The color change occurs when starch interacts with iodine. Given the.pdf
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The color change occurs when starch interacts with iodine. Given the location of the color change, which particles (starch, iodine, or both) must have moved across the dialysis membrane, and why? Solution Starch Test: Adding iodine to a solution or directly on a material which contains starch it gives blue black colour.If the Amylose starch is not present, it give orange or yellow. Iodine Test: iodine might be used as indicator to follow the changes of iodide ion and iodine element. Iodine element will give blue-black colour in the presence of Iodide ion. EXPLANATION Starch is a polysaccharide contains molecules of amylose and amylopectin. Amylose in starch is responsible for the formation of a deep blue color in the presence of iodine. In the Iodine test , Iodine dissolved in aqueous solution of Potassium Iodide which reacts with starch to produce bluish colour. transfer of CHARGE between Starch &Iodine changes the energy level spaces in the Iodine or Starch complex that are for absorbing the visible light giving the complex its intense blue colour. the dialysis membrane does not allow any kind of large substances to pass through the pores of the membrane.the dialysis membrane was permeable to iodine and not to starch,because starch has larger molecules. the colour change occurs inside the dialysis membrane with iodine,because iodine has small molecules,selective permeability is a property of cell membrane that allows the small molecules to pass through them...
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Describe in detail the specimen of a potato with a drop of iodine in.pdf
Describe in detail the specimen of a potato with a drop of iodine in comparison with a potatoe
with just a drop of water under the microscope.
Solution
Answer:
Starch (a carbohydrate) can be identified by a positive iodine test. Few drops of iodine solution
are added to a raw or boiled sample of the substance suspected of containing starch. If starch is
present, the substance will turn blue-black in the area where the iodine solution touches it.
Substances with very small amounts of starch will turn faint blue or greenish instead of blue-
black. Many materials of biological origin have high starch content. These materials can be
identified with the iodine test.
The sample specimen with potato and a drop of iodine will turn blue black concluding for the
presence of large amount of starch.
The sample specimen with potato and a drop of water will not undergo a change in colour..
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Till that time, it was not known that these \'factors\' were actually genes which are very active at
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Describe in detail the specimen of a potato with a drop of iodine in.pdf
Describe in detail the specimen of a potato with a drop of iodine in comparison with a potatoe
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Solution
Answer:
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I need help with two things. First I\'d like someone to check over my code for the 3 //TODO
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public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
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for(int i=0; i doLinkedHashSetRemoveMedian(int numItems) {
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long endTime = getTimestamp();
long totalTime = endTime - startTime;
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}
Solution
PART 1
Hi friend, first part you have done correctly except a small mistake. In first TODO you missed
\'=\' sign:
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
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set.add(i);
}
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set.remove(median);
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}
PART 2:
public static LinkedHashSet doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
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for(int i=0; i<=numitems-1; i++)
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int current_size = set.size();
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Explain Internal Controls and Corporate Governance.SolutionInt.pdf
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Enforcement: 6. Migratory Bird Act (1918) a. Amendment Year: b. International or National?: c.
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Please find the answers below as tabulated:
Answer 2:
Lacey act
Answer 3
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Answer 4:
Public Health Service Act (1912)
Governed by Service Commission Corps and United States Department of Health and Human
Services and enacted by Office of the Assistant Secretary for Preparedness and Response along
with Office of Global AffairsActDrafted yearAmendment
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the residential areas.Formulated and executed by the United States congress
Answer 2:
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products etc.The United States Federal Environmental Legislation
Answer 3
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antiquitiesNationalThis law was formulated in order to create monuments from the land of the
country in.
discuss the drawbacks of programmed and interrupt driven io and des.pdf
discuss the drawbacks of programmed and interrupt driven i/o and describe in general the
functionality of the DNA
Solution
Programmed I/O
Programmed I/O (PIO) refers to data transfers initiated by a CPU under driver software control
to access registers or memory on a device.
The CPU issues a command then waits for I/O operations to be complete. As the CPU is faster
than the I/O module, the problem with programmed I/O is that the CPU has to wait a long time
for the I/O module of concern to be ready for either reception or transmission of data. The CPU,
while waiting, must repeatedly check the status of the I/O module, and this process is known as
Polling. As a result, the level of the performance of the entire system is severely degraded.
Programmed I/O basically works in these ways:
Interrupt
The CPU issues commands to the I/O module then proceeds with its normal work until
interrupted by I/O device on completion of its work.
For input, the device interrupts the CPU when new data has arrived and is ready to be retrieved
by the system processor. The actual actions to perform depend on whether the device uses I/O
ports, memory mapping.
For output, the device delivers an interrupt either when it is ready to accept new data or to
acknowledge a successful data transfer. Memory-mapped and DMA-capable devices usually
generate interrupts to tell the system they are done with the buffer.
Although Interrupt relieves the CPU of having to wait for the devices, but it is still inefficient in
data transfer of large amount because the CPU has to transfer the data word by word between I/O
module and memory.
The main limitation of programmed I/O and interrupt driven I/O is given below:
Programmed I/O
Each instructions selects one I/O device (by number) and transfers a single character (byte)
Example: microprocessor controlled video terminal.
Four registers: input status and character, output status and character.
Interrupt-driven I/O
Primary disadvantage of programmed I/O is that CPU spends most of its time in a tight loop
waiting for the device to become ready. This is called busy waiting.
With interrupt-driven I/O, the CPU starts the device and tells it to generate an interrupt when it is
finished.
Done by setting interrupt-enable bit in status register.
Still requires an interrupt for every character read or written.
Interrupting a running process is an expensive business (requires saving context).
Requires extra hardware (DMA controller chip).
All these limitation can be overcome by the Introduction of DMA (Direct Memory Access)
To write block of 32 bytes from memory address 100 to device 4
1. CPU writes 32, 100, 4 into the first three DMA registers (memory address, count, device
number)
2. CPU puts code for WRITE (say 1) into fourth (direction) DMA register, which signals DMA
controller to begin operation
3. Controller reads (via bus request as CPU would) byte 100 from memory
4. Controller makes I/O request to write to device 4
5. Controller increments m.
do it on excel and please kindly upload the excel file. or explain m.pdf
do it on excel and please kindly upload the excel file. or explain me how to do this question on
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Solution
In numerical analysis, Newton\'s method (also known as the Newton–Raphson method), named
after IsaacNewton and Joseph Raphson, is a method for finding successively better
approximations to the roots (or zeroes) of a real-valued function. x : f ( x ) = 0 . {\\displaystyle
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Describe the structure and location of the parathyroid glands.Desc.pdf
Describe the structure and location of the parathyroid glands.
Describe the structure and location of the adrenal glands. Include the organization of the
following parts and their major role: Cortex and Medulla
Explain how the adrenal medulla and the sympathetic nervous system interact.
Describe the circadian cycle of melatonin.
Describe how the HPA axis controls sex-hormone release.
Describe what happens in the body in diabetes mellitus and explain the difference between Type
1 and Type 2 diabetes.
Solution
1.Structure and location of parathyroid glands:
Structure :
Loaction :These glands are situated behind the left and right lobes of the thyroid.
2.Structure and location of adrenal glands:
Structure :
Cortex :It is the outermost layer and is made up of three zones:
Zona glomerulosa :The outermost layer of the adrenal cortex is the zona glomerulosa and is
made up of oval cells that separated by thin strands of connective tissue.They secrete aldosterone
that function to regulate blood pressure.
Zona fasciculata:Lies next to zona glomerulosa.It is the largest layer made up of columnar
cells.They secrete hormone cortisol that is essential in maintaining homeostasis.
Zona reticularis:Lies next to zona fasciculata and near to medulla.Cells are in form of irregular
cords and they secrete androgen hormones for male reproductive functions.
Medulla :Present in the centre of the adrenal gland.Presence of chromaffin cells that secrete
hormones catecholamines adrenaline and noradrenaline. in response to fright and fear.
Location of adrenal gland:The adrenal glands are located on both sides of the body in the
retroperitoneum above the kidneys medially.
3.Adrenal and sympathetic nervous sytem interaction : In response to fear and fright , the
sympathetic nervous sysyem is activated that in turn activates the adrenal medulla to secrete
norepinephrine.Nor epinephrine prepares the body to respond to the situation immediately by
increasing the blood flow to our skeletal muscles to act accordingly.
4.Circadian cycle of melatonin: Melatonin is secreted by the pineal gland controls our sleep and
wake up cycle in a day.The internal clock of our body known as circadian rhythm regulates the
amount of melatonin produced by pineal gland.Melatonin levels rise in the dark after the sunset
and are at elevated levels at the night tht makes us asleep.In the morning,with sunrise the levels
drop making us awake.
5.HPA axis controls sex hormones: Sex difference is the result of interactions between the
gonadal hormone system and the HPA axis.In adulthood,androgen levels increase HPA activity
while and estrogen level decreases HPA activity.Monoamine neurotransmitters norepinephrine
and serotonin are released from HPA axis that determine differences in sex.Signals that originate
from the brain and periphery are integrated together to influence the level of HPA axis activity,
and these are sex-specific.
6.Diabetes Mellitus : In diabetes mellitus, the blood sugar level increase.
Describe how the composition of MALT tissue bypasses the need for.pdf
Describe how the composition of MALT tissue bypasses the need for dendritic cells to travel to
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mucosal membranes are a major avenue for pathogen entry? Is this benefit provided during the
first or subsequent encounters with a specific pathogen? Why?
Solution
Ans Problem 3. Problem Type: Comprehension:
MALT (mucosa-associated lymphoid tissue) is scattered along all mucosal linings of body and is
responsible for initiating immune responses to antigens that are specifically encountered along
these mucosal linings.
Mucosal epithelium contains special flat epithelial cells called M (which are membrane-like)
cells or FAE (also known as follicle associated epithelial cells). These special cells take small
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that tend to cross mucosa, and make IgA type of antibodies, which are then secreted directly onto
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Solution
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Solution
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Year d) interactions with other species for space or resources-classic experiment with Paramecia
ook in at pop: Grown separately One tends \"Or O\' hod P to have a Place or o P kaudatum higer
tenders aurelia than the owner to hide They can Grown in mixed culture lire together o o a. 2 4 6
8 10 12 14 16 18 Days Other considerations: a) interactions from competition Paramecia exa c)
age ofreproduction Guvenile to senescence) reproductive value d) seasonal effects a weather
(regular and catastrophic) thrips example e) effects of habitat degradation fKocarrying capacity)
is effectively unknown for natural pops of all species
Solution
d) Interaction with other species for space or resources-classical example of paramecia
1st diagram: Here both the species of paramecium P. aurelia and P. caudatumare are gwown
separately. In this case there is no competition between both species. Therefore these are
growing in number and number is increasing with time (days).
2nd diagram: Here both the species of paramecium P. aurelia and P. caudatumare gwown are
grown together. Here you can see that one species population is increasing (P. aurelia) and other
is decreasing (P. caudatumare). Here ‘Principle of Competitive Exclusion’ works. When two
species are too similar, they cannot co-exsits because one of them will prove superior (dominant)
competitor. In other words we can say that two species cannot co-exist when they are actively
dependent on same resources.
d) Seasonal effect of weather (regular or catastrophic) strips example
Change in weather also effects the growth of a particular species. One species grow well in
winters whereas other in summer or autumn. As the favourable season changes the regular
growth is inhibited and whole population is destroyed (catastrophe).
Example: Agricultural crops.
Write a program that creates a linked list of families. Each family .pdf
Write a program that creates a linked list of families. Each family contains a husband, his wife
and a set of children. Each family is linked to another by a pointer called nextFamily. Husband is
linked to his wife by a pointer called myWife. The children are linked to their mother through a
pointer called myChildren. Children are linked by a pointer called mySibling.
Therefore, there are three types of nodes. The Husband node (class) contains the following
information:
SSN of type long
firstName of type string
lastName of type string
nextFamily of Husband*
myWife of type Wife*
The Wife class contains the following information:
SSN of type long
firstName of type string
lastName of type string
myChildren of Child*
The Child class contains the following information:
SSN of type long
firstName of type string
lastName of type string
mySibling of Child*
Your program reads a set of commands from the transaction file and acts accordingly. The
commands in the transaction file include
AddHusband
RemoveHusband
AddWife
RemoveWife
AddAChild
RemoveAChild
PrintAFamily
PrintAllFamilies
AddHusbad comes with three more information: his SSN, his first name and his last name. When
you get this command, you need to create a node of type Husband and insert it to the top of the
Link list of the families
RemveHusband comes with one more information: the SSN. This node should be deleted. In
order to delete this node, first you need to remove the children (if any), then remove the wife (if
he is married) and then remove the node itself
AddWife comes with 4 more information, the SSN of the wife, her first name, her last name and
finally the SSN of her husband. Based on the SSN of her husband, her husband must be found in
the linked list and the wife node should be linked to it.
RemoveWife, comes with the SSN of her husband. In this case, the husband should be searched
and his wife node should be deleted. If the wife node is attached to the children, first remove all
the children and then remove the wide node
AddAchild comes with 4 more information, the SSN of the child, child first name, child last
name and finally the SSN of the father. Based on the SSN of the father, the father must be found
in the linked list and the child should be attached to the mother.
RemoveAchild, comes with the SSN of the father and SSN of the child. In this case, the father
should be searched and his wife node should be followed to find the child that should be deleted.
If you find the child, you need to remove it from the family
PrintAFamily comes with the SSN of the father. You need to search that family in the linked list
and print the information of the father. The mother (if any) and the children (if any) on the screen
PrintAllFamilies, goes through the entire linked list of the family and prints the information of
each family one by one.
Test your program with the following transaction file
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AddHusband 200200200 Joe Brown
AddHusband 300300300 Kevin Tarr
A.
Why do Radicals maintain capitalism will always tend to enter into c.pdf
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Solution
Keynes challenged Say’s law that supply will create its own demand, and argues it is
incompetent to tackle the problems of unemployment and of the trade cycle. The radicals believe
that capitalism has reduced the need for work, blurred the edges between work and free time and
loosened the relationship between work and wages. Information is corroding the market’s ability
to form prices correctly because markets are based on scarcity while information is abundant.
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of lowercase letters of the alphabet. Then have it be able to determine whether strings in a string
file are in the language or not. The program must take the two files as command line arguments
(do not prompt the user).
Solution
#include
#include
#define STATES 256
#define SYMBOLS 20
int N_symbols;
int NFA_states;
char *NFAtab[STATES][SYMBOLS];
int DFA_states; /* number of DFA states */
int DFAtab[STATES][SYMBOLS];
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void put_dfa_table(
int tab[][SYMBOLS], /* DFA table */
int nstates, /* number of states */
int nsymbols) /* number of input symbols */
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int i, j;
puts(\"STATE TRANSITION TABLE\");
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printf(\" | \");
for (i = 0; i < nsymbols; i++) printf(\" %c \", \'0\'+i);
printf(\"\ -----+--\");
for (i = 0; i < nsymbols; i++) printf(\"-----\");
printf(\"\ \");
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printf(\" %c | \", \'A\'+i); /* state */
for (j = 0; j < nsymbols; j++)
printf(\" %c \", \'A\'+tab[i][j]);
printf(\"\ \");
}
}
/*Initialize NFA table.*/
void init_NFA_table()
{
NFAtab[0][0] = \"12\";
NFAtab[0][1] = \"13\";
NFAtab[1][0] = \"12\";
NFAtab[1][1] = \"13\";
NFAtab[2][0] = \"4\";
NFAtab[2][1] = \"\";
NFAtab[3][0] = \"\";
NFAtab[3][1] = \"4\";
NFAtab[4][0] = \"4\";
NFAtab[4][1] = \"4\";
NFA_states = 5;
DFA_states = 0;
N_symbols = 2;
}
/*String \'t\' is merged into \'s\' in an alphabetical order.*/
void string_merge(char *s, char *t)
{
char temp[STATES], *r=temp, *p=s;
while (*p && *t) {
if (*p == *t) {
*r++ = *p++; t++;
} else if (*p < *t) {
*r++ = *p++;
} else
*r++ = *t++;
}
*r = \'\\0\';
if (*p) strcat(r, p);
else if (*t) strcat(r, t);
strcpy(s, temp);
}
/*Get next-state string for current-state string.*/
void get_next_state(char *nextstates, char *cur_states,
char *nfa[STATES][SYMBOLS], int n_nfa, int symbol)
{
int i;
char temp[STATES];
temp[0] = \'\\0\';
for (i = 0; i < strlen(cur_states); i++)
string_merge(temp, nfa[cur_states[i]-\'0\'][symbol]);
strcpy(nextstates, temp);
}
int state_index(char *state, char statename[][STATES], int *pn)
{
int i;
if (!*state) return -1; /* no next state */
for (i = 0; i < *pn; i++)
if (!strcmp(state, statename[i])) return i;
strcpy(statename[i], state); /* new state-name */
return (*pn)++;
}
/*
Convert NFA table to DFA table.
Return value: number of DFA states.
*/
int nfa_to_dfa(char *nfa[STATES][SYMBOLS], int n_nfa,
int n_sym, int dfa[][SYMBOLS])
{
char statename[STATES][STATES];
int i = 0; /* current index of DFA */
int n = 1; /* number of DFA states */
char nextstate[STATES];
int j;
strcpy(statename[0], \"0\"); /* start state */
for (i = 0; i < n; i++) { /* for each DFA state */
for (j = 0; j < n_sym; j++) { /* for each input symbol */
get_next_state(nextstate, statename[i], nfa, n_nfa, j);
dfa[i][j] = state_index(nextstate, statename, &n);
}
}
return n; /* number of DFA states */
}
void main()
{
init_NFA_table();
DFA_states = nfa_to_dfa(NFAtab, N.
Which of these hypotheses cannot be tested by an experiment Biologi.pdf
Which of these hypotheses cannot be tested by an experiment? Biological interactions between
barnacle species are responsible for their distribution patterns. Night-feeding fishes rely on their
sense of smell to locate prey. The growth of salt marsh gross is limited by nitrogen availability.
Certain humpback whale behaviors are important for attracting a mate. Goldfish prefer to eat
brightly-colored foods. TRUE/FALSE The oceans cover less than half of the earth\'s surface.
Ocean productivity is the amount of food produced by marine organisms and the number of
organisms this food supports. Aristotle developed one of the first taxonomical classification
schemes. Today it is generally believed that animals cannot live below the lit regions of the
ocean. Ecology is: a type of life style. the study of the interaction of organisms and their
environment. a place where organisms live. what happens when a predator consumes a prey
item. the study of the physical environment in an ecosystem. Organisms that live under similar
environmental conditions and location, and interact directly or indirectly are part of the same:
ecosystem. biosystem. abiotic habitat. biotic habitat. population. An example of abiotic
factors influencing marine organisms is: predation. competition for food. parasitism. salinity.
bacterial disease. Habitat refers to where an organism: lives. feeds. reproduces. forages. All
of the above.
Solution
Answers
15. False
Oceans cover about 70%of the surface of the earth
16. True
17. True
Aristotle developed the first classification system which divided all known organisms into two
groups plants and animals
18. False
Multiple choice
19. b study of interaction of organism\'s their environment
Ecology is the scientific analysis and study of interactions among organisms and their
environment . It is an interdisciplinary field that includes biology, geography and science .
20. a ecosystem
Interactions describes the different ways organisms obtain their food and energy . Some
organisms can make their own food and other organisms have to their food by eating other
organism
22. a lives
A habitat is an ecological or environmental area that is inhabited by a particular species of
animals plants or other type of organism . It is the natural environment in which an organism
lives ..
Which is the mode of inheritance that is most likely [That is the mo.pdf
Which is the mode of inheritance that is most likely? [That is the mode of inheritance that would
mean that the pattern would HAVE to be as described, not just that is could have come out this
way.] sex-limited, autosomal dominant Y-linked X-linked recessive autosomal recessive X-
linked dominant autosomal dominant autosomal sex-influenced
Solution
Non-autosomal mode of inheritance is the one related to sex-chromosomes. The alleles
associated with sex-chromosomes represent a differential kind of inheritance pattern and display
a differential phenotype response in offsprings based upon their sex. Majority of the human
diseases related to sex-chromosomes are X-linked in nature such as color blindness, hemophilia,
Duchenne muscular dystrophy, Vitamin D resistant rickets or X-linked hypophosphatemia, Rett
syndrome, Incontinentia pigmention and Goltz syndrome.
Since both parents, a male and a female carry an X chromosome, they both are potential donors
of X-linked allele of a disease. Since the offsprings can acquire a differential X-chromosome
profile either from mother or father, their genetic make up will be dependent upon that of their
parents and thus, their disease condition (carrier, affected, unaffected).
Thus, X-linked (dominant or recessive) mode of sex-chromosome related diseases need to be
explained theoratically and observationally, and not only be described as general procedures of
genetic segregation. Since X-linked recessive diseases are more likely to cause ambiguity among
inheritance pattern, a more detailed explanation would be requried for X-linked recessive alleles..
Which of the following is a good example of homoplasy (convergent ev.pdf
Which of the following is a good example of homoplasy (convergent evolution)?
a. Dolphins and humans
b. Apes and monkeys
c. Dolphins and fish
d. Dolphins and whales
e. Cats and bats
Wings of bats are:
a. Plesiomorphic (ancestral) feature for mammals
b. Plesiomorphic feature for bats and birds
c. Synapomorphic (shared-derived) feature for vertebrates
d. Synapomorphic feature for bats
e. Synapomorphic feature for bats and birds
Solution
1). c. Dolphins and fish
Analogous structures (convergent evolution) are not found in the common ancestors but the same
structure or similar structure with the same function can be found in one or more groups. In such
cases, it is not clear to tract the ancestry but the homologous structures are helpful to track the
ancestry. For example, both the dolphins and sharks share the fins but the dolphins are mammals,
while the shark are fish..
What happens at the mid-blastula transition in Xenopus Which part o.pdf
What happens at the mid-blastula transition in Xenopus? Which part of the Xenopus blastula
becomes the mesoderm? Where does the dorsal lip tissue end up when, Xenopus gastrulation is
finished? Where does the vegetal pole end up when gastrulation is finished? How is the central
nervous system formed? Where does the neural crest come from and what does it become?
What are the main tissue types formed from the mesoderm? What happens during the cortical
rotation? What is the principal determinant for endoderm formation in Xenopus? What are the
main mesoderm-inducing factors in Xenopus?
Solution
2. Midblastula transition refers to a number of dramatic changes that occur at this stage in
Xenopus. The important event that happens during this stage is activation of zygotic genes. The
Cell cycle is lengthened, the asynchronous cycle of different cells. Development of cellular
motility.
3. The blastula is composed of three different regions out of which the equatorial or marginal
zone, which compose the walls of the blastocoel differentiate primarily into mesodermal tissue.
4. The lip of the blastopore nearer to the animal pole is termed the dorsal lip of the blastopore.
The dorsal lip of the blastopore forms due to the contraction of bottle cells. It becomes the head
and the dorsal midline structures all along the body. It forms notochord, flanking parts of
somites, floor plate of neural tube.
5. The vegetal plate undergoes primary invagination to produce the archenteron or primitive gut.
Primary invagination is thought to result from changes in the shape of cells in the vegetal plate.
6. During the development, the dorsal ectoderm rolls up into a tube, which becomes the brain at
the anterior end and the spinal cord at the posterior end.
7. The neural crest is a population of migratory cells, derived from the neural folds. In the head,
it forms much of the skeleton. It forms dorsal root ganglia, enteric ganglia and pigment cells.
8. Mesoderm develops into notochord, somites, kidneys, gonads, limbs, lateral plate, blood
islands, heart and blood vessels during the Xenopus development.
9. On the surface of the dorsal egg cortex egg toward the animal pole and internally components
associated with cortex move from a vegetal to a dorsal position..
What is an example of Mendel’s first hypothesis regarding patterns o.pdf
What is an example of Mendel’s first hypothesis regarding patterns of inheritance? What
important vocabulary word do we learn in this hypothesis?
Solution
Gregor Johann Mendel studied the pea plant in his garden. He selected characters that were
easily identifiable. He grew the plants he selected to make sure they were pure – i.e. they were
pure line. A pure line is a plant that does not show any variation for the particular character
under study. Based on his studies, and the careful records he maintained, Mendel proposed his
first hypothesis as;
Vocabulary
Trait
Factors
Genes
Alleles
Homozygous
Heterozygous
Dominant
Recessive
F1.
MATATAG CURRICULUM: ASSESSING THE READINESS OF ELEM. PUBLIC SCHOOL TEACHERS I...
In this research, it concludes that while the readiness of teachers in Caloocan City to implement the MATATAG Curriculum is generally positive, targeted efforts in professional development, resource distribution, support networks, and comprehensive preparation can address the existing gaps and ensure successful curriculum implementation.
MASS MEDIA STUDIES-835-CLASS XI Resource Material.pdf
mass media studies
CBSE
Full Book
for class 11th
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I need an inttroduction for my paper on Conducting and Analyzing Sta.pdf
I need an inttroduction for my paper on Conducting and Analyzing Statistcal Tests
Solution
Statistical tests are of various types, depending upon the nature of the study. Statistical tests
provide a method for making quantitative decisions about a particular sample. Statistical tests
mainly test the hypothesis that is made about the significance of an observed sample.
Statistics.
Identify the contributions to Microbiology made by Redi, Tyndall, an.pdf
Identify the contributions to Microbiology made by Redi, Tyndall, and Pasteur.
Solution
Francesco Redi (1626-1697) was an Italian physician, who first refused the theory of
spontaneous generation. He refused the idea of spontaneous generation of maggots (larva of
flies) from the meat.
John Tyndall (1820-1893) is a physicist of 19th century. He studied diamagnetism and also the
made discoveries regarding the physical properties of air and the infrared radiation. He studied
the effect of radiant energy on the air.
Louis Pasteur was a French chemist during the 19th century. He disproved the theory of
spontaneous generation and won the debate on this topic and Academy’s prize. He filtered the air
through a gun-cotton filter and observed the presence of microbes on the filter under the
microscope..
I need help with two things. First Id like someone to check over m.pdf
I need help with two things. First I\'d like someone to check over my code for the 3 //TODO
statements below to verify that they\'re correct
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
Solution
PART 1
Hi friend, first part you have done correctly except a small mistake. In first TODO you missed
\'=\' sign:
public static HashSet doHashSetInsertMedian(int numItems) {
System.out.print(\"doHashSetInsertMedian: \");
HashSet set = new HashSet<>();
// TODO Write code that adds integers 0 through (numItems - 1)
// to set, inside a loop.
for(int i=0; i<=numItems-1; i++){
set.add(i);
}
// TODO Write code that removes integer (numItems / 2)
// from set.
int median = numItems / 2;
set.remove(median);
long startTime = getTimestamp();
// TODO Write code that adds integer (numItems / 2)
// to set.
set.add(median);
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}
PART 2:
public static LinkedHashSet doLinkedHashSetRemoveMedian(int numItems) {
System.out.print(\"doLinkedHashSetRemoveMedian: \");
LinkedHashSet set = new LinkedHashSet<>();
// TODO Write code that adds integers 0 through (numitems - 1)
// to set.
for(int i=0; i<=numitems-1; i++)
set.add(i);
long startTime = getTimestamp();
// TODO Write code that removes the median element
// from set, repeatedly until the set is empty.
int median;
int current_size = set.size();
while(set.size() > 0){
current_size = set.size(); //getting current size of set
median = current_size/2; // finding current median
set.remove(median); // removing median
}
long endTime = getTimestamp();
long totalTime = endTime - startTime;
System.out.println(totalTime);
return set;
}.
Explain Internal Controls and Corporate Governance.SolutionInt.pdf
Explain Internal Controls and Corporate Governance.
Solution
Internal Control is generally defined as \" the process designed , implemented and maintained by
those charged with governance ,management to provide reasonable assurance with regard to the
financial reporting ,effectiveness and efficiency of operations ,protection of assets and
compliance with prevalent laws and regulations.
Objectives of Internal Control :
(i) To ensure that the financial transactions are preformed within the general or specific
authorisation limits set by the management of the company
(ii) The accounting transactions are properly recorded with regards to correct amounts, as per
applicable financial reporting framework, in the right accounting period and per accounting
policies and principles
(iii) Assets are safeguarded against unauthorised use, disposition and physically protected against
damage
Corporate Governance :
Corporate governance is defined as \" the system by which an enterprise is directed and
controlled to achieve the objective of increasing the shareholders value by enhancing economic
performance.\" Corporate governance typically involves relationship among the board of
directors, controlling shareholders and other stakeholders of the company. Issues that corporate
governance ideally deals with issues which include roles of the CEO and board of directors,
controls assurance and risk management. Corporate governance is a key aspect to enhance
shareholders value, drive the enterprise to do best business and achieve the company\'s
objectives. Corporate governance also serves as an mechanism to oversee the company\'s
performance in financial and operational terms. The Sarbanes Oxley Act 2002 provides various
sections and clauses for an effective working of the Corporate Governance in US companies..
Environmental Laws and Acts Project Research all of the following law.pdf
Environmental Laws and Acts Project Research all of the following laws, acts, and/or treaties
and answer the questions below. Forest Reserve Act (1891) a. Amendment Year: b. International
or National?: Description of Function/Environmental Issues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 2. Lacey Act Draft Year: b. Amendment Year:
International or National?: Description of Function/Environmental Issues Affected d. e.
Agency/Group Responsible for Regulation and Enforcement: 3. Antiquities Act (1906) ent Year:
nternational or National?: c. Description of Function/Environmental Issues Affected: d.
Responsible for Regulation and Enforcement: 4, Public Health Service Act Amendment Year: b.
International or National? c, Description of Function/Environmental ssues Affected: d.
Agency/Group Responsible for Regulation and Enforcement: 5. National Park Service Act
(1916) a. Amendment Year: b. International or National? c. Description of
Function/Environmental lssues Affected: d. Agency/Group Responsible for Regulation and
Enforcement: 6. Migratory Bird Act (1918) a. Amendment Year: b. International or National?: c.
Description of Function/Environmentallssues Affected: d. Agency/Group Responsible for
Regulation and Enforcement: 7. Taylor Grazing Act (1934) a. Amendment Year: b. nternational
or National? c. Description of Function/Environmental lssues Affected: d. Agency/Group
Responsible for Regulation and Enforcement: 8, Soil Conservation Act (1935) a. Amendment
Year: b, International or National? c. Description of Function/E Issues Affected nvironmental d.
Agency/Group Responsible for Regulation and Enforcement: 9. Federal Insecticide, Fungicide,
and Rodenticide Act (1947) a. Amendment Year: b, International or National? c. Description of
Function/Environmental lssues Affected:
Solution
Please find the answers below as tabulated:
Answer 2:
Lacey act
Answer 3
:Antiquities act(1906)
Answer 4:
Public Health Service Act (1912)
Governed by Service Commission Corps and United States Department of Health and Human
Services and enacted by Office of the Assistant Secretary for Preparedness and Response along
with Office of Global AffairsActDrafted yearAmendment
yearNatiaonl/internationalDescriptionagency/groupAnswer 1: Forest reserve
act1891NoneNational (for U.S.)This law gave the President the power to keep the forest area
and residential area separate. This helped in keeping the forest reserves intact while managing
the residential areas.Formulated and executed by the United States congress
Answer 2:
Lacey act1990Amended in 2008 and entitled as Food, Conservation, and Energy ActNationalA
law of conservation of illegal trade of fauna and flora and parts thereof such as fisheries, plant
products etc.The United States Federal Environmental Legislation
Answer 3
:Antiquities act(1906)1906Amended in 1950 to incorporate some domains of
antiquitiesNationalThis law was formulated in order to create monuments from the land of the
country in.
discuss the drawbacks of programmed and interrupt driven io and des.pdf
discuss the drawbacks of programmed and interrupt driven i/o and describe in general the
functionality of the DNA
Solution
Programmed I/O
Programmed I/O (PIO) refers to data transfers initiated by a CPU under driver software control
to access registers or memory on a device.
The CPU issues a command then waits for I/O operations to be complete. As the CPU is faster
than the I/O module, the problem with programmed I/O is that the CPU has to wait a long time
for the I/O module of concern to be ready for either reception or transmission of data. The CPU,
while waiting, must repeatedly check the status of the I/O module, and this process is known as
Polling. As a result, the level of the performance of the entire system is severely degraded.
Programmed I/O basically works in these ways:
Interrupt
The CPU issues commands to the I/O module then proceeds with its normal work until
interrupted by I/O device on completion of its work.
For input, the device interrupts the CPU when new data has arrived and is ready to be retrieved
by the system processor. The actual actions to perform depend on whether the device uses I/O
ports, memory mapping.
For output, the device delivers an interrupt either when it is ready to accept new data or to
acknowledge a successful data transfer. Memory-mapped and DMA-capable devices usually
generate interrupts to tell the system they are done with the buffer.
Although Interrupt relieves the CPU of having to wait for the devices, but it is still inefficient in
data transfer of large amount because the CPU has to transfer the data word by word between I/O
module and memory.
The main limitation of programmed I/O and interrupt driven I/O is given below:
Programmed I/O
Each instructions selects one I/O device (by number) and transfers a single character (byte)
Example: microprocessor controlled video terminal.
Four registers: input status and character, output status and character.
Interrupt-driven I/O
Primary disadvantage of programmed I/O is that CPU spends most of its time in a tight loop
waiting for the device to become ready. This is called busy waiting.
With interrupt-driven I/O, the CPU starts the device and tells it to generate an interrupt when it is
finished.
Done by setting interrupt-enable bit in status register.
Still requires an interrupt for every character read or written.
Interrupting a running process is an expensive business (requires saving context).
Requires extra hardware (DMA controller chip).
All these limitation can be overcome by the Introduction of DMA (Direct Memory Access)
To write block of 32 bytes from memory address 100 to device 4
1. CPU writes 32, 100, 4 into the first three DMA registers (memory address, count, device
number)
2. CPU puts code for WRITE (say 1) into fourth (direction) DMA register, which signals DMA
controller to begin operation
3. Controller reads (via bus request as CPU would) byte 100 from memory
4. Controller makes I/O request to write to device 4
5. Controller increments m.
do it on excel and please kindly upload the excel file. or explain m.pdf
do it on excel and please kindly upload the excel file. or explain me how to do this question on
excel. I really need help.
Solution
In numerical analysis, Newton\'s method (also known as the Newton–Raphson method), named
after IsaacNewton and Joseph Raphson, is a method for finding successively better
approximations to the roots (or zeroes) of a real-valued function. x : f ( x ) = 0 . {\\displaystyle
x:f(x)=0\\,.}.
Describe the structure and location of the parathyroid glands.Desc.pdf
Describe the structure and location of the parathyroid glands.
Describe the structure and location of the adrenal glands. Include the organization of the
following parts and their major role: Cortex and Medulla
Explain how the adrenal medulla and the sympathetic nervous system interact.
Describe the circadian cycle of melatonin.
Describe how the HPA axis controls sex-hormone release.
Describe what happens in the body in diabetes mellitus and explain the difference between Type
1 and Type 2 diabetes.
Solution
1.Structure and location of parathyroid glands:
Structure :
Loaction :These glands are situated behind the left and right lobes of the thyroid.
2.Structure and location of adrenal glands:
Structure :
Cortex :It is the outermost layer and is made up of three zones:
Zona glomerulosa :The outermost layer of the adrenal cortex is the zona glomerulosa and is
made up of oval cells that separated by thin strands of connective tissue.They secrete aldosterone
that function to regulate blood pressure.
Zona fasciculata:Lies next to zona glomerulosa.It is the largest layer made up of columnar
cells.They secrete hormone cortisol that is essential in maintaining homeostasis.
Zona reticularis:Lies next to zona fasciculata and near to medulla.Cells are in form of irregular
cords and they secrete androgen hormones for male reproductive functions.
Medulla :Present in the centre of the adrenal gland.Presence of chromaffin cells that secrete
hormones catecholamines adrenaline and noradrenaline. in response to fright and fear.
Location of adrenal gland:The adrenal glands are located on both sides of the body in the
retroperitoneum above the kidneys medially.
3.Adrenal and sympathetic nervous sytem interaction : In response to fear and fright , the
sympathetic nervous sysyem is activated that in turn activates the adrenal medulla to secrete
norepinephrine.Nor epinephrine prepares the body to respond to the situation immediately by
increasing the blood flow to our skeletal muscles to act accordingly.
4.Circadian cycle of melatonin: Melatonin is secreted by the pineal gland controls our sleep and
wake up cycle in a day.The internal clock of our body known as circadian rhythm regulates the
amount of melatonin produced by pineal gland.Melatonin levels rise in the dark after the sunset
and are at elevated levels at the night tht makes us asleep.In the morning,with sunrise the levels
drop making us awake.
5.HPA axis controls sex hormones: Sex difference is the result of interactions between the
gonadal hormone system and the HPA axis.In adulthood,androgen levels increase HPA activity
while and estrogen level decreases HPA activity.Monoamine neurotransmitters norepinephrine
and serotonin are released from HPA axis that determine differences in sex.Signals that originate
from the brain and periphery are integrated together to influence the level of HPA axis activity,
and these are sex-specific.
6.Diabetes Mellitus : In diabetes mellitus, the blood sugar level increase.
Describe how the composition of MALT tissue bypasses the need for.pdf
Describe how the composition of MALT tissue bypasses the need for dendritic cells to travel to
the lymph node to present antigen to B and T cells. What benefit does this provide, given that gut
mucosal membranes are a major avenue for pathogen entry? Is this benefit provided during the
first or subsequent encounters with a specific pathogen? Why?
Solution
Ans Problem 3. Problem Type: Comprehension:
MALT (mucosa-associated lymphoid tissue) is scattered along all mucosal linings of body and is
responsible for initiating immune responses to antigens that are specifically encountered along
these mucosal linings.
Mucosal epithelium contains special flat epithelial cells called M (which are membrane-like)
cells or FAE (also known as follicle associated epithelial cells). These special cells take small
amounts of gut antigens that enter the lumen of gut, and pass them to antigen producing cells and
lymphocytes of the MALT. M Cells also produce B-lymphocytes in gut react to micro-organisms
that tend to cross mucosa, and make IgA type of antibodies, which are then secreted directly onto
the gut lumen. These antibodies help in prevention of micro-organisms entering the gut by
sticking to the epithelium of gut, and can neutralise toxins and viruses both first and
subsequently also due to presence of memory B cells and T cells that then move to mucosal
effector sites..
C. wye D. delta E. not defined Question 6 What is the configuration o.pdf
C. wye D. delta E. not defined Question 6 What is the configuration of R3, R4 and R6? O A.
wye B. not defined O C. delta D. parallel E. Series Question 7 What is the configuration of R5,
R7 and R8? R2 LR3
Solution
its a delta connection
Standard 3-phase circuits or networks take on two major forms with names that represent the
way in which the resistances are connected, a Star connected network which has the symbol of
the letter, (wye) and a Delta connected network which has the symbol of a triangle, (delta).
A bullet that lodges in the heart would a. be located in the ventra.pdf
A bullet that lodges in the heart would: a. be located in the ventral body cavity. b. penetrate the
visceral peritoneum c. be located in the vertebral cavity. d. penetrate the parietal pleura. e. be
located laterally to a bullet that lodges in the lung.
Solution
3. d. A bullet that lodges in the heart would pentrate the parietal pluera to reach the heart..
Year d) interactions with other species for space or resources-classi.pdf
Year d) interactions with other species for space or resources-classic experiment with Paramecia
ook in at pop: Grown separately One tends \"Or O\' hod P to have a Place or o P kaudatum higer
tenders aurelia than the owner to hide They can Grown in mixed culture lire together o o a. 2 4 6
8 10 12 14 16 18 Days Other considerations: a) interactions from competition Paramecia exa c)
age ofreproduction Guvenile to senescence) reproductive value d) seasonal effects a weather
(regular and catastrophic) thrips example e) effects of habitat degradation fKocarrying capacity)
is effectively unknown for natural pops of all species
Solution
d) Interaction with other species for space or resources-classical example of paramecia
1st diagram: Here both the species of paramecium P. aurelia and P. caudatumare are gwown
separately. In this case there is no competition between both species. Therefore these are
growing in number and number is increasing with time (days).
2nd diagram: Here both the species of paramecium P. aurelia and P. caudatumare gwown are
grown together. Here you can see that one species population is increasing (P. aurelia) and other
is decreasing (P. caudatumare). Here ‘Principle of Competitive Exclusion’ works. When two
species are too similar, they cannot co-exsits because one of them will prove superior (dominant)
competitor. In other words we can say that two species cannot co-exist when they are actively
dependent on same resources.
d) Seasonal effect of weather (regular or catastrophic) strips example
Change in weather also effects the growth of a particular species. One species grow well in
winters whereas other in summer or autumn. As the favourable season changes the regular
growth is inhibited and whole population is destroyed (catastrophe).
Example: Agricultural crops.
Write a program that creates a linked list of families. Each family .pdf
Write a program that creates a linked list of families. Each family contains a husband, his wife
and a set of children. Each family is linked to another by a pointer called nextFamily. Husband is
linked to his wife by a pointer called myWife. The children are linked to their mother through a
pointer called myChildren. Children are linked by a pointer called mySibling.
Therefore, there are three types of nodes. The Husband node (class) contains the following
information:
SSN of type long
firstName of type string
lastName of type string
nextFamily of Husband*
myWife of type Wife*
The Wife class contains the following information:
SSN of type long
firstName of type string
lastName of type string
myChildren of Child*
The Child class contains the following information:
SSN of type long
firstName of type string
lastName of type string
mySibling of Child*
Your program reads a set of commands from the transaction file and acts accordingly. The
commands in the transaction file include
AddHusband
RemoveHusband
AddWife
RemoveWife
AddAChild
RemoveAChild
PrintAFamily
PrintAllFamilies
AddHusbad comes with three more information: his SSN, his first name and his last name. When
you get this command, you need to create a node of type Husband and insert it to the top of the
Link list of the families
RemveHusband comes with one more information: the SSN. This node should be deleted. In
order to delete this node, first you need to remove the children (if any), then remove the wife (if
he is married) and then remove the node itself
AddWife comes with 4 more information, the SSN of the wife, her first name, her last name and
finally the SSN of her husband. Based on the SSN of her husband, her husband must be found in
the linked list and the wife node should be linked to it.
RemoveWife, comes with the SSN of her husband. In this case, the husband should be searched
and his wife node should be deleted. If the wife node is attached to the children, first remove all
the children and then remove the wide node
AddAchild comes with 4 more information, the SSN of the child, child first name, child last
name and finally the SSN of the father. Based on the SSN of the father, the father must be found
in the linked list and the child should be attached to the mother.
RemoveAchild, comes with the SSN of the father and SSN of the child. In this case, the father
should be searched and his wife node should be followed to find the child that should be deleted.
If you find the child, you need to remove it from the family
PrintAFamily comes with the SSN of the father. You need to search that family in the linked list
and print the information of the father. The mother (if any) and the children (if any) on the screen
PrintAllFamilies, goes through the entire linked list of the family and prints the information of
each family one by one.
Test your program with the following transaction file
AddHusband 100100100 Jim Smith
AddHusband 200200200 Joe Brown
AddHusband 300300300 Kevin Tarr
A.
Why do Radicals maintain capitalism will always tend to enter into c.pdf
Why do Radicals maintain capitalism will always tend to enter into crisis?
Solution
Keynes challenged Say’s law that supply will create its own demand, and argues it is
incompetent to tackle the problems of unemployment and of the trade cycle. The radicals believe
that capitalism has reduced the need for work, blurred the edges between work and free time and
loosened the relationship between work and wages. Information is corroding the market’s ability
to form prices correctly because markets are based on scarcity while information is abundant.
They feel the system’s defence mechanism is to form monopolies. Goods and services are
appearing no longer responsive to the dictates of the market and the managerial hierarchy.
Parallel currencies, time banks, and cooperatives have been proliferated and the whole swaths of
economic life are beginning to move to a different rhythm.
*****.
Write a C program that converts an NFA file to a DFA, where the lang.pdf
Write a C program that converts an NFA file to a DFA, where the language of the NFA consists
of lowercase letters of the alphabet. Then have it be able to determine whether strings in a string
file are in the language or not. The program must take the two files as command line arguments
(do not prompt the user).
Solution
#include
#include
#define STATES 256
#define SYMBOLS 20
int N_symbols;
int NFA_states;
char *NFAtab[STATES][SYMBOLS];
int DFA_states; /* number of DFA states */
int DFAtab[STATES][SYMBOLS];
/*Print state-transition table.*/
void put_dfa_table(
int tab[][SYMBOLS], /* DFA table */
int nstates, /* number of states */
int nsymbols) /* number of input symbols */
{
int i, j;
puts(\"STATE TRANSITION TABLE\");
/* input symbols: \'0\', \'1\', ... */
printf(\" | \");
for (i = 0; i < nsymbols; i++) printf(\" %c \", \'0\'+i);
printf(\"\ -----+--\");
for (i = 0; i < nsymbols; i++) printf(\"-----\");
printf(\"\ \");
for (i = 0; i < nstates; i++) {
printf(\" %c | \", \'A\'+i); /* state */
for (j = 0; j < nsymbols; j++)
printf(\" %c \", \'A\'+tab[i][j]);
printf(\"\ \");
}
}
/*Initialize NFA table.*/
void init_NFA_table()
{
NFAtab[0][0] = \"12\";
NFAtab[0][1] = \"13\";
NFAtab[1][0] = \"12\";
NFAtab[1][1] = \"13\";
NFAtab[2][0] = \"4\";
NFAtab[2][1] = \"\";
NFAtab[3][0] = \"\";
NFAtab[3][1] = \"4\";
NFAtab[4][0] = \"4\";
NFAtab[4][1] = \"4\";
NFA_states = 5;
DFA_states = 0;
N_symbols = 2;
}
/*String \'t\' is merged into \'s\' in an alphabetical order.*/
void string_merge(char *s, char *t)
{
char temp[STATES], *r=temp, *p=s;
while (*p && *t) {
if (*p == *t) {
*r++ = *p++; t++;
} else if (*p < *t) {
*r++ = *p++;
} else
*r++ = *t++;
}
*r = \'\\0\';
if (*p) strcat(r, p);
else if (*t) strcat(r, t);
strcpy(s, temp);
}
/*Get next-state string for current-state string.*/
void get_next_state(char *nextstates, char *cur_states,
char *nfa[STATES][SYMBOLS], int n_nfa, int symbol)
{
int i;
char temp[STATES];
temp[0] = \'\\0\';
for (i = 0; i < strlen(cur_states); i++)
string_merge(temp, nfa[cur_states[i]-\'0\'][symbol]);
strcpy(nextstates, temp);
}
int state_index(char *state, char statename[][STATES], int *pn)
{
int i;
if (!*state) return -1; /* no next state */
for (i = 0; i < *pn; i++)
if (!strcmp(state, statename[i])) return i;
strcpy(statename[i], state); /* new state-name */
return (*pn)++;
}
/*
Convert NFA table to DFA table.
Return value: number of DFA states.
*/
int nfa_to_dfa(char *nfa[STATES][SYMBOLS], int n_nfa,
int n_sym, int dfa[][SYMBOLS])
{
char statename[STATES][STATES];
int i = 0; /* current index of DFA */
int n = 1; /* number of DFA states */
char nextstate[STATES];
int j;
strcpy(statename[0], \"0\"); /* start state */
for (i = 0; i < n; i++) { /* for each DFA state */
for (j = 0; j < n_sym; j++) { /* for each input symbol */
get_next_state(nextstate, statename[i], nfa, n_nfa, j);
dfa[i][j] = state_index(nextstate, statename, &n);
}
}
return n; /* number of DFA states */
}
void main()
{
init_NFA_table();
DFA_states = nfa_to_dfa(NFAtab, N.
Which of these hypotheses cannot be tested by an experiment Biologi.pdf
Which of these hypotheses cannot be tested by an experiment? Biological interactions between
barnacle species are responsible for their distribution patterns. Night-feeding fishes rely on their
sense of smell to locate prey. The growth of salt marsh gross is limited by nitrogen availability.
Certain humpback whale behaviors are important for attracting a mate. Goldfish prefer to eat
brightly-colored foods. TRUE/FALSE The oceans cover less than half of the earth\'s surface.
Ocean productivity is the amount of food produced by marine organisms and the number of
organisms this food supports. Aristotle developed one of the first taxonomical classification
schemes. Today it is generally believed that animals cannot live below the lit regions of the
ocean. Ecology is: a type of life style. the study of the interaction of organisms and their
environment. a place where organisms live. what happens when a predator consumes a prey
item. the study of the physical environment in an ecosystem. Organisms that live under similar
environmental conditions and location, and interact directly or indirectly are part of the same:
ecosystem. biosystem. abiotic habitat. biotic habitat. population. An example of abiotic
factors influencing marine organisms is: predation. competition for food. parasitism. salinity.
bacterial disease. Habitat refers to where an organism: lives. feeds. reproduces. forages. All
of the above.
Solution
Answers
15. False
Oceans cover about 70%of the surface of the earth
16. True
17. True
Aristotle developed the first classification system which divided all known organisms into two
groups plants and animals
18. False
Multiple choice
19. b study of interaction of organism\'s their environment
Ecology is the scientific analysis and study of interactions among organisms and their
environment . It is an interdisciplinary field that includes biology, geography and science .
20. a ecosystem
Interactions describes the different ways organisms obtain their food and energy . Some
organisms can make their own food and other organisms have to their food by eating other
organism
22. a lives
A habitat is an ecological or environmental area that is inhabited by a particular species of
animals plants or other type of organism . It is the natural environment in which an organism
lives ..
Which is the mode of inheritance that is most likely [That is the mo.pdf
Which is the mode of inheritance that is most likely? [That is the mode of inheritance that would
mean that the pattern would HAVE to be as described, not just that is could have come out this
way.] sex-limited, autosomal dominant Y-linked X-linked recessive autosomal recessive X-
linked dominant autosomal dominant autosomal sex-influenced
Solution
Non-autosomal mode of inheritance is the one related to sex-chromosomes. The alleles
associated with sex-chromosomes represent a differential kind of inheritance pattern and display
a differential phenotype response in offsprings based upon their sex. Majority of the human
diseases related to sex-chromosomes are X-linked in nature such as color blindness, hemophilia,
Duchenne muscular dystrophy, Vitamin D resistant rickets or X-linked hypophosphatemia, Rett
syndrome, Incontinentia pigmention and Goltz syndrome.
Since both parents, a male and a female carry an X chromosome, they both are potential donors
of X-linked allele of a disease. Since the offsprings can acquire a differential X-chromosome
profile either from mother or father, their genetic make up will be dependent upon that of their
parents and thus, their disease condition (carrier, affected, unaffected).
Thus, X-linked (dominant or recessive) mode of sex-chromosome related diseases need to be
explained theoratically and observationally, and not only be described as general procedures of
genetic segregation. Since X-linked recessive diseases are more likely to cause ambiguity among
inheritance pattern, a more detailed explanation would be requried for X-linked recessive alleles..
Which of the following is a good example of homoplasy (convergent ev.pdf
Which of the following is a good example of homoplasy (convergent evolution)?
a. Dolphins and humans
b. Apes and monkeys
c. Dolphins and fish
d. Dolphins and whales
e. Cats and bats
Wings of bats are:
a. Plesiomorphic (ancestral) feature for mammals
b. Plesiomorphic feature for bats and birds
c. Synapomorphic (shared-derived) feature for vertebrates
d. Synapomorphic feature for bats
e. Synapomorphic feature for bats and birds
Solution
1). c. Dolphins and fish
Analogous structures (convergent evolution) are not found in the common ancestors but the same
structure or similar structure with the same function can be found in one or more groups. In such
cases, it is not clear to tract the ancestry but the homologous structures are helpful to track the
ancestry. For example, both the dolphins and sharks share the fins but the dolphins are mammals,
while the shark are fish..
What happens at the mid-blastula transition in Xenopus Which part o.pdf
What happens at the mid-blastula transition in Xenopus? Which part of the Xenopus blastula
becomes the mesoderm? Where does the dorsal lip tissue end up when, Xenopus gastrulation is
finished? Where does the vegetal pole end up when gastrulation is finished? How is the central
nervous system formed? Where does the neural crest come from and what does it become?
What are the main tissue types formed from the mesoderm? What happens during the cortical
rotation? What is the principal determinant for endoderm formation in Xenopus? What are the
main mesoderm-inducing factors in Xenopus?
Solution
2. Midblastula transition refers to a number of dramatic changes that occur at this stage in
Xenopus. The important event that happens during this stage is activation of zygotic genes. The
Cell cycle is lengthened, the asynchronous cycle of different cells. Development of cellular
motility.
3. The blastula is composed of three different regions out of which the equatorial or marginal
zone, which compose the walls of the blastocoel differentiate primarily into mesodermal tissue.
4. The lip of the blastopore nearer to the animal pole is termed the dorsal lip of the blastopore.
The dorsal lip of the blastopore forms due to the contraction of bottle cells. It becomes the head
and the dorsal midline structures all along the body. It forms notochord, flanking parts of
somites, floor plate of neural tube.
5. The vegetal plate undergoes primary invagination to produce the archenteron or primitive gut.
Primary invagination is thought to result from changes in the shape of cells in the vegetal plate.
6. During the development, the dorsal ectoderm rolls up into a tube, which becomes the brain at
the anterior end and the spinal cord at the posterior end.
7. The neural crest is a population of migratory cells, derived from the neural folds. In the head,
it forms much of the skeleton. It forms dorsal root ganglia, enteric ganglia and pigment cells.
8. Mesoderm develops into notochord, somites, kidneys, gonads, limbs, lateral plate, blood
islands, heart and blood vessels during the Xenopus development.
9. On the surface of the dorsal egg cortex egg toward the animal pole and internally components
associated with cortex move from a vegetal to a dorsal position..
What is an example of Mendel’s first hypothesis regarding patterns o.pdf
What is an example of Mendel’s first hypothesis regarding patterns of inheritance? What
important vocabulary word do we learn in this hypothesis?
Solution
Gregor Johann Mendel studied the pea plant in his garden. He selected characters that were
easily identifiable. He grew the plants he selected to make sure they were pure – i.e. they were
pure line. A pure line is a plant that does not show any variation for the particular character
under study. Based on his studies, and the careful records he maintained, Mendel proposed his
first hypothesis as;
Vocabulary
Trait
Factors
Genes
Alleles
Homozygous
Heterozygous
Dominant
Recessive
F1.
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