Ans.C.Cysteinein assimilatory sulfate reduction the sulfate is tak.pdf
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Ans.C.Cysteine in assimilatory sulfate reduction the sulfate is taken up by the roots of the plant is transported to the shoots. and in the shoots these sulfate is reduced by a series of enzymes. in the shoots the sulfate is activated to APS (adenosine 5\'-Phosphosulfate) then reduced to sulfite by the enzyme APS reductase, and this sulfite is reduced further to sulfide and the sulfide reacts with O- acetylserine and forms Cysteine. Cysteine further forms methionine. the cysteine and methionine have importance and involves in structure and functions of various enzymes. Solution Ans.C.Cysteine in assimilatory sulfate reduction the sulfate is taken up by the roots of the plant is transported to the shoots. and in the shoots these sulfate is reduced by a series of enzymes. in the shoots the sulfate is activated to APS (adenosine 5\'-Phosphosulfate) then reduced to sulfite by the enzyme APS reductase, and this sulfite is reduced further to sulfide and the sulfide reacts with O- acetylserine and forms Cysteine. Cysteine further forms methionine. the cysteine and methionine have importance and involves in structure and functions of various enzymes..
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Sulphate Assimilation
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Biochemistry of sulphur reduction
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plants and other microorganisms
Sulphur metabolism
This document provides information on sulfur metabolism in plants. It discusses sulfate uptake and transport, sulfur activation through ATP sulfurylase and APS kinase, and sulfate reduction via APS sulfotransferase. Key regulation points are the enzymes ATP sulfurylase, APS reductase, and serine acetyltransferase which can limit the pathways when overexpressed, leading to increased sulfur levels in plants. The document also outlines subcellular compartmentation of sulfur processes between the plasma membrane, cytoplasm and chloroplast.
Sulfate assimilation
Sulfate assimilation is the process by which plants convert inorganic sulfate into organic sulfur compounds. There are two pathways for sulfate assimilation: reductive and sulfation. The reductive pathway involves three steps - activation of sulfate to APS, reduction of APS to sulfide, and incorporation of sulfide into cysteine. Cysteine can then be used to synthesize other organic sulfur compounds like methionine, glutathione, and S-adenosyl methionine. The sulfation pathway involves phosphorylation of APS to PAPS, which can then add sulfate to organic molecules via sulfotransferases. Sulfur-containing compounds are exported throughout the plant via phloem.
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Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
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cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are
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Part4. The Anaphase II is responsible for the separation of sister chromatids to form
chromosomes. Sister chromatids are chromosomes of either maternal or paternal.
Part5. During Prophase I homologous chromosomes physically pair up. This phase includes
paring up of chromosomes and homologous recombination.
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chromosomes. Sister chromatids are chromosomes of either maternal or paternal.
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paring up of chromosomes and homologous recombination..
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
Performance Ratios
Market Value Added (MVA):
MVA=(Company’s market value) minus(Invested capital)
Market to book=Market value of shares/Book value of shares
Data on market value are not available.
Hence the above performance ratios cannot be calculated
Profitability Ratios
Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity)
Dividend information is not available
Return on capital(ROC)=(Net Income)/Average (Debt+Equity)
Return on asset(ROA)=Net Income/(Average total asset)
Return on Equity(ROE)=Net Income/(Average Equity)
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B
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Dec 31 2015
Dec 31 2014
2015Average
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D
Debt
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Equity
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F
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G
Total asset
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Return on capital(ROC)
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Return on asset(ROA)
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Operating Profit margin=Operating Profit/Sales
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Days in inventory=365/inventory turnover(cogs)
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A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Operating Income
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E
Sales
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F
Total assets
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25509
25608
G
Cost of goods sold
15383
H
Total inventory
3518
3706
3612
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Profit margin
0.16974962
J=D/E
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0.229437801
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16.97%
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22.94%
Asset turnover
1.18220868
Inventory turnover(cogs)
4.25885936
Days in inventory
85.7036989
Days
Inventory turnover (Sales)
8.38150609
Days sales in inventory
43.5482592
Days
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Debt
9246
8432
8839
E
Equity
11747
13142
12444.5
F
Debt+Equity
20993
21574
21283.5
G
Total asset
25707
25509
25608
H=C/F
Return on capital(ROC)
0.241454648
I=C/G
Return on asset(ROA)
0.200679475
J=C/E
Return on Equity(ROE)
0.412953514
Solution
Performance Ratios
Market Value Added (MVA):
MVA=(Company’s market value) minus(Invested capital)
Market to book=Market value of shares/Book value of shares
Data on market value are not available.
Hence the above performance ratios cannot be calculated
Profitability Ratios
Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity)
Dividend information is not available
Return on capital(ROC)=(Net Income)/Average (Debt+Equity)
Return on asset(ROA)=Net Income/(Average total asset)
Return on Equity(ROE)=Net Income/(Average Equity)
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Debt
9246
8432
8839
E
Equity
11747
13142
124.
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P = 212option ASolutionP = 212option A.pdf
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Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
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Solution
Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains
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Solution
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= 2.416
Solution
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Solution
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hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
hi Q is nearly done,just give me dia of shaft
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hi Q is nearly done,just give me dia of shaft.
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O2 is simply oxygen gas. Gases are usually made up of two of the same elements.
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Solution
O2 is simply oxygen gas. Gases are usually made up of two of the same elements.
(Cl2 = chlorine gas, N2=nitrogen gas).
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
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repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage
virus or plasmid.
Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the
CRISPR/Cpf1 system of Francisella novicida.
Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded
DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent
motif and requiring only a CRISPR RNA (crRNA) for successful targeting.
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system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response
in human host. This immune system in bacteria play role in the regulation of those genes, which
are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR
immune system also helps bacterium to bypass the human immune system.
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complex at the correct spot on the double helix to cleave target DNA. CRISPR/Cpf1 systems
activity has three stages:
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the CRISPR array.
2. Formation of crRNAs: processing of pre-cr-RNAs producing of mature crRNAs to guide the
Cas protein.
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target DNA sequence.
Solution
CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each
repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage
virus or plasmid.
Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the
CRISPR/Cpf1 system of Francisella novicida.
Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded
DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent
motif and requiring only a CRISPR RNA (crRNA) for successful targeting.
A recent study says that this system has the ability to regulate endogenous genes, responsible for
pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR
system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response
in human host. This immune system in bacteria play role in the regulation of those genes, which
are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR
immune system also helps bacterium to bypass the human immune system.
The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the
complex at the correct spot on the double helix to cleave target DNA. CRISP.
A Survey of Techniques for Maximizing LLM Performance.pptx
A Survey of Techniques for Maximizing LLM Performance
Natural birth techniques - Mrs.Akanksha Trivedi Rama University
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More from anwarsadath111
There are several ways in which the gene regulation in eukaryotes di.pdf
Gene regulation differs between prokaryotes and eukaryotes in several key ways. In prokaryotes, DNA is not enclosed in a nucleus and transcription and translation occur simultaneously in the cytoplasm. In eukaryotes, DNA is packaged into nucleosomes and contained within the nucleus, where transcription occurs to produce RNA that is then transported to the cytoplasm for translation. Additionally, eukaryotes have multiple RNA polymerases and levels of gene expression regulation, whereas prokaryotes have a single RNA polymerase and regulate gene expression primarily at transcription.
The water must be boiled to remove traces of dissolved carbon dioxid.pdf
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S
Solution
The water must be boiled to remove traces of dissolved carbon dioxide (CO2).
CO2 dissolves in water to form carbonic acid (H2CO3):
CO2 + H2O => H2CO3
H2CO3 is a weak acid and partially ionizes to give H+ ions:
H2CO3 => H+ + HCO3-
HCO3- => H+ + CO32-
The H+ ions can decompose thiosulfate ions and reduce the concentration of sodium thiosulfate
according to the reaction:
2 S2O32- + H+ => HSO3- + S.
adsorb onto silica gel Solu.pdf
The document discusses adsorbing a solution onto silica gel. Silica gel is used to adsorb or absorb a solution. Adsorption of the solution onto the silica gel is mentioned.
Solution Plants show different types of symptoms when there is an.pdf
Solution
:
Plants show different types of symptoms when there is any nutritional vacancy in
plants.Nitrogen is one of the most common nutrient deficiency in plants.Nitrogen deficiency
leads to decrease in growth.Plants are mainly short.Some leaves not able to produce chlorophyll
which leads to green color to them.So it give rise to yellow color in leaves..
SF4SolutionSF4.pdf
This very short document contains the acronym "SF4" repeated three times without any other context or explanation. It does not provide enough information to generate a multi-sentence summary.
Quick ratio = (Cash + accounts receivable)Accounts payableFor 200.pdf
Quick ratio = (Cash + accounts receivable)/Accounts payable
For 2008 the quick ratio = (140 + 780)/1120
= 920/1120
= .82
A .82
Solution
Quick ratio = (Cash + accounts receivable)/Accounts payable
For 2008 the quick ratio = (140 + 780)/1120
= 920/1120
= .82
A .82.
Ques-1) What is the most likely cause of a throat infection in a 6-y.pdf
Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she
require treatment?
Throat infections caused by streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given).
The infection is strep throat and it is caused by streptococcal bacteria (group B). These bacteria
infect the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting,
body aches. Throat infections caused by streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Apart from antibiotics, the major future intervention to
treat strep throat is to implement vaccination to boost innate immune levels in the infant because
antibiotics such as macrolides have potential inhibitor effects on the growth of the useful gutfolra
in infants & toddlers. Probiotics, gicing more Vitamin D & C and natural remedies are major
future interventions to avoid strep throat in children. Immunity stimulation is specific future
intervention to combat strep throat in infants as vaccination promotes immune factors to reduce
the bacteria that cause inflammation of the blood vessels near throat.
The infection is strep throat and streptococcal bacteria (group B) cause it. These bacteria infect
the upper respiratory tract and causes fever, sore throat, abdominal pain, fatigue, vomiting, body
aches. In a Gram stain of sputum specimen of beta hemolytic streptococci, \"squamous epithelial
cell type\" of body cell provides an indication that the sputum specimen represents material from
an active infection. Other cells are present in the sputum during the active infection are
polymorphonuclear leukocytes (PMNs)
Ques-2) What is the significance of facial swelling? List the possible causes of facial swelling?
1). Throat infections caused by streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms
(antibiotics such as penicillin, corticosteroids and diuretics are to be given).
2). The bacteria causes inflammation of the blood vessels supplying glomeruli, which impairs the
glomerular filtration. Decreased urine filtration causes edema (swelling of body organs).
Ques-3) What are the possible causes of dark urine?
Dark urine is mainly due to \"presence of RBC\" in the blood. Urine was positive for 3+ protein
and for3+ blood by dipstick. The examination of urinary sediment revealed numerous RBCs,
many of which were dismorphic. Red blood cell casts were also present.
Solution
Ques-1) What is the most likely cause of a throat infection in a 6-year old child? Does she
require treatment?
Throat infections caused by streptococcal bacteria may cause \"post-streptococcal
glomerulonephritis\" if left untreated. Yes, the child needs treatment to treat these symptoms
(antibioti.
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic
cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are
undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So
body’s maximum cells like somatic cells undergo mitosis rather than meiosis.
Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this
section by multiple sites of duplication.
Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division.
These are non-identical sister chromatids.
Part4. The Anaphase II is responsible for the separation of sister chromatids to form
chromosomes. Sister chromatids are chromosomes of either maternal or paternal.
Part5. During Prophase I homologous chromosomes physically pair up. This phase includes
paring up of chromosomes and homologous recombination.
Solution
Part1. Option 3rd; Somatic cells such as liver cells cannot undergo meiosis cell division. Mitotic
cells are required to be diploid in number of chromosomes. Sperm cells and germ cells are
undergo fusion with other reproductive cells, thus they get diploid chromosomes after fusion. So
body’s maximum cells like somatic cells undergo mitosis rather than meiosis.
Part2. The S phase includes the duplication of each chromosome. All DNA has duplicated in this
section by multiple sites of duplication.
Part3. Maternal and parental chromosomes are separated in Anaphase I of meiosis cell division.
These are non-identical sister chromatids.
Part4. The Anaphase II is responsible for the separation of sister chromatids to form
chromosomes. Sister chromatids are chromosomes of either maternal or paternal.
Part5. During Prophase I homologous chromosomes physically pair up. This phase includes
paring up of chromosomes and homologous recombination..
Performance RatiosMarket Value Added (MVA)MVA=(Company’s market.pdf
Performance Ratios
Market Value Added (MVA):
MVA=(Company’s market value) minus(Invested capital)
Market to book=Market value of shares/Book value of shares
Data on market value are not available.
Hence the above performance ratios cannot be calculated
Profitability Ratios
Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity)
Dividend information is not available
Return on capital(ROC)=(Net Income)/Average (Debt+Equity)
Return on asset(ROA)=Net Income/(Average total asset)
Return on Equity(ROE)=Net Income/(Average Equity)
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Debt
9246
8432
8839
E
Equity
11747
13142
12444.5
F
Debt+Equity
20993
21574
21283.5
G
Total asset
25707
25509
25608
H=C/F
Return on capital(ROC)
0.241454648
I=C/G
Return on asset(ROA)
0.200679475
J=C/E
Return on Equity(ROE)
0.412953514
Return on capital(ROC)
24.15%
Return on asset(ROA)
20.07%
Return on Equity(ROE)
41.30%
Efficiency Ratios
Profit margin=Net Profit/Sales
Operating Profit margin=Operating Profit/Sales
Asset turnover=Sales/Average Total Assets
Inventory turnover(cogs)=Cost of goods sold/Average total inventory
Days in inventory=365/inventory turnover(cogs)
Inventory turnover is also calculated from sales:
Inventory turnover (Sales)=Sales/average total inventory
Days sales in inventory=365/inventory turnover (sales)
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Operating Income
6946
E
Sales
30274
F
Total assets
25707
25509
25608
G
Cost of goods sold
15383
H
Total inventory
3518
3706
3612
I=C/E
Profit margin
0.16974962
J=D/E
Operating Profit margin
0.229437801
K=E/F
Asset turnover
1.182208685
L=G/H
Inventory turnover(cogs)
4.258859358
M=365/L
Days in inventory
85.70369889
N=E/H
Inventory turnover (Sales)
8.381506091
P=365/N
Days sales in inventory
43.54825923
Profit margin
16.97%
Operating Profit margin
22.94%
Asset turnover
1.18220868
Inventory turnover(cogs)
4.25885936
Days in inventory
85.7036989
Days
Inventory turnover (Sales)
8.38150609
Days sales in inventory
43.5482592
Days
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Debt
9246
8432
8839
E
Equity
11747
13142
12444.5
F
Debt+Equity
20993
21574
21283.5
G
Total asset
25707
25509
25608
H=C/F
Return on capital(ROC)
0.241454648
I=C/G
Return on asset(ROA)
0.200679475
J=C/E
Return on Equity(ROE)
0.412953514
Solution
Performance Ratios
Market Value Added (MVA):
MVA=(Company’s market value) minus(Invested capital)
Market to book=Market value of shares/Book value of shares
Data on market value are not available.
Hence the above performance ratios cannot be calculated
Profitability Ratios
Return on capital(ROC)=(Net Income –Dividend)/(Debt+Equity)
Dividend information is not available
Return on capital(ROC)=(Net Income)/Average (Debt+Equity)
Return on asset(ROA)=Net Income/(Average total asset)
Return on Equity(ROE)=Net Income/(Average Equity)
A
B
(A+B)/2
Dec 31 2015
Dec 31 2014
2015Average
C
Net income
5139
D
Debt
9246
8432
8839
E
Equity
11747
13142
124.
package employeeType.employee;public class Employee { private .pdf
The document defines an Employee class and subclasses for Hourly, Salary, and Commission employees. It includes methods to calculate pay, apply raises, and reset tracking periods. The main method creates sample employees, alters values, calculates pay and bonuses, applies raises, and outputs the results.
P = 212option ASolutionP = 212option A.pdf
This short document discusses a mathematical problem where P equals 2/12. The solution provided is option A. No other context or explanation is given about the nature of the problem or what option A represents as the solution.
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
OSI MODEL:
It has 7 layers
INTERNET MODEL:
It has5 Layers
DOD MODEL:
It has 4 Layers
Solution
OSI MODEL:
It has 7 layers
INTERNET MODEL:
It has5 Layers
DOD MODEL:
It has 4 Layers.
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains
pyrimidine ring which is linked to thiazole ring
Solution
Option 2 namely TPP contains a thiazolium ring is correct. This is because TPP contains
pyrimidine ring which is linked to thiazole ring.
NH4NO3 = N2O + 2 H2OMoles of NH4NO3 = 12 x moles of H2O= 12 x .pdf
NH4NO3 = N2O + 2 H2O
Moles of NH4NO3 = 1/2 x moles of H2O
= 1/2 x 0.625
= 0.3125 mol
Solution
NH4NO3 = N2O + 2 H2O
Moles of NH4NO3 = 1/2 x moles of H2O
= 1/2 x 0.625
= 0.3125 mol.
No. of shares outstanding = Total assets x weight of equity price .pdf
No. of shares outstanding = Total assets x weight of equity / price per share
= $12,020,000 x50% / $8
= 751250
Net Income = Total assets x Return on assets
= $12,020,000 x 15.10%
= 1,815,020
Earnings per share = Net Income/ No. of shares outstanding
= 1,815,020 / 751250
= 2.416
Solution
No. of shares outstanding = Total assets x weight of equity / price per share
= $12,020,000 x50% / $8
= 751250
Net Income = Total assets x Return on assets
= $12,020,000 x 15.10%
= 1,815,020
Earnings per share = Net Income/ No. of shares outstanding
= 1,815,020 / 751250
= 2.416.
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M
Solution
molarity = moles / volume = 2.0 x 10-3 mol / (18.5/1000) L = 0.11 M.
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;
public class Main {
public static int gridArray1[][] = null;
public static int gridArray2[][] = null;
public static boolean playerTurn = true;
public static int count = 0;
public static void setGridDimensions(int x_max, int y_max) {
Main.gridArray1 = new int[x_max][y_max];
Main.gridArray2 = new int[x_max][y_max];
for (int i = 0; i < x_max; i++) {
for (int j = 0; j < y_max; j++) {
gridArray1[i][j] = -1;
gridArray2[i][j] = -1;
}
}
}
static void placeShip(int starting_x, int starting_y, int length, int direction) {
count++;
int ships[] = new int[10];
int battleshipArrayPositions1[] = new int[4];
int cruiserArrayPositions1[] = new int[3];
int submarineArrayPositions1[] = new int[3];
int destroyerArrayPositions1[] = new int[2];
int carrierArrayPositions2[] = new int[5];
int battleshipArrayPositions2[] = new int[4];
int cruiserArrayPositions2[] = new int[3];
int submarineArrayPositions2[] = new int[3];
int destroyerArrayPositions2[] = new int[2];
if (playerTurn) {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray1[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray1[starting_x][i] = 0;
}
}
} else {
if (direction == 0) {
for (int i = starting_x; i < starting_x + length; i++) {
Main.gridArray2[i][starting_y] = 0;
}
} else {
for (int i = starting_y; i < starting_y + length; i++) {
Main.gridArray2[starting_x][i] = 0;
}
}
}
}
boolean isConflictingShipPlacement(int starting_x, int starting_y, int length, int direction) {
return false;
}
static int shoot(int x, int y) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (Main.gridArray2[i][j] == 0) {
return 0;
}
}
}
} else {
}
return 0;
}
boolean hasBeenAttempted(int x, int y) {
return false;
}
static void displayGrid(boolean showShips) {
int count = 0;
System.out.print(\" 0 1 2 3 4 5 6 7 8 9\");
if (showShips) {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray1[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
} else {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray2[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray2[i][j] == 0) {
System.out.print(\"@ \");
} else if (Main.gridArray2[i][j] == 1) {
System.out.print(\"+ \");
} else {
System.out.print(\"X \");
}
}
}
}
} else {
if (playerTurn) {
for (int i = 0; i < 10; i++) {
for (int j = 0; j < 10; j++) {
if (j % 10 == 0) {
System.out.print(\"\ \" + count + \" \");
count++;
}
if (Main.gridArray1[i][j] == -1) {
System.out.print(\"- \");
} else if (Main.gridArray1[i][j] == 0) {
System.out.print(\"- \");
} el.
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
hi Q is nearly done,just give me dia of shaft
Solution
hi Q is nearly done,just give me dia of shaft.
O2 is simply oxygen gas. Gases are usually made .pdf
O2 is simply oxygen gas. Gases are usually made up of two of the same elements.
(Cl2 = chlorine gas, N2=nitrogen gas)
Solution
O2 is simply oxygen gas. Gases are usually made up of two of the same elements.
(Cl2 = chlorine gas, N2=nitrogen gas).
CRISPR are segments of prokaryotic DNA containing short repetitions .pdf
CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each
repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage
virus or plasmid.
Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the
CRISPR/Cpf1 system of Francisella novicida.
Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded
DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent
motif and requiring only a CRISPR RNA (crRNA) for successful targeting.
A recent study says that this system has the ability to regulate endogenous genes, responsible for
pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR
system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response
in human host. This immune system in bacteria play role in the regulation of those genes, which
are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR
immune system also helps bacterium to bypass the human immune system.
The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the
complex at the correct spot on the double helix to cleave target DNA. CRISPR/Cpf1 systems
activity has three stages:
1. Adaptation: Cas1 and Cas2 proteins facilitate the adaptation of small fragments of DNA into
the CRISPR array.
2. Formation of crRNAs: processing of pre-cr-RNAs producing of mature crRNAs to guide the
Cas protein.
3. Interference: the Cpf1 is bound to a crRNA to form a binary complex to identify and cleave a
target DNA sequence.
Solution
CRISPR are segments of prokaryotic DNA containing short repetitions of base sequences. Each
repetition is followed by short segments of spacer DNA due to exposures to a bacteriophage
virus or plasmid.
Cpf1 was the second nuclease discovered after Cas9 nuclease,it was discovered in the
CRISPR/Cpf1 system of Francisella novicida.
Cpf1 showed several differences to Cas9 including causing a \'staggered\' cut in double stranded
DNA as opposed to the \'blunt\' cut produced by Cas9, relying on a \'T rich\' Protospacer adjacent
motif and requiring only a CRISPR RNA (crRNA) for successful targeting.
A recent study says that this system has the ability to regulate endogenous genes, responsible for
pathogenesis and virulence of Francisella novicida. It was also shown that type II CRISPR
system can repress bacterial lipoprotein transcription that leads to the pro-inflammatory response
in human host. This immune system in bacteria play role in the regulation of those genes, which
are responsible for encoding of factors that have impact in pathogenesis of bacteria. CRISPR
immune system also helps bacterium to bypass the human immune system.
The Cpf1 system consist of a Cpf1 enzyme and a guide RNA that finds and positions the
complex at the correct spot on the double helix to cleave target DNA. CRISP.
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Solution Plants show different types of symptoms when there is an.pdf
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package employeeType.employee;public class Employee { private .pdf
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OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
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molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
molarity = moles volume = 2.0 x 10-3 mol (18.51000) L = 0.11 M.pdf
import java.util.Scanner;public class Main { public static in.pdf
import java.util.Scanner;public class Main { public static in.pdf
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
O2 is simply oxygen gas. Gases are usually made .pdf
O2 is simply oxygen gas. Gases are usually made .pdf
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Strategies for Effective Upskilling is a presentation by Chinwendu Peace in a Your Skill Boost Masterclass organisation by the Excellence Foundation for South Sudan on 08th and 09th June 2024 from 1 PM to 3 PM on each day.A Strategic Approach: GenAI in Education
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
DRUGS AND ITS classification slide share
Any substance (other than food) that is used to prevent, diagnose, treat, or relieve symptoms of a
disease or abnormal condition
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