Answer A) Subsidiary cells lose turgorExplanationPotassium ent.pdf
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Answer: A) Subsidiary cells lose turgor Explanation: Potassium enters the cell sap of the guard cells thereby reducing the water potential while increasing the osmotic concentration (and the O.P.) of the cell sap. Hence, endosmosis occurs (water moves from subsidiary cells to guard cells by loss of tugor in the subsidiary cells), guard cells become turgid and kidney-shaped and the stomata open. A. Opening of stomata depends upon following conditions: (a) Presence of light. (b) Decrease in starch contents of guard cells. (c) Increased concentration of malic acid in guard cells. (d) Influx of K+ ions in guard cells. (e) Efflux of H+ ions from guard cells. (f) Intake of CI ions by guard cells. (g) Low CO2 concentration in an around guard cells. (h) High pH (more than 7) in guard cells (hence, alkaline medium of the cell sap in guard cells). (i) High T.P. in guard cells due to endosmosis, (turgidity of cells). (j) TP more towards thin wall of guard cell & stomata open. B. Closing of stomata depends on following conditions: (a) Absence of light. (b) Decreased concentration of malic acid in guard cells. (c) Efflux of K+ ions from guard cells. (d) Influx of H+ ions in guard cells. (e) Acidic medium of the cell sap in guard cells. (f) Loss of Cl– ions from guard cells. (g) Increases CO2 concentration in and around guard cell due to release of CO2 in respiration combined with the absence of photosynthetic activity in dark. (h) Presence of plant growth inhibiting hormone abscissic acid (ABA), (i) Loss of turgidity and loss of kidney-shape by guard cells. All these conditions represent the reversal of the daytime events. Under these conditions, the guard cells lose water by exosmosis and become flaccid. This causes closing of the stomata. Solution Answer: A) Subsidiary cells lose turgor Explanation: Potassium enters the cell sap of the guard cells thereby reducing the water potential while increasing the osmotic concentration (and the O.P.) of the cell sap. Hence, endosmosis occurs (water moves from subsidiary cells to guard cells by loss of tugor in the subsidiary cells), guard cells become turgid and kidney-shaped and the stomata open. A. Opening of stomata depends upon following conditions: (a) Presence of light. (b) Decrease in starch contents of guard cells. (c) Increased concentration of malic acid in guard cells. (d) Influx of K+ ions in guard cells. (e) Efflux of H+ ions from guard cells. (f) Intake of CI ions by guard cells. (g) Low CO2 concentration in an around guard cells. (h) High pH (more than 7) in guard cells (hence, alkaline medium of the cell sap in guard cells). (i) High T.P. in guard cells due to endosmosis, (turgidity of cells). (j) TP more towards thin wall of guard cell & stomata open. B. Closing of stomata depends on following conditions: (a) Absence of light. (b) Decreased concentration of malic acid in guard cells. (c) Efflux of K+ ions from guard cells. (d) Influx of H+ ions in guard cells. (e) Acidic medium of the cell sap in.
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This document contains a practice test with multiple choice questions about cell organelles and their functions. It tests knowledge of the structures and roles of organelles like the nucleus, mitochondria, chloroplasts, endoplasmic reticulum, Golgi apparatus, lysosomes, cell membrane, and vacuoles. It also contains short answer and multiple choice questions about cellular transport mechanisms, like diffusion, osmosis, and active transport, and how they relate to movement across the cell membrane.
stomatalregulation-141211094949-conversion-gate01.pdf
This document discusses stomatal regulators in plant physiology. It describes the structure and function of stomata, guard cells, and the leaf cuticle. It explains that guard cells open and close the stomata to control gas exchange and transpiration. The key factors that influence stomatal aperture are described as light, temperature, carbon dioxide levels, and water content. The mechanisms of stomatal opening and closing are summarized, involving ion transport and changes in turgor pressure in the guard cells. The roles of the plant hormone ABA and environmental stresses like drought in triggering stomatal closure are also covered.
BSC1005 Biology General Chapter 6 1 Chapter 6 Cel.docx
BSC1005 Biology General
Chapter 6
1
Chapter 6: Cell Energy, Transport and Enzymes
1) Many of the enzymes that control a deep-sea firefly squid's ability to produce light energy from chemical energy are located
A) in membranes.
B) in the nucleus.
C) within chloroplasts.
D) outside of cells.
E) within mitochondria.
2) The fluid mosaic model describes the plasma membrane as consisting of
A) a phospholipid bilayer with embedded carbohydrates.
B) two layers of phospholipids with protein sandwiched between them.
C) a protein bilayer with embedded phospholipids.
D) carbohydrates, proteins, and phospholipids that can drift in the membrane.
E) individual proteins and phospholipids that can drift in a phospholipid bilayer.
3) Membrane phospholipids
A) have hydrophobic heads that face the center of the membrane and are shielded from water.
B) have hydrophilic tails that face outward and are exposed to water.
C) often have "kinks" in their tails caused by the presence of a double bond between carbons.
D) remain fluid because they are tightly packed against one another.
E) form impermeable layers for cells.
4) The cholesterol associated with animal cell membranes
A) is attached to membrane proteins and extends into the watery environment surrounding the cell.
B) helps to stabilize the cell membrane at body temperature.
C) makes the cell membrane fluid at room temperature.
D) is an abnormality resulting from a diet high in cholesterol.
E) helps solidify the membranes when the room temperature is below freezing.
5) A major function of glycoproteins and glycolipids in the cell membrane is to
A) glue cells together to form tissues.
B) allow the cells of an embryo to sort themselves into tissues and organs.
C) attach the cell membrane to the cytoskeleton.
D) help the cell resist swelling.
E) help the cell retain its shape.
6) When physicians perform an organ transplant, they choose a donor whose tissues match those of the recipient as closely as
possible. Which of the following cell components are being matched?
A) plasma membrane phospholipids
B) plasma membrane proteins
C) cell-surface carbohydrates
D) plasma membrane cholesterols
E) cytoskeletal elements
7) Most of the functions of a cell membrane are performed by
A) glycolipids.
B) proteins.
C) phospholipids.
D) cholesterol.
E) nucleotides.
BSC1005 Biology General
Chapter 6
2
9) Relaying a message from a membrane receptor to a molecule that performs a specific function within a cell is called
A) signal transduction.
B) inhibition.
C) competition.
D) self-recognition.
E) selective permeability.
10) Plasma membranes are selectively permeable. This means that
A) anything can pass into or out of a cell as long as the membrane is intact and the cell is healthy.
B) the plasma membrane allows some substances to enter or leave a cell more easily than others.
C) glucose cannot ...
MECHANISM OF TRANSPIRATION IN PLANTS [THEORIES]
Transpiration is the loss of water vapor from plant leaves through stomata. Stomata are small openings in leaves that allow for gas exchange. Almost 98-99% of water absorbed by plants is lost through transpiration. The rate of transpiration varies by plant - for example, sunflowers transpire 2 liters per day while apple trees transpire 36-45 liters per day. Transpiration occurs through four main processes - stomatal, cuticular, bark, and lenticular transpiration. Stomatal transpiration is the most important type and makes up 50-97% of total transpiration. Stomata open and close in response to turgor changes in guard cells that surround the
1BIOLOGY 101 Fall 2016FINAL EXAMINATIONPlease copy and.docx
1
BIOLOGY 101 Fall 2016
FINAL EXAMINATION
Please copy and paste the final examination into a Word file. Complete it in this form (do not make any structural changes!) and submit it as an attachment into your Assignment Folder.Do not forget to put your name on top of the exam!
The absolute deadline for submission is Sunday, October 16, NOON ET.
I cannot accept any later submissions.
YOUR NAME:
_______________________________________________________________
Total possible points: 100
I. Multiple choice questions. Please bold or underline the correct answer (1point each=50 points)
1. Several features, or properties, of living things may also be found in non-living things, such as fire. Of the following, which is the least likely to be found in non-living things?
A. Consumption of energy-containing molecules
B. Growth
C. Reproduction
D. Homeostasis
E. Response to external stimuli
2. Suppose you conduct an experiment which simulates glacial recession over time. What is the dependent variable in this experiment?
A. Glacial mass
B. Sunlight
C. The season
D. Time
3. If life on another planet were fundamentally like life on Earth, the macromolecules of living things would be based on which element?
A. Carbon
B. Hydrogen
C. Nitrogen
D. Oxygen
E. Phosphorus
4. The effectiveness of a medication containing growth hormones is tested on a group of young male rabbits 3 weeks of age. The best control group would be:
A. Any group of rabbits
B. A group of male rabbits, three weeks old, not given the medication
C. A group of female rabbits, three weeks old, not given the medication
D. A mixed group of male/female rabbits, three weeks old, not given the medication
E. No control is required; just measure whether the rabbits grew
5. Temperature is a measure of
A. The potential energy in a substance
B. How fast the molecule in a substance are moving
C. The energy lost from a substance to its surroundings.
D. How much heat is being absorbed by a substance
E. All of the above
6. Which of the following refers to a substance that has ‘lost’ a carbon dioxide molecule?
A. Denatured
B. Phosphorilated
C. Decarboxylated
D. Carbonated
E. Oxidated
7. Which of the following reactions or pathways is catabolic?
A. Converting glucose to carbon dioxide and water (cellular respiration)
B. Making starch from many glucose monomers
C. Photosynthesis, which builds glucose from carbon dioxide using energy from light
D. Making ATP from ADP and phosphate
8. One human disease is caused by a change in the DNA from GAA to GUA. This change is an example of:
A. Crossing-over
B. A meiosis error
C. A mitosis error
D. A mutation
9. During which part of the cell cycle is DNA polymerase most active?
A. Cytokinesis
B. G1 phase
C. G2 phase
D. S phase
E. M phase
10. Which of the following correctly describes a buffer?
A. A buffer converts an alkaline solution to neutral
B. A buffer converts an acid solution to neutral.
C. A buffer converts alkaline solutions to acid solutions.
D. A b ...
The Membrane Of A Eukaryotic Cell
The cell membrane surrounds the cytoplasm of the cell and separates its contents from the external environment. It is a semi-permeable bilayer that regulates what enters and exits the cell through membrane proteins. The nucleus contains the cell's DNA and controls its metabolism and reproduction. Mitochondria have a double membrane and produce energy for the cell in the form of ATP through aerobic respiration. Ribosomes are sites of protein synthesis and consist of large and small subunits that can float freely in the cell or attach to the endoplasmic reticulum.
Layers of GIT
The document summarizes the layers of the gastrointestinal tract and the mechanism of gastric acid secretion. It discusses four layers of the GI tract: the mucosa, submucosa, muscularis, and serosa. It then describes the four-step mechanism of gastric acid secretion by parietal cells: 1) active transport of ions creates a charge that allows diffusion of ions into canaliculi, 2) hydrogen ions are actively secreted in exchange for potassium ions, 3) water passes into canaliculi by osmosis, 4) carbon dioxide forms bicarbonate ions that diffuse out in exchange for chloride ions.
Diffusion and osmosis
1. The document is a biology worksheet about diffusion and osmosis containing multiple choice and short answer questions.
2. Some of the questions ask about the processes of diffusion and osmosis, such as how oxygen moves into root hair cells and examples of osmosis.
3. Other questions provide diagrams and experiments involving plant and animal cells in different solutions to be analyzed in terms of diffusion and osmosis.
Biology Diffusion
- Molecules move freely in gases and liquids which allows substances to diffuse and become evenly distributed.
- Diffusion is the movement of molecules from an area of higher concentration to lower concentration and allows substances like oxygen to enter cells.
- While diffusion works for single-celled organisms and transporting oxygen in the human body, it is too slow for large organisms so systems like blood circulation evolved.
diffusion ppt for concepts
- Molecules in gases, liquids, and solids move through diffusion. Diffusion occurs naturally as molecules move from areas of higher concentration to lower concentration.
- Diffusion allows oxygen and carbon dioxide to move freely across cell membranes in organisms. Larger organisms have developed transport systems like circulatory systems to move gases and nutrients over greater distances via diffusion.
- Plants, fish, and humans all rely on diffusion, with gases diffusing through leaves in plants, across gill surfaces in fish, and across lung air sacs in humans. The structures in each have large surface areas to increase diffusion rates.
Water Relations In Plants
This document discusses plant water relations and transport. It covers several topics:
1. It describes how turgor pressure in plant cells is sensitive to small changes in cell volume due to the rigid cell wall. As cell volume decreases, turgor pressure and water potential decrease sharply at first and then more gradually.
2. It explains how water and minerals are transported through the root system, entering through root hairs, moving through the apoplast and symplast of root cells, and being restricted by the casparian strip in endodermal cells before entering the xylem vessels for transport upward.
3. It discusses the mechanisms of cohesion and adhesion that allow for the ascent of water through
MH-CET 2014 Practice Test - Biology
MH-CET 2014 Practice Test - Biology, MH-CET 2014, Practice Test - Biology, MH-CET 2014 Biology Preparation
Ged Science Test
This document contains a 50 question science practice test with multiple choice answers. The test covers topics in biology, chemistry, physics and earth science. Questions assess understanding of concepts like photosynthesis, cellular structure, genetics, the scientific method, and ecosystems. The test is designed to help students prepare for science exams.
Key a. Diffusion b. Osmosis c. Active Transport Into the Cells d. Ac.pdf
Key: a. Diffusion b. Osmosis c. Active Transport Into the Cells d. Active Transport Out of the
Cells 1. Magnesium in Brackish water 2. Sodium in Sea water 3. Chlorine in Brackish water 4.
Calcium in Seawater 5. Chlorine in Seawater 6. Potassium in Brackish water 7. Sulphate in
Seawater 8. Chlorine in Brackish water 9. Magnesium in Seawater 10. Calcium in Brackish
water
ACTIVE TRANSPORT INTO AND OUT OF CELLS Active Transport makes it possible for
cells to maintain intemal conditions that are chemically different from the surrounding medium.
For cxample, some seaweed accumulate mineral ions such as potassium in their cells in
concentrations much higher than the surrounding seawater by Active Transport of these ions into
their cells. Problem: The following data shows concentrations of various ions inside the cells of
seaweed that can grow in both seawater and estuarine (brackish) water. Note: Freshwater
contains less than 1% dissolved substances. Salt water contains about 3.5% dissolved substances.
Brackish (estuarine) water contains 1.8% dissolved substances. Key: a. Diffusion b. Osmosis c.
Active Transport Into the Cells d. Active Transport Out of the Cells 1. Magnesium in Brackish
water 2. Sodium in Sea water 3. Chlorine in Brackish water 4. Calcium in Seawater 5. Chlorine
in Seawater 6. Potassium in Brackish water 7. Sulphate in Seawater 8. Chlorine in Brackish
water 9. Magnesium in Seawater 10. Calcium in Brackish water.
Neet 2018 Biology Sample Question Paper
Download free papers and solutions at www.misostudy.com. NEET 2018 Sample Paper of Biology. For appearing in the competitive exam of the country students need to practice more sample papers and previous papers.
Biological science Comprehensive LET Reviewer
1. The kidney maintains osmotic balance by filtering excess material from the blood.
2. At the end of mitosis, the cell divides through cytokinesis where a new cell wall is formed.
3. A boring tool would first encounter xylem, then vascular cambium, and then phloem when drilling into a tree.
Similar to Answer A) Subsidiary cells lose turgorExplanationPotassium ent.pdf (20)
Cell Physiology Spring 2016 Name Page 1 This test is .docx
Cell Physiology Spring 2016 Name Page 1 This test is .docx
stomatalregulation-141211094949-conversion-gate01.pdf
stomatalregulation-141211094949-conversion-gate01.pdf
BSC1005 Biology General Chapter 6 1 Chapter 6 Cel.docx
BSC1005 Biology General Chapter 6 1 Chapter 6 Cel.docx
1BIOLOGY 101 Fall 2016FINAL EXAMINATIONPlease copy and.docx
1BIOLOGY 101 Fall 2016FINAL EXAMINATIONPlease copy and.docx
Key a. Diffusion b. Osmosis c. Active Transport Into the Cells d. Ac.pdf
Key a. Diffusion b. Osmosis c. Active Transport Into the Cells d. Ac.pdf
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yes, roasting essentially means heating the ore, .pdf
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides.
Solution
yes, roasting essentially means heating the ore, in excess of air, to a temperature
below the melting point of the metal. this expels moisture, and converts the impurities to oxides..
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The adition is slow because the benzyne formed is.pdf
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
solvent oranother compound present in the medium.
Solution
The adition is slow because the benzyne formed is unstable, so itcould react with
the reagent and form other species. The driving force of the reaction is the formation of
stableproducts (N2 and CO2), but you are right,benzyne is so unstable that it can react with the
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Yes, I do agree, that some eukaryotic cells survive only on the reso.pdf
Yes, I do agree, that some eukaryotic cells survive only on the resources provided by other
eukaryotic cells and that prokaryotic cells are not involved in this type of \"supply for life\".
Symbiotic root colonization by Eukaryotic cells by Mycorrhizal mycelia:
Mycorrhizal mycelia can only survive in a symbiotic relationship with other plant root \"cells
(eukaryotic)\" by supplying nitrogen materials to the plants followed by obtaining its nutrients
for survival. Mycorrhizal mycelia cannot survive outside the environment as prokaryotes do
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants. The relative number of competitors, the relative availability of symbiotic partners
such as Mycorrhizal mycelia are living on the roots surface of the host plant as mutualistic
symbionts (supply of life) and enabling the host root colonization of land plants with increased
absorption of nutrients such as phosphates, nitrates, water etc. by providing the surface area for
absorption.
Fungi and plant symbiosis represents one of the most important interspecific eukaryotic
partnerships to have occurred on the planet. Fungal mycelia and ancient archaic prokaryotes are
living on the roots surface of the host land plants as mutualistic symbionts in 500 millions of
years ago and enabling the host root colonization with increased abruption of nutrients such as
phosphates, nitrates, water etc. by providing the surface area for absorption. This process
produced a vital photosynthesis rate by increasing total gross primary productivity finally
enhancing the release of exclusive amounts of oxygen in the atmosphere result in higher oxygen
concentration together to form ozone layer to protect against harmful UV radiation. The oxygen
gas is conductive to air-breathing animals on the land whereas dissolved oxygen is conductive to
all aquatic animals. These land plants with differential reproductive ability to produce
gametophytes enabled their sexual reproduction (embryophytes, all vascular plants including
angiosperms and gymnosperms) to produce more progeny to generate higher rate of
photosynthesis associated with increased rate of colonization to produce oxygen gas to air-
breathing animals.
The traits that allowed them to maintain symbiotic relationships are as explained in case of fungi,
in which hydrolytic enzymes have capability to assimilate \"nitrogenous material and supply\"
them to plants.
The altitudinal increase of root-to-shoot (C/N) ratio is meticulously due to low nutrient
availability. This is because of “low Michorrizha availability for symbiotic relationship” to
supply nutrient as elevated altitudes lowered temperatures. Therefore, low temperature and low
soil microbial decomposition with organic matter with water logging & low nitrogen availability
make the plants to increase their C/N ratio as an ev.
What are the key components of an internal audit What are the impli.pdf
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report issues and challenges identified and negotiate action plans with management to address
the problems.
Regular follow-up on reported findings at appropriate intervals. Internal audit departments
maintain a follow-up database for this purpose.
Solution
What are the key components of an internal audit? What are the implications of having a
thorough understanding of each? Using your firm (or a firm that you familiar with), select a
significant company resource (such as engineering capability, or supply chain, or customer
service) and apply the Four Tests of a Resources Compeititve Power and determine if provides a
sustainable competitive advantage. Support your input.
Answer:
Key Components & Implications of Internal Audit:
Components:
Implications:
Establish and communicate the scope and objectives for the audit to appropriate management.
Developing an understanding of the business area under review. This includes objectives,
measurements, and key transaction types. This involves review of documents and interviews.
Flowcharts and narratives may be created if necessary.
So that we can concentrate on the key business activities which are having in Risk of Fraud or
Misappropriation.
Identify management practices in the five components of control used to ensure each key risk is
properly controlled and monitored. Internal Audit check can be a helpful tool to identify
common risks and desired controls in the specific process or industry being audited.
Develop and execute a risk-based sampling and testing approach to determine whether the most
important management controls are operating as intended.
Report .
IV I II III Solution .pdf
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To differentiate between the mutated and normal genes, the DNA of th.pdf
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence.
Solution
To differentiate between the mutated and normal genes, the DNA of the person is isolated and
amplified by the PCR (polymerase chain reaction). This genome is subjected to the DNA
microarrays, i.e. DNA microchips that contain the single strand sequences of normal genes
including the sequences of its possible mutations.
The DNA is cut into small fragments and these fragments are labelled with different fluorescent
dyes (the sample from the patient and the normal DNA sample). Both the DNA are allowed to
hybridize and bind to the single strands of the DNA chip. The difference in the percent
hybridization indicates the presence of mutations. if the person does not have any mutations,
both the samples hybridize equivalents and produce the same fluorescence.
Thus, if the person has the retinoblastoma mutant gene, two different fluorescences will appear,
indicating the hybridizations of the mutant and the normal genes whereas an unaffected person
produces unifom fluorescence..
The most critical shortcomings of these assumptions are1.The flow .pdf
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients.
Solution
The most critical shortcomings of these assumptions are
1.The flow of water is considered to be one dimensional but in reality it is three dimensional.
2.Darcy\'s law does not seem to hold good at high hydraulic gradients..
The following statements are true1. The tendency of water to mini.pdf
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
molecules and water molecules.
3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
Cellwalls)
Solution
The following statements are true:
1. The tendency of water to minimize it\'s interaction with non-polar substances is known as
hydrophobic effect. ( in hydrophobic, hydro refers to water and phobic refers to fear. )
2.Hydrophobic molecules donot dissolve in water due to lack of Hydrogen bonding between it\'s
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3. Amphipathic substances are the structural basis of biological bilayer membranes. ( Ex. Animal
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The data is Quantitative as there are numbers , and its an observati.pdf
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination.
Solution
The data is Quantitative as there are numbers , and its an observational study, where they studies
about the H.pylori in the cause of ulceration.
2. From table -II, we see that among the 22 patients having Ulcer, 18 have infection of H.pylori.
Out of 16, 8 were with infection.
Gastritis and Duodenitis.
Out of 22, 18 patients with gastric ulcers had bacteria, and four had both the ulcers, and all were
Helicobacter positive. With gastric ulcer 77% had bacteria, and with duodenal ulcer 100% had
bacteria. 4 patients with ulcers became negative for bacteria is significant as they had taken
NSAIS drugs which are known to cause ulcers.
Although gastric colonization with H.pylori, induces histologic gastritis in all infected
individuals, only a small portion of them develop any significant sign and symptom for this
colonization. It sestimated that H. pylori, infected patients have a 10-20% lifetime risk of
developing ulcer disease. The risk of development of these diseases in the presence of H.pylori,
depends on a variety of bacteria, host, and environmental factors, that relate to the pattern and
severity of gastritis. Thus a patient can show normal examination..
Thank you everyone! The answer was O(n!)SolutionThank you ever.pdf
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Solution1 ) The purpose of criminal law is to help maintain social.pdf
Solution
1 ) The purpose of criminal law is to help maintain social order and stability . The purpose of
criminal law is to establish a system of prohibitions , penalties and correctional measures to deal
with conduct that unjustifiably or inexcusably causes or threatens harm to those individual or
public interests for which state protection is appropriate .
2) Classification according to the nature of crimes
Mala in se : These are acts that are immoral or wrong in themselves or acts that are naturally evil
. Ex : Murder, rape , arson , burglary etc
Mala prohibita : These crimes are not naturally evil , but are prohibited by statute because they
infringe upon the right of the others . This type of act is not wrong in some societies but wrong in
other societies . Example : Traffic offenses , gambling etc .
Classification according to degree :
a) Felonies : Crimes that are punishable by death or imprisonment in a state prision or
penitentiary are felonies .
b) Misdemeanors : Crimes or offenses not amounting to felonies are Misdemeanors . They are
offenses for which the punishment is other than death or imprisonment in a state prison , or that
have not been designated felonies by statute.
c) Treason : Treason is the only crime described in the constitution and it is given a higher
classification than felony .
3) Example of white-collar crimes are bank fraud, blackmail , bribery , computer fraud,
counterfeiting , credit card fraud , currency schemes, embezzelement , environmental schemes .
The punishment for these crimes should be recovery of all the money made by white collar
crimes and to put the white collar criminals in prision ..
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num.txt(Save this file under D Drive.Then the path of the file is D.pdf
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
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CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<
Solution
num.txt(Save this file under D Drive.Then the path of the file is D:\\\ um.txt)
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CPP Code
#include
#include
using namespace std;
//Declaring Global variable
int sum=0;
//Function declarations
void readFromFile();
void writeToFile();
/* Function implementation which will
* read the numbers from the file and calculates the sum of
* the numbers which are multiples of 2 or 5
*/
void readFromFile()
{
int number,sum=0;
//fileInput reference
ifstream dataIn;
//Opening input file
dataIn.open(\"D:\\\ um.txt\");
while(dataIn>>number)
{
if(number%2==0 || number%5==0)
{
sum+=number;
}
}
//closing the input stream
dataIn.close();
}
/* Function implementation which will
* display the sum into another text file
*/
void writeToFile()
{
//fileoutput reference
ofstream myfile;
//Opening the file
myfile.open (\"D:\\\ um1.txt\");
//Writing the contents to the file
myfile <<\"The Sum of Multiples of 2 or 5 is :\"<.
n = 7 which is odd so median is the middle value of te data after ar.pdf
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4.
Solution
n = 7 which is odd so median is the middle value of te data after arranging in ascending order.
1,1,2,4,5,6,7
median = (7+1)/2 th observation = 4th observation = 4
So median of the data is 4..
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