Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
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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..
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Cell Division
Cell division occurs through mitosis and meiosis. Mitosis produces two identical daughter cells and is used for growth and tissue repair. Meiosis produces gametes like sperm and eggs with half the number of chromosomes and is used for sexual reproduction. The cell cycle includes interphase, where the cell grows and DNA replicates, and the M phase where mitosis occurs. Mitosis involves prophase, metaphase, anaphase and telophase stages to divide the cell. Meiosis involves two cell divisions, meiosis I and meiosis II, to produce four haploid cells from one diploid cell. This ensures genetic variation in offspring.
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This document provides an overview of cell cycle, cell division, and meiosis. It describes the stages of the cell cycle including interphase and mitosis. It then explains the two main types of cell division - mitosis, which produces identical daughter cells, and meiosis, which reduces the chromosome number by half to produce gametes like sperm and egg cells. The stages of meiosis, including meiosis I and meiosis II, are outlined in detail. Key events in meiosis I like homologous chromosome pairing, crossing over, and separation of homologous chromosomes are discussed. The phases of each meiotic division - prophase, metaphase, anaphase, and telophase - are also summarized.
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Ques-1Part-a Mitosis it is a somatic cell division in an organi.pdf
Ques-1:
Part-a: Mitosis: it is a somatic cell division in an organism & this division is useful to maintain
body growth
Part-b: Meiosis: it is the reproductive reduction cell division & it is useful to produce gametes for
fertilization finally to generate a mature embryo
Part-c: cytokinesis: It is defined as the process of division of cytoplasm during cell division
either in mitosis or meiosis & this is useful for carrying cell organelles
Part-d: chromosome: it is the hereditary material that carries genomic information for gene
expression & useful for inheritance of parental traits to offspring
Mitosis ---> somatic cell division & produces two identical daughter cells with equal number of
chromosomes (cytokinesis & karyokinesis).
Meiosis ----> a reduction division occurs in reproductive germ cells in which allosomes
determine sex determination finally generates half of the chromosomes in daughter cells
Part-a: Mitosis: it is a somatic cell division in an organism & this division is useful to maintain
body growth
Part-b: Meiosis: it is the reproductive reduction cell division & it is useful to produce gametes for
fertilization finally to generate a mature embryo
Part-c: cytokinesis: It is defined as the process of division of cytoplasm during cell division
either in mitosis or meiosis & this is useful for carrying cell organelles
Part-d: chromosome: it is the hereditary material that carries genomic information for gene
expression & useful for inheritance of parental traits to offspring
Cell plate is absent during animal cell division instead a complete cleavage of the cells during
division can be observed in animal cells. Cell plate formation during cell division observed only
in plants.
Haploid cells (n) are normally produced by reduction division of meiosis from diploid parent
cells (2n). This type of division occurs in germ cells of male and female species. The
chromosomes of the diploid germ cell undergo reduction in meiosis I and generate half of the
chromosomes when compared to parent diploid cell.
Meiosis division in a reproductive cell often results in four haploid cells with only single set of
chromosomes.
Meiosis-I prophase-I In this process double stranded homologous chromosomes line up each
other, and result in forming a tetrad; this is termed as Synapsis.
Terminal chaismata hold the homologous chromosomes together in metaphase I finally these are
going to align on the metaphase plate. Synapsis allows homologous chromosomes to travel to the
center by forming spindle formation through microtubules attaching to the centromeres finally to
the opposite poles of the cell. Later, crossing over takes place to exchanges paternal and maternal
genes to form tetrad where genetic variation occurs. Crossing over takes place to exchange equal
amounts of DNA generates genetic variability in prophase -I of meiosis.
Crossing over occurs in which lined up chromosomal ends exchange genetic information by
switching their places. Crossing over can un.
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Ques-1Part-a Mitosis it is a somatic cell division in an organi.pdf
Ques-1:
Part-a: Mitosis: it is a somatic cell division in an organism & this division is useful to maintain
body growth
Part-b: Meiosis: it is the reproductive reduction cell division & it is useful to produce gametes for
fertilization finally to generate a mature embryo
Part-c: cytokinesis: It is defined as the process of division of cytoplasm during cell division
either in mitosis or meiosis & this is useful for carrying cell organelles
Part-d: chromosome: it is the hereditary material that carries genomic information for gene
expression & useful for inheritance of parental traits to offspring
Mitosis ---> somatic cell division & produces two identical daughter cells with equal number of
chromosomes (cytokinesis & karyokinesis).
Meiosis ----> a reduction division occurs in reproductive germ cells in which allosomes
determine sex determination finally generates half of the chromosomes in daughter cells
Part-a: Mitosis: it is a somatic cell division in an organism & this division is useful to maintain
body growth
Part-b: Meiosis: it is the reproductive reduction cell division & it is useful to produce gametes for
fertilization finally to generate a mature embryo
Part-c: cytokinesis: It is defined as the process of division of cytoplasm during cell division
either in mitosis or meiosis & this is useful for carrying cell organelles
Part-d: chromosome: it is the hereditary material that carries genomic information for gene
expression & useful for inheritance of parental traits to offspring
Cell plate is absent during animal cell division instead a complete cleavage of the cells during
division can be observed in animal cells. Cell plate formation during cell division observed only
in plants.
Haploid cells (n) are normally produced by reduction division of meiosis from diploid parent
cells (2n). This type of division occurs in germ cells of male and female species. The
chromosomes of the diploid germ cell undergo reduction in meiosis I and generate half of the
chromosomes when compared to parent diploid cell.
Meiosis division in a reproductive cell often results in four haploid cells with only single set of
chromosomes.
Meiosis-I prophase-I In this process double stranded homologous chromosomes line up each
other, and result in forming a tetrad; this is termed as Synapsis.
Terminal chaismata hold the homologous chromosomes together in metaphase I finally these are
going to align on the metaphase plate. Synapsis allows homologous chromosomes to travel to the
center by forming spindle formation through microtubules attaching to the centromeres finally to
the opposite poles of the cell. Later, crossing over takes place to exchanges paternal and maternal
genes to form tetrad where genetic variation occurs. Crossing over takes place to exchange equal
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Crossing over occurs in which lined up chromosomal ends exchange genetic information by
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- A reminder for parents to pick up and return student modules/outputs while following COVID protocols.
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(1) c. Two-tailed test
(2)d. Ho: Men and women are the same in their political understanding.
(3)b. Ho:
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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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2. Mitosis and meiosis are described as two types of cell division. Mitosis produces two identical daughter cells and occurs in body cells, while meiosis produces gametes and reduces chromosome number.
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Cell division occurs through mitosis and meiosis. Mitosis produces two identical daughter cells from one parent cell during growth and repair. Meiosis reduces the chromosome number by half and produces genetic variation through independent assortment and crossing over during gamete formation for sexual reproduction. The cell cycle is tightly regulated and consists of interphase, mitosis, and cytokinesis. Errors in meiosis can result in genetic disorders.
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The document provides an overview of mitosis and meiosis. It describes the cell cycle and its two main phases: interphase and M-phase. Interphase consists of G1, S, and G2 phases where the cell grows and prepares for division. M-phase involves either mitosis or meiosis. Mitosis divides somatic cells and results in two identical daughter cells. Meiosis produces gametes through two cell divisions, resulting in four daughter cells each with half the number of chromosomes. The stages of each process including prophase, metaphase, anaphase and telophase are defined in detail.
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Similarities:
- Both are types of cell division that involve dividing up chromosomes.
- Both involve the separation of duplicated chromosomes into new daughter cells.
Differences:
- Mitosis produces two identical diploid daughter cells used for growth and repair. Meiosis produces four haploid daughter cells (eggs or sperm).
- Mitosis involves one division, meiosis involves two divisions (Meiosis I and Meiosis II).
- During mitosis, sister chromatids separate but remain identical. During meiosis, homologous chromosomes pair up and recombine, producing genetic diversity.
- Mitosis is asexual reproduction, meiosis is a part
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Chapter 3 cell division
The document summarizes cell division and the two main types: mitosis and meiosis. Mitosis produces two identical daughter cells and is used for growth, repair, and replacement of somatic cells. It involves the phases of interphase, prophase, metaphase, anaphase, telophase, and cytokinesis. Meiosis produces four non-identical gamete cells through two rounds of division and is involved in sexual reproduction.
Chapter 8
Chromosomes contain DNA and come in pairs, with one chromosome from each parent. Cells undergo either mitosis or meiosis to divide. Mitosis produces two identical daughter cells and occurs in body cells, while meiosis produces four unique haploid cells and occurs in sex cells. During meiosis, homologous chromosomes can crossover and recombine, introducing genetic variation in offspring. Sexual reproduction uses meiosis to create gametes from two parents, while asexual reproduction clones the single parent.
Meiosis presentation for PFS 2014
The document summarizes the process of meiosis in producing gametes or sex cells. It explains that meiosis involves two cell divisions, Meiosis I and Meiosis II, starting with one diploid cell that undergoes DNA replication and ends with four haploid cells. The key stages of each division (prophase, metaphase, anaphase, telophase) are described. Meiosis results in genetic variation through independent assortment and crossing over during prophase I. The differences between mitosis and meiosis are highlighted, with meiosis producing haploid cells through two divisions while mitosis results in two identical diploid daughter cells.
Meiosis.ppt..
It will give idea about Meiosis with excellent picture so, anyone can easily understand what actually meiosis is.
Cell cycles and its regulation presented by dr.sibi p ittiyavirah, professor,...
Cell cycles and its regulation presented by dr.sibi p ittiyavirah, professor,...Dr.SIBI P ITTIYAVIRAH
The document summarizes the cell cycle and its key stages. It discusses:
- The cell cycle consists of interphase and the mitotic phase. Interphase includes G1, S, and G2 phases where the cell grows and duplicates its DNA.
- The mitotic phase includes mitosis, where the cell separates copied DNA and divides its cytoplasm, forming two daughter cells. Mitosis consists of prophase, prometaphase, metaphase, anaphase and telophase.
- The stages carefully coordinate cell growth and division to duplicate the cell and its contents in an organized manner.Similar to Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf (20)
Cell cycles and its regulation presented by dr.sibi p ittiyavirah, professor,...
Cell cycles and its regulation presented by dr.sibi p ittiyavirah, professor,...
More from anwarsadath111
(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.
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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hi Q is nearly done,just give me dia of shaftSolutionhi Q is n.pdf
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Part1. Option 3rd; Somatic cells such as liver cells cannot undergo .pdf
- 1. 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.