well known, in several areas, NOT
including:
A. Sutton, who proposed that genes were on chromosomes.
B. Morgan, who identified fruit fly mutations.
C. Haldane and others who used populations to connect genetics to evolution.
D. Watson and Crick\'s determination of the structure of DNA.
E. Actually, Mendel\'s work was critical in each of these areas.
6. Prokaryotes and eukaryotes differ in that:
A. prokaryotes have a nucleus.
B. they decode DNA to different amino acids.
C. prokaryotes tend to be larger.
D. prokaryotes tend to be less complex.
E. None of the above.
7. Genes:
A. are carried on chromosomes.
B. come in pairs in eukaryotic cells.
C. come in alternative forms, known as alleles.
D. are encoded by DNA.
E. All of the above.
8. Phenotypes are:
A. measurable traits of an organism.
B. not inherited directly.
C. not changeable during the life of an organism.
D. A and B.
E. All of the above.
9. Evolution:
A. is genetic change.
B. starts with variation.
C. does not require change of DNA.
D. A and B.
E. All of the above.
10. The flow of information is as follows:
A. protein to DNA to RNA.
B. protein to RNA to DNA.
C. DNA to RNA to protein.
D. RNA to protein to DNA.
E. None of the above.
11. Mutations are changes of DNA:
A. that can affect a single gene.
B. that can affect large regions of chromosomes.
C. that can be passed on to daughter cells.
D. that can change amino acid sequences.
E. All of the above.
Solution
1). C. the albinism seems to have conferred social advantages.
2). D. urban design and layout
3). D. is a consequence of origin of species.
4). B. acquisition genetics.
5). C. Haldane and others who used populations to connect genetics to evolution.
6). E. None of the above.
7). A. are carried on ch.
1. Which of the following is correct about the protease inhibitor dr.pdf
1. Which of the following is correct about the protease inhibitor drugs designed to treat HIV
infection?
d. They cure the disease
2. Viruses like Ebola, Marburg, and Hanta are known as \'zoonotic\' because...
3. HIV, Ebola, and Marburg are best described as...
a. double-stranded DNA viruses
b. single-stranded RNA viruses
c. double-stranded DNA viruses
d. single-stranded DNA virusesa. They result in lysis of virus particles
Solution
Answer:
1). c. They slow the progress of the disease
Protease is an enzyme in your body that’s important for HIV replication. Protease inhibitor drugs
block the action of protease. That means they prevent the protease enzyme from doing its part in
the steps that allow HIV to multiply. In this way, protease inhibitors can interrupt the HIV
lifecycle. This can stop the virus from multiplying in your body.
2). c. they are animal-borne.
Zoonotic: Pertaining to a zoonosis: a disease that can be transmitted from animals to people or,
more specifically, a disease that normally exists in animals but that can infect humans.
3). b. single-stranded RNA viruses
Ebola, and Marburg are negative-sense single standed RNA viruses whereas HIV is positive-
sense single standed RNA viruses..
1. Match the molecule to the chemical used to test for it to the col.pdf
1. Match the molecule to the chemical used to test for it to the color that the chemical would
change for a positive test.
Macromolecule Reagent Color Change
Sugar Biuret Reagent Green, Orange, or Red
Color
Protein Sudan III Blue-black Color
Lipid Iodine Red or Orange Color
Starch Benedict\'s
Solution
Blue-Violet Color.
2024.06.01 Introducing a competency framework for languag learning materials ...
http://sandymillin.wordpress.com/iateflwebinar2024
Published classroom materials form the basis of syllabuses, drive teacher professional development, and have a potentially huge influence on learners, teachers and education systems. All teachers also create their own materials, whether a few sentences on a blackboard, a highly-structured fully-realised online course, or anything in between. Despite this, the knowledge and skills needed to create effective language learning materials are rarely part of teacher training, and are mostly learnt by trial and error.
Knowledge and skills frameworks, generally called competency frameworks, for ELT teachers, trainers and managers have existed for a few years now. However, until I created one for my MA dissertation, there wasn’t one drawing together what we need to know and do to be able to effectively produce language learning materials.
This webinar will introduce you to my framework, highlighting the key competencies I identified from my research. It will also show how anybody involved in language teaching (any language, not just English!), teacher training, managing schools or developing language learning materials can benefit from using the framework.
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
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If the proportion of an African population that is susceptible to Ma.pdf
If the proportion of an African population that is susceptible to Malaria is 0.49, what proportion
of the population is affected with Sickle Cell Anemia? A. 0.68. B. 0.09. C. 0.45. D. 0.34.
Solution
B 0.09
In literature it was mentioned that 40% of popullation in africa suffering from malaria are
sucessiptible to sickel cell anemia..
Identify nursing diagnoses goal and outcome criteria and interven.pdf
Identify nursing diagnoses goal and outcome criteria and interventions for the postoperative
patient
Identify nursing diagnoses goal and outcome criteria and interventions for the postoperative
patient
Solution
ASSESSMENT: - After the surgery, when the operation procedure is completed, the patient is
usually shifted to intensive care unit. When patient is admitted to PACU determine the medical
diagnosis, the surgery procedure, pain, consciousness, vital signs, medical history, how patient
tolerated the procedure? what was the medication given, and other assessment data listed earlier.
Check and set up equipment like oxygen, suction devices, urinary drainage etc. A more detailed
assessment should be performed when a patient returns to the nursing unit. It is actually
collecting, organizing, validating, and documenting patient data and its purpose is to establish a
database about the patient’s response to health concerns or illness and the ability to manage
health care needs.
DIAGNOSIS: - Medical diagnoses are important factors influencing patient outcomes during
hospitalization. For example, ICU patients with infectious diseases at admission had higher
mortality than the patients with gastrointestinal diseases. After surgery patients are at an
increased risk of severe complication. Nursing diagnosis is the nurse’s clinical judgment about
the patient’s response to potential health conditions and comorbid medical conditions. Several
studies show that these diagnosis influence 31 different types of patient outcomes (e.g. hospital
mortality, the length of stay, and ICU readmission) and interventions. A standard nursing
diagnosis includes three structural components: the problem, the etiology and supporting data. Its
purpose is to identify patient’s strengths and health problems that can be prevented or resolved
by collaborative and independent nursing interventions. and to develop a list of nursing
diagnoses and collaborative problems.
GOAL AND OUTCOME: - The goal is aimed at the medical diagnosis. The outcomes are aimed
at meeting the goals. A patient goal represents a predicted resolution of a diagnosis, evidence of
progress toward resolution, progress toward improved health, or continued maintenance of good
health. Each goal is limited to a time so the health care team has a common time limit for solving
the problem. For example, the goal of “patient will achieve pain relief. A short-term goal is an
objective behavior or response that you expect a patient to achieve in a short time, usually in a
week. A long-term goal is an objective behavior or response that you expect a patient to achieve
over a longer period, usually over several weeks, or even several months (for example “Patient
will leave nicotine products within two months”). Determine how to prevent, reduce, or resolve
the identified patient problems; how to support patient strengths; and how to implement nursing
interventions in a well organized, individualized, and goal-directed way, .
How would you arrange these concepts in terms of concreteness (from .pdf
How would you arrange these concepts in terms of concreteness (from most to least concrete)?
Justify your arrangement.
Asset
Common stock
Equity
Investment
Security
Stock
Solution
Below is the arrangement of concepts in terms of concreteness (most to least):
1. Stock 2. Common Stock 3. Security 4. Investment 5. Asset 6. Equity
Justification : As per my understanding, I have arranged all concepts based on how easily can we
do the valuation and the scope of a particular concept.
Stock is a current asset i.e. the valuation of stock can be done on an yearly basis. Since, it is short
term in nature and can be valued at cost using methods like FIFO, LIFO, etc. Cost is easily
available by looking at the invoices, purchase orders, etc.
Common Stock is a long term liability also called as equity shares. They can be valued using
different techniques like Dividend Discount model. Stock prices are easily available in the
exchange markets.
Securities comes next as they are wider than Common stock. They are the tradeable financial
instruments. They are divided into three parts - equity securities, debt securities and derivatives.
We can value securities from the market which is determined by the demand and supply forces.
Investments includes short term and long term investments. They are valued at the lesser of cost
or fair market value. The market value of investemnts can not be seen on stock exchanges and
hence lesser concrete than securities.
Assets includes both short term and long term assets. Fixed assets like Long term investments,
building, factory, machinery and current assets like Stock, short term investments, account
receivables, cash are included in the broad defination of assets. Hence, investments are a subset
of assets and hence more concrete than assets. Again, there is no trade market for assets to do the
asset valuation.
Equity is difference between assets and liabilities. Equity is a wider term which includes
common stock, preffered stock, retained earnings (calculated from Income statement), security
premium, revaluation reserves (calculated as the difference between the fair market value and
book value of assets). Hence, they can be regarded as least concrete of the given list..
For 12 pages randomly selected from War and Peace by Leo Tolstoy, th.pdf
For 12 pages randomly selected from War and Peace by Leo Tolstoy, the data set lists the mean
number of words per sentence, the mean number of characters per word.
Words/sentence
Characters/word
20.6
4.3
28.0
4.5
12.0
4.5
11.5
4.5
17.4
4.5
19.7
4.8
20.3
4.3
17.8
4.2
22.1
4.7
31.4
4.3
18.3
4.4
11.7
4.5
1. ( 15 points) Prepare a frequency distribution of words/sentence with a class width of 3 words
and another with class width of 4 words.
Words/sentence
Characters/word
20.6
4.3
28.0
4.5
12.0
4.5
11.5
4.5
17.4
4.5
19.7
4.8
20.3
4.3
17.8
4.2
22.1
4.7
31.4
4.3
18.3
4.4
11.7
4.5
Solution
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-14.4
13.0
3.0
3
25.0
3
25.0
14.5-17.4
16.0
3.0
1
8.3
4
33.3
17.5-20.4
19.0
3.0
4
33.3
8
66.7
20.5-23.4
22.0
3.0
2
16.7
10
83.3
23.5-26.4
25.0
3.0
0
0.0
10
83.3
26.5-29.4
28.0
3.0
1
8.3
11
91.7
29.5-32.4
31.0
3.0
1
8.3
12
100.0
12
100.0
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-15.4
13.5
4.0
3
25.0
3
25.0
15.5-19.4
17.5
4.0
3
25.0
6
50.0
19.5-23.4
21.5
4.0
4
33.3
10
83.3
23.5-27.4
25.5
4.0
0
0.0
10
83.3
27.5-31.4
29.5
4.0
2
16.7
12
100.0
12
100.0
Words/sentence
cumulative
midpoint
width
frequency
percent
frequency
percent
11.5-14.4
13.0
3.0
3
25.0
3
25.0
14.5-17.4
16.0
3.0
1
8.3
4
33.3
17.5-20.4
19.0
3.0
4
33.3
8
66.7
20.5-23.4
22.0
3.0
2
16.7
10
83.3
23.5-26.4
25.0
3.0
0
0.0
10
83.3
26.5-29.4
28.0
3.0
1
8.3
11
91.7
29.5-32.4
31.0
3.0
1
8.3
12
100.0
12
100.0.
Explain what an enumerated type isSolutionEnum is also called.pdf
Explain what an enumerated type is?
Solution
Enum is also called enumerated Data type
Enum is a user defined data type
Syntax: of enumerated data type
Enum is a data type which contains fixed set of constants
identifier is a user defined datatype variable name
val1,val2,val3.....valn is a set of enum values
enum is just used for generation index values.
index starts from 0 to n
example
solution.cpp
#include//header for input output function
using namespace std;//it tells the compiler to link std namespace
int main()
{//main function
int i;
enum season {winter,summer,spring,fall};//enum data type
for(i=winter;i<=fall;i++)//for loop
cout<.
Explain how the three germ layers participate in the formation of ext.pdf
Explain how the three germ layers participate in the formation of extraembryonic membranes,
and discuss the importance of the placenta as an endocrine organ.
Solution
The three germ layers are not only responsible for the differentiation of specialized tissue rather
are also responsible for the formation of the extraembryonic membrane, that protects and
nourishes the embryo. The three extraembryonic membranes are formed from the primitive germ
layers:
The yolk sac is the membrane that forms from the endoderm and splanchnic mesoderm that
grows from the area opaca to form the sac. the yolk sac surrounds the yolk and draws the. the
endodermal gets more specialized and digest the yolk and pass its nutrients to the vitelline
circulation and thus to the membrane.
The amnion is formed from the ectoderm and somatic mesoderm that surrounds the embryo,
forming a protective cover for the embryo. the inner layer of the main secretes amniotic fluid in
which the embryo floats and prevents it from drying.
The chorion-allantoic membrane in which the chorion is made from the ectoderm and somatic
mesoderm while the allantoic membrane is formed from the endoderm and splanchnic
mesoderm. the allantoic membrane works with the chorion to exchange gasses between the
embryo and the outside environment. The allantois also stores the wastes of the embryo.
The placenta is considered as an endocrine gland present temporarily during pregnancy. the
placenta is responsible for the production of hormones that are responsible for the healthy
pregnancy. It secretes placental lactogen, placental growth relaxin as these hormones are
important for the healthy propagation of the pregnancy..
Ethical issue What would you doTony, a data analyst for a major c.pdf
Ethical issue: What would you do
Tony, a data analyst for a major casino, is working after normal business hours to finish an
important project. He realizes that he is missing data that had been sent to his coworker Robert.
Tony had inadvertently observed Robert typing his password several days ago and decides to log
into Robert\'s computer and resend the data to himself. Upon doing so, Tony sees an open email
regarding gambling bets Robert placed over the last several days with a local sports book. All
employees of the casino are forbidden to engage in gambling activities to avoid any hint of
conflict of interest.
Tony knows he should report this but would have to admit to violating the company\'s
information technology regulations by logging into Robert\'s computer. If he warns Robert to
stop his betting, he would also have to reveal the source of his information. What does Tony do
in this situation?
Solution
In this situation, Tony should warn Robert to stop his betting. He should clearly specify Robert
that he has accidently seen his password few days back. Due to the urgency of data, he logged
onto his computer and seen the mails. He should apologize for his mistake for logging into his
account without his permission. However, at the same time he should warn Robert that if he
found his further involvement in gambling activities through any source, then he would report it
to the management..
Compare and contrast the structure of a nephron with that of an alve.pdf
Compare and contrast the structure of a nephron with that of an alveolus in the lung. Identify
TWO (2) similarities and TWO (2) differences. (2 marks)
Solution
DIFFERENCES:1=Each alveolas is a very small bag_like structure bearing a component known
as Surfactant while as each nephron is a curled tube_like arrangement with a cup_shaped
structure called Bowman\'s capsule.2=Alveoli are present inside the lungs while as nephrons are
lying in kidneys. SIMILARITIES:1=Both Alveoli and nephrons are present in larger quantity in
their respective organs.2=Both the structures act as functional units..
Classless addressing was introduced to extend the life of the IPv ad.pdf
Classless addressing was introduced to extend the life of the IPv address scheme:
a. True b. False
Solution
True.
Explanation:
IP addresses are described as consisting of two groups of bits in the address: the most significant
bits are the network address (or network prefix or network block), which identifies a whole
network or subnet, and the least significant set forms the host identifier, which specifies a
particular interface of a host on that network. This division is used as the basis of traffic routing
between IP networks and for address allocation policies. Classful network design for IPv4 sized
the network address as one or more 8-bit groups, resulting in the blocks of Class A, B, or C
addresses. Classless Inter-Domain Routing allocates address space to Internet service providers
and end users on any address bit boundary, instead of on 8-bit segments. In IPv6, however, the
interface identifier has a fixed size of 64 bits by convention, and smaller subnets are never
allocated to end users..
C++ Complexity analysis Represent the time complexity of the follow.pdf
C++ Complexity analysis: Represent the time complexity of the following recursive algorithm,
T(n), as a recurrence equation:
int pow_7( int n ){
if ( n==1)
return 7;
if ( n > 1)
return ( 7 * pow_7( n-1 ) );
}
Solution
T(n) = T(n-1) + 1
T(n) = T(n-2)+2
suppose after k iteration
T(n) = T(n-k)+k
then T(n) = T(n-k-1)+k+1....T(n) = T(n-(k+1))+(k+1)
=> true for k+1 also.
T(n) = O(n).
Can someone provide a solution for this assignmentPurpose of This.pdf
Can someone provide a solution for this assignment?
Purpose of This Assignment
The purpose of this assignment is to develop your abilities to write a larger application involving
arrays, structures and linked lists. It also introduces switchable tracing.
Graphs and Terminology
This assignment uses weighted graphs, as described in assignment 4. Be sure that you are
familiar with weighted graphs.
Two vertices are said to be adjacent if there is an edge that connects them directly to one another.
A given edge is incident on each of the vertices that it connects.
Think of the vertices of a weighted graph as towns and the edges as roads. The weight of an edge
is the length of the road. One thing that you might like to know is how to get from one town to
another by the shortest possible route. For example, in the following weighted graph, the shortest
route from vertex 1 to vertex 5 is to go from 1 to 3 and then from 3 to 5, and the length of that
route is 27. You add the weights of the edges that you use.
For this assignment, the weights are real numbers (type double). All of the weights are required
to be nonnegative. Edges of weight 0 are allowed.
Assignment Requirements
Functional Requirements
Write a program that reads information about a weighted graph from the standard input. The
input format is described in detail below. After the description of the graph, the input has two
vertex numbers, s and t.
Your program should print a description of the graph, followed by the shortest path from s to t
and the distance from s to t via that path, on the standard output.
For example, the input might look like this.
That says that there are five vertices. There is an edge from vertex 1 to vertex 2 with weight 9.0,
an edge from vertex 1 to vertex 3 with weight 12.0, etc. The start vertex s is 1, and the end vertex
t is 5. The output for this input would be as follows.
Nonfunctional Requirements
Create a directory to hold assignment 6 and call your program dijkstra.cpp. Start with the
standard template.
Below is a description of an algorithm, based on Dijkstra\'s algorithm, for solving this problem.
You are required to use that algorithm, and to follow the guidelines for its implementation. It is
not acceptable to rely on a different approach to the problem. Follow the design.
Make variable and function names sensible. Use sensible terminology. If something is a graph,
do not call it an edge. If something is an edge, do not call it a vertex. Keep functions short and
simple.
As always, you must follow the coding standards for this course.
Representing Graphs: Types Vertex, Edge and Graph
This assignment uses a different graph representation from assignment 4. Here, we use the
adjacency list representation.
1. Type Vertex
Create and document type Vertex. Each vertex v has the following pieces of information.
A pointer to a linked list of edges listing all edges that are incident on v. This list is called an
adjacency list.
A real number indicating v\'s shor.
Before first law of thermodynamics, there is zeroth law of thermodyn.pdf
Before first law of thermodynamics, there is zeroth law of thermodynamics. The thermometer
works on the principle of zeroth law of thermodynamics. Can you explain what is a zeroth law of
thermodynamics?
Solution
Zeroth law states that Objects into contact will reach a thermal equilibrium (and heat transfers
from hotter to colder), this means one can use a device that can measure this difference of this
quantity as follows:
We have a device (called a thermometer) which has an indication. When the \"thermometer
device\" comes into contact with another object, 0th law states that they wil reach thermal
equilibrium . Then the indication of the device will change to reflect the difference of this
quantity that changed or transfered and refered to as temperature .
One such device is the thermometer based on mercury (which is used as the indicator due to its
special heat expansion factor)..
An enzyme that has lost its shape a. is denatured and no longer func.pdf
An enzyme that has lost its shape a. is denatured and no longer functional. b. is denatured but
still functional. c. is not different from a regular enzyme as shape is not important. d. can be
restored by extreme temperatures or pH swings. Predict whether each of the following has ionic,
polar covalent, or not O_2
Solution
7) Each enzyme has specific active sites. When an enzyme loses its shape then its active sites are
no longer functional. As a result, the properties of the enzyme are destroyed. This process is
called denaturation.
Therefore, an enzyme that has lost its shape is denatured and no longer functional.
(Answer) (a)
8) In oxygen molecule (O2), two oxygen atoms are joined by double covalent bond. The bonds
are formed by sharing of electron pairs. The electron pairs are equally shared between two
oxygen atoms. For this reason, oxygen molecule has non-polar covalent bonds..
A valley fog(ground fog) is an example of a _______ fog type.Fill .pdf
A valley fog(ground fog) is an example of a _______ fog type.Fill in the correct answer in the
blank.
A valley fog(ground fog) is an example of a _______ fog type.Fill in the correct answer in the
blank.
Solution
Valley fog forms in mountain valleys, often during winter. It is essentially a radiation fog
confined by local topography,
So, Radiation fog.
A simple pendulum makes 115 complete oscillations in 2.40 min at a lo.pdf
A simple pendulum makes 115 complete oscillations in 2.40 min at a location where g = 9.80
m/s^2. Find the period of the pendulum. Find the length of the pendulum.
Solution
time period = time taken in one oscillation
=> T = 2.4*60)/(115 = 1.25 sec
b)
T = 2pi*sqrt(L/g)
=> L = (1.25/(2*3.14))^2*9.8 = 0.388 m.
A large group of people is to be checked for two common symptoms of .pdf
A large group of people is to be checked for two common symptoms of a certain disease. It is
thought that
20% possess symptom A alone;
30% possess symptom B alone
10% possess both
the remainder have neither symptoms
If one person chosen at random from this group, find the conditional probability P(A | B).
Solution
P(A and B) = 0.1
P(B) = 0.3
so P(A/B) = P(A and B) / P(b) = 0.1/0.3 = 1/3.
1.) What does “exonuclease” activity mean Which enzyme important fo.pdf
1.) What does “exonuclease” activity mean? Which enzyme important for DNA polymerization
has both 5’ – 3’ AND 3’ – 5’ exonuclease activity? Which is important for proofreading
function?
2.) How are the site(s) where replication begin(s) is/are characterized as in both prokaryotes and
eukaryotes? What is beneficial about this? Are the sites the same in prokaryotes and eukaryotes?
3.) Arrange the following proteins in the proper order in which they participate in DNA
replication.
1 = Primase
2 = Helicase
3 = Single-strand binding proteins
4 = DNA polymerase I
5 =DNA polymerase III
4.) When does DNA replicate in the cell cycle of eukaryotes and prokaryotes?
5.) Why are Okazaki fragments only found on the lagging strand, but not on the leading strand of
DNA?
6.) What protein transports the histones of Nucleosomes into the nucleus?
7.) Describe the differences between Topoisomerase I and Topoisomerase II.
8.) During DNA replication, ___________ adds an RNA primer to naked template strands of
DNA. Why is this necessary during replication? What fills the gap after removal of this RNA
primer?
9.) What is the problem with the ends of linear eukaryotic chromosomes? How does telomerase
help the issue?
Solution
1. Exonuclease activity means that enzymes work by cleaving nucleotides one at a time from the
end (exo) of a polynucleotide chain. A hydrolyzing reaction breaks phosphodiester bonds at
either the 3’ or the 5’ ends.
DNA polymerase I is the enzyme important for DNA polymerization and has both 5’ – 3’ and 3’
– 5’ exonuclease activity, which is used in editing and proofreading DNA for errors. The 3\' to 5\'
can remove only one mononucleotide at a time, whereas the 5\' to 3\' activity can remove up to
10 nucleotides at a time.
2. Sites from which the process of replication starts are different in prokaryotes and Eukaryotes.
Prokaryotes have only one origin of replication per DNA molecule, whereas in ekaryotes, there
are multiple sites for origin of replication.
To initiate DNA replication in Eukaryotes, multiple replicative proteins assemble on and
dissociate from these replicative origins. DNA double helix is unwound and an initial priming
event by DNA polymerase occurs on the leading strand.To initiate the process of DNA
replication, multiple replicative proteins assemble on and dissociate from these replicative
origins. The individual factors work together to direct the formation of the pre-replication
complex (pre-RC), a key intermediate in the replication initiation process.
In case of prokaryotes, Helicase separates the DNA to form a replication fork at the origin of
replication where DNA replication begins. Replication forks extend bi-directionally as
replication continues. The initiation of DNA replication is mediated by DnaA, a protein that
binds to a region of the origin known as the DnaA box. E. coli, has 4 DnaA boxes, each of which
contains a highly conserved 9 bp consensus sequence 5\' - TTATCCACA - 3\'.
3. Enzymes that participate in DNa.
1. The case of albinism among the Hopi Indians of Black M.pdf
1. The case of albinism among the Hopi Indians of Black Mesa is different from other
cases of albinism because of all of the following EXCEPT:
A. the incidence is much higher among the Hopis.
B. the albinism seems to have provided a reproductive advantage.
C. the albinism seems to have conferred social advantages.
D. the albinism did not involve vision defects.
E. Actually, all of the above are accurate.
2. Genetics has NOT played a key role in which of the following areas?
A. agriculture
B. medicine
C. pharmaceutical production
D. urban design and layout
E. Actually, it has played a key role in all of these areas.
3. Genetic variation in populations:
A. demonstrates descent from a common ancestor.
B. provides the basis of adaptation.
C. explains the use of similar genetic systems in all organisms.
D. is a consequence of origin of species.
E. All of the above.
4. Genetics as a discipline can be subdivided into all of the following EXCEPT:
A. transmission genetics.
B. acquisition genetics.
C. molecular genetics.
D. population genetics.
E. Actually, all of these are found in the discipline.
5. While modern genetics starts with the experiments of Gregor Mendel in 1866, it wasn\'t
until the start of the 20th century that those studies became well known, in several areas, NOT
including:
A. Sutton, who proposed that genes were on chromosomes.
B. Morgan, who identified fruit fly mutations.
C. Haldane and others who used populations to connect genetics to evolution.
D. Watson and Crick\'s determination of the structure of DNA.
E. Actually, Mendel\'s work was critical in each of these areas.
6. Prokaryotes and eukaryotes differ in that:
A. prokaryotes have a nucleus.
B. they decode DNA to different amino acids.
C. prokaryotes tend to be larger.
D. prokaryotes tend to be less complex.
E. None of the above.
7. Genes:
A. are carried on chromosomes.
B. come in pairs in eukaryotic cells.
C. come in alternative forms, known as alleles.
D. are encoded by DNA.
E. All of the above.
8. Phenotypes are:
A. measurable traits of an organism.
B. not inherited directly.
C. not changeable during the life of an organism.
D. A and B.
E. All of the above.
9. Evolution:
A. is genetic change.
B. starts with variation.
C. does not require change of DNA.
D. A and B.
E. All of the above.
10. The flow of information is as follows:
A. protein to DNA to RNA.
B. protein to RNA to DNA.
C. DNA to RNA to protein.
D. RNA to protein to DNA.
E. None of the above.
11. Mutations are changes of DNA:
A. that can affect a single gene.
B. that can affect large regions of chromosomes.
C. that can be passed on to daughter cells.
D. that can change amino acid sequences.
E. All of the above.
Solution
1). C. the albinism seems to have conferred social advantages.
2). D. urban design and layout
3). D. is a consequence of origin of species.
4). B. acquisition genetics.
5). C. Haldane and others who used populations to connect genetics to evolution.
6). E. None of the above.
7). A. are carried on ch.
1. Which of the following is correct about the protease inhibitor dr.pdf
1. Which of the following is correct about the protease inhibitor drugs designed to treat HIV
infection?
d. They cure the disease
2. Viruses like Ebola, Marburg, and Hanta are known as \'zoonotic\' because...
3. HIV, Ebola, and Marburg are best described as...
a. double-stranded DNA viruses
b. single-stranded RNA viruses
c. double-stranded DNA viruses
d. single-stranded DNA virusesa. They result in lysis of virus particles
Solution
Answer:
1). c. They slow the progress of the disease
Protease is an enzyme in your body that’s important for HIV replication. Protease inhibitor drugs
block the action of protease. That means they prevent the protease enzyme from doing its part in
the steps that allow HIV to multiply. In this way, protease inhibitors can interrupt the HIV
lifecycle. This can stop the virus from multiplying in your body.
2). c. they are animal-borne.
Zoonotic: Pertaining to a zoonosis: a disease that can be transmitted from animals to people or,
more specifically, a disease that normally exists in animals but that can infect humans.
3). b. single-stranded RNA viruses
Ebola, and Marburg are negative-sense single standed RNA viruses whereas HIV is positive-
sense single standed RNA viruses..
1. Match the molecule to the chemical used to test for it to the col.pdf
1. Match the molecule to the chemical used to test for it to the color that the chemical would
change for a positive test.
Macromolecule Reagent Color Change
Sugar Biuret Reagent Green, Orange, or Red
Color
Protein Sudan III Blue-black Color
Lipid Iodine Red or Orange Color
Starch Benedict\'s
Solution
Blue-Violet Color.
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Compare and contrast the structure of a nephron with that of an alve.pdf
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Before first law of thermodynamics, there is zeroth law of thermodyn.pdf
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Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
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For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
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Priya Shaw - https://www.linkedin.com/in/priya-shaw
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