Alice Agent is an employee of Patti Principal. She negligently runs .pdf
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Alice Agent is an employee of Patti Principal. She negligently runs a stop sign in the course of her employment and injures Bart Bikerider. Can Bart sue Patti for his damages from Alice’s Negligence? Why or why not? Solution Yes Bart can sue Patti for his damages because the employer is responsible for actions of employees during the course of their employment. The doctrine is called respondeat superior..
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Note the structures through which light passes to reach the retina. W.pdf
Note the structures through which light passes to reach the retina. Write these structures in
order.
Solution
The order of light passage from outside till to retina:
Initially light enters cornea & sclera layers ---> pupil ---> circular Iris that surrounds pupil using
pupillary sphincter muscle --> light enters lens---> retinal layers ---> rods & cones
Explanation:
Neural layer is the inner layer; it is in contact with the vitreous humour. The neural layer
contains three distinct populations of cells namely,
From the outer (pigmented epithelial layer) to the inner (towards the vitreous humour), these
cells are arranged in the order of, photoreceptors, bipolar cells, ganglion cells.
Photoreceptor cells (rods and cones) are communicated with the nerve cells called bipolar cells.
These bipolar cells synapse with the ganglion cells, which transmit the impulses to the brain via
axons in the optic nerve.
Retina is the innermost nervous coat of eyeball, and it contains some special photo receptor
structures called “rods and cones.” These rods and cones receive the light and send impulses to
the optic nerve.
The inner nuclear layer of the retina contains oval shaped bipolar cells. The dendrites of bipolar
cells synapse with the photoreceptor cells, and axons of bipolar cells synapse with the ganglionic
cells.
Ganglion cells are multipolar cells present in the ganglion cell layer of retina; their axons are
present in the inner surface of the retina, forming the optic nerve. The dendrites of ganglion cells
synapse with the axons of bipolar cells. These ganglion cells transmit the impulses to the brain
via axons in the optic nerve..
Muriel recently stopped taking her medication resulting in flulike s.pdf
Muriel recently stopped taking her medication resulting in flulike symptoms. She was most
likely taking a(n):
A. atypical second-generation antidepressant.
B. SSRI.
C. MAOI.
D. tricyclic antidepressant.
Solution
1. The above all 4 drugs belong to antidepressant drugs.
2. Atypical second-generation antidepressants are antipsychotic drugs. These drugs have less
side effects.
3. The remaining three SSRI, MAOI and Tricyclic antidepressant are antidepressant mostly used
in panic disorder or social phobia.
4. MAOI is used as last line treatment when SSRI and Tricyclic antidepressant have failed.
5. Side effects of tricyclic antidepressant causes symptoms like flu such as dry mounth, dry nose,
blurry vision, and increasing body temperature.
6. So she was most likely taking tricyclic antidepressant..
managerial reporting systems areStandardizedRigidFlexible.pdf
managerial reporting systems are:
Standardized
Rigid
Flexible
managerial reporting systems are:
Standardized
Rigid
Flexible
managerial reporting systems are:
Standardized
Rigid
Flexible
Solution
Managerial accounting systems are Flexible. The budgets, standards are revised according to
the situations and to adapt to changes.
Lot X_1, ..., X_n be independent and identically distributed continuo.pdf
Lot X_1, ..., X_n be independent and identically distributed continuous random variables having
cumulative distribution function F(x). and let X_(i) denote the ordered sequence of the (realized
values) of these random variables, i.e. X_(1)
Solution
a)
P( Y > X) = X1 + X2 + X3 ... Xn
b)
P( X < Y < X ) = X1 + X2 +X3 + ... +X6.
Let Y follows a Poisson distribution with mean lambda that is, Y app.pdf
Let Y follows a Poisson distribution with mean lambda: that is, Y approx equal
Poisson(lambda). Show that E(Y) = lambda. Hint: you may use one of the following two
methods Method 1: use exponential series e^x = 1 + x + x^2/2! + x^3/3! + .,.. Method 2: Use the
2nd property of a valid pmf that summation y p(y) = 1
Solution
P(x; ?) = (e-?) (?x) / x!
xP(x) = x (e-?) (?x) / x! = (e-?) (?x) / (x-1)!= (e-?) (?) (?x-1) / (x-1)!
Mean = sum of all xP(X)
= (e-?) (?) e??
= ?
Thus mean of Poisson distribution is its parameter.
Or E(Y) = Lemda.
In humans, the HOXD homeotic gene cluster plays a critical role in l.pdf
In humans, the HOXD homeotic gene cluster plays a critical role in limb development. In one
large family, 16 of 36 members expressed one of two dominantly inherited malformations of the
feet known as rocker bottom foot (CVT) or claw foot (CMT). One individual had one foot with
CVT and the other with CMT. Genomic analysis identified a single missense mutation in the
HOXD10 gene, resulting in a single amino acid substitution in the homeodomain of the encoded
transcription factor. This region is crucial for making contact and binding to the target genes
controlled by this protein. All family members with the foot malformations were heterzygotes;
all unaffected members were homozygous for the normal allele.
1) Given that affected heterozygotes carry one normal allele of the HOX10D gene, how mught a
dominant mutation in a gene encoding a transcription factor lead to a developmental
malformation?
2) How can two clinically different phenotypes result from the same mutation?
3) What might we learn about the control of developmental processes from an understanding of
how this mutation works?
Solution
1)
If the wild type allele is dominant, it would not allow the expression of the dominant mutation. If
the wild type allele is recessive, it would allow the expression of the dominant mutation.
Here, the heterozygotes possess wild type allele that is recessive and hence cannot prevent the
expression of the mutation. Homozygotes possess wild type allele that is dominant and hence can
prevent the mutation expression.
2)
If the recessive alleles present at a given locus are different from each other in having two
mutations at different loci, then two different phenotypes result from a single mutation. Here,
one allele represents a dominant mutation and the other is a recessive allele of normal type.
Often two unrelated allele inheritance might result in the two different phenotypes, one due to a
classic mutation and other due to a rare and a new mutation.
3)
If a weak or recessive allele exists along with a dominant mutation at the same locus, the
dominant mutation might have negative impact on the normal wild type recessive allele as well
during the inheritance. This impact will show a lot of difference in the expression of the gene and
in turn in the control of the developmental processes..
I am sorry but my major does not cover programming in depth (ICT) an.pdf
I am sorry but my major does not cover programming in depth (ICT) and we are expected to
know advanced java programming. I am looking for help. I have purchased a book from Oracle
but I will not be up to speed this semester. Please Help me!!!
1. Write an app called viewer that will have a Label at the top saying \"My Viewer\" (or
something like that)
2. Will have JButtons at the bottom that will do Next, Previous, and Quit
3. Have the whole middle be a JLabel in which you will display Images stored in a directory.
4. The directory can be named Resource.
5. When you run the program (java viewer) it will read all the names in the Resource Directory.
Then, when you click Next or Previous it will display an Image.
6. Note: you will need to find a java method that exists for reading a whole directory of
filenames. You can store all those names in a String Array when run the program.
7. You will use a counter or index that is an int an when you click Next it will increment the
counter until it reach some maximum value and then you will set it to 0. Previous will decrement
the counter until it goes negative and then it will set the counter to the Maximum index ( which
is how many filenames you have in the Image names array)
8. Submit the program viewer.java I should be able to use it with my own Resource directory
Solution
Compilation process
javac Viewer.java
Run
java Viewer
Click on next or previous a file chooser will be opened and you can select your file
The Code is in four file.
// File Dg.java
import java.io.File;
public class Dg {
public File nextFile(File file) {
File f[] = new File[5000];
f = file.getParentFile().listFiles(new ImageFileFilter());
int max = f.length;
max = max - 1;
// System.out.println(file);
for (int i = 0; i < f.length; i++) {
if (f[i].equals(file)) {
if (i + 1 > max) {
return f[0];
} else {
return f[i + 1];
}
} else {
// System.out.println(0);
// return file;
}
}
return file;
}
public File priFile(File file) {
File f[] = new File[5000];
f = file.getParentFile().listFiles(new ImageFileFilter());
int min = 0;
// System.out.println(file);
for (int i = 0; i < f.length; i++) {
if (f[i].equals(file)) {
min = i - 1;
if (min < 0) {
return f[f.length-1];
} else {
return f[min];
}
} else {
// System.out.println(0);
// return file;
}
}
return file;
}
}
/// Image.java
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.geom.AffineTransform;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;
import javax.swing.JPanel;
import java.awt.Color;
import javax.swing.JLabel;
import java.awt.Font;
public class Image extends JPanel {
/**
*
*/
private static final long serialVersionUID = 1L;
public BufferedImage img = null;
/**
* Create the panel.
*/
@SuppressWarnings(\"static-access\")
public Image() {
setBackground(new Color(255, 255, 255));
setLayout(null);
try {
Viewer ocr = new Viewer();
img = ImageIO.read(ocr.file);
JLabel lblNewLabel = new JLabel(Viewer.f.
explain why blockage or removal of the lymphatic vessels can result .pdf
explain why blockage or removal of the lymphatic vessels can result in significant edema
Solution
LYMPH: Lymph is the second largest circulatory system in human body.It is made up of
lymphatic vessels or capillaries.
FUNCTION OF LYMPHATIC VESSELS: The main function of lymphatic vessels or lymphatic
capillaries is pick up the fluid that leaks in to the tissues from the blood stream and return into
the circulatory system.it absorbs the lymphatic fluid from the tissues.
The blockage ir removal of the lymphatic vessels leads to the significant edema.
EDEMA: Edema means swellimg of a perticular body part due to the accumulation of fluid.or
inflamation of a body part.other wise it can happens due to the injury of a body part.
The obstruction of the lymphatic capillaries or lymphatic vessels leads to its disfunction..so that
the absorption of accumulatic lymphatic fluid from the tissues does not takes place in the body..
this leads to the edema..This may occurs at a perticular region of the body or somtimes entire
body can be affected..
Recommended
Note the structures through which light passes to reach the retina. W.pdf
Note the structures through which light passes to reach the retina. Write these structures in
order.
Solution
The order of light passage from outside till to retina:
Initially light enters cornea & sclera layers ---> pupil ---> circular Iris that surrounds pupil using
pupillary sphincter muscle --> light enters lens---> retinal layers ---> rods & cones
Explanation:
Neural layer is the inner layer; it is in contact with the vitreous humour. The neural layer
contains three distinct populations of cells namely,
From the outer (pigmented epithelial layer) to the inner (towards the vitreous humour), these
cells are arranged in the order of, photoreceptors, bipolar cells, ganglion cells.
Photoreceptor cells (rods and cones) are communicated with the nerve cells called bipolar cells.
These bipolar cells synapse with the ganglion cells, which transmit the impulses to the brain via
axons in the optic nerve.
Retina is the innermost nervous coat of eyeball, and it contains some special photo receptor
structures called “rods and cones.” These rods and cones receive the light and send impulses to
the optic nerve.
The inner nuclear layer of the retina contains oval shaped bipolar cells. The dendrites of bipolar
cells synapse with the photoreceptor cells, and axons of bipolar cells synapse with the ganglionic
cells.
Ganglion cells are multipolar cells present in the ganglion cell layer of retina; their axons are
present in the inner surface of the retina, forming the optic nerve. The dendrites of ganglion cells
synapse with the axons of bipolar cells. These ganglion cells transmit the impulses to the brain
via axons in the optic nerve..
Muriel recently stopped taking her medication resulting in flulike s.pdf
Muriel recently stopped taking her medication resulting in flulike symptoms. She was most
likely taking a(n):
A. atypical second-generation antidepressant.
B. SSRI.
C. MAOI.
D. tricyclic antidepressant.
Solution
1. The above all 4 drugs belong to antidepressant drugs.
2. Atypical second-generation antidepressants are antipsychotic drugs. These drugs have less
side effects.
3. The remaining three SSRI, MAOI and Tricyclic antidepressant are antidepressant mostly used
in panic disorder or social phobia.
4. MAOI is used as last line treatment when SSRI and Tricyclic antidepressant have failed.
5. Side effects of tricyclic antidepressant causes symptoms like flu such as dry mounth, dry nose,
blurry vision, and increasing body temperature.
6. So she was most likely taking tricyclic antidepressant..
managerial reporting systems areStandardizedRigidFlexible.pdf
managerial reporting systems are:
Standardized
Rigid
Flexible
managerial reporting systems are:
Standardized
Rigid
Flexible
managerial reporting systems are:
Standardized
Rigid
Flexible
Solution
Managerial accounting systems are Flexible. The budgets, standards are revised according to
the situations and to adapt to changes.
Lot X_1, ..., X_n be independent and identically distributed continuo.pdf
Lot X_1, ..., X_n be independent and identically distributed continuous random variables having
cumulative distribution function F(x). and let X_(i) denote the ordered sequence of the (realized
values) of these random variables, i.e. X_(1)
Solution
a)
P( Y > X) = X1 + X2 + X3 ... Xn
b)
P( X < Y < X ) = X1 + X2 +X3 + ... +X6.
Let Y follows a Poisson distribution with mean lambda that is, Y app.pdf
Let Y follows a Poisson distribution with mean lambda: that is, Y approx equal
Poisson(lambda). Show that E(Y) = lambda. Hint: you may use one of the following two
methods Method 1: use exponential series e^x = 1 + x + x^2/2! + x^3/3! + .,.. Method 2: Use the
2nd property of a valid pmf that summation y p(y) = 1
Solution
P(x; ?) = (e-?) (?x) / x!
xP(x) = x (e-?) (?x) / x! = (e-?) (?x) / (x-1)!= (e-?) (?) (?x-1) / (x-1)!
Mean = sum of all xP(X)
= (e-?) (?) e??
= ?
Thus mean of Poisson distribution is its parameter.
Or E(Y) = Lemda.
In humans, the HOXD homeotic gene cluster plays a critical role in l.pdf
In humans, the HOXD homeotic gene cluster plays a critical role in limb development. In one
large family, 16 of 36 members expressed one of two dominantly inherited malformations of the
feet known as rocker bottom foot (CVT) or claw foot (CMT). One individual had one foot with
CVT and the other with CMT. Genomic analysis identified a single missense mutation in the
HOXD10 gene, resulting in a single amino acid substitution in the homeodomain of the encoded
transcription factor. This region is crucial for making contact and binding to the target genes
controlled by this protein. All family members with the foot malformations were heterzygotes;
all unaffected members were homozygous for the normal allele.
1) Given that affected heterozygotes carry one normal allele of the HOX10D gene, how mught a
dominant mutation in a gene encoding a transcription factor lead to a developmental
malformation?
2) How can two clinically different phenotypes result from the same mutation?
3) What might we learn about the control of developmental processes from an understanding of
how this mutation works?
Solution
1)
If the wild type allele is dominant, it would not allow the expression of the dominant mutation. If
the wild type allele is recessive, it would allow the expression of the dominant mutation.
Here, the heterozygotes possess wild type allele that is recessive and hence cannot prevent the
expression of the mutation. Homozygotes possess wild type allele that is dominant and hence can
prevent the mutation expression.
2)
If the recessive alleles present at a given locus are different from each other in having two
mutations at different loci, then two different phenotypes result from a single mutation. Here,
one allele represents a dominant mutation and the other is a recessive allele of normal type.
Often two unrelated allele inheritance might result in the two different phenotypes, one due to a
classic mutation and other due to a rare and a new mutation.
3)
If a weak or recessive allele exists along with a dominant mutation at the same locus, the
dominant mutation might have negative impact on the normal wild type recessive allele as well
during the inheritance. This impact will show a lot of difference in the expression of the gene and
in turn in the control of the developmental processes..
I am sorry but my major does not cover programming in depth (ICT) an.pdf
I am sorry but my major does not cover programming in depth (ICT) and we are expected to
know advanced java programming. I am looking for help. I have purchased a book from Oracle
but I will not be up to speed this semester. Please Help me!!!
1. Write an app called viewer that will have a Label at the top saying \"My Viewer\" (or
something like that)
2. Will have JButtons at the bottom that will do Next, Previous, and Quit
3. Have the whole middle be a JLabel in which you will display Images stored in a directory.
4. The directory can be named Resource.
5. When you run the program (java viewer) it will read all the names in the Resource Directory.
Then, when you click Next or Previous it will display an Image.
6. Note: you will need to find a java method that exists for reading a whole directory of
filenames. You can store all those names in a String Array when run the program.
7. You will use a counter or index that is an int an when you click Next it will increment the
counter until it reach some maximum value and then you will set it to 0. Previous will decrement
the counter until it goes negative and then it will set the counter to the Maximum index ( which
is how many filenames you have in the Image names array)
8. Submit the program viewer.java I should be able to use it with my own Resource directory
Solution
Compilation process
javac Viewer.java
Run
java Viewer
Click on next or previous a file chooser will be opened and you can select your file
The Code is in four file.
// File Dg.java
import java.io.File;
public class Dg {
public File nextFile(File file) {
File f[] = new File[5000];
f = file.getParentFile().listFiles(new ImageFileFilter());
int max = f.length;
max = max - 1;
// System.out.println(file);
for (int i = 0; i < f.length; i++) {
if (f[i].equals(file)) {
if (i + 1 > max) {
return f[0];
} else {
return f[i + 1];
}
} else {
// System.out.println(0);
// return file;
}
}
return file;
}
public File priFile(File file) {
File f[] = new File[5000];
f = file.getParentFile().listFiles(new ImageFileFilter());
int min = 0;
// System.out.println(file);
for (int i = 0; i < f.length; i++) {
if (f[i].equals(file)) {
min = i - 1;
if (min < 0) {
return f[f.length-1];
} else {
return f[min];
}
} else {
// System.out.println(0);
// return file;
}
}
return file;
}
}
/// Image.java
import java.awt.Graphics;
import java.awt.Graphics2D;
import java.awt.geom.AffineTransform;
import java.awt.image.BufferedImage;
import java.io.File;
import java.io.IOException;
import javax.imageio.ImageIO;
import javax.swing.JPanel;
import java.awt.Color;
import javax.swing.JLabel;
import java.awt.Font;
public class Image extends JPanel {
/**
*
*/
private static final long serialVersionUID = 1L;
public BufferedImage img = null;
/**
* Create the panel.
*/
@SuppressWarnings(\"static-access\")
public Image() {
setBackground(new Color(255, 255, 255));
setLayout(null);
try {
Viewer ocr = new Viewer();
img = ImageIO.read(ocr.file);
JLabel lblNewLabel = new JLabel(Viewer.f.
explain why blockage or removal of the lymphatic vessels can result .pdf
explain why blockage or removal of the lymphatic vessels can result in significant edema
Solution
LYMPH: Lymph is the second largest circulatory system in human body.It is made up of
lymphatic vessels or capillaries.
FUNCTION OF LYMPHATIC VESSELS: The main function of lymphatic vessels or lymphatic
capillaries is pick up the fluid that leaks in to the tissues from the blood stream and return into
the circulatory system.it absorbs the lymphatic fluid from the tissues.
The blockage ir removal of the lymphatic vessels leads to the significant edema.
EDEMA: Edema means swellimg of a perticular body part due to the accumulation of fluid.or
inflamation of a body part.other wise it can happens due to the injury of a body part.
The obstruction of the lymphatic capillaries or lymphatic vessels leads to its disfunction..so that
the absorption of accumulatic lymphatic fluid from the tissues does not takes place in the body..
this leads to the edema..This may occurs at a perticular region of the body or somtimes entire
body can be affected..
Describe three different symbiotic relationships, with examples. Whi.pdf
Describe three different symbiotic relationships, with examples. Which relationship produces the
most coevolution? Explain.
Please type the answer. Type the answer in your own words please. no copying from the internet
has to be in own words
Solution
Three different types of symbiotic relatioships are :
1. Mutualism
2. Commensalism
3. Parasitism
1.Mutualism is the relationship between organisms where both are benefitted from each other.
Examples of mutualism:
a. Oxpecker on zebra eats pest. Oxpecker gets food and zebra gets rid of pest.
b. Shape of flower and structure of pollinator is also an example of mutulism as shape of the
flower is specific for the pollinator to allow pollination.
c. Bacteria in gut of human intestine benefit human by partially digesting the food difficult to
digest. Bacteria gets food through this and human is able to digest tougher to dogest food.
d. Mycorhhiza is association between fungus and gymnosperm root. Plant gets benefitted by
collection of nutrients from soil and fungus gets benefit by plant in form of food and shelter.
2. Commensalism: In commensalism association, one organism is benefitted whereas other is not
affected at all.
examples:
a. Cattle Egret and cattle: as cattle grazes and uproots plants, insects that escape are eaten by
egret. Egret gets the benefit wheras cattle is not affected.
b Sucker fish attaches itself to the shark and travels along. Shark doesnt get affected whereas the
sucker fish gets left over food of shark.
c.Epiphytes are plants that grow on larger trees. They just climb the trees for receiving sunlight
and do not cause any affect to the plant.
d. CLown fish lives in tentacles of Sea anemones. Sea anemone doesnt get affected whereas
clown fish gets the advantage of protection from predators.
3. Parasitism: In this relationship, the parasite harms the host to derive the benefit in terms of
food or shelter or both. In this type of relationship one species is harmed whereas other is
benefitted.
examples of parasitism:
a. Lice on Human body: Lice sucks blood can causes harm to human.
b. Cuscuta causes harm to higher plants by living on it and taking the benefit of food and shelter.
c. Ascaris lives in human intestine . It benefits from receiving food and shelter and causes harm
to human.
d. tapeworm in Human intestine as they get partially digested food and deprive human of
nutrients.
Out of these relationships, coevolution is most important for species showing mutualistic
behaviour as:
1. They both have large dependency on each other for their survival.
2. Any change in one species can lead to extinction of the other. Coevolution is the only strategy
for their survival..
Describe the differences seen in the hydra, planarian, clam, grasshop.pdf
Describe the differences seen in the hydra, planarian, clam, grasshopper, and pig for symmetry,
tissue organization, body cavity, digestive system, circulatory system, respiratory system,
excretory system, support system and nervous system.
Solution
Hydra
Planarian:
Clam:
Symmetry
radial
bilateral
bilateral
Tissue Organization
diploblastic
triploblastic
triploblastic
Type of Body Cavity:
gastrovascular cavity
acoelomate
coelomate:
pericardial cavity
Digestive Openings
one
mouth
(pharynx extended during feeding)
mouth, a nu s
Circulatory System
none
none
heart, open c.s.
Respiratory Organs:
none
none
incurrent and excurrent siphon, gills
Excretory System
none
protonephridia, flame bulb
nephridia
Support System
: none
none
nephridia
Nervous System
Organization: nerve net
Organization: brain, ventral nerve cords
Hydra
Planarian:
Clam:
Symmetry
radial
bilateral
bilateral
Tissue Organization
diploblastic
triploblastic
triploblastic
Type of Body Cavity:
gastrovascular cavity
acoelomate
coelomate:
pericardial cavity
Digestive Openings
one
mouth
(pharynx extended during feeding)
mouth, a nu s
Circulatory System
none
none
heart, open c.s.
Respiratory Organs:
none
none
incurrent and excurrent siphon, gills
Excretory System
none
protonephridia, flame bulb
nephridia
Support System
: none
none
nephridia
Nervous System
Organization: nerve net
Organization: brain, ventral nerve cordsNervous System Organization: three ganglia.
Companyhas prot s Price per unit Variable cost per unit Fixed costs p.pdf
Companyhas prot s Price per unit Variable cost per unit Fixed costs per month $%4 $17,00
What is the contribution margin ratio? A) 18% B) 70% C) 48% D) 33%
Solution
Contribution margin=Sales-Variable costs
=(40-12)=$28/unit
Hence Contribution margin ratio=Contribution margin/Sales
=(28/40)
which is equal to
=70%.
at log kHz SolutionYou can have several problems 1) if you a.pdf
at log kHz
Solution
You can have several problems:
1) if you are working with transistors sure you\'re not saturating.
2) if diodes are sure they are doing the right contact or try another.
3) Make sure your input source this good.
4) for me the most certain is that you are using an active elemnto and this is in court and why
that image, check input sources, the currents that these manipulated are optimal for circuit
elements you use or application you\'re doing..
Bark, which is thicken, toughen periderm, arises from this lateral m.pdf
Bark, which is thicken, toughen periderm, arises from this lateral meristem?
Vascular cambium
Cork cambium
QUESTION 2
Seeds with edible fruit are most likely dispersed by
wind
animals
water
bursting
QUESTION 3
Seeds with wings are most likely dispersed by
wind
animals
water
bursting
QUESTION 4
The arrangement of flowers on the stem with the newest flowers near the end of the shoot is
Determinate inflorescence
Indeterminate inflorescence
QUESTION 5
The arrangement of flowers on the stem with the oldest flowers near the end of the shoot is
Determinate inflorescence
Indeterminate inflorescence
QUESTION 6
The cortex in stems is made of this tissue
Dermal
Ground
Vascular
QUESTION 7
The epidermis, cuticle, guard cells, root hairs and trichomes arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 8
The epidermis, cuticle, guard cells, root hairs and trichomes arise from this type of tissue?
Dermal
Ground
Vascular
QUESTION 9
The parenchyma, collenchyma and sclerenchyma arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 10
The parenchyma, collenchyma and sclerenchyma arise from this type of tissue?
Dermal
Ground
Vascular
QUESTION 11
The xylem and phloem in leaves are contained in which structure?
blade
petiole
stipules
vein
QUESTION 12
This meristem is located in the root and shoot tips.
Apical meristem
Lateral meristem
QUESTION 13
This meristem is located in the stems of plants that live longer than one year.
Apical meristem
Lateral meristem
QUESTION 14
Xylem and phloem elements arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 15
Xylem and phloem elements arise from this type of tissue?
Dermal
Ground
Vascular
Vascular cambium
Cork cambium
Solution
1. Cork cambium. It palys a major role in secondary growth like wood (bark) in dicotyledons.
Vascular cambium is responsible for Xylem and Pholem formation.
2.Animals. Edible part of fruits are eaten by animals and remaining part (seeds) dropped in other
locations.
3. Winds. It has a smaller seeds with feather - like or wings it helps easily transfer in winds.
4. InDeterminate inflorescence.
Acropetal sucession - The new flowers arranged at the top and old flowers are at bottom is
Racemose or Indeterminate inflorescence.
5. Determinate inflorescence
Basipetal sucession - The new flowers arranged at the bottom and old flower are at tip is cymose
or determinate inflorescence.
6.Ground Tissue
It contains simple tissuse are parenchyma, collenchyma and sclerenchyma. Parenchyma cells are
usually present in cortex.
7. Protoderm
It helps in develop epidermal tissues.
8. Dermal tissue
Epidermal tissue system present in outer most layer of plant system. It contains epidermal,
stomato, trichomes and hairs etc.,
9. Ground meristem.
Ground meristem give rise to ground tissues and contains parenchyma, collenchyma and
sclerenchyma.
10. Ground Tissues.
11. Vein
Lamina - Green expanded part of leaf. Petiole - It attach leaf and stem. Stip.
Why do you think it is usually preferable to start with the side con.pdf
Why do you think it is usually preferable to start with the side containing the more complicated
expression when establishing a trigonometric identity?
Solution
1. Human brains are accustomed to simplify complicated things rather than complicating a
simpler one.
2. A simpler/simplest conclusion of a complicated problem in trigonometric identities is only a
handful;mere 2 or 3.
But if I want to complicate a simpler side, outcomes are numerous.
For eg: prove cosx*tanx =sinx
Now I know that if I want to simplify LHS, I will try to convert it is terms of Sin or Cos (which
we consider simple in the world of trigonometry)
So cosx*(sinx/cosx)= cosx which is RHS.
But if I want to complicate sinx ,I can write it as
1/cscx or 1-cos^2x or (1-cos2x)/2 . . . . And the list goes on!
So, by and large, it is customary simplify complicated things rather than the other way.
Hope it clarifies your doubt!.
Why is the mucus thicker than normal in CF patients A good answer w.pdf
Why is the mucus thicker than normal in CF patients? A good answer will include a description
of how osmosis is affected by CF.
Solution
Mucus is a substance that coats our Epithelial surface.It keeps surfaces lubricated and helps
prevent disease by trapping potential invaders.Invading microbes can become trapped in the
mucus.The epithelial cells are lined with cilia that are constantly beating and moving mucus with
trapped bacteria up and out of the respiratory tract.Mucus is amixture of predominantly water
and glycproteins.The mucins are produced by goblet cell that are interspersed among the
respiratory epithelial cell.The composition,water content and volume of mucus are controlled by
the secretion and absorption of ions and water across the respiratory epithelium.The salt
imbalance resulting from a disfuctional chloride channel could through osmosis lead to the thick
mucus characterstic of cystic fibrosis mutation could lead to disease.cystic fibrosis is a thick that
lines both the airways and the intestinal tract of affected individual.In the respiratory tract this
thick mucus is not removed by the cillia that usually beat and push mucus up into the throat.This
thick mucus builds up in the airways and bacteria trapped in the mucus are not removed.Thus
bacterial respiratory infections are extremely common and dangerous in CF individuals..
Which of the following primordial reproductive structures is NOT ext.pdf
Which of the following primordial reproductive structures is NOT external?
A) Urogenital sinus
B) Genital tubercle
C) Urogenital folds
D) Labioscrotal swelling
Solution
Urogenital sinus is the promordial structure not external. It is present only in the develovement
of Reproductive organs.It is divided in to three main regions upper , phallic and pelvic..
what kind of accessory protein would you classify IRS1 and 2 as.pdf
what kind of accessory protein would you classify IRS1 and 2 as?
Docking protein
Adaptor protein?
Transcription factor?
SH2 domain protein
Solution
Docking protein - The robosomes are of two types based on their occurence in the cell. Free
ribosomes and membrane bound (on ER surface). Ever wondered why and how they are of two
types? Yes!! Some specific proteins which are found on the surface of certain specific ribosomal
particles making them \"dock\" (stick) on to the ER surface are called docking proteins.
Adapter protein - An adapter protein is one which assists in signal transduction across
membranes. They aid receptor mechanisms in the cells.
Transcription factor - The process of copying cell DNA codons in to an mRNA strand in called
transcription. This is assisted by certain types of protein called Transcription factors.
SH2 domain - These are specific proteins that mediate and fecilitate protein-protein inteactions
in the cell. SH2 domain (a structurally special part within a protein molecule) is found within Src
oncoprotein (cancer protein) which makes the Src oncoprotein attach to other protein containing
phosphorylated tyrosin residues.
**IRS1 and IRS2 proteins are classified as adapter proteins.**.
What is the main difference between prokaryotic and eukaryotic riboso.pdf
What is the main difference between prokaryotic and eukaryotic ribosomes? Prokaryotic
ribosomes synthesize RNA whereas eukaryotic ribosomes synthesize proteins Eukaryotic
ribosomes are composed of RNA and protons whereas prokaryote ribosomes are
Solution
None of the above
The main difference is that Prokaryotic ribosomes(70S) having two subunits- 30S and 50S
and Eukaryotic ribosomes(80S) also having two subunits- 40S and 60S
Both are composed of RNA & proteins. 70S having 60% RNA & 40% proteins while 80S having
40%RNA & 60% proteins..
what is the cardinality of a sample spaceSolutionA sample spa.pdf
what is the cardinality of a sample space?
Solution
A sample space is a collection of all possible outcomes of a random experiment. A random
variable is a function defined on a sample space. We shall consider several examples shortly.
Later on we shall introduce probability functions on the sample spaces. A sample space may be
finite or infinite. Infinite sample spaces may be discrete or continuous.
What is computer information systems in three paragraphs.Solutio.pdf
What is computer information systems in three paragraphs.
Solution
A computer information system be a system collected of people and computer that process or
interpret information. The word is also from time to time used in more limited senses to refer to
only the software used to sprint a automated database or to refer to only a computer system.
A computer(based) information system, or CBIS, uses computers to gather, procedure, store,
examine and allocate information for a specific reason, such as gathering a trade objective. The
main mechanism of a CBIS comprise hardware, software, data, events and public.
A computer(based) information system be basically an IS using computer knowledge to take out
some or all of its intended tasks. The basic mechanism of computer based information system
are:-
· Hardware these are the plans like the check, processor, printer and keyboard, all of which
work jointly to accept, procedure, show information and in sequence.
· Software are the program that let the hardware to procedure the data.
· Databases are the meeting of connected files or tables contain related data.
· Networks are a between system that allow diverse computer to hand out resources.
· Procedures are the instructions for combine the mechanism above to development in
sequence and produce the ideal output..
Van der Waals bonding is due tothe sharing of electrons between at.pdf
Van der Waals bonding is due to
the sharing of electrons between atoms.
the transfer of electrons between atoms.
unequal charge distribution around neutral molecules.
atoms bonding to hydrogen molecules.
Solution
Covalent bond = the sharing of electrons between atoms.
Ionic bond = the transfer of electrons between atoms.
Hydrogen bond = atoms bonding to hydrogen molecules.
Vanderwaals bonding = unequal charge distribution around neutral molecules..
USE TORA TODetermine the maximum flow and the optimum flow in eac.pdf
USE TORA TO:
Determine the maximum flow and the optimum flow in each arc for the network in the following
figure.
Clearly show all steps and data input.
Exterior Graph Data, Nodes listed outside paranthesis
Node 1 (14) to Node 3 (0)
Node 3 (10) to Node 5 (0)
Node 5(0) to Node 4 (5)
Node 4(6) to Node 2 (7)
Node 2(0) to Node 1 (8)
Interior Graph (Triangle ) Data
Node3 (10) to Node2(5)
Node3 (9) to Node 4(7)
Second Interior Graph Data
Node5 (0) to Node2(6)
Solution
The most successful and simple model is Erlang\'s loss system where the blocking probability is
given by Erlang\'s B-formula. The traffic is described by the offered traffic A, the system (only
one link) by the number of channels, and the strategy is full accessibility with lost calls cleared.
The above-mentioned three elements of the model are each described by only one parameter.
Only single-channel calls are considered. The network performance is described by the blocking
probability E1,n(A), i.e. the probability that a call attempts is blocked because all n channels are
busy. For Erlang\'s loss system time, call, and traffic congestions are equal. The state
probabilities are given by the truncated Poisson distribution, and when the number of channels is
very large this becomes a Poisson distribution. This model has been very successful for traffic
engineering. The background for this success is that the traffic is very well modeled by one
parameter only. The underlying mathematical assumption is a Poisson arrival process. This is
fulfilled when the traffic is generated by many independent users, which is the case for
telephony. If the arrival process is a Poisson process, then the model is insensitive to the holding
time distribution, which means that only the mean holding time is of importance. So the model is
very robust to the traffic and models the real world extremely well. Improvement function: This
denotes the increase in carried traffic when the number of channels is increased by one from n to
n + 1: 64 Attention: This is not an ITU publication made available to the public, but an internal
ITU Document intended only for use by the Member States of the ITU and by its Sector
Members and their respective staff and collaborators in their ITU related work. It shall not be
made available to, and used by, any other persons or entities without the prior written consent of
the ITU. ITU Telecom Network Planning Reference Manual - Draft version 4.1 January 2007
4.8.4.3 Engset\'s loss system The Poisson arrival process is the most random process, and the
calls are generated by a very large number of independent sources, each having an infinitesimal
calling rate. In many real systems the number of users is limited, and the arrival process is more
regular or smooth than random traffic. This is modeled by Engset\'s loss system where we have a
finite number S of users (traffic sources) which alternates between the states off (= idle) and on
(= busy). When a source is idle it generates cal.
The three ways of promoting ACTIVE transport area) Channels, pori.pdf
The three ways of promoting ACTIVE transport are:
a) Channels, porins and coupled transporters.
b) Porins, channels and ATP driven transport.
c) Coupled transporters, ATP and LIGHT driven transport.
d) Antiporter, symporter and coupled transporters.
Solution
c. Coupled transporters, ATP and LIGHT driven transport
ATP use the energy derived from hydrolysis of ATP to transport solutes from the
electrochemical gradient.
Coupled transporters use the energy stored in the electrochemical gradient of one solute to
transport another.
Light-driven transport use energy derived from gentle to drive translocation..
The picture below describes three aspects of using a variable. Identi.pdf
The picture below describes three aspects of using a variable. Identify each. Name the three
types of loops and briefly describe the main use of each.
Solution
1)
A)This one describes the datatype of the varibles, datatType are int,float,char,short,long,double
etc
B)This one describes the identifier/variable, which is sequence of symbols, which is
representation of a memory block.
C)value, this is value of the variable, this value is stored in the memory block that is associated
to the variable
2)
A)for loop: syntax is for(i=0;i.
The idea of the principle of superposition in stratigraphy is that t.pdf
The idea of the principle of superposition in stratigraphy is that the layer on the bottom in a
stratigraphic sequence is the youngest layer. the layer on the bottom in a stratigraphic sequence
is the oldest layer. the decay at a known rate of the radioactive isotope in the stratigraphy makes
it possible to determine the date. all layers in a stratigraphic sequence date to the same time. An
artifact is a. an object made by humans, such as a stone tool. a person who excavates and studies
artifacts. A location with evidence of past human activity, including artifacts and other remains.
an object used by humans, but not modified by them, such as plant remains. An event that
occurred at the end of the Mesolithic was the first appearance of life on earth. first appearance
of reptiles. first appearance of the therapsids. extinction of the dinosaurs.
Solution
31.b.The principle of superposition states that rock forms horizontally with the oldest layers of
rock lying at the bottom of a formation and newer rock resting on top.
32.a.Any object made by human beings,especially with a view to subsequent use.It is a
handmade object.
33.4.extinction of dinosaurs.
The Force is strong among the inhabitants of Tatooine, the birthplac.pdf
The Force is strong among the inhabitants of Tatooine, the birthplace of Luke Skywalker. Not all
of its inhabitants are capable of becoming a Jedi, but the probability is higher than among the rest
of the members of the Alliance. Information was collected regarding this phenomena and it was
determined that 54% of all of the members of the Alliance were born on Tatooine. It was also
found that of all the members of the Alliance, 44% had the potential to become a Jedi. Finally, it
was determined that 35% of all members of the Alliance were both from Tatooine and capable of
becoming a Jedi. Based on this information calculate the following probabilities:
1. What is the probability that a member of the Alliance is randomly selected and found to be
from Tatooine and not a Jedi?
2. What is the probability that a member of the Alliance is randomly selected and found to be
from Tatooine or a Jedi?
3. What is the probability that a member of the Alliance is randomly selected and found to be
neither from Tatooine nor a Jedi?
4. Given that a member of the Alliance is a Jedi, what is the probability that he/she is from
Tatooine?
Solution
a Jedi
1.
P(from Tatooine and not a jedi) = P(from Tatooine) - P(from Tatooine and a Jedi) = 0.54-0.35 =
0.19
2.
P(from Tatooine or a jedi) = 0.35+0.19+0.09 = 0.63
3.
P(neither from Tatooine nor a jedi) = 1 - 0.63 = 0.37
4.
P(from Tatooine | a jedi) = 0.35/0.44 = 0.80
a Jedinot a jedifrom Tatooine0.350.54-0.35 = 0.19not from Tatooine0.44-0.35 = 0.091-
(0.35+0.19+0.09)= 0.37.
Suppose that you are given a short fragment ofDNA to sequence. You am.pdf
Suppose that you are given a short fragment ofDNA to sequence. You amplify the fragment
with PCR and set up a series of four dideoxy reactions. You then separate the products of the
reactions by gel electrophoresis and obtain the following banding pattern: Write out the base
sequence of the original fragment that you were given. Original sequence: 5\'-__3^r
Solution
The sequence of the DNA fragment from 5\'-3\' will be:
TACTGATGC
these method of DNA sequencing is known as Sanger method or dideoxynucleotide chain
termination method of DNA sequencing. In this method first of all the DNA fragment to be
sequenced is first of all is denatured by denaturing agents and then is annealed with primers to
initiate the copying of DNA, after which it is extended by DNA polymerase enzyme using 1)
normal deoxynucleotide triphosphate(dNTPs) and 2)dideoxynucleotide triphosphates (ddNTPs)
which are chain terminating nucleotide.
Normally in copying of DNA the new dNTP is added to the 3’OH group of the previous dNTP
thus the extension of DNA continues. The ddNTPS lacks these 3’OH group thus the next dNTP
cannot be added thus the DNA polymerase fails in its work and cannot copy the DNA and thus
these leds in the termination of the chain.
Thus the copied strand lenth depends on when the ddNTP is incorporated,as these incorporation
leads to stopping of chain elongation. as in the above question 4 different testubes have been
prepared all of them containing denatured DNA, annealed primer , DNA polymerase and dNTP
but they vary in ddNTP containing different types of ddNTP when incorporated stops the chain
at that particular ddNTP containing aliquot. thus when such testtubes are run in electrophoresis
give varying bands of different length which can be read from 5\'-3\' direction. here, starting
from down i.e. T will be the first nucleotidethen the next band to the level just above it here it is
A and in the same way other can be read..
Suppose that species I has 2n=10 and species II has 2n=12. How many .pdf
Suppose that species I has 2n=10 and species II has 2n=12. How many chromosomes would be
present in the allotetraploid?
Solution
Allotetraploid is the condition where hybrid must have chromosomes from both species and a
total of 4n(chromosome set four times of haploid organism)
Species I has 2n=10 (n = 5) and species II has 2n=12 (n = 6)
For allotetraploid there are 3 possible combination -
2n from both species I and species II - 10 + 12 = 22 chromosomes
3n from species I + 1n from species II : 3(5) + 1(6) = 21 chromosomes
1n from species I + 3n from species II : 1(5) + 3(6) = 23 chromosomes..
The approach at University of Liverpool.pptx
How libraries can support authors with open access requirements for UKRI funded books
Wednesday 22 May 2024, 14:00-15:00.
Model Attribute Check Company Auto Property
In Odoo, the multi-company feature allows you to manage multiple companies within a single Odoo database instance. Each company can have its own configurations while still sharing common resources such as products, customers, and suppliers.
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Describe three different symbiotic relationships, with examples. Whi.pdf
Describe three different symbiotic relationships, with examples. Which relationship produces the
most coevolution? Explain.
Please type the answer. Type the answer in your own words please. no copying from the internet
has to be in own words
Solution
Three different types of symbiotic relatioships are :
1. Mutualism
2. Commensalism
3. Parasitism
1.Mutualism is the relationship between organisms where both are benefitted from each other.
Examples of mutualism:
a. Oxpecker on zebra eats pest. Oxpecker gets food and zebra gets rid of pest.
b. Shape of flower and structure of pollinator is also an example of mutulism as shape of the
flower is specific for the pollinator to allow pollination.
c. Bacteria in gut of human intestine benefit human by partially digesting the food difficult to
digest. Bacteria gets food through this and human is able to digest tougher to dogest food.
d. Mycorhhiza is association between fungus and gymnosperm root. Plant gets benefitted by
collection of nutrients from soil and fungus gets benefit by plant in form of food and shelter.
2. Commensalism: In commensalism association, one organism is benefitted whereas other is not
affected at all.
examples:
a. Cattle Egret and cattle: as cattle grazes and uproots plants, insects that escape are eaten by
egret. Egret gets the benefit wheras cattle is not affected.
b Sucker fish attaches itself to the shark and travels along. Shark doesnt get affected whereas the
sucker fish gets left over food of shark.
c.Epiphytes are plants that grow on larger trees. They just climb the trees for receiving sunlight
and do not cause any affect to the plant.
d. CLown fish lives in tentacles of Sea anemones. Sea anemone doesnt get affected whereas
clown fish gets the advantage of protection from predators.
3. Parasitism: In this relationship, the parasite harms the host to derive the benefit in terms of
food or shelter or both. In this type of relationship one species is harmed whereas other is
benefitted.
examples of parasitism:
a. Lice on Human body: Lice sucks blood can causes harm to human.
b. Cuscuta causes harm to higher plants by living on it and taking the benefit of food and shelter.
c. Ascaris lives in human intestine . It benefits from receiving food and shelter and causes harm
to human.
d. tapeworm in Human intestine as they get partially digested food and deprive human of
nutrients.
Out of these relationships, coevolution is most important for species showing mutualistic
behaviour as:
1. They both have large dependency on each other for their survival.
2. Any change in one species can lead to extinction of the other. Coevolution is the only strategy
for their survival..
Describe the differences seen in the hydra, planarian, clam, grasshop.pdf
Describe the differences seen in the hydra, planarian, clam, grasshopper, and pig for symmetry,
tissue organization, body cavity, digestive system, circulatory system, respiratory system,
excretory system, support system and nervous system.
Solution
Hydra
Planarian:
Clam:
Symmetry
radial
bilateral
bilateral
Tissue Organization
diploblastic
triploblastic
triploblastic
Type of Body Cavity:
gastrovascular cavity
acoelomate
coelomate:
pericardial cavity
Digestive Openings
one
mouth
(pharynx extended during feeding)
mouth, a nu s
Circulatory System
none
none
heart, open c.s.
Respiratory Organs:
none
none
incurrent and excurrent siphon, gills
Excretory System
none
protonephridia, flame bulb
nephridia
Support System
: none
none
nephridia
Nervous System
Organization: nerve net
Organization: brain, ventral nerve cords
Hydra
Planarian:
Clam:
Symmetry
radial
bilateral
bilateral
Tissue Organization
diploblastic
triploblastic
triploblastic
Type of Body Cavity:
gastrovascular cavity
acoelomate
coelomate:
pericardial cavity
Digestive Openings
one
mouth
(pharynx extended during feeding)
mouth, a nu s
Circulatory System
none
none
heart, open c.s.
Respiratory Organs:
none
none
incurrent and excurrent siphon, gills
Excretory System
none
protonephridia, flame bulb
nephridia
Support System
: none
none
nephridia
Nervous System
Organization: nerve net
Organization: brain, ventral nerve cordsNervous System Organization: three ganglia.
Companyhas prot s Price per unit Variable cost per unit Fixed costs p.pdf
Companyhas prot s Price per unit Variable cost per unit Fixed costs per month $%4 $17,00
What is the contribution margin ratio? A) 18% B) 70% C) 48% D) 33%
Solution
Contribution margin=Sales-Variable costs
=(40-12)=$28/unit
Hence Contribution margin ratio=Contribution margin/Sales
=(28/40)
which is equal to
=70%.
at log kHz SolutionYou can have several problems 1) if you a.pdf
at log kHz
Solution
You can have several problems:
1) if you are working with transistors sure you\'re not saturating.
2) if diodes are sure they are doing the right contact or try another.
3) Make sure your input source this good.
4) for me the most certain is that you are using an active elemnto and this is in court and why
that image, check input sources, the currents that these manipulated are optimal for circuit
elements you use or application you\'re doing..
Bark, which is thicken, toughen periderm, arises from this lateral m.pdf
Bark, which is thicken, toughen periderm, arises from this lateral meristem?
Vascular cambium
Cork cambium
QUESTION 2
Seeds with edible fruit are most likely dispersed by
wind
animals
water
bursting
QUESTION 3
Seeds with wings are most likely dispersed by
wind
animals
water
bursting
QUESTION 4
The arrangement of flowers on the stem with the newest flowers near the end of the shoot is
Determinate inflorescence
Indeterminate inflorescence
QUESTION 5
The arrangement of flowers on the stem with the oldest flowers near the end of the shoot is
Determinate inflorescence
Indeterminate inflorescence
QUESTION 6
The cortex in stems is made of this tissue
Dermal
Ground
Vascular
QUESTION 7
The epidermis, cuticle, guard cells, root hairs and trichomes arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 8
The epidermis, cuticle, guard cells, root hairs and trichomes arise from this type of tissue?
Dermal
Ground
Vascular
QUESTION 9
The parenchyma, collenchyma and sclerenchyma arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 10
The parenchyma, collenchyma and sclerenchyma arise from this type of tissue?
Dermal
Ground
Vascular
QUESTION 11
The xylem and phloem in leaves are contained in which structure?
blade
petiole
stipules
vein
QUESTION 12
This meristem is located in the root and shoot tips.
Apical meristem
Lateral meristem
QUESTION 13
This meristem is located in the stems of plants that live longer than one year.
Apical meristem
Lateral meristem
QUESTION 14
Xylem and phloem elements arise from this type of meristem?
Protoderm
Ground meristem
Procambium
QUESTION 15
Xylem and phloem elements arise from this type of tissue?
Dermal
Ground
Vascular
Vascular cambium
Cork cambium
Solution
1. Cork cambium. It palys a major role in secondary growth like wood (bark) in dicotyledons.
Vascular cambium is responsible for Xylem and Pholem formation.
2.Animals. Edible part of fruits are eaten by animals and remaining part (seeds) dropped in other
locations.
3. Winds. It has a smaller seeds with feather - like or wings it helps easily transfer in winds.
4. InDeterminate inflorescence.
Acropetal sucession - The new flowers arranged at the top and old flowers are at bottom is
Racemose or Indeterminate inflorescence.
5. Determinate inflorescence
Basipetal sucession - The new flowers arranged at the bottom and old flower are at tip is cymose
or determinate inflorescence.
6.Ground Tissue
It contains simple tissuse are parenchyma, collenchyma and sclerenchyma. Parenchyma cells are
usually present in cortex.
7. Protoderm
It helps in develop epidermal tissues.
8. Dermal tissue
Epidermal tissue system present in outer most layer of plant system. It contains epidermal,
stomato, trichomes and hairs etc.,
9. Ground meristem.
Ground meristem give rise to ground tissues and contains parenchyma, collenchyma and
sclerenchyma.
10. Ground Tissues.
11. Vein
Lamina - Green expanded part of leaf. Petiole - It attach leaf and stem. Stip.
Why do you think it is usually preferable to start with the side con.pdf
Why do you think it is usually preferable to start with the side containing the more complicated
expression when establishing a trigonometric identity?
Solution
1. Human brains are accustomed to simplify complicated things rather than complicating a
simpler one.
2. A simpler/simplest conclusion of a complicated problem in trigonometric identities is only a
handful;mere 2 or 3.
But if I want to complicate a simpler side, outcomes are numerous.
For eg: prove cosx*tanx =sinx
Now I know that if I want to simplify LHS, I will try to convert it is terms of Sin or Cos (which
we consider simple in the world of trigonometry)
So cosx*(sinx/cosx)= cosx which is RHS.
But if I want to complicate sinx ,I can write it as
1/cscx or 1-cos^2x or (1-cos2x)/2 . . . . And the list goes on!
So, by and large, it is customary simplify complicated things rather than the other way.
Hope it clarifies your doubt!.
Why is the mucus thicker than normal in CF patients A good answer w.pdf
Why is the mucus thicker than normal in CF patients? A good answer will include a description
of how osmosis is affected by CF.
Solution
Mucus is a substance that coats our Epithelial surface.It keeps surfaces lubricated and helps
prevent disease by trapping potential invaders.Invading microbes can become trapped in the
mucus.The epithelial cells are lined with cilia that are constantly beating and moving mucus with
trapped bacteria up and out of the respiratory tract.Mucus is amixture of predominantly water
and glycproteins.The mucins are produced by goblet cell that are interspersed among the
respiratory epithelial cell.The composition,water content and volume of mucus are controlled by
the secretion and absorption of ions and water across the respiratory epithelium.The salt
imbalance resulting from a disfuctional chloride channel could through osmosis lead to the thick
mucus characterstic of cystic fibrosis mutation could lead to disease.cystic fibrosis is a thick that
lines both the airways and the intestinal tract of affected individual.In the respiratory tract this
thick mucus is not removed by the cillia that usually beat and push mucus up into the throat.This
thick mucus builds up in the airways and bacteria trapped in the mucus are not removed.Thus
bacterial respiratory infections are extremely common and dangerous in CF individuals..
Which of the following primordial reproductive structures is NOT ext.pdf
Which of the following primordial reproductive structures is NOT external?
A) Urogenital sinus
B) Genital tubercle
C) Urogenital folds
D) Labioscrotal swelling
Solution
Urogenital sinus is the promordial structure not external. It is present only in the develovement
of Reproductive organs.It is divided in to three main regions upper , phallic and pelvic..
what kind of accessory protein would you classify IRS1 and 2 as.pdf
what kind of accessory protein would you classify IRS1 and 2 as?
Docking protein
Adaptor protein?
Transcription factor?
SH2 domain protein
Solution
Docking protein - The robosomes are of two types based on their occurence in the cell. Free
ribosomes and membrane bound (on ER surface). Ever wondered why and how they are of two
types? Yes!! Some specific proteins which are found on the surface of certain specific ribosomal
particles making them \"dock\" (stick) on to the ER surface are called docking proteins.
Adapter protein - An adapter protein is one which assists in signal transduction across
membranes. They aid receptor mechanisms in the cells.
Transcription factor - The process of copying cell DNA codons in to an mRNA strand in called
transcription. This is assisted by certain types of protein called Transcription factors.
SH2 domain - These are specific proteins that mediate and fecilitate protein-protein inteactions
in the cell. SH2 domain (a structurally special part within a protein molecule) is found within Src
oncoprotein (cancer protein) which makes the Src oncoprotein attach to other protein containing
phosphorylated tyrosin residues.
**IRS1 and IRS2 proteins are classified as adapter proteins.**.
What is the main difference between prokaryotic and eukaryotic riboso.pdf
What is the main difference between prokaryotic and eukaryotic ribosomes? Prokaryotic
ribosomes synthesize RNA whereas eukaryotic ribosomes synthesize proteins Eukaryotic
ribosomes are composed of RNA and protons whereas prokaryote ribosomes are
Solution
None of the above
The main difference is that Prokaryotic ribosomes(70S) having two subunits- 30S and 50S
and Eukaryotic ribosomes(80S) also having two subunits- 40S and 60S
Both are composed of RNA & proteins. 70S having 60% RNA & 40% proteins while 80S having
40%RNA & 60% proteins..
what is the cardinality of a sample spaceSolutionA sample spa.pdf
what is the cardinality of a sample space?
Solution
A sample space is a collection of all possible outcomes of a random experiment. A random
variable is a function defined on a sample space. We shall consider several examples shortly.
Later on we shall introduce probability functions on the sample spaces. A sample space may be
finite or infinite. Infinite sample spaces may be discrete or continuous.
What is computer information systems in three paragraphs.Solutio.pdf
What is computer information systems in three paragraphs.
Solution
A computer information system be a system collected of people and computer that process or
interpret information. The word is also from time to time used in more limited senses to refer to
only the software used to sprint a automated database or to refer to only a computer system.
A computer(based) information system, or CBIS, uses computers to gather, procedure, store,
examine and allocate information for a specific reason, such as gathering a trade objective. The
main mechanism of a CBIS comprise hardware, software, data, events and public.
A computer(based) information system be basically an IS using computer knowledge to take out
some or all of its intended tasks. The basic mechanism of computer based information system
are:-
· Hardware these are the plans like the check, processor, printer and keyboard, all of which
work jointly to accept, procedure, show information and in sequence.
· Software are the program that let the hardware to procedure the data.
· Databases are the meeting of connected files or tables contain related data.
· Networks are a between system that allow diverse computer to hand out resources.
· Procedures are the instructions for combine the mechanism above to development in
sequence and produce the ideal output..
Van der Waals bonding is due tothe sharing of electrons between at.pdf
Van der Waals bonding is due to
the sharing of electrons between atoms.
the transfer of electrons between atoms.
unequal charge distribution around neutral molecules.
atoms bonding to hydrogen molecules.
Solution
Covalent bond = the sharing of electrons between atoms.
Ionic bond = the transfer of electrons between atoms.
Hydrogen bond = atoms bonding to hydrogen molecules.
Vanderwaals bonding = unequal charge distribution around neutral molecules..
USE TORA TODetermine the maximum flow and the optimum flow in eac.pdf
USE TORA TO:
Determine the maximum flow and the optimum flow in each arc for the network in the following
figure.
Clearly show all steps and data input.
Exterior Graph Data, Nodes listed outside paranthesis
Node 1 (14) to Node 3 (0)
Node 3 (10) to Node 5 (0)
Node 5(0) to Node 4 (5)
Node 4(6) to Node 2 (7)
Node 2(0) to Node 1 (8)
Interior Graph (Triangle ) Data
Node3 (10) to Node2(5)
Node3 (9) to Node 4(7)
Second Interior Graph Data
Node5 (0) to Node2(6)
Solution
The most successful and simple model is Erlang\'s loss system where the blocking probability is
given by Erlang\'s B-formula. The traffic is described by the offered traffic A, the system (only
one link) by the number of channels, and the strategy is full accessibility with lost calls cleared.
The above-mentioned three elements of the model are each described by only one parameter.
Only single-channel calls are considered. The network performance is described by the blocking
probability E1,n(A), i.e. the probability that a call attempts is blocked because all n channels are
busy. For Erlang\'s loss system time, call, and traffic congestions are equal. The state
probabilities are given by the truncated Poisson distribution, and when the number of channels is
very large this becomes a Poisson distribution. This model has been very successful for traffic
engineering. The background for this success is that the traffic is very well modeled by one
parameter only. The underlying mathematical assumption is a Poisson arrival process. This is
fulfilled when the traffic is generated by many independent users, which is the case for
telephony. If the arrival process is a Poisson process, then the model is insensitive to the holding
time distribution, which means that only the mean holding time is of importance. So the model is
very robust to the traffic and models the real world extremely well. Improvement function: This
denotes the increase in carried traffic when the number of channels is increased by one from n to
n + 1: 64 Attention: This is not an ITU publication made available to the public, but an internal
ITU Document intended only for use by the Member States of the ITU and by its Sector
Members and their respective staff and collaborators in their ITU related work. It shall not be
made available to, and used by, any other persons or entities without the prior written consent of
the ITU. ITU Telecom Network Planning Reference Manual - Draft version 4.1 January 2007
4.8.4.3 Engset\'s loss system The Poisson arrival process is the most random process, and the
calls are generated by a very large number of independent sources, each having an infinitesimal
calling rate. In many real systems the number of users is limited, and the arrival process is more
regular or smooth than random traffic. This is modeled by Engset\'s loss system where we have a
finite number S of users (traffic sources) which alternates between the states off (= idle) and on
(= busy). When a source is idle it generates cal.
The three ways of promoting ACTIVE transport area) Channels, pori.pdf
The three ways of promoting ACTIVE transport are:
a) Channels, porins and coupled transporters.
b) Porins, channels and ATP driven transport.
c) Coupled transporters, ATP and LIGHT driven transport.
d) Antiporter, symporter and coupled transporters.
Solution
c. Coupled transporters, ATP and LIGHT driven transport
ATP use the energy derived from hydrolysis of ATP to transport solutes from the
electrochemical gradient.
Coupled transporters use the energy stored in the electrochemical gradient of one solute to
transport another.
Light-driven transport use energy derived from gentle to drive translocation..
The picture below describes three aspects of using a variable. Identi.pdf
The picture below describes three aspects of using a variable. Identify each. Name the three
types of loops and briefly describe the main use of each.
Solution
1)
A)This one describes the datatype of the varibles, datatType are int,float,char,short,long,double
etc
B)This one describes the identifier/variable, which is sequence of symbols, which is
representation of a memory block.
C)value, this is value of the variable, this value is stored in the memory block that is associated
to the variable
2)
A)for loop: syntax is for(i=0;i.
The idea of the principle of superposition in stratigraphy is that t.pdf
The idea of the principle of superposition in stratigraphy is that the layer on the bottom in a
stratigraphic sequence is the youngest layer. the layer on the bottom in a stratigraphic sequence
is the oldest layer. the decay at a known rate of the radioactive isotope in the stratigraphy makes
it possible to determine the date. all layers in a stratigraphic sequence date to the same time. An
artifact is a. an object made by humans, such as a stone tool. a person who excavates and studies
artifacts. A location with evidence of past human activity, including artifacts and other remains.
an object used by humans, but not modified by them, such as plant remains. An event that
occurred at the end of the Mesolithic was the first appearance of life on earth. first appearance
of reptiles. first appearance of the therapsids. extinction of the dinosaurs.
Solution
31.b.The principle of superposition states that rock forms horizontally with the oldest layers of
rock lying at the bottom of a formation and newer rock resting on top.
32.a.Any object made by human beings,especially with a view to subsequent use.It is a
handmade object.
33.4.extinction of dinosaurs.
The Force is strong among the inhabitants of Tatooine, the birthplac.pdf
The Force is strong among the inhabitants of Tatooine, the birthplace of Luke Skywalker. Not all
of its inhabitants are capable of becoming a Jedi, but the probability is higher than among the rest
of the members of the Alliance. Information was collected regarding this phenomena and it was
determined that 54% of all of the members of the Alliance were born on Tatooine. It was also
found that of all the members of the Alliance, 44% had the potential to become a Jedi. Finally, it
was determined that 35% of all members of the Alliance were both from Tatooine and capable of
becoming a Jedi. Based on this information calculate the following probabilities:
1. What is the probability that a member of the Alliance is randomly selected and found to be
from Tatooine and not a Jedi?
2. What is the probability that a member of the Alliance is randomly selected and found to be
from Tatooine or a Jedi?
3. What is the probability that a member of the Alliance is randomly selected and found to be
neither from Tatooine nor a Jedi?
4. Given that a member of the Alliance is a Jedi, what is the probability that he/she is from
Tatooine?
Solution
a Jedi
1.
P(from Tatooine and not a jedi) = P(from Tatooine) - P(from Tatooine and a Jedi) = 0.54-0.35 =
0.19
2.
P(from Tatooine or a jedi) = 0.35+0.19+0.09 = 0.63
3.
P(neither from Tatooine nor a jedi) = 1 - 0.63 = 0.37
4.
P(from Tatooine | a jedi) = 0.35/0.44 = 0.80
a Jedinot a jedifrom Tatooine0.350.54-0.35 = 0.19not from Tatooine0.44-0.35 = 0.091-
(0.35+0.19+0.09)= 0.37.
Suppose that you are given a short fragment ofDNA to sequence. You am.pdf
Suppose that you are given a short fragment ofDNA to sequence. You amplify the fragment
with PCR and set up a series of four dideoxy reactions. You then separate the products of the
reactions by gel electrophoresis and obtain the following banding pattern: Write out the base
sequence of the original fragment that you were given. Original sequence: 5\'-__3^r
Solution
The sequence of the DNA fragment from 5\'-3\' will be:
TACTGATGC
these method of DNA sequencing is known as Sanger method or dideoxynucleotide chain
termination method of DNA sequencing. In this method first of all the DNA fragment to be
sequenced is first of all is denatured by denaturing agents and then is annealed with primers to
initiate the copying of DNA, after which it is extended by DNA polymerase enzyme using 1)
normal deoxynucleotide triphosphate(dNTPs) and 2)dideoxynucleotide triphosphates (ddNTPs)
which are chain terminating nucleotide.
Normally in copying of DNA the new dNTP is added to the 3’OH group of the previous dNTP
thus the extension of DNA continues. The ddNTPS lacks these 3’OH group thus the next dNTP
cannot be added thus the DNA polymerase fails in its work and cannot copy the DNA and thus
these leds in the termination of the chain.
Thus the copied strand lenth depends on when the ddNTP is incorporated,as these incorporation
leads to stopping of chain elongation. as in the above question 4 different testubes have been
prepared all of them containing denatured DNA, annealed primer , DNA polymerase and dNTP
but they vary in ddNTP containing different types of ddNTP when incorporated stops the chain
at that particular ddNTP containing aliquot. thus when such testtubes are run in electrophoresis
give varying bands of different length which can be read from 5\'-3\' direction. here, starting
from down i.e. T will be the first nucleotidethen the next band to the level just above it here it is
A and in the same way other can be read..
Suppose that species I has 2n=10 and species II has 2n=12. How many .pdf
Suppose that species I has 2n=10 and species II has 2n=12. How many chromosomes would be
present in the allotetraploid?
Solution
Allotetraploid is the condition where hybrid must have chromosomes from both species and a
total of 4n(chromosome set four times of haploid organism)
Species I has 2n=10 (n = 5) and species II has 2n=12 (n = 6)
For allotetraploid there are 3 possible combination -
2n from both species I and species II - 10 + 12 = 22 chromosomes
3n from species I + 1n from species II : 3(5) + 1(6) = 21 chromosomes
1n from species I + 3n from species II : 1(5) + 3(6) = 23 chromosomes..
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