D is correct. Solution D.pdf
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1. The TextBox control allows you to capture input the user has type.pdf
1. The TextBox control allows you to capture input the user has typed on the keyboard.
2. txt is commonly used prefix for TextBox control names.
3. String concatenation means that one string is appended to another.
4. A variable is a storage location in the computer\'s memory, used for holding information while
the program is running.
5. A variable declaration is a statement that causes Visual Basic to create a variable in memory.
6. A variable\'s datatype determines whether a variable can hold a string, an integer, a date, or
some other kind of data.
7. A local variable is declared inside a procedure.
8. When Option Strict is set to ON only widening conversions are permitted.
9. A function is a specialized routine that performs a specific operation and then returns a value.
10. A CInt function converts an expression to an integer.
Solution
1. The TextBox control allows you to capture input the user has typed on the keyboard.
2. txt is commonly used prefix for TextBox control names.
3. String concatenation means that one string is appended to another.
4. A variable is a storage location in the computer\'s memory, used for holding information while
the program is running.
5. A variable declaration is a statement that causes Visual Basic to create a variable in memory.
6. A variable\'s datatype determines whether a variable can hold a string, an integer, a date, or
some other kind of data.
7. A local variable is declared inside a procedure.
8. When Option Strict is set to ON only widening conversions are permitted.
9. A function is a specialized routine that performs a specific operation and then returns a value.
10. A CInt function converts an expression to an integer..
1. In the above diagram C arteriol blood which carries oxyginated bl.pdf
1. In the above diagram C arteriol blood which carries oxyginated blood from left vetricle to
body parts, The partial pressure PO2 is 80-100 and partial pressure of PCO2 is 40.
2.E represents veinus blood, which carries deoxyginated blood from tissues to heart. The partial
pressure PO2 is 40-50 and partial pressure of CO2 PCO2 is 45.
3. Hyberbaric setting are used in medical treatment in which ambient pressure is greater than sea
level. It is used in hyberbaric oxygen theraphy.
4. This helps to increase the higher of concentration of oxygen in tissues. But higher pressure of
xygen may cause toxicity. It causes temporary blurred vision.
B. Differece in Partial pressure gradient between alveolar air and pulmonary capillary blood, and
systemic capillary blood and surrounding tissue.
Ability of oxygen to bind haemoglibin is called affinity for oxygen. Fetal haemoglobin has more
greater affinity to oxygen.
Solution
1. In the above diagram C arteriol blood which carries oxyginated blood from left vetricle to
body parts, The partial pressure PO2 is 80-100 and partial pressure of PCO2 is 40.
2.E represents veinus blood, which carries deoxyginated blood from tissues to heart. The partial
pressure PO2 is 40-50 and partial pressure of CO2 PCO2 is 45.
3. Hyberbaric setting are used in medical treatment in which ambient pressure is greater than sea
level. It is used in hyberbaric oxygen theraphy.
4. This helps to increase the higher of concentration of oxygen in tissues. But higher pressure of
xygen may cause toxicity. It causes temporary blurred vision.
B. Differece in Partial pressure gradient between alveolar air and pulmonary capillary blood, and
systemic capillary blood and surrounding tissue.
Ability of oxygen to bind haemoglibin is called affinity for oxygen. Fetal haemoglobin has more
greater affinity to oxygen..
1. Activation of helicase and phosphorylation of iniator proteins2.pdf
1. Activation of helicase and phosphorylation of iniator proteins
2. origin of replication is bound by ORC
(ORC serves as the foundation for the assembly of pre-replicative complex)
3. S-cdk is activated
(S-cdk is a part of pre replicative complex)
4. helicase and helicase loader bind to origin, forming pre-replicative complex.
(DnaA hydrolyzes ATP, DnaB-helicase, DnaC-helicase loader, SSBP- stabilize single strands,
DnaG- primase- recruits polymerase)
5. Dna replication begins
6. pre-RC disassembles
(CDKs prevent formation of the pre-RC in late G1, S, and G2 phases by transporting MCM2-7
and Cdt1 from the nucleus, degrading Cdc6)
Solution
1. Activation of helicase and phosphorylation of iniator proteins
2. origin of replication is bound by ORC
(ORC serves as the foundation for the assembly of pre-replicative complex)
3. S-cdk is activated
(S-cdk is a part of pre replicative complex)
4. helicase and helicase loader bind to origin, forming pre-replicative complex.
(DnaA hydrolyzes ATP, DnaB-helicase, DnaC-helicase loader, SSBP- stabilize single strands,
DnaG- primase- recruits polymerase)
5. Dna replication begins
6. pre-RC disassembles
(CDKs prevent formation of the pre-RC in late G1, S, and G2 phases by transporting MCM2-7
and Cdt1 from the nucleus, degrading Cdc6).
The link to constants is dead. .pdf
The link to constants is dead.
Solution
The link to constants is dead..
O has a lone pair of electrons .pdf
O has a lone pair of electrons
Solution
O has a lone pair of electrons.
Magnesium has 12 electrons in its ground state. W.pdf
Magnesium has 12 electrons in its ground state. When it forms anionic compound,
it loses its 2 valence electrons to nitrogen. Itnow has 10 electrons, and it has the electron
configuration ofneon. Thus, the highest energy level is 2, and there are 8electrons there.
Solution
Magnesium has 12 electrons in its ground state. When it forms anionic compound,
it loses its 2 valence electrons to nitrogen. Itnow has 10 electrons, and it has the electron
configuration ofneon. Thus, the highest energy level is 2, and there are 8electrons there..
d. sugar only sugar is a compound, the others are.pdf
d. sugar only sugar is a compound, the others are mixtures
Solution
d. sugar only sugar is a compound, the others are mixtures.
Recommended
1. The TextBox control allows you to capture input the user has type.pdf
1. The TextBox control allows you to capture input the user has typed on the keyboard.
2. txt is commonly used prefix for TextBox control names.
3. String concatenation means that one string is appended to another.
4. A variable is a storage location in the computer\'s memory, used for holding information while
the program is running.
5. A variable declaration is a statement that causes Visual Basic to create a variable in memory.
6. A variable\'s datatype determines whether a variable can hold a string, an integer, a date, or
some other kind of data.
7. A local variable is declared inside a procedure.
8. When Option Strict is set to ON only widening conversions are permitted.
9. A function is a specialized routine that performs a specific operation and then returns a value.
10. A CInt function converts an expression to an integer.
Solution
1. The TextBox control allows you to capture input the user has typed on the keyboard.
2. txt is commonly used prefix for TextBox control names.
3. String concatenation means that one string is appended to another.
4. A variable is a storage location in the computer\'s memory, used for holding information while
the program is running.
5. A variable declaration is a statement that causes Visual Basic to create a variable in memory.
6. A variable\'s datatype determines whether a variable can hold a string, an integer, a date, or
some other kind of data.
7. A local variable is declared inside a procedure.
8. When Option Strict is set to ON only widening conversions are permitted.
9. A function is a specialized routine that performs a specific operation and then returns a value.
10. A CInt function converts an expression to an integer..
1. In the above diagram C arteriol blood which carries oxyginated bl.pdf
1. In the above diagram C arteriol blood which carries oxyginated blood from left vetricle to
body parts, The partial pressure PO2 is 80-100 and partial pressure of PCO2 is 40.
2.E represents veinus blood, which carries deoxyginated blood from tissues to heart. The partial
pressure PO2 is 40-50 and partial pressure of CO2 PCO2 is 45.
3. Hyberbaric setting are used in medical treatment in which ambient pressure is greater than sea
level. It is used in hyberbaric oxygen theraphy.
4. This helps to increase the higher of concentration of oxygen in tissues. But higher pressure of
xygen may cause toxicity. It causes temporary blurred vision.
B. Differece in Partial pressure gradient between alveolar air and pulmonary capillary blood, and
systemic capillary blood and surrounding tissue.
Ability of oxygen to bind haemoglibin is called affinity for oxygen. Fetal haemoglobin has more
greater affinity to oxygen.
Solution
1. In the above diagram C arteriol blood which carries oxyginated blood from left vetricle to
body parts, The partial pressure PO2 is 80-100 and partial pressure of PCO2 is 40.
2.E represents veinus blood, which carries deoxyginated blood from tissues to heart. The partial
pressure PO2 is 40-50 and partial pressure of CO2 PCO2 is 45.
3. Hyberbaric setting are used in medical treatment in which ambient pressure is greater than sea
level. It is used in hyberbaric oxygen theraphy.
4. This helps to increase the higher of concentration of oxygen in tissues. But higher pressure of
xygen may cause toxicity. It causes temporary blurred vision.
B. Differece in Partial pressure gradient between alveolar air and pulmonary capillary blood, and
systemic capillary blood and surrounding tissue.
Ability of oxygen to bind haemoglibin is called affinity for oxygen. Fetal haemoglobin has more
greater affinity to oxygen..
1. Activation of helicase and phosphorylation of iniator proteins2.pdf
1. Activation of helicase and phosphorylation of iniator proteins
2. origin of replication is bound by ORC
(ORC serves as the foundation for the assembly of pre-replicative complex)
3. S-cdk is activated
(S-cdk is a part of pre replicative complex)
4. helicase and helicase loader bind to origin, forming pre-replicative complex.
(DnaA hydrolyzes ATP, DnaB-helicase, DnaC-helicase loader, SSBP- stabilize single strands,
DnaG- primase- recruits polymerase)
5. Dna replication begins
6. pre-RC disassembles
(CDKs prevent formation of the pre-RC in late G1, S, and G2 phases by transporting MCM2-7
and Cdt1 from the nucleus, degrading Cdc6)
Solution
1. Activation of helicase and phosphorylation of iniator proteins
2. origin of replication is bound by ORC
(ORC serves as the foundation for the assembly of pre-replicative complex)
3. S-cdk is activated
(S-cdk is a part of pre replicative complex)
4. helicase and helicase loader bind to origin, forming pre-replicative complex.
(DnaA hydrolyzes ATP, DnaB-helicase, DnaC-helicase loader, SSBP- stabilize single strands,
DnaG- primase- recruits polymerase)
5. Dna replication begins
6. pre-RC disassembles
(CDKs prevent formation of the pre-RC in late G1, S, and G2 phases by transporting MCM2-7
and Cdt1 from the nucleus, degrading Cdc6).
The link to constants is dead. .pdf
The link to constants is dead.
Solution
The link to constants is dead..
O has a lone pair of electrons .pdf
O has a lone pair of electrons
Solution
O has a lone pair of electrons.
Magnesium has 12 electrons in its ground state. W.pdf
Magnesium has 12 electrons in its ground state. When it forms anionic compound,
it loses its 2 valence electrons to nitrogen. Itnow has 10 electrons, and it has the electron
configuration ofneon. Thus, the highest energy level is 2, and there are 8electrons there.
Solution
Magnesium has 12 electrons in its ground state. When it forms anionic compound,
it loses its 2 valence electrons to nitrogen. Itnow has 10 electrons, and it has the electron
configuration ofneon. Thus, the highest energy level is 2, and there are 8electrons there..
d. sugar only sugar is a compound, the others are.pdf
d. sugar only sugar is a compound, the others are mixtures
Solution
d. sugar only sugar is a compound, the others are mixtures.
True.Gases have a larger entropy than liquids as they have a lower.pdf
True.
Gases have a larger entropy than liquids as they have a lower density (hence larger specific
volume).
So the gas molecules have more possible energy configurations and a higher entropy.
Solution
True.
Gases have a larger entropy than liquids as they have a lower density (hence larger specific
volume).
So the gas molecules have more possible energy configurations and a higher entropy..
The balance of payments of a country is a systematic record of all e.pdf
The balance of payments of a country is a systematic record of all economic transactions
between the residents of a country and the rest of the world. BOP presents a classified record of
all receipts on account of goods exported, services rendered and capital received by residents and
payments made by them on account of goods imported and services received from the capital
transferred to non-residents or foreigners.
It adopts a double-entry book-keeping system. It has two sides: credit side and debit side.
Receipts are recorded on the credit side and payments on the debit side.
Current account of BOP
BOP on current account is a statement of actual receipts and payments in short period. It includes
the value of export and imports of both visible and invisible goods. There can be either surplus or
deficit in current account.The current account includes:- export & import of services, interests,
profits, dividends and unilateral receipts/payments from/to abroad. BOP on current account
refers to the inclusion of three balances of namely – Merchandise balance, Services balance and
Unilateral Transfer balance.
Capital Account of BOP
The capital account records all international transactions that involve a resident of the country
concerned changing either his assets with or his liabilities to a resident of another country.
Transactions in the capital account reflect a change in a stock – either assets or liabilities. It is
difference between the receipts and payments on account of capital account. It refers to all
financial transactions. The capital account involves inflows and outflows relating to investments,
short term borrowings/lending, and medium term to long term borrowing/lending.
Solution
The balance of payments of a country is a systematic record of all economic transactions
between the residents of a country and the rest of the world. BOP presents a classified record of
all receipts on account of goods exported, services rendered and capital received by residents and
payments made by them on account of goods imported and services received from the capital
transferred to non-residents or foreigners.
It adopts a double-entry book-keeping system. It has two sides: credit side and debit side.
Receipts are recorded on the credit side and payments on the debit side.
Current account of BOP
BOP on current account is a statement of actual receipts and payments in short period. It includes
the value of export and imports of both visible and invisible goods. There can be either surplus or
deficit in current account.The current account includes:- export & import of services, interests,
profits, dividends and unilateral receipts/payments from/to abroad. BOP on current account
refers to the inclusion of three balances of namely – Merchandise balance, Services balance and
Unilateral Transfer balance.
Capital Account of BOP
The capital account records all international transactions that involve a resident of the country
concerned changing either his asset.
REITs are Real Estate Investment Trusts. REITs are considered as a u.pdf
REITs are Real Estate Investment Trusts. REITs are considered as a unique vehicle for studying
capital structure decisions due to the fact that REITs do not use the conventional models and
approaches like the net income approach or the net operating income approach or the Modigliani
and Miller approach to determine their capital structure. Their capital structure usually exhibits
use of low leverage with use of debt that has longer maturity. REITs maintain a much higher
proportion of fixed rate debt and tend to avoid or reduce its dependence on secured debts. REIT
shuns the use of variable rate debt and there exists an inverse relationship between the value of
REIT and leverage.
Research and empirical studies suggests that the optimal capital structure of REITs is one that
has the highest “q ratio”, low leverage, and higher proportion of fixed rate debts.
Solution
REITs are Real Estate Investment Trusts. REITs are considered as a unique vehicle for studying
capital structure decisions due to the fact that REITs do not use the conventional models and
approaches like the net income approach or the net operating income approach or the Modigliani
and Miller approach to determine their capital structure. Their capital structure usually exhibits
use of low leverage with use of debt that has longer maturity. REITs maintain a much higher
proportion of fixed rate debt and tend to avoid or reduce its dependence on secured debts. REIT
shuns the use of variable rate debt and there exists an inverse relationship between the value of
REIT and leverage.
Research and empirical studies suggests that the optimal capital structure of REITs is one that
has the highest “q ratio”, low leverage, and higher proportion of fixed rate debts..
Please follow the code and description Software Test Manager T.pdf
Please follow the code and description :
Software Test Manager :
The role of Test manager commonly refers to the activity of managing and checking the
computer software adn its components testing process.They generallu use some management
tools so as to manage the tests that have been implemented in a test procedure.
Roles and Responsibilities :
1) The effective software test managers need to understand the discipline of testing and the
process respectively.
2) They are also need to be able to manage and implement a effective testing process in their
organizations. So thei basically needs the team leading skills, with effective communication
skills.
3) One needs to understand the testing efforts and the implementations by analyzing the
requirements of project before hand.
4) This step ivolves the estimatation of the management support for the time required, witht the
available resources and the budget required to perform the testing to complete the process
successfully.
5) Need to Organize the meetings regularly so as to get the clear picture of the present situation
of the testing process.
6) Build a testing team of professionals with appropriate skills, attitudes and motivation.
7) Identification of the training requirements and forward it to the Project Manager.
8) Develop the test plan for the tasks, dependencies and participants required to mitigate the risks
to system quality and obtain stakeholder support for this plan.
9) Arrange the Hardware and software requirement for the Test Setup. Assignment of the tasks to
all Testing Team members and ensure that all of them have sufficient work in the project.
10) Document, implement, monitor, and enforce all processes for testing as per standards defined
by the organization. Check / Review the Test Cases documents frequently.
11) Keep track of the new requirements / change in the requirements of the Project. Escalate the
issues about project requirements to the higher authorities.
12) Organize the status meetings and send the Status Report to the Client. Attend the regular
client call and discuss the weekly status with the client.
13) Tracking and preparing the report of testing activities like test testing results, test case
coverage, required resources, defects discovered and their status, performance baselines.
14) Ensure the timely delivery of different testing milestones in the near future.
15) Prepares / updates the metrics dashboard at the end of a phase or at the completion of project.
Quality Assurance Manager :
In general Quality assurance is a way of preventing mistakes or defects persisted in the
manufactured products/softwares and avoiding the problems when delivering solutions or
services to the respecitve customers. A QA Manager basically looks after the assurance and the
perfect delivery of the product/software in the stipulated and to maintain the customer
satisfaction.
Roles and Responsibilities :
1) Determining, negotiating and agreeing the available in-house quality proced.
Part DGASTRULA----answerAmid gastrulation cells move to the insi.pdf
Part D
GASTRULA----answer
Amid gastrulation cells move to the inside of the blastula, hence framing two (in diploblastic
creatures) or three (triploblastic) germ layers. The developing life amid this procedure is known
as a gastrula. The germ layers are alluded to as the ectoderm, mesoderm and endoderm. In
diploblastic creatures just the ectoderm and the endoderm are available .
Among various creatures, distinctive blends of the accompanying procedures jump out at place
the cells in the inside of the developing life:
Epiboly ,Ingression ,Invagination archenteron ,Delamination and Involution
Polar expansion - Cells at the polar closures of the blastula/gastrula multiply, for the most part at
the creature shaft.
===============================================================
Part E
What happens to a cell during the process of differentiation
Answer : option a
It undergoes a change towards a more specialised form or function
===========================================================
Part F
child is born without a fully developed pancreas. He is normal in all other respects & thus
doctors believe the problem began in early development. Which part of development did the
problem probably occur?
endoderm formation-answer
pancreas:
It is a glandular organ in the stomach related framework and endocrine arrangement of
vertebrates. In people, it is situated in the stomach cavity behind the stomach. It is an endocrine
organ creating a few imperative hormones, including insulin, glucagon, somatostatin, and
pancreatic polypeptide which course in the blood. The pancreas is additionally a stomach related
organ, discharging pancreatic juice containing stomach related chemicals that help assimilation
and ingestion
Solution
Part D
GASTRULA----answer
Amid gastrulation cells move to the inside of the blastula, hence framing two (in diploblastic
creatures) or three (triploblastic) germ layers. The developing life amid this procedure is known
as a gastrula. The germ layers are alluded to as the ectoderm, mesoderm and endoderm. In
diploblastic creatures just the ectoderm and the endoderm are available .
Among various creatures, distinctive blends of the accompanying procedures jump out at place
the cells in the inside of the developing life:
Epiboly ,Ingression ,Invagination archenteron ,Delamination and Involution
Polar expansion - Cells at the polar closures of the blastula/gastrula multiply, for the most part at
the creature shaft.
===============================================================
Part E
What happens to a cell during the process of differentiation
Answer : option a
It undergoes a change towards a more specialised form or function
===========================================================
Part F
child is born without a fully developed pancreas. He is normal in all other respects & thus
doctors believe the problem began in early development. Which part of development did the
problem probably occur?
endoderm formation-answer
pancreas:
It is a glandul.
AnswerThe given experiment is related to the bacterial transforma.pdf
Answer:
The given experiment is related to the bacterial transformation. Bacterial transformation is the
transfer of foreign genetic material (either DNA or RNA) into the recipient cell through direct
uptake from the environment. It is a process involves exchange of genetic material between of
two organisms through non-reproductive mode in contrast to the traditional reproduction
(vertical transfer through sexual or asexual reproduction). It results to the change in the genotype
of the recipient cell due to incorporation of newly acquired DNA into the recipient genome
through either recombination or insertion.
The transformants can be identified using a selective medium. The identification of the
transformants (containing the vector/foreign molecule) require the presence of a suitable marker
genes on the vector molecule, whose expression provides a means of identifying cells containing
it (transformants). The most popular marker used for it are: a) Antibiotic resistant gene, b) colour
substance developing gene, c) galactosidase gene complementation.
In the given experiment 3 markers were used:
a) The vector (pGLO plasmid) contain the gene for Green Fluorescent Protein (GFP) and the
ampicillin resistance gene (which codes for beta-lactamase). Because it shares a bidirectional
promoter with a gene for metabolizing arabinose, the GFP gene is expressed in the presence of
arabinose, which makes the transgenic organism express its fluorescence under UV light. GFP
can be induced in bacteria containing the pGLO plasmid by growing them on +arabinose plates.
b) Ampicillin antibiotic, inhibits the cell wall synthesis. It acts as a competitive inhibitor of the
enzyme transpeptidase.
c) Arabinose (as a carbon source for bacteria).
The genetic transformation resulting to the change the phenotype of the organism (E. coli in case
of the given experiment). The results found after the transformation experiment would be in
observed in 4 type of plates.
1) Control plate (Growth medium only)
2) Control plate (Growth medium+ampicillin)
3) Transformant plate (Growth medium+ampicillin+pGLO)
4) Transformant plate (Growth medium+ampicillin+pGLO+arabinose)
The observations of control plate (E. coli cells before transformantion experiment) might provide
reference data to determine if any genetic transformation has occurred. Thus, we have to record
the a) Number of colonies, b) Size of the colony, c) Color of the colonies, d) Distribution of the
colonies on the plate, e) Visible appearance when viewed with ultraviolet (UV) light), f) The
ability of the cells to survive and multiply in the presence of an antibiotic ampicillin.
Expected Results:
*Plate 1: The results would obtained that countable (between 30-300), small, circular, shiny,
colourless colonies of E. coli were distributed throught out the plate. They would be non-
fluorescent when obsered under UV-light.
*Plate 2: No colonies will be developed in the presence of ampicillin because the E. coli is non
resistant .
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part.pdf
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part will be together in the
solution
NaClO =================> Na+ + ClO-
[It is similar with sodium sulphate, it is also an oxy acid salt. so the sulphur and O will be
together]
Na2SO4 =========> 2 Na+ + SO42- ]
Solution
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part will be together in the
solution
NaClO =================> Na+ + ClO-
[It is similar with sodium sulphate, it is also an oxy acid salt. so the sulphur and O will be
together]
Na2SO4 =========> 2 Na+ + SO42- ].
Answer A] Water solventates any substances that .pdf
Answer: A] Water solventates any substances that may dissociate into charged ions.
Taking salt for example (NaCl), the water will allow the solid salt to dissociate into Na+ and Cl-
ions. How does it do this chemically? Water molecules are polar. Hence, they surround the ions
in something called a hydration shell. Water is H2O, the hydrogens have a positive dipole
whereas the oxygen has a negative dipole. The negative sides of the water molecule (oxygen)
will all surround and face the positive charged Na+ ions. The positive sides of the water
molecule (hydrogens) will surround the negatively charged Cl- ions. B] A hydrogen bond is
formed between an hydrogen atom on one molecule and an electronegative atom on another
molecule (oxygen, flourine, nitrogen, etc.). The hydrogen atom on the first molecule must be
covalently binded to an electronegative atom as well. Overall hydrogen bond: O-H...O (dotted
line is the hydrogen bond). In water, these bonds are rapidly formed and destroyed. It is not a
covalent bond, but sort of an interaction force between water molecules. Physically, this may be
seen with ice. Ice forms crystal-like structures due to the neat organization of these hydrogen
interactions. C] Water will tend to isolate hydrophobic substances. Water is polar and hence
likes to mix with other polar substances. Since hydrophobic substances are non-polar, the water
will physically separate out the hydrophobic substances in solution. In chemical terms, the water
will just form clatherates (organized shells of water) around any hydrophobic substance. This can
be seen if you add oil to water. It just doesn\'t mix.
Solution
Answer: A] Water solventates any substances that may dissociate into charged ions.
Taking salt for example (NaCl), the water will allow the solid salt to dissociate into Na+ and Cl-
ions. How does it do this chemically? Water molecules are polar. Hence, they surround the ions
in something called a hydration shell. Water is H2O, the hydrogens have a positive dipole
whereas the oxygen has a negative dipole. The negative sides of the water molecule (oxygen)
will all surround and face the positive charged Na+ ions. The positive sides of the water
molecule (hydrogens) will surround the negatively charged Cl- ions. B] A hydrogen bond is
formed between an hydrogen atom on one molecule and an electronegative atom on another
molecule (oxygen, flourine, nitrogen, etc.). The hydrogen atom on the first molecule must be
covalently binded to an electronegative atom as well. Overall hydrogen bond: O-H...O (dotted
line is the hydrogen bond). In water, these bonds are rapidly formed and destroyed. It is not a
covalent bond, but sort of an interaction force between water molecules. Physically, this may be
seen with ice. Ice forms crystal-like structures due to the neat organization of these hydrogen
interactions. C] Water will tend to isolate hydrophobic substances. Water is polar and hence
likes to mix with other polar substances. Since hydr.
In this case HF is the most ionic because H and F have thelarges.pdf
In this case HF is the most ionic because H and F have thelargest difference in electronegativity.
The value for H is2.1; the value for F is 4.0. The difference is 1.9. All of the other options given
have a lower difference.
Solution
In this case HF is the most ionic because H and F have thelargest difference in electronegativity.
The value for H is2.1; the value for F is 4.0. The difference is 1.9. All of the other options given
have a lower difference..
Here some of the reasons why list instead of a vector(i) Storing .pdf
Here some of the reasons why list instead of a vector:
(i) Storing large data, that is if few elements contains more data then it is inefficient in storing
into
vectors. But this can easily done using lists.
(ii) Iterations, insertning and deleting will take more time in vectors comparitively to lists.
(iii) In lists searching and traversing are inconsequential.
(iv) Splice and merge methods are not available in vectors. If we try to do using vectors then
operation will be more expensive.
Solution
Here some of the reasons why list instead of a vector:
(i) Storing large data, that is if few elements contains more data then it is inefficient in storing
into
vectors. But this can easily done using lists.
(ii) Iterations, insertning and deleting will take more time in vectors comparitively to lists.
(iii) In lists searching and traversing are inconsequential.
(iv) Splice and merge methods are not available in vectors. If we try to do using vectors then
operation will be more expensive..
(1)Through antibiotic resistance, bacteria will resist the action of.pdf
(1)Through antibiotic resistance, bacteria will resist the action of antibiotic and there are
different ways through which bacteria will become antibiotic resistance.
Conjugation: Here the resistant bacteria will transfer genetic material responsible for antibiotic
resistance to another bacterium by forming direct contact with them through sex pilus.
Transduction: The resistance traits of one bacterium will get packed into the head portion of the
virus (bacteriophage) and the release phage progeny from these antibiotic resistant bacteria will
infect susceptible bacteria and transfer the antibiotic resistance genes to them.
Transformation: Here bacteria capable of undergoing transformation will take up short fragments
of naked DNA responsible for antibiotic resistance from the surrounding environment by direct
uptake.
By undergoing genetic mutation: By undergoing genetic mutations, the bacteria will produce
chemicals which will eliminate the action of antibiotics and in some cases the target will be
modified to escape from antibiotic action.
(2) The genetic code is universal and bacterial cell will try to translate the gene to produce the
protein, but the proteins will not similar to human proteins because
Bacteria are lacking splicing mechanism it cannot splice out the introns from exons and thus it
will try to transcribe and translate the gene with both introns and exons giving different proteins.
In addition to this the proteins produced by bacteria will be different form human cells because,
bacteria used 70S ribosomes and the translation it does is different from the human ribosome and
another thing is codon usage is different in bacteria and we will have different genomic tRNA
pool and the post-translation modification present in human cells is missing in bacteria.
(3) Not all bacteria present in environment can be culturable, what you can see through
microscope will far outweigh the number of colonies that grow on a Petri plate and this shows
that you cannot culture all the bacteria. Through sophisticated microscopic techniques you can
identify the unculturable bacteria and through molecular techniques, such as metagenomic
sequencing will help you to get more details about this.
Solution
(1)Through antibiotic resistance, bacteria will resist the action of antibiotic and there are
different ways through which bacteria will become antibiotic resistance.
Conjugation: Here the resistant bacteria will transfer genetic material responsible for antibiotic
resistance to another bacterium by forming direct contact with them through sex pilus.
Transduction: The resistance traits of one bacterium will get packed into the head portion of the
virus (bacteriophage) and the release phage progeny from these antibiotic resistant bacteria will
infect susceptible bacteria and transfer the antibiotic resistance genes to them.
Transformation: Here bacteria capable of undergoing transformation will take up short fragments
of naked DNA responsible for antibio.
D. Role Based Access Control (RBAC) allows specific people to be ass.pdf
D. Role Based Access Control (RBAC) allows specific people to be assigned to specific roles
with specific privileges. It allows access to be assigned to groups and works well where there are
high levels of turnover.
Solution
D. Role Based Access Control (RBAC) allows specific people to be assigned to specific roles
with specific privileges. It allows access to be assigned to groups and works well where there are
high levels of turnover..
CreateTable.m filefunction multiplicationTable = CreateTable(baseV.pdf
CreateTable.m file
function multiplicationTable = CreateTable(baseValue)
multiplicationTable = zeros(1,6);
for i=0:5
multiplicationTable(i+1) = 2*i;
end
end
Solution
CreateTable.m file
function multiplicationTable = CreateTable(baseValue)
multiplicationTable = zeros(1,6);
for i=0:5
multiplicationTable(i+1) = 2*i;
end
end.
Capacity of the program memory is 0.248 kb.SolutionCapacity of.pdf
Capacity of the program memory is 0.248 kb.
Solution
Capacity of the program memory is 0.248 kb..
BrO- could not disproportionateSolutionBrO- could not dispropo.pdf
BrO- could not disproportionate
Solution
BrO- could not disproportionate.
Binary Search works on the concept of calculating the mid index valu.pdf
Binary Search works on the concept of calculating the mid index value, and will divide the actual
list to half of its size, if not found at mid position.
If the element is found in the first comparison itself, therefore, the situation is considered the best
case, and the number of comparisons required in this scenario is just ONE. And the time
complexity of binary search in best case O(1).
If the element is found in the last comparison, or if the element is not found in the list, the
situation is considered the worst case, and the number of comparisons required in this scenario is
log n, where n is the number of elements in the list. And the time complexity of binary search in
worst case O(log n).
Therefore, the amounts of steps in a binary search change its time complexity. And the amount
of steps required is directly proportional to the complexity(number of comparisons) of the binary
search algorithm.
If you need any refinements, just get back to me.
Solution
Binary Search works on the concept of calculating the mid index value, and will divide the actual
list to half of its size, if not found at mid position.
If the element is found in the first comparison itself, therefore, the situation is considered the best
case, and the number of comparisons required in this scenario is just ONE. And the time
complexity of binary search in best case O(1).
If the element is found in the last comparison, or if the element is not found in the list, the
situation is considered the worst case, and the number of comparisons required in this scenario is
log n, where n is the number of elements in the list. And the time complexity of binary search in
worst case O(log n).
Therefore, the amounts of steps in a binary search change its time complexity. And the amount
of steps required is directly proportional to the complexity(number of comparisons) of the binary
search algorithm.
If you need any refinements, just get back to me..
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; .pdf
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; oxide ion d 14 C6 ; Carbon
Solution
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; oxide ion d 14 C6 ; Carbon.
Answer A. H2S serves as electron donorChemolithotrophs are difined.pdf
Answer A. H2S serves as electron donor
Chemolithotrophs are difined as the organisms that can produce metabolic useful energy by
oxidatio of inorganic compounds. Chemolithotrophic life is possible in the presence as well as in
the absence of molecular oxygen.
Thiobacillus and Chlorobium bacteria are chemolithotrops. The chemilithotrops a use
H2S(inorganic compounds) as electron donor rather that organic compounds. As
Solution
Answer A. H2S serves as electron donor
Chemolithotrophs are difined as the organisms that can produce metabolic useful energy by
oxidatio of inorganic compounds. Chemolithotrophic life is possible in the presence as well as in
the absence of molecular oxygen.
Thiobacillus and Chlorobium bacteria are chemolithotrops. The chemilithotrops a use
H2S(inorganic compounds) as electron donor rather that organic compounds. As.
Answer (1)Pros of investing directly in individual stocks- If inves.pdf
Answer (1):Pros of investing directly in individual stocks- If investors directly invest into growth
stocks, their capital grows very fast as the stocks go up. They also get direct dividend from the
company. They do not have to pay any entry and exit load. They can directly expose to the stock
market.
Cons of investing directly in individual stocks- Investors are directly exposed to risk, pertaining
to particular stock, some stocks are more risky and some are less. Investing directly into
individual stocks does not provide benefit of diversification. Bonds\' prices get fluctuated over
the period of time. It is a risky move.
Pros of Investing indirectly through a mutual fund- Investing into mutual funds gives these
benefits-
Cons of Investing indirectly through a mutual fund- Investing into mutual funds has its own
disadvantage also. It charges some hidden fees and entry and exit loan.
Answer (2): Difference between alpha and beta risk- is as following-
Answer (4): After the slow down in economy and financial markets, new act were being passed
and investors are getting back the confidence in the stock markets. People invested into growth
mutual funds as the market was at lower level. Their mindset was bullish towards the market so
they invested into equities as well as mutual funds for getting future returns. Expenses were also
reduced in 2009.
Answer (5): Difference between the SEC and FINRA as regulators-
Solution
Answer (1):Pros of investing directly in individual stocks- If investors directly invest into growth
stocks, their capital grows very fast as the stocks go up. They also get direct dividend from the
company. They do not have to pay any entry and exit load. They can directly expose to the stock
market.
Cons of investing directly in individual stocks- Investors are directly exposed to risk, pertaining
to particular stock, some stocks are more risky and some are less. Investing directly into
individual stocks does not provide benefit of diversification. Bonds\' prices get fluctuated over
the period of time. It is a risky move.
Pros of Investing indirectly through a mutual fund- Investing into mutual funds gives these
benefits-
Cons of Investing indirectly through a mutual fund- Investing into mutual funds has its own
disadvantage also. It charges some hidden fees and entry and exit loan.
Answer (2): Difference between alpha and beta risk- is as following-
Answer (4): After the slow down in economy and financial markets, new act were being passed
and investors are getting back the confidence in the stock markets. People invested into growth
mutual funds as the market was at lower level. Their mindset was bullish towards the market so
they invested into equities as well as mutual funds for getting future returns. Expenses were also
reduced in 2009.
Answer (5): Difference between the SEC and FINRA as regulators-.
Ans BIOSBIOS verifies if theoptical drive is recognised by the sys.pdf
Ans BIOS
BIOS verifies if theoptical drive is recognised by the system
Device Manager is used to uninstall/reinstall the device driver for optical drive
Solution
Ans BIOS
BIOS verifies if theoptical drive is recognised by the system
Device Manager is used to uninstall/reinstall the device driver for optical drive.
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/
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
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True.Gases have a larger entropy than liquids as they have a lower.pdf
True.
Gases have a larger entropy than liquids as they have a lower density (hence larger specific
volume).
So the gas molecules have more possible energy configurations and a higher entropy.
Solution
True.
Gases have a larger entropy than liquids as they have a lower density (hence larger specific
volume).
So the gas molecules have more possible energy configurations and a higher entropy..
The balance of payments of a country is a systematic record of all e.pdf
The balance of payments of a country is a systematic record of all economic transactions
between the residents of a country and the rest of the world. BOP presents a classified record of
all receipts on account of goods exported, services rendered and capital received by residents and
payments made by them on account of goods imported and services received from the capital
transferred to non-residents or foreigners.
It adopts a double-entry book-keeping system. It has two sides: credit side and debit side.
Receipts are recorded on the credit side and payments on the debit side.
Current account of BOP
BOP on current account is a statement of actual receipts and payments in short period. It includes
the value of export and imports of both visible and invisible goods. There can be either surplus or
deficit in current account.The current account includes:- export & import of services, interests,
profits, dividends and unilateral receipts/payments from/to abroad. BOP on current account
refers to the inclusion of three balances of namely – Merchandise balance, Services balance and
Unilateral Transfer balance.
Capital Account of BOP
The capital account records all international transactions that involve a resident of the country
concerned changing either his assets with or his liabilities to a resident of another country.
Transactions in the capital account reflect a change in a stock – either assets or liabilities. It is
difference between the receipts and payments on account of capital account. It refers to all
financial transactions. The capital account involves inflows and outflows relating to investments,
short term borrowings/lending, and medium term to long term borrowing/lending.
Solution
The balance of payments of a country is a systematic record of all economic transactions
between the residents of a country and the rest of the world. BOP presents a classified record of
all receipts on account of goods exported, services rendered and capital received by residents and
payments made by them on account of goods imported and services received from the capital
transferred to non-residents or foreigners.
It adopts a double-entry book-keeping system. It has two sides: credit side and debit side.
Receipts are recorded on the credit side and payments on the debit side.
Current account of BOP
BOP on current account is a statement of actual receipts and payments in short period. It includes
the value of export and imports of both visible and invisible goods. There can be either surplus or
deficit in current account.The current account includes:- export & import of services, interests,
profits, dividends and unilateral receipts/payments from/to abroad. BOP on current account
refers to the inclusion of three balances of namely – Merchandise balance, Services balance and
Unilateral Transfer balance.
Capital Account of BOP
The capital account records all international transactions that involve a resident of the country
concerned changing either his asset.
REITs are Real Estate Investment Trusts. REITs are considered as a u.pdf
REITs are Real Estate Investment Trusts. REITs are considered as a unique vehicle for studying
capital structure decisions due to the fact that REITs do not use the conventional models and
approaches like the net income approach or the net operating income approach or the Modigliani
and Miller approach to determine their capital structure. Their capital structure usually exhibits
use of low leverage with use of debt that has longer maturity. REITs maintain a much higher
proportion of fixed rate debt and tend to avoid or reduce its dependence on secured debts. REIT
shuns the use of variable rate debt and there exists an inverse relationship between the value of
REIT and leverage.
Research and empirical studies suggests that the optimal capital structure of REITs is one that
has the highest “q ratio”, low leverage, and higher proportion of fixed rate debts.
Solution
REITs are Real Estate Investment Trusts. REITs are considered as a unique vehicle for studying
capital structure decisions due to the fact that REITs do not use the conventional models and
approaches like the net income approach or the net operating income approach or the Modigliani
and Miller approach to determine their capital structure. Their capital structure usually exhibits
use of low leverage with use of debt that has longer maturity. REITs maintain a much higher
proportion of fixed rate debt and tend to avoid or reduce its dependence on secured debts. REIT
shuns the use of variable rate debt and there exists an inverse relationship between the value of
REIT and leverage.
Research and empirical studies suggests that the optimal capital structure of REITs is one that
has the highest “q ratio”, low leverage, and higher proportion of fixed rate debts..
Please follow the code and description Software Test Manager T.pdf
Please follow the code and description :
Software Test Manager :
The role of Test manager commonly refers to the activity of managing and checking the
computer software adn its components testing process.They generallu use some management
tools so as to manage the tests that have been implemented in a test procedure.
Roles and Responsibilities :
1) The effective software test managers need to understand the discipline of testing and the
process respectively.
2) They are also need to be able to manage and implement a effective testing process in their
organizations. So thei basically needs the team leading skills, with effective communication
skills.
3) One needs to understand the testing efforts and the implementations by analyzing the
requirements of project before hand.
4) This step ivolves the estimatation of the management support for the time required, witht the
available resources and the budget required to perform the testing to complete the process
successfully.
5) Need to Organize the meetings regularly so as to get the clear picture of the present situation
of the testing process.
6) Build a testing team of professionals with appropriate skills, attitudes and motivation.
7) Identification of the training requirements and forward it to the Project Manager.
8) Develop the test plan for the tasks, dependencies and participants required to mitigate the risks
to system quality and obtain stakeholder support for this plan.
9) Arrange the Hardware and software requirement for the Test Setup. Assignment of the tasks to
all Testing Team members and ensure that all of them have sufficient work in the project.
10) Document, implement, monitor, and enforce all processes for testing as per standards defined
by the organization. Check / Review the Test Cases documents frequently.
11) Keep track of the new requirements / change in the requirements of the Project. Escalate the
issues about project requirements to the higher authorities.
12) Organize the status meetings and send the Status Report to the Client. Attend the regular
client call and discuss the weekly status with the client.
13) Tracking and preparing the report of testing activities like test testing results, test case
coverage, required resources, defects discovered and their status, performance baselines.
14) Ensure the timely delivery of different testing milestones in the near future.
15) Prepares / updates the metrics dashboard at the end of a phase or at the completion of project.
Quality Assurance Manager :
In general Quality assurance is a way of preventing mistakes or defects persisted in the
manufactured products/softwares and avoiding the problems when delivering solutions or
services to the respecitve customers. A QA Manager basically looks after the assurance and the
perfect delivery of the product/software in the stipulated and to maintain the customer
satisfaction.
Roles and Responsibilities :
1) Determining, negotiating and agreeing the available in-house quality proced.
Part DGASTRULA----answerAmid gastrulation cells move to the insi.pdf
Part D
GASTRULA----answer
Amid gastrulation cells move to the inside of the blastula, hence framing two (in diploblastic
creatures) or three (triploblastic) germ layers. The developing life amid this procedure is known
as a gastrula. The germ layers are alluded to as the ectoderm, mesoderm and endoderm. In
diploblastic creatures just the ectoderm and the endoderm are available .
Among various creatures, distinctive blends of the accompanying procedures jump out at place
the cells in the inside of the developing life:
Epiboly ,Ingression ,Invagination archenteron ,Delamination and Involution
Polar expansion - Cells at the polar closures of the blastula/gastrula multiply, for the most part at
the creature shaft.
===============================================================
Part E
What happens to a cell during the process of differentiation
Answer : option a
It undergoes a change towards a more specialised form or function
===========================================================
Part F
child is born without a fully developed pancreas. He is normal in all other respects & thus
doctors believe the problem began in early development. Which part of development did the
problem probably occur?
endoderm formation-answer
pancreas:
It is a glandular organ in the stomach related framework and endocrine arrangement of
vertebrates. In people, it is situated in the stomach cavity behind the stomach. It is an endocrine
organ creating a few imperative hormones, including insulin, glucagon, somatostatin, and
pancreatic polypeptide which course in the blood. The pancreas is additionally a stomach related
organ, discharging pancreatic juice containing stomach related chemicals that help assimilation
and ingestion
Solution
Part D
GASTRULA----answer
Amid gastrulation cells move to the inside of the blastula, hence framing two (in diploblastic
creatures) or three (triploblastic) germ layers. The developing life amid this procedure is known
as a gastrula. The germ layers are alluded to as the ectoderm, mesoderm and endoderm. In
diploblastic creatures just the ectoderm and the endoderm are available .
Among various creatures, distinctive blends of the accompanying procedures jump out at place
the cells in the inside of the developing life:
Epiboly ,Ingression ,Invagination archenteron ,Delamination and Involution
Polar expansion - Cells at the polar closures of the blastula/gastrula multiply, for the most part at
the creature shaft.
===============================================================
Part E
What happens to a cell during the process of differentiation
Answer : option a
It undergoes a change towards a more specialised form or function
===========================================================
Part F
child is born without a fully developed pancreas. He is normal in all other respects & thus
doctors believe the problem began in early development. Which part of development did the
problem probably occur?
endoderm formation-answer
pancreas:
It is a glandul.
AnswerThe given experiment is related to the bacterial transforma.pdf
Answer:
The given experiment is related to the bacterial transformation. Bacterial transformation is the
transfer of foreign genetic material (either DNA or RNA) into the recipient cell through direct
uptake from the environment. It is a process involves exchange of genetic material between of
two organisms through non-reproductive mode in contrast to the traditional reproduction
(vertical transfer through sexual or asexual reproduction). It results to the change in the genotype
of the recipient cell due to incorporation of newly acquired DNA into the recipient genome
through either recombination or insertion.
The transformants can be identified using a selective medium. The identification of the
transformants (containing the vector/foreign molecule) require the presence of a suitable marker
genes on the vector molecule, whose expression provides a means of identifying cells containing
it (transformants). The most popular marker used for it are: a) Antibiotic resistant gene, b) colour
substance developing gene, c) galactosidase gene complementation.
In the given experiment 3 markers were used:
a) The vector (pGLO plasmid) contain the gene for Green Fluorescent Protein (GFP) and the
ampicillin resistance gene (which codes for beta-lactamase). Because it shares a bidirectional
promoter with a gene for metabolizing arabinose, the GFP gene is expressed in the presence of
arabinose, which makes the transgenic organism express its fluorescence under UV light. GFP
can be induced in bacteria containing the pGLO plasmid by growing them on +arabinose plates.
b) Ampicillin antibiotic, inhibits the cell wall synthesis. It acts as a competitive inhibitor of the
enzyme transpeptidase.
c) Arabinose (as a carbon source for bacteria).
The genetic transformation resulting to the change the phenotype of the organism (E. coli in case
of the given experiment). The results found after the transformation experiment would be in
observed in 4 type of plates.
1) Control plate (Growth medium only)
2) Control plate (Growth medium+ampicillin)
3) Transformant plate (Growth medium+ampicillin+pGLO)
4) Transformant plate (Growth medium+ampicillin+pGLO+arabinose)
The observations of control plate (E. coli cells before transformantion experiment) might provide
reference data to determine if any genetic transformation has occurred. Thus, we have to record
the a) Number of colonies, b) Size of the colony, c) Color of the colonies, d) Distribution of the
colonies on the plate, e) Visible appearance when viewed with ultraviolet (UV) light), f) The
ability of the cells to survive and multiply in the presence of an antibiotic ampicillin.
Expected Results:
*Plate 1: The results would obtained that countable (between 30-300), small, circular, shiny,
colourless colonies of E. coli were distributed throught out the plate. They would be non-
fluorescent when obsered under UV-light.
*Plate 2: No colonies will be developed in the presence of ampicillin because the E. coli is non
resistant .
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part.pdf
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part will be together in the
solution
NaClO =================> Na+ + ClO-
[It is similar with sodium sulphate, it is also an oxy acid salt. so the sulphur and O will be
together]
Na2SO4 =========> 2 Na+ + SO42- ]
Solution
NaClO is an oxyacid salt. So when it disassociate, the oxy acid part will be together in the
solution
NaClO =================> Na+ + ClO-
[It is similar with sodium sulphate, it is also an oxy acid salt. so the sulphur and O will be
together]
Na2SO4 =========> 2 Na+ + SO42- ].
Answer A] Water solventates any substances that .pdf
Answer: A] Water solventates any substances that may dissociate into charged ions.
Taking salt for example (NaCl), the water will allow the solid salt to dissociate into Na+ and Cl-
ions. How does it do this chemically? Water molecules are polar. Hence, they surround the ions
in something called a hydration shell. Water is H2O, the hydrogens have a positive dipole
whereas the oxygen has a negative dipole. The negative sides of the water molecule (oxygen)
will all surround and face the positive charged Na+ ions. The positive sides of the water
molecule (hydrogens) will surround the negatively charged Cl- ions. B] A hydrogen bond is
formed between an hydrogen atom on one molecule and an electronegative atom on another
molecule (oxygen, flourine, nitrogen, etc.). The hydrogen atom on the first molecule must be
covalently binded to an electronegative atom as well. Overall hydrogen bond: O-H...O (dotted
line is the hydrogen bond). In water, these bonds are rapidly formed and destroyed. It is not a
covalent bond, but sort of an interaction force between water molecules. Physically, this may be
seen with ice. Ice forms crystal-like structures due to the neat organization of these hydrogen
interactions. C] Water will tend to isolate hydrophobic substances. Water is polar and hence
likes to mix with other polar substances. Since hydrophobic substances are non-polar, the water
will physically separate out the hydrophobic substances in solution. In chemical terms, the water
will just form clatherates (organized shells of water) around any hydrophobic substance. This can
be seen if you add oil to water. It just doesn\'t mix.
Solution
Answer: A] Water solventates any substances that may dissociate into charged ions.
Taking salt for example (NaCl), the water will allow the solid salt to dissociate into Na+ and Cl-
ions. How does it do this chemically? Water molecules are polar. Hence, they surround the ions
in something called a hydration shell. Water is H2O, the hydrogens have a positive dipole
whereas the oxygen has a negative dipole. The negative sides of the water molecule (oxygen)
will all surround and face the positive charged Na+ ions. The positive sides of the water
molecule (hydrogens) will surround the negatively charged Cl- ions. B] A hydrogen bond is
formed between an hydrogen atom on one molecule and an electronegative atom on another
molecule (oxygen, flourine, nitrogen, etc.). The hydrogen atom on the first molecule must be
covalently binded to an electronegative atom as well. Overall hydrogen bond: O-H...O (dotted
line is the hydrogen bond). In water, these bonds are rapidly formed and destroyed. It is not a
covalent bond, but sort of an interaction force between water molecules. Physically, this may be
seen with ice. Ice forms crystal-like structures due to the neat organization of these hydrogen
interactions. C] Water will tend to isolate hydrophobic substances. Water is polar and hence
likes to mix with other polar substances. Since hydr.
In this case HF is the most ionic because H and F have thelarges.pdf
In this case HF is the most ionic because H and F have thelargest difference in electronegativity.
The value for H is2.1; the value for F is 4.0. The difference is 1.9. All of the other options given
have a lower difference.
Solution
In this case HF is the most ionic because H and F have thelargest difference in electronegativity.
The value for H is2.1; the value for F is 4.0. The difference is 1.9. All of the other options given
have a lower difference..
Here some of the reasons why list instead of a vector(i) Storing .pdf
Here some of the reasons why list instead of a vector:
(i) Storing large data, that is if few elements contains more data then it is inefficient in storing
into
vectors. But this can easily done using lists.
(ii) Iterations, insertning and deleting will take more time in vectors comparitively to lists.
(iii) In lists searching and traversing are inconsequential.
(iv) Splice and merge methods are not available in vectors. If we try to do using vectors then
operation will be more expensive.
Solution
Here some of the reasons why list instead of a vector:
(i) Storing large data, that is if few elements contains more data then it is inefficient in storing
into
vectors. But this can easily done using lists.
(ii) Iterations, insertning and deleting will take more time in vectors comparitively to lists.
(iii) In lists searching and traversing are inconsequential.
(iv) Splice and merge methods are not available in vectors. If we try to do using vectors then
operation will be more expensive..
(1)Through antibiotic resistance, bacteria will resist the action of.pdf
(1)Through antibiotic resistance, bacteria will resist the action of antibiotic and there are
different ways through which bacteria will become antibiotic resistance.
Conjugation: Here the resistant bacteria will transfer genetic material responsible for antibiotic
resistance to another bacterium by forming direct contact with them through sex pilus.
Transduction: The resistance traits of one bacterium will get packed into the head portion of the
virus (bacteriophage) and the release phage progeny from these antibiotic resistant bacteria will
infect susceptible bacteria and transfer the antibiotic resistance genes to them.
Transformation: Here bacteria capable of undergoing transformation will take up short fragments
of naked DNA responsible for antibiotic resistance from the surrounding environment by direct
uptake.
By undergoing genetic mutation: By undergoing genetic mutations, the bacteria will produce
chemicals which will eliminate the action of antibiotics and in some cases the target will be
modified to escape from antibiotic action.
(2) The genetic code is universal and bacterial cell will try to translate the gene to produce the
protein, but the proteins will not similar to human proteins because
Bacteria are lacking splicing mechanism it cannot splice out the introns from exons and thus it
will try to transcribe and translate the gene with both introns and exons giving different proteins.
In addition to this the proteins produced by bacteria will be different form human cells because,
bacteria used 70S ribosomes and the translation it does is different from the human ribosome and
another thing is codon usage is different in bacteria and we will have different genomic tRNA
pool and the post-translation modification present in human cells is missing in bacteria.
(3) Not all bacteria present in environment can be culturable, what you can see through
microscope will far outweigh the number of colonies that grow on a Petri plate and this shows
that you cannot culture all the bacteria. Through sophisticated microscopic techniques you can
identify the unculturable bacteria and through molecular techniques, such as metagenomic
sequencing will help you to get more details about this.
Solution
(1)Through antibiotic resistance, bacteria will resist the action of antibiotic and there are
different ways through which bacteria will become antibiotic resistance.
Conjugation: Here the resistant bacteria will transfer genetic material responsible for antibiotic
resistance to another bacterium by forming direct contact with them through sex pilus.
Transduction: The resistance traits of one bacterium will get packed into the head portion of the
virus (bacteriophage) and the release phage progeny from these antibiotic resistant bacteria will
infect susceptible bacteria and transfer the antibiotic resistance genes to them.
Transformation: Here bacteria capable of undergoing transformation will take up short fragments
of naked DNA responsible for antibio.
D. Role Based Access Control (RBAC) allows specific people to be ass.pdf
D. Role Based Access Control (RBAC) allows specific people to be assigned to specific roles
with specific privileges. It allows access to be assigned to groups and works well where there are
high levels of turnover.
Solution
D. Role Based Access Control (RBAC) allows specific people to be assigned to specific roles
with specific privileges. It allows access to be assigned to groups and works well where there are
high levels of turnover..
CreateTable.m filefunction multiplicationTable = CreateTable(baseV.pdf
CreateTable.m file
function multiplicationTable = CreateTable(baseValue)
multiplicationTable = zeros(1,6);
for i=0:5
multiplicationTable(i+1) = 2*i;
end
end
Solution
CreateTable.m file
function multiplicationTable = CreateTable(baseValue)
multiplicationTable = zeros(1,6);
for i=0:5
multiplicationTable(i+1) = 2*i;
end
end.
Capacity of the program memory is 0.248 kb.SolutionCapacity of.pdf
Capacity of the program memory is 0.248 kb.
Solution
Capacity of the program memory is 0.248 kb..
BrO- could not disproportionateSolutionBrO- could not dispropo.pdf
BrO- could not disproportionate
Solution
BrO- could not disproportionate.
Binary Search works on the concept of calculating the mid index valu.pdf
Binary Search works on the concept of calculating the mid index value, and will divide the actual
list to half of its size, if not found at mid position.
If the element is found in the first comparison itself, therefore, the situation is considered the best
case, and the number of comparisons required in this scenario is just ONE. And the time
complexity of binary search in best case O(1).
If the element is found in the last comparison, or if the element is not found in the list, the
situation is considered the worst case, and the number of comparisons required in this scenario is
log n, where n is the number of elements in the list. And the time complexity of binary search in
worst case O(log n).
Therefore, the amounts of steps in a binary search change its time complexity. And the amount
of steps required is directly proportional to the complexity(number of comparisons) of the binary
search algorithm.
If you need any refinements, just get back to me.
Solution
Binary Search works on the concept of calculating the mid index value, and will divide the actual
list to half of its size, if not found at mid position.
If the element is found in the first comparison itself, therefore, the situation is considered the best
case, and the number of comparisons required in this scenario is just ONE. And the time
complexity of binary search in best case O(1).
If the element is found in the last comparison, or if the element is not found in the list, the
situation is considered the worst case, and the number of comparisons required in this scenario is
log n, where n is the number of elements in the list. And the time complexity of binary search in
worst case O(log n).
Therefore, the amounts of steps in a binary search change its time complexity. And the amount
of steps required is directly proportional to the complexity(number of comparisons) of the binary
search algorithm.
If you need any refinements, just get back to me..
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; .pdf
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; oxide ion d 14 C6 ; Carbon
Solution
a Lithium ; 7Li3 b Calcium +2 ion c O -2 ion ; oxide ion d 14 C6 ; Carbon.
Answer A. H2S serves as electron donorChemolithotrophs are difined.pdf
Answer A. H2S serves as electron donor
Chemolithotrophs are difined as the organisms that can produce metabolic useful energy by
oxidatio of inorganic compounds. Chemolithotrophic life is possible in the presence as well as in
the absence of molecular oxygen.
Thiobacillus and Chlorobium bacteria are chemolithotrops. The chemilithotrops a use
H2S(inorganic compounds) as electron donor rather that organic compounds. As
Solution
Answer A. H2S serves as electron donor
Chemolithotrophs are difined as the organisms that can produce metabolic useful energy by
oxidatio of inorganic compounds. Chemolithotrophic life is possible in the presence as well as in
the absence of molecular oxygen.
Thiobacillus and Chlorobium bacteria are chemolithotrops. The chemilithotrops a use
H2S(inorganic compounds) as electron donor rather that organic compounds. As.
Answer (1)Pros of investing directly in individual stocks- If inves.pdf
Answer (1):Pros of investing directly in individual stocks- If investors directly invest into growth
stocks, their capital grows very fast as the stocks go up. They also get direct dividend from the
company. They do not have to pay any entry and exit load. They can directly expose to the stock
market.
Cons of investing directly in individual stocks- Investors are directly exposed to risk, pertaining
to particular stock, some stocks are more risky and some are less. Investing directly into
individual stocks does not provide benefit of diversification. Bonds\' prices get fluctuated over
the period of time. It is a risky move.
Pros of Investing indirectly through a mutual fund- Investing into mutual funds gives these
benefits-
Cons of Investing indirectly through a mutual fund- Investing into mutual funds has its own
disadvantage also. It charges some hidden fees and entry and exit loan.
Answer (2): Difference between alpha and beta risk- is as following-
Answer (4): After the slow down in economy and financial markets, new act were being passed
and investors are getting back the confidence in the stock markets. People invested into growth
mutual funds as the market was at lower level. Their mindset was bullish towards the market so
they invested into equities as well as mutual funds for getting future returns. Expenses were also
reduced in 2009.
Answer (5): Difference between the SEC and FINRA as regulators-
Solution
Answer (1):Pros of investing directly in individual stocks- If investors directly invest into growth
stocks, their capital grows very fast as the stocks go up. They also get direct dividend from the
company. They do not have to pay any entry and exit load. They can directly expose to the stock
market.
Cons of investing directly in individual stocks- Investors are directly exposed to risk, pertaining
to particular stock, some stocks are more risky and some are less. Investing directly into
individual stocks does not provide benefit of diversification. Bonds\' prices get fluctuated over
the period of time. It is a risky move.
Pros of Investing indirectly through a mutual fund- Investing into mutual funds gives these
benefits-
Cons of Investing indirectly through a mutual fund- Investing into mutual funds has its own
disadvantage also. It charges some hidden fees and entry and exit loan.
Answer (2): Difference between alpha and beta risk- is as following-
Answer (4): After the slow down in economy and financial markets, new act were being passed
and investors are getting back the confidence in the stock markets. People invested into growth
mutual funds as the market was at lower level. Their mindset was bullish towards the market so
they invested into equities as well as mutual funds for getting future returns. Expenses were also
reduced in 2009.
Answer (5): Difference between the SEC and FINRA as regulators-.
Ans BIOSBIOS verifies if theoptical drive is recognised by the sys.pdf
Ans BIOS
BIOS verifies if theoptical drive is recognised by the system
Device Manager is used to uninstall/reinstall the device driver for optical drive
Solution
Ans BIOS
BIOS verifies if theoptical drive is recognised by the system
Device Manager is used to uninstall/reinstall the device driver for optical drive.
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