There are several ways in which the gene regulation in eukaryotes di.pdf
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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.
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There are two main types of cells - prokaryotic and eukaryotic cells. Prokaryotic cells were the earliest form of life and lack a nucleus and organelles. Eukaryotic cells are larger and more complex, with a nucleus that contains DNA organized into chromosomes and membrane-bound organelles. The key differences are that prokaryotes lack a nucleus and organelles while eukaryotes have these structures.
Types of cells notes ppt
There are two main types of cells - prokaryotic and eukaryotic cells. Prokaryotic cells were the earliest form of life and lack a nucleus and organelles. Eukaryotic cells are larger and more complex, with a nucleus that contains DNA organized into chromosomes and membrane-bound organelles. The key differences are that prokaryotes lack a nucleus and organelles while eukaryotes have these structures.
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1. What major biomolecule has the most number of functions and why?
2. How does DNA coil or become more compact? Give two reasons why it must be compacted?
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specifically coiled which is required to enter cell division.
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2. The prokaryotic genome is generally a circular piece of DNA that can exist in multiple copies and ranges in length but is at least a few million base pairs. It is packaged into the nucleoid through supercoiling facilitated by nucleoid-associated proteins.
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We use a base (in this case sodium bicarbonate) during the separation of the non-aqueous layer
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1. What major biomolecule has the most number of functions and why?
2. How does DNA coil or become more compact? Give two reasons why it must be compacted?
3. Explain what crossing over and independent assortment are. Where does each happen in
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1. What major biomolecule has the most number of functions and why?
2. How does DNA coil or become more compact? Give two reasons why it must be compacted?
3. Explain what crossing over and independent assortment are. Where does each happen in
meiosis?
2. How does DNA coil or become more compact? Give two reasons why it must be compacted?
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meiosis?
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1. There are four major types of biomolecules. Carbohydrates, proteins, lipids and nucleic acid.
Out of these, the nucleic acid has the most number of function. Nucleic acid are polymers present
in the living cells and are of two types: Ribonucleic acid (RNA) and Deoxyribonucleic acid
(DNA). DNA carry genes which are responsible for storing and transmitting the hereditary
information from one generation to the next. RNA help in protein formation and also have
functional activity as enzymes.
2. The entire length of DNA in each cell could reach up to 2 meters if measured end to end. The
DNA is coiled with the help of proteins called histones to form nucleosomes which are
compacted together to form chromatin fibre. The reason for it to be compacted : to accomodate
in the small size of the cell which has the size in micro meteres. It is also required to be
specifically coiled which is required to enter cell division.
3. Crossing Over is the process of exchange of genes between two homologous chromosomes
during sexual reproduction. This results in non-identical chromatids called recombinant
chromosomes. It happens in the pachytene stage of Prophase I of meiosis.
Independent assortment is the independent separation of homologous chromosomes. The alleles
of different genes get sorted into gametes randomly. This gives different hereditary traits equal
probability to occur together. in the offspring. It occurs in Metaphase I of meiosis..
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1. Mutations are changes in the nucleotide sequence of DNA that can arise spontaneously during DNA replication or due to damage from mutagens.
2. DNA repair enzymes work to minimize mutations by correcting errors during replication or reacting to damaged DNA.
3. If a mismatch introduced during replication is not repaired, it will become a permanent mutation when that region is replicated again.
Prokaryote genome
This document summarizes key aspects of genome organization in prokaryotes. It notes that prokaryotes like E. coli have a single circular chromosome composed of DNA that is compacted into a nucleoid structure. The DNA is highly condensed via supercoiling facilitated by enzymes like topoisomerases. Prokaryotic genomes are also typically smaller than eukaryotic ones and can contain extra DNA on plasmids that are exchanged between bacteria.
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1. In prokaryotes, the genome is located in an irregularly shaped region within the cell called the nucleoid, which is not surrounded by a membrane like the eukaryotic nucleus.
2. The prokaryotic genome is generally a circular piece of DNA that can exist in multiple copies and ranges in length but is at least a few million base pairs. It is packaged into the nucleoid through supercoiling facilitated by nucleoid-associated proteins.
3. DNA supercoiling allows for very long strands of DNA to be tightly packaged into a prokaryotic cell. This involves the introduction of plectonemic supercoils that twist the DNA into loops and wind it around nucle
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- Section 1 : Cell
-- What is a Cell?
-- What is DNA?
-- What is mitochondrial DNA?
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-- What is a chromosome?
-- How many chromosomes do people have?
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-- What are proteins and what do they do?
-- How do genes direct the production of proteins?
-- Can genes be turned on and off in cells?
-- What is epigenome?
-- How do cells divide?
-- How do genes control the growth and division of cells?
-- How do genetics indicate the location of a gene?
- Section 3: Genetic Mapping
-- What is genetic mapping?
-- How do researchers create a genetic map?
-- What are genetic markers?
Lesson 13.4
The document discusses gene regulation and expression in prokaryotes and eukaryotes. In prokaryotes like bacteria, genes are organized into operons and regulated by DNA-binding proteins that control transcription. The lac operon in E. coli regulates genes needed to break down lactose and is turned on or off by the lac repressor binding in the presence or absence of lactose. In eukaryotes, gene expression is more complex and regulated by transcription factors binding regulatory regions and controlling transcription. RNA interference is another mechanism where small RNAs regulate genes by interfering with mRNA. Master control genes regulate development by controlling gene expression and cell differentiation.
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This document contains the answers to an activity on biochemistry and cell organization submitted by a student named Maricris P. Nebiar. It defines several key terms related to cells and organelles. It also lists 5 differences between prokaryotic and eukaryotic cells, identifies which organelles contain DNA, are sites of energy production, and are surrounded by a double membrane. Finally, it summarizes the advantages of eukaryotic cells over prokaryotic cells and explains why a claim of discovering mitochondria in bacteria is unlikely.
Nucleic Acid Based Therapeutic Delivery System
This topic is covered for M Pharm Pharmaceutics students as per PCI syllabus. Hope this will be helpful.
Genome organisation
Genetic Organisation:
All cellular activities are encoded within a cell’s DNA.
The sequence of bases within a DNA molecule represents the genetic information of the cell.
Segments of DNA molecules are called genes, and individual genes contain the instructional code necessary for synthesizing various proteins, enzymes, or stable RNA molecules.
Genetic control
Covers the flow of information from DNA to Protein synthesis, Transcription, Types of RNA, Genetic code, Protein Synthesis, Cell Function and cell reproduction
Genetic control
This document discusses genetic control and regulation at the molecular level. It covers DNA and RNA structure, transcription, translation, gene regulation, and cell division. The central dogma of molecular genetics is described involving DNA transcription to RNA and translation to protein. The types of RNA are defined including mRNA, tRNA, rRNA. Transcription and translation processes are explained in detail. Gene regulation mechanisms like promoters, transcription factors, and feedback systems are covered. Finally, the document discusses cell reproduction through DNA replication and cell mitosis.
Animal Cell Structure
Animal cells are eukaryotic cells or cells with a membrane-bound nucleus.
DNA in animal cells is housed within the nucleus.
In addition to having nucleus animal cells also contain other membrane-bound organelles.
Organelles have a wide range of responsibilities that include everything from producing hormones and enzymes to providing energy for animal cells.
All living things are made up of cells that make up their body structure. Some of these living things are single-celled and other organisms are made up of more than one cell.
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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.
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There are several ways in which the gene regulation in eukaryotes di.pdf
- 1. There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protein · Post-translation: After protein synthesis. Solution There are several ways in which the gene regulation in eukaryotes differ from that in prokaryotes, as follows: 1. Prokaryotes are unicellular organisms and do not have defined nucleus, therefore the DNA resides in the cytoplasm. The gene regulation takes place in cytoplasm only i.e. simultaneous transcription and translation. Whereas, eukaryotes are Complex cell structure with a defined nucleus. The DNA is present in nucleus where its transcription takes place into RNA. RNA passes into cytoplasm where its translation takes place into protein. 2 Prokaryotes have naked DNA which are not associated with any protein or other substances to affect gene regulation.
- 2. Whereas, in eukaryotes DNA forms nucleosomes by wounding around the histone octamers. They are packed within chromatins. This complex structure controls the gene regulation. 3. In prokaryotes all genes are transcribed be same type of RNA polymerase. Whereas, in eukaryotes, there are three different kind of RNA polymerase (I, II and III) involved in gene regulation which are much bigger and complex than prokaryotic RNA polymerase. 4. In prokaryotes the gene expression regulation occurs primarily at transcription stage only. Whereas, in eukaryotes gene expression regulation may occurs at several levels: · Epigenetics: When DNA uncoils for transcription · Transcription: When RNA is transcribed · Post-transcription: When RNA passes into the cytoplasm · Translation: When RNA translates into protein · Post-translation: After protein synthesis.