double the amount of water you are dissolving it .pdf
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double the amount of water you are dissolving it into Solution double the amount of water you are dissolving it into.
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(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
(1) c. Two-tailed test
(2)d. Ho: Men and women are the same in their political understanding.
(3)b. Ho:
Solution
(1) c. Two-tailed test
(2)d. Ho: Men and women are the same in their political understanding.
(3)b. Ho:.
We use a base (in this case sodium bicarbonate) during the separatio.pdf
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer
from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer;
when deprotonated the compounds become more soluble in the aqueous layer then the non-
aqueous and \"jump\" into the aqueous layer.
Solution
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer
from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer;
when deprotonated the compounds become more soluble in the aqueous layer then the non-
aqueous and \"jump\" into the aqueous layer..
While computing the dilluted earnings per share Dividend paid on Pre.pdf
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted
from the calculation.
Solution
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted
from the calculation..
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.
Recommended
(1) c. Two-tailed test(2)d. Ho Men and women are the same in .pdf
(1) c. Two-tailed test
(2)d. Ho: Men and women are the same in their political understanding.
(3)b. Ho:
Solution
(1) c. Two-tailed test
(2)d. Ho: Men and women are the same in their political understanding.
(3)b. Ho:.
We use a base (in this case sodium bicarbonate) during the separatio.pdf
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer
from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer;
when deprotonated the compounds become more soluble in the aqueous layer then the non-
aqueous and \"jump\" into the aqueous layer.
Solution
We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer
from the aqueous. The base deprotonates unwanted compounds found in the non-aqueous layer;
when deprotonated the compounds become more soluble in the aqueous layer then the non-
aqueous and \"jump\" into the aqueous layer..
While computing the dilluted earnings per share Dividend paid on Pre.pdf
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted
from the calculation.
Solution
While computing the dilluted earnings per share Dividend paid on Preferred stock is omitted
from the calculation..
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.
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
C. Router(config-if)#ip access-group 110 in is the right answer.
ip-access-group is a command in interface configuration mode which is used to place an access
list on an interface.
Solution
C. Router(config-if)#ip access-group 110 in is the right answer.
ip-access-group is a command in interface configuration mode which is used to place an access
list on an interface..
C is falseSolutionC is false.pdf
The document states that the statement "C is false" is false, meaning that C must be true. It provides no other information or context.
B). 12SolutionB). 12.pdf
El documento no contiene información sustancial para resumir. Consiste en un encabezado "Solution" seguido de "B). 1/2", lo que sugiere que es una parte de una solución más larga o un problema, pero no proporciona detalles sobre el tema o contenido.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
The document provides code for implementing a linked list data structure in C++. It includes class definitions for a ListInterface, Node, and LinkedList class. The Node class represents individual nodes in the linked list. The LinkedList class implements the linked list and inherits from ListInterface. It uses Node objects to store and connect list elements. Helper functions and exceptions are also defined to support linked list operations like insertion, removal and accessing elements.
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
Answer:
The nucleus is separated from the cytoplasm by a phospholipid bilayer known as nuclear
envelope. Transport of macromolecules across the nuclear membrane takes place through
specialized nuclear pores. Proteins found in the nucleus are synthesized in the cytoplasm and
imported into the nucleus through nuclear pore complexes. Such proteins contain a nuclear-
localization signal (NLS) that directs their selective transport into the nucleus.4 proteins are
required for nuclear localization- importin , importin , NTF-2 & Ran.
Answer-A:
It has been previously said that nuclear localization signal(NLS) is located within the amino acid
sequence of a protein which needs to be localized in the nucleus. Importins actually identify this
signal within the protein and interacts with it and facilitates the transfer of nuclear protein from
cytosol to nucleus. If the NLS is not present and has already been deleted, the importins won\'t
be able to identify the protein and transport them into the nucleus. In this present case, deletion
mutation of the protein has been performed to identify the part of the amino acid sequence acting
as NLS. In addition leptomycin B has also been added which prevents nuclear export and is not
relevant with respect to nuclear localization and NLS.
NLS is located in the amino acid range- 240-357. It is so because it can be observed from the
available results that localization into the nucleus takes place for deletion mutants having amino
acids - 1-576, 115- 576, 240-576 , 240- 357 & 115-357.
Conclusions:
115-576 - NLS is not located in amino acid sequence 1-115
240-576 - NLS is not located within the amino acid sequence - 1-239
240-357 - NLS is not located in amino acid sequence 1-240 & 358-576
115-357 - NLS is not located within 1-115 & 358-576
Localization takes place in all the above cases and the deletion mutants are always found in the
nucleus and they may or may not be found in the cytoplasm. From assessment of these
conclusions, it can only be inferred that NLS is located in the range 240-357.
Answer B:
There are some “shuttling” proteins that contain a nuclear-export signal (NES) that stimulates
their export from the nucleus to the cytoplasm through nuclear pores, in addition to an NLS that
results in their reuptake into the nucleus. Export of proteins from the nucleus is mediated by
proteins known as Exportins which first forms a complex with Ran·GTP and then binds the NES
in a cargo protein. Once it crosses the nuclear pore, the Ran GAP associated with the NPC
cytoplasmic filaments stimulates conversion of Ran·GTP to Ran·GDP. The accompanying
conformational change in Ran leads to dissociation of the complex. The NES-containing cargo
protein is released into the cytosol, while exportin 1 and Ran·GDP are transported back into the
nucleus through NPCs.
In this case, NES is located in the amino acid region 115-240. It is speculated to be so because:
The deletion mutant bearing amino acids 115-576 possess the NES since i.
The History of Stoke Newington Street Names
Presented at the Stoke Newington Literary Festival on 9th June 2024
www.StokeNewingtonHistory.com
Introduction to AI for Nonprofits with Tapp Network
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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.
C. Router(config-if)#ip access-group 110 in is the right answer.ip.pdf
C. Router(config-if)#ip access-group 110 in is the right answer.
ip-access-group is a command in interface configuration mode which is used to place an access
list on an interface.
Solution
C. Router(config-if)#ip access-group 110 in is the right answer.
ip-access-group is a command in interface configuration mode which is used to place an access
list on an interface..
C is falseSolutionC is false.pdf
The document states that the statement "C is false" is false, meaning that C must be true. It provides no other information or context.
B). 12SolutionB). 12.pdf
El documento no contiene información sustancial para resumir. Consiste en un encabezado "Solution" seguido de "B). 1/2", lo que sugiere que es una parte de una solución más larga o un problema, pero no proporciona detalles sobre el tema o contenido.
AnswerNote LinkedList.cpp is written and driver program main.cpp.pdf
The document provides code for implementing a linked list data structure in C++. It includes class definitions for a ListInterface, Node, and LinkedList class. The Node class represents individual nodes in the linked list. The LinkedList class implements the linked list and inherits from ListInterface. It uses Node objects to store and connect list elements. Helper functions and exceptions are also defined to support linked list operations like insertion, removal and accessing elements.
AnswerThe nucleus is separated from the cytoplasm by a phospholip.pdf
Answer:
The nucleus is separated from the cytoplasm by a phospholipid bilayer known as nuclear
envelope. Transport of macromolecules across the nuclear membrane takes place through
specialized nuclear pores. Proteins found in the nucleus are synthesized in the cytoplasm and
imported into the nucleus through nuclear pore complexes. Such proteins contain a nuclear-
localization signal (NLS) that directs their selective transport into the nucleus.4 proteins are
required for nuclear localization- importin , importin , NTF-2 & Ran.
Answer-A:
It has been previously said that nuclear localization signal(NLS) is located within the amino acid
sequence of a protein which needs to be localized in the nucleus. Importins actually identify this
signal within the protein and interacts with it and facilitates the transfer of nuclear protein from
cytosol to nucleus. If the NLS is not present and has already been deleted, the importins won\'t
be able to identify the protein and transport them into the nucleus. In this present case, deletion
mutation of the protein has been performed to identify the part of the amino acid sequence acting
as NLS. In addition leptomycin B has also been added which prevents nuclear export and is not
relevant with respect to nuclear localization and NLS.
NLS is located in the amino acid range- 240-357. It is so because it can be observed from the
available results that localization into the nucleus takes place for deletion mutants having amino
acids - 1-576, 115- 576, 240-576 , 240- 357 & 115-357.
Conclusions:
115-576 - NLS is not located in amino acid sequence 1-115
240-576 - NLS is not located within the amino acid sequence - 1-239
240-357 - NLS is not located in amino acid sequence 1-240 & 358-576
115-357 - NLS is not located within 1-115 & 358-576
Localization takes place in all the above cases and the deletion mutants are always found in the
nucleus and they may or may not be found in the cytoplasm. From assessment of these
conclusions, it can only be inferred that NLS is located in the range 240-357.
Answer B:
There are some “shuttling” proteins that contain a nuclear-export signal (NES) that stimulates
their export from the nucleus to the cytoplasm through nuclear pores, in addition to an NLS that
results in their reuptake into the nucleus. Export of proteins from the nucleus is mediated by
proteins known as Exportins which first forms a complex with Ran·GTP and then binds the NES
in a cargo protein. Once it crosses the nuclear pore, the Ran GAP associated with the NPC
cytoplasmic filaments stimulates conversion of Ran·GTP to Ran·GDP. The accompanying
conformational change in Ran leads to dissociation of the complex. The NES-containing cargo
protein is released into the cytosol, while exportin 1 and Ran·GDP are transported back into the
nucleus through NPCs.
In this case, NES is located in the amino acid region 115-240. It is speculated to be so because:
The deletion mutant bearing amino acids 115-576 possess the NES since i.
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OSI MODELIt has 7 layersINTERNET MODELIt has5 LayersDOD MO.pdf
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Option 2 namely TPP contains a thiazolium ring is correct. This is b.pdf
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