What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
Solution
Role of environmental factors on transpiration
1. Relative humidity
The loss of water vapour from stomata occurs by simple diffusion. When the atmosphere is very
humid, the rate of loss of water vapour is lowered.
Wind:
When the wind velocity is high, rate of transpiration also increases.
Temperature:
Warm air hold more water vapour so the rate of loss of water increases with increase in
temperature.
Light:
Opening and closing of stomata is regulated by light. More light result in high degree of
transpiration.
Soil water content:
When there is less water content in the soil , the leaf droops / get dehydrated making the stomata
to close resulting in less rate of transpiration.
Carbon dioxide concentration
When the concentration is high, stomata closes resulting in low rate of transpiration and vice
versa.
Plants defense mechanism against water loss
Plants require water for it\'s survival. So it will have to regulate water loss ( especially during
drought) .
Several leaf adaptations help to combat water loss.
Leaf arrangements play important role, transversely arranged leaves loss more water than
perpendicularly arranged ones.
Other anatomical features of the leaves such as sunken/vestigial stomata and closely arranged
mesophyll cells reduces transpiration.
Presence of hair, cuticle(waxy) , gums, mucilage etc on leaf surface help reducing transpiration.
During drought, plant hormone abscisic acid play an important role by closing the stomata
reducing water loss..
This assignment is comprised of the dtails about the process of transpiration in plants, when was its study first conducted,who discovered the actual phenomenon at first, what's the actual process, its types, its importance for the plants. Studying this u will get almost all your points cleared.
This assignment is comprised of the dtails about the process of transpiration in plants, when was its study first conducted,who discovered the actual phenomenon at first, what's the actual process, its types, its importance for the plants. Studying this u will get almost all your points cleared.
Water is hydrosphere is made up of all the water on Earth. This includes all of the rivers, lakes, streams, oceans, groundwater, polar ice caps, glaciers and moisture in the air (like rain and snow). The hydrosphere is found on the surface of Earth, but also extends down several miles below, as well as several miles up into the atmosphere. So, there is a need for study of water as a scarce resource.
WHAT IS HYDROLOGICAL CYCLE
SYSTEM APPROACH IN HYDROLOGY
HYDROLOGIC INPUT & OUTPUT
VARIATION IN HYDROLOGICAL CYCLE
COMPONENTS
EVAPORATION
EVAPOTRANSPIRATION
PRECIPITATION
INTERCEPTION
INFILTRATION
GROUND WATER
RUN-OFF
HUMAN IMPACT
EARTH SURFACE
CLIMATE CHANGE
ATMOSPHERIC POLLUTION
MULTI PURPOSE PROJECTS
WATER WITHDRAWAL
MANAGEMENT AND CONTROL
Water as an Ecological Factor by Salman Saeed Lecturer BotanySalman Saeed
Water as an Ecological Factor
lecture for Biology, Botany, Zoology, and Chemistry Students by Salman Saeed lecturer Botany University College of Management and Sciences Khanewal, Pakistan.
About Author: Salman Saeed
Qualification: M.SC (Botany), M. Phil (Biotechnology) from BZU Multan.
M. Ed & B. Ed from GCU Faisalabad, Pakistan.
Email: Salmanbotanist@gmail.com
Liquid water is converted to gaseous water (water vapor) by the process of evaporation. Water travels from the Earth’s surface to the atmosphere via evaporation. Evaporation results from the dissolution of the bonds holding the water molecules together as a result of heat energy.
Osmoregulation, and adaptation in plants against abiotic factors plant stres...Raheel Hayat Rahee
Osmoregulation in plants and adaptation in plants against abiotic factors
Follow to get more updated information.You wil get all types of information According to your study and if you want to order any ppt formation according to your topic I can also provide you. Hope so you will not be disappointed 😃. Be happy stay blessed
Translocation and Absorption of water
Absorption of water
Functions of water
Active and Passive absorption
Factors affecting absorption of water
Effective root zones
Moisture extraction pattern
Translocation
Consider a binary search tree T with nodes containing the four fields.pdfSIGMATAX1
Consider a binary search tree T with nodes containing the four fields key, left, right, p. and ht,
where key is an integer, left, and right are pointers to the two subtrees, p is a pointer to the parent
of the node, and ht is the height of the node (the maximum length of a path from this node to a
leaf). Design an algorithm for a procedure insert (T, x) that inserts to a binary searches tree T a
new node x (whose key is not in the tree yet), and updates lit fields accordingly. The function
should work in O(h) time, where h is the length of the access path.
Solution
Algorithm is implemented in C:
struct node
{
int key;
struct node *left, *right;
};
//function to create a new BST node
struct node *new_Node(int item)
{
struct node *tem = (struct node *)malloc(sizeof(struct node));
tem->key = item;
tem->left = tem->right = NULL;
return tem;
}
/INSERT PROCEDURE
/* Given key in BST,function to insert a new node. */
struct node* insert(struct node* T, int x)
{
/* If tree is empty then return a new node */
if (T == NULL) return new_Node(x);
/* else, repeat down the tree */
if (x < T->x)
T->left = insert(T->left, x);
else if (key > node->key)
T->right = insert(T->right, x);
return T;
}.
Based on the articleWhy It Is Time to Look Beyond Algal Genes in P.pdfSIGMATAX1
Based on the article
Why It Is Time to Look Beyond Algal Genes in Photosynthetic Slugs
By: Cessa Rauch1, Jan de Vries 1, Sophie Rommel2, Laura E. Rose2, Christian Woehle3,
Gregor Christa1, Elise M. Laetz4, Heike Wa ¨gele4, Aloysius G.M. Tielens5,6, Jo ¨rg
Nickelsen7, Tobias Schumann8, Peter Jahns8, and Sven B. Gould1,*
Answer this:
--> It has been argued that the reason for detection of such a small amount of mRNA (the read
count) of lateral gene transfer origin in the slug cells is due to only very few cells containing
plastids. Do you agree with this argument? Why or why not?
--> The authors present several alternative reasons for the stability of plastids within the slug
cells. Describe two of these reasons.
Solution
1. No, we donot agree with the argument.
Through LGT from algae to slugs has similar RNA sequence,supported by genome sequence of
slug DNA, failed to give an evidence for such LGT events later on. So by this, it is clear that
there is no proof that the evolution of kleptoplasty in photosynthetic slugs involves LGT events,
where they depend purely on nuclear genes.
In some of the photosynthetic slugs, plastids sequestered from algae can continue to
photosynthesize for weeks, or even months, in the absence of algal nuclei. Therefore, in
conclusion, some isolated algal plastids are by nature more robust than those of land plants and
there is no evidence that any of these require the expression of slug genes of algal origin.
2. Plastids have effective photoprotection mechanisms.
Slugs acquiring robust plastids will not automatically retain them long-term and endure
starvation.
Slugs are physiologically adapted and require to retain them functionally..
More Related Content
Similar to What are the role of environmental factors on transpiration and h.pdf
Water is hydrosphere is made up of all the water on Earth. This includes all of the rivers, lakes, streams, oceans, groundwater, polar ice caps, glaciers and moisture in the air (like rain and snow). The hydrosphere is found on the surface of Earth, but also extends down several miles below, as well as several miles up into the atmosphere. So, there is a need for study of water as a scarce resource.
WHAT IS HYDROLOGICAL CYCLE
SYSTEM APPROACH IN HYDROLOGY
HYDROLOGIC INPUT & OUTPUT
VARIATION IN HYDROLOGICAL CYCLE
COMPONENTS
EVAPORATION
EVAPOTRANSPIRATION
PRECIPITATION
INTERCEPTION
INFILTRATION
GROUND WATER
RUN-OFF
HUMAN IMPACT
EARTH SURFACE
CLIMATE CHANGE
ATMOSPHERIC POLLUTION
MULTI PURPOSE PROJECTS
WATER WITHDRAWAL
MANAGEMENT AND CONTROL
Water as an Ecological Factor by Salman Saeed Lecturer BotanySalman Saeed
Water as an Ecological Factor
lecture for Biology, Botany, Zoology, and Chemistry Students by Salman Saeed lecturer Botany University College of Management and Sciences Khanewal, Pakistan.
About Author: Salman Saeed
Qualification: M.SC (Botany), M. Phil (Biotechnology) from BZU Multan.
M. Ed & B. Ed from GCU Faisalabad, Pakistan.
Email: Salmanbotanist@gmail.com
Liquid water is converted to gaseous water (water vapor) by the process of evaporation. Water travels from the Earth’s surface to the atmosphere via evaporation. Evaporation results from the dissolution of the bonds holding the water molecules together as a result of heat energy.
Osmoregulation, and adaptation in plants against abiotic factors plant stres...Raheel Hayat Rahee
Osmoregulation in plants and adaptation in plants against abiotic factors
Follow to get more updated information.You wil get all types of information According to your study and if you want to order any ppt formation according to your topic I can also provide you. Hope so you will not be disappointed 😃. Be happy stay blessed
Translocation and Absorption of water
Absorption of water
Functions of water
Active and Passive absorption
Factors affecting absorption of water
Effective root zones
Moisture extraction pattern
Translocation
Consider a binary search tree T with nodes containing the four fields.pdfSIGMATAX1
Consider a binary search tree T with nodes containing the four fields key, left, right, p. and ht,
where key is an integer, left, and right are pointers to the two subtrees, p is a pointer to the parent
of the node, and ht is the height of the node (the maximum length of a path from this node to a
leaf). Design an algorithm for a procedure insert (T, x) that inserts to a binary searches tree T a
new node x (whose key is not in the tree yet), and updates lit fields accordingly. The function
should work in O(h) time, where h is the length of the access path.
Solution
Algorithm is implemented in C:
struct node
{
int key;
struct node *left, *right;
};
//function to create a new BST node
struct node *new_Node(int item)
{
struct node *tem = (struct node *)malloc(sizeof(struct node));
tem->key = item;
tem->left = tem->right = NULL;
return tem;
}
/INSERT PROCEDURE
/* Given key in BST,function to insert a new node. */
struct node* insert(struct node* T, int x)
{
/* If tree is empty then return a new node */
if (T == NULL) return new_Node(x);
/* else, repeat down the tree */
if (x < T->x)
T->left = insert(T->left, x);
else if (key > node->key)
T->right = insert(T->right, x);
return T;
}.
Based on the articleWhy It Is Time to Look Beyond Algal Genes in P.pdfSIGMATAX1
Based on the article
Why It Is Time to Look Beyond Algal Genes in Photosynthetic Slugs
By: Cessa Rauch1, Jan de Vries 1, Sophie Rommel2, Laura E. Rose2, Christian Woehle3,
Gregor Christa1, Elise M. Laetz4, Heike Wa ¨gele4, Aloysius G.M. Tielens5,6, Jo ¨rg
Nickelsen7, Tobias Schumann8, Peter Jahns8, and Sven B. Gould1,*
Answer this:
--> It has been argued that the reason for detection of such a small amount of mRNA (the read
count) of lateral gene transfer origin in the slug cells is due to only very few cells containing
plastids. Do you agree with this argument? Why or why not?
--> The authors present several alternative reasons for the stability of plastids within the slug
cells. Describe two of these reasons.
Solution
1. No, we donot agree with the argument.
Through LGT from algae to slugs has similar RNA sequence,supported by genome sequence of
slug DNA, failed to give an evidence for such LGT events later on. So by this, it is clear that
there is no proof that the evolution of kleptoplasty in photosynthetic slugs involves LGT events,
where they depend purely on nuclear genes.
In some of the photosynthetic slugs, plastids sequestered from algae can continue to
photosynthesize for weeks, or even months, in the absence of algal nuclei. Therefore, in
conclusion, some isolated algal plastids are by nature more robust than those of land plants and
there is no evidence that any of these require the expression of slug genes of algal origin.
2. Plastids have effective photoprotection mechanisms.
Slugs acquiring robust plastids will not automatically retain them long-term and endure
starvation.
Slugs are physiologically adapted and require to retain them functionally..
Comparecontrast the following classes of hormones by identifying thr.pdfSIGMATAX1
Compare/contrast the following classes of hormones by identifying three similarities and four
differences among/between them. In addition, be able to list several specific examples of each.
AMINE vs. PEPTIDE vs. STEROID
Solution
Three similarities between the three hormones-
1. Being hormones, the site of synthesis and the site of action of all the three hormones are
different.
2. They are all organic molecules and rich in carbon and hydrogen.
3. They are all released directly into the blood stream.
Differences
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine
Peptide
Steroid
Amine
Composition
Made of few to many amino acids(3-200), synthesized from mRNA
Derived from cholesterol (from pregnenolone, a steroid precursor)
Derived from amino acids tyrosine or tryptophan
Solubility in lipid bilayer
Not soluble
soluble
Not soluble
Nature
Hydrophillic with cell surface receptors( except- thyroid hormones are lipid soluble)
Lipophillic with intracellular receptors
Hydrophillic with cell surface receptors
Mode of Action
Through secondary messengers like cAMP, cGMP etc by cascading effect
Once inside the cell it binds DNA and modifies transcription
Through binding to beta adrenergic receptors, by secondary messengers (signal transduction) or
calcium or phosphoinositides
Examples
ADH, calcitonin, Insulin, Pituitary Hormones (LH,FSH, ACTH, TSH,GH)
Estrogen, Progesterone, testosterone, cortisol
Epinephrine, nor epinephrine, melatonin, dopamine.
Why does the incidence of acute disease decrease with ageSoluti.pdfSIGMATAX1
Why does the incidence of acute disease decrease with age?
Solution
Incidence of acute diseases decrease with age because the body which is through all the worst
times have gained enough resistance though the immunity decreases with age still the body will
be ready to fight with acute infections and go over that due to the resistance developed in course
of lifetime. i hope this is a brief answer and would help you. Thank you.
White eye color in Drosophila was discovered to be sex linked by Tho.pdfSIGMATAX1
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:
White eye color in Drosophila was discovered to be sex linked by Thomas Hunt Morgan, who
observed that:A. reciprocal crosses produced different results, which is not true for most
autosomal traits.B. the inheritance pattern depended on the sex of the individuals.C. the white-
eyed flies were sterile, in contrast to the red-eyed flies.D. A and B.E. All of the above
statements apply.
Solution
The answer is D. A and B.
A. reciprocal crosses produced different results, which is not true for most autosomal traits.
B. the inheritance pattern depended on the sex of the individuals..
Which of the following protons would you NOT expect to find embedded .pdfSIGMATAX1
Which of the following protons would you NOT expect to find embedded in the cytoplasmic
membrane of any bacteria? Primary active transport proteins. ATP synthase. Enzyme I complex
of the Group translocation system. Basel body protein rings of flagella. Which of the following is
true? Both photosystem I and photosystem II rely on light bombarding chlorophyll. Photosystem
II donates electrons to photosystem I in a linear electron transport chain. Photosystem I only (not
photosystem II) directly produces NADPH utilized for the Calvin cycle. All of the above are
true. None of the above are true. Which one of these prokaryotic characteristics can\'t co-exist in
a single cell? Mesospheric and acidophil. Aero-tolerant and basophilic. Thermophiles and
psychrophilic. Fermentative and Anaerobic. If a population of 4 bacterial cells in log phase, how
many cells can we expect to see in an hour\'s time if the cell has a generation time of 30 minutes?
8 cells. 16 cells. 64 cells. 128 cells. The classification of ribosomes in bacteria would be:
Ribosomes are proteins and bacteria contain 70s versions. Ribosomes are proteins and bacteria
contain 80s versions. Ribosomes are membrane lipids and bacteria contain 70s versions.
Ribosomes are membrane lipids and bacteria contain 80s versions. Where do The Calvin Cycle
and Glycolysis converge? Only after the \"prep step\" once acetyl CoA is formed, leading to the
Calvin Cycle Only when NADH is utilized in the electron transport chain At Glyceraldehyde 3
Phosphate. None of the above are correct.
Solution
8.a primary active transport proteins
The basal body of a bacterial flagellum is a rod and a system of rings embedded in the cell
envelope.Gram-negative flagella usually have an L ring in the plane of the lipopolysaccharide in
the outer membrane,a periplasmic P ring in the plane of the peptidoglycan around the flagellar
rod and a M S ring that is located within and above the cytoplasmic membrane.A ,C ring extends
into the cytoplasm.
Enzyme I transfers phosphate from PEP to HPr.PEP group translocation is a distinct method
used by bacteria for sugar uptake where the source of energy is from phosphoenolpyruvate
(PEP).It is known as multicomponent system that always involves enzymes of the plasma
membrane and those in the cytoplasm.
ATP synthase is located within the thylakoid membrane and the inner mitochondrial membrane
in bacteria.
9.d all are true.
In photosystem I energy is absorbed by a pair of P700 chlrophyll a molecules raise to an excited
energy level.From there they pass onto FeS4,then onto ferrodoxin,and finally onto ferrodoxin-
NADP reductase.After 2 electrons have reduced ferrodoxin-NADP reductase they are transferred
to NADP+ reducing it to NADPH and a H+.Photosystem II works with Photosystem I and two
series of enzymes imbedded in the thylakoid membrane to transfer energy from the form of light
to that stored in chemical bands and gradients which the plant can use in a process called
noncyclic photophosph.
Which of these characteristics distinguishes angiosperms from other .pdfSIGMATAX1
Which of these characteristics distinguishes angiosperms from other seed plants?
roots
Which of these characteristics distinguishes angiosperms from other seed
plants?flowersseedstracheids
roots
Solution
1. Only angiosperms bear true fruit and have flowers.
2.sporophyte and gametophyte generations, swimming sperm and desiccation tolerance
3. monocots (ie cereals)
4. mechanical support (in form of stem and root anchored to the soil)
5. the diploid gametophyte
6. Pollen grains contain the male gametes of gymnosperms, and seeds contain the female
gametes called ovules..
Which of the following is MOST responsible for maintaining the doubl.pdfSIGMATAX1
Which of the following is MOST responsible for maintaining the double-stranded structure of
DNA?
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
B. Hydrogen bonding between the 3\'-OH of deoxyribose and purine and pyrimidine ring
nitrogens.
C. Hydrophobic interactions between contiguous base pairs and hydrogen bonding between
complementary bases on opposite strands.
D. Interactions between the phosphate groups in separate chains.
E. Hydrophilic interactions between the pentoses in separate chains.
Solution
A. Attractive interactions between positively charged purines and negatively charged
pyrimidines.
Explanation- Attractive interactions between partial positive charge on purines and partial
negative charge on pyrimidines is responsible for maintaining stability of DNA.
Hydrogen bonds in the DNA strands are individually weak but are collectively strong. These
bonds between complementary base pairs stabilize the helical structure of DNA.
Adenine is paired through two hydrogen bonds with thymine. Guanidine is paired through three
hydrogen bonds with cytosine..
What will be quantization step size in numbers and in voltage for th.pdfSIGMATAX1
What will be quantization step size in numbers and in voltage for this Arduino Code? Using 5V
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
const int led1 = 2;
const int led2 = 3;
const int led3 = 4;
void setup() {
pinMode(led1, OUTPUT);
pinMode(led2, OUTPUT);
pinMode(led3, OUTPUT);
// initialize serial communication at 9600 bits per second:
Serial.begin(9600);
}
// the loop routine runs over and over again forever:
void loop() {
// read the input on analog pin 0:
int sensorValue = analogRead(A0);
int dataConv = sensorValue*(8.0/1024);
//write analog equvivalant data on led pins
switch(dataConv)
{
case 0: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 1: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,LOW);
break;
}
case 2: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 3: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,LOW);
break;
}
case 4: {
digitalWrite(led1,LOW);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 5: {
digitalWrite(led1,HIGH);
digitalWrite(led2,LOW);
digitalWrite(led3,HIGH);
break;
}
case 6: {
digitalWrite(led1,LOW);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
case 7: {
digitalWrite(led1,HIGH);
digitalWrite(led2,HIGH);
digitalWrite(led3,HIGH);
break;
}
}
Serial.println(sensorValue); // print out the value you read:
Serial.println(dataConv);
delay(1000); // delay in between reads for stability
}
Solution
// Random LED Dots - from noise source
// Ed Nisley - KE4ANU - September 2015
//----------
// Pin assignments
const byte PIN_HEARTBEAT = 8; // DO - heartbeat LED
const byte PIN_SYNC = A3; // DO.
What is the need for a shadow page table (One or two brief sentences.pdfSIGMATAX1
What is the need for a shadow page table? (One or two brief sentences please)
Solution
Virtualization allows multiple operating system instances to run concurrently on a single
computer. Virtualization is the creation of a virtual rather than actual version of something, such
as an operating system, a server, a storage device or network resources.
Virtualization is where the framework divides the resource into one or more execution
environments Even something as simple as partitioning a hard drive is considered virtualization
because you take one drive and partition it to create two separate hard drives.
Types of virtualization:
Storage virtualization: is the pooling of physical storage from multiple network storage devices
into what appears to be a single storage device
Data virtualization: is abstracting the traditional technical details of data and data management,
such as location, performance or format..
Network virtualization: is a method of combining the available resources in a network by
splitting up the available bandwidth into channels, each of which is independent from the others.
Server virtualization: the partitioning a physical server into smaller virtual servers.
Shadow Tables are objects used to improve the way machines, networks and programs handle
information.
Shadow page tables are used by the hypervisor to keep track of the state in which the guest
thinks its page tables should be.
The guest can\'t be allowed access to the hardware page tables because then it would essentially
have control of the machine. So, the hypervisor keeps the \"real\" mappings guest virtual -> host
physical in the hardware when the relevant guest is executing, and keeps a representation of the
page table..
What step of protein synthesis is shown in the figure elongation st.pdfSIGMATAX1
What step of protein synthesis is shown in the figure? elongation stage of transcription
termination stage of translation mRNA processing elongation stage of translation initiation
stage of transcription
Solution
elongation stage of transcription
Explanation: newly synthesized mRNA separates from the DNA template as the RNA plymerase
continues elongation. multiple RNA plymerases transcribe simultaneously..
Water is forced through a contraction causing low pressure. The wat.pdfSIGMATAX1
Water is forced through a contraction causing low pressure. The water is observed to boil at a
pressure of -13.01 psi. If atmospheric pressure is 14.70 psi, what is the temperature, in °F, of the
water?
Solution
the water is start boiling at 212 °F
it means the temperature of water at 14.70psi is 212 °F +239.538 °F =451.53 °F.
Using a PIC 18 Write a time overflow interrupt routine that detects .pdfSIGMATAX1
Using a PIC 18 Write a time overflow interrupt routine that detects the interrupt source if it
overflows, increments the timer overflow counter, and finally clears the interrupt flag
Solution
ORG 00H
GOTO MAIN
ORG 0X08H
MOVLW 01H
ADDWF TMR0L
BCF INTCON,TMR0IF
RETIF
MAIN
ORG 100H
BCF INTCON,TMR0IF
BSF T0CON,TMR0ON
BSF INTCON,TMR0IE
BSF INTC0N,GIE
HERE GOTO HERE
END.
The three types of mixtures are and The weak bond forming a bridge .pdfSIGMATAX1
The three types of mixtures are and The weak bond forming a bridge between a hydrogen atom
and nitrogen or oxygen is a bond. A is an inorganic compound which has a bitter taste, feels
slipper and is a proton acceptor.
Solution
10. Three types of mixtures are suspensions, solutions and colloids
11.answer is hydrogen bond.
A hydrogen bond is the electrostatic attraction between two polar groups that occurs when a
hydrogen (H) atom covalently bound to a highly electronegative atom such as nitrogen (N),
oxygen (O), or fluorine (F) experiences the electrostatic field of another highly electronegative
atom nearby.
12.answer is Base..
Some commercial airplanes recirculate approximately 50 of the cabin .pdfSIGMATAX1
Some commercial airplanes recirculate approximately 50% of the cabin air in order to increase
fuel efficiency. The researchers studied 1096 airline passengers, among which some traveled on
airplanes that recirculated air and others traveled on planes that did not recirculate air. Of the 515
passengers who flew on planes that did not recirculate air, 106 reported post-flight respiratory
symptoms, while 112 of the 581 passengers on planes that did recirculate air reported such
symptoms. Is there sufficient evidence to conclude that the proportion of passengers with post-
flight respiratory symptoms differs for planes that do and do not recirculate air? Test the
appropriate hypotheses using a = 0.05. You may assume that it is reasonable to regard these two
samples as being independently selected and as representative of the two populations of interest.
(Use a statistical computer package to calculate the P-value. Use Pdo not recirculate - Pdo
recirculate- Round your test statistic to two decimal places and your P-value to four decimal
places.)
Solution
Null , There Is No Significance between them Ho: p1 = p2
Alternate , There Is Significance between them H1: p1 != p2
Test Statistic
Sample 1 : X1 =106, n1 =515, P1= X1/n1=0.206
Sample 2 : X2 =112, n2 =581, P2= X2/n2=0.193
Finding a P^ value For Proportion P^=(X1 + X2 ) / (n1+n2)
P^=0.199
Q^ Value For Proportion= 1-P^=0.801
we use Test Statistic (Z) = (P1-P2)/(P^Q^(1/n1+1/n2))
Zo =(0.206-0.193)/Sqrt((0.199*0.801(1/515+1/581))
Zo =0.54
| Zo | =0.54
Critical Value
The Value of |Z | at LOS 0.05% is 1.96
We got |Zo| =0.54 & | Z | =1.96
Make Decision
Hence Value of |Zo | < | Z | and Here we Do not Reject Ho
P-Value: Two Tailed ( double the one tail ) -Ha : ( P != 0.5403 ) = 0.589
Hence Value of P0.05 < 0.589,Here We Do not Reject Ho
No sufficient evidence to condude that the proportion of passengers
with post-flight respratory symptoms differs for planes that do and do not recirculate air
ANS:
Z = 0.54
P = 0.589.
Solve for 02 sin 2+sin 5=0. Separate your answers with commas.pdfSIGMATAX1
Solve for 0??<2?: sin 2?+sin 5?=0. Separate your answers with commas.
?= Solve for 0 lessthanorequalto theta
Solution
Use formula : sinx +siny = 2sin(x+y)/2cos(x-y)/2
So, sin 2+sin 5=0.
2sin(7)/2cos(3/2) =0
sin(7)/2 =0
= 0, 2pi/7 , 4pi/7, 6pi/7 , 8pi/7 , 10pi/7 , 12pi/7 ,
cos(3)/2 =0
= pi/3 , pi , 5pi/3.
Required 1. Briefly describe different computer controls available .pdfSIGMATAX1
Required: 1. Briefly describe different computer controls available to an organization.
a. Identify how these controls help to maintain system security and reduce risk.
b. Do companies need a formal internal policy that addresses security violations? Why/why not?
c. Consider what type of actions a company should take against employees that circumvent
computer
Solution
Security Controls:
Administrative Controls, Technical or Logical Controls, Physical Controls
The constituents of Security:
Vulnerability, Threat, Risk, Exposure, Countermeasure or Safeguard
The data vendor grants access needs or may choose to give this function to business unit
managers. And it is the data owner who will contract with security violations pertaining to the
data he is liable for defending.
Subjects that should be measured when developing a security policy:
Classification Controls:
Qualitative Risk Analysis Techniques:.
a. What is Snort [1]b. According to “Snort Users Manual” (downloa.pdfSIGMATAX1
a. What is Snort? [1]
b. According to “Snort Users Manual” (downloadable under “Docs” menu), what are the three
modes in which Snort operates and how it behaves under each mode? Show an example
command to put Snort in each mode. [6]
c. Explain the difference between the rule headers and rule options [2]
d. List 8 different types of actions Snort may take upon a rule match. Describe what each action
type does. [2]
Solution
a. Snort: Snort is an open source network capable of performing real-time traffic study and
packet-logging on IP networks. It can transmit the protocol analysis, content searching &
matching and is used to detect a range of attacks and investigates, such as buffer run over,
cautious port scans, OS fingerprinting efforts.
d. Types of snort actions:.
Proponents of decentralized forms of socialism usually supportA co.pdfSIGMATAX1
Proponents of decentralized forms of socialism usually support
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
A command economy
Control of the economy by a vanguard party of intellectuals
Control of the economy by various local groups of workers
State control of the economy
The Phoenix Suns
Solution
Proponents of decentralized forms of socialism usually support the following things such a
1.A command economy.
2.Control of the economy by a vanguard party of intellectuals.
3.Control of the economy by various local groups of workers.
4.State control of the economy.
5.The phoenix suns..
Unit 8 - Information and Communication Technology (Paper I).pdfThiyagu K
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
What are the role of environmental factors on transpiration and h.pdf
1. What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
What are the role of environmental factors on transpiration and how plants defend themselves
against water loss?
Solution
Role of environmental factors on transpiration
1. Relative humidity
The loss of water vapour from stomata occurs by simple diffusion. When the atmosphere is very
humid, the rate of loss of water vapour is lowered.
Wind:
When the wind velocity is high, rate of transpiration also increases.
Temperature:
Warm air hold more water vapour so the rate of loss of water increases with increase in
temperature.
Light:
Opening and closing of stomata is regulated by light. More light result in high degree of
transpiration.
Soil water content:
When there is less water content in the soil , the leaf droops / get dehydrated making the stomata
to close resulting in less rate of transpiration.
Carbon dioxide concentration
When the concentration is high, stomata closes resulting in low rate of transpiration and vice
versa.
Plants defense mechanism against water loss
Plants require water for it's survival. So it will have to regulate water loss ( especially during
drought) .
Several leaf adaptations help to combat water loss.
Leaf arrangements play important role, transversely arranged leaves loss more water than
perpendicularly arranged ones.
Other anatomical features of the leaves such as sunken/vestigial stomata and closely arranged
mesophyll cells reduces transpiration.
Presence of hair, cuticle(waxy) , gums, mucilage etc on leaf surface help reducing transpiration.
2. During drought, plant hormone abscisic acid play an important role by closing the stomata
reducing water loss.
