Determine whether the following pair of events are dependent of indep.pdf
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Determine whether the following pair of events are dependent of independent. Event E is the event that a resident of Colorado lives in Denver, and F is the event that a resident of Colorado lives in either Denver or Boulder. Are the events dependent or independent? Independent Dependent Solution Both the events E & F are Independent.
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Effects of protease inhibitors in the treatment of secondary infecti.pdf
Effects of protease inhibitors in the treatment of secondary infections in HIV-positive patients
Find the average complication rate in the two groups of HIV-positive patients. Protease
inhibitors (PIs) arc used in the treatment of infection by human immunodeficiency virus (HIV).
A military hospital provides unlimited access to a full spectrum of HIV medications. Has this
access reduced admissions caused by secondary infections? In particular, has unlimited access to
PIs reduced the annual admission rate of patients with pulmonary complications? The number of
admissions for HIV. separated into patients with and those without pulmonary complications,
was obtained for the 4 years (1992-1995) before access to PIs and for the 2 years (1996-1997)
after PIs became available.
Solution
The complication rate before PIs are available is: number of complications in the four years/total
number of patients in those four years
=194/485=0.4
The complication rate before PIs are available is: number of complications in the two years/total
number of patients in those two years
=25/92=0.27
Thus average complication rate is: (0.4+0.27)/2=0.335.
Describe the process (step by step) how a glucose molecule is absorb.pdf
Describe the process (step by step) how a glucose molecule is absorbed from the lumen of the
intestine to the blood. Include what energy drives this process. Then explain how researchers like
Pia Roder were able to use radioactivity to determine where and how the transporters functioned.
(The answer must be in a long paragraph form)
Solution
Carbohydrates initially undergo chemical digestion by salivary amylases result in formation of
simple monomer sugars of \"glucose\" etc finally absorbed directly into the sublingual
circulation. Later mixed food material enters into the pharynx from mouth due to the peristaltic
movement of the visceral smooth muscles finally enters into the stomach
Disaccharidases are the enzymes secreted from the brush border cells of intestine. These
enzymes are sucrase, maltase, and lactase. These enzymes acts on disachraides such as maltose,
sucrose and lactose and digests them into monomer sugar units of glucose & galactose. Simple,
soluble and absorbable monomer glucose molecules \"diffused\" (diffusion) into the blood
portals finally food chyme enters into the large intestine for reabsorption.
It has clearly observed the \"sodium -glucose symport\" enable the transport of glucose
monomers after digestion into the intestinal epithelial cells via microvilli (improve absorption)
finally into the interstitial space followed by movement of glucose into the blood down the
concentration gradient (crossing basilar membrane). Primary active transport using Na+-K+-
ATPase improves movement sodium inside and potassium outside using energy followed by
building up of interstitial sodium concentration. Sodium –glucose symport has considered as
secondary active transport as it is a co-transporter.
P V Roder analysis of \"glucose transporter functions\" via radioactivity principle:
The sodium-glucose linked transporter (SGLTs, a co-transporter) uses the energy generated by
the ATPase pump through the downhill sodium ion gradient for the glucose transportation across
the apical membrane (against the glucose gradient). SGLTs are an example of secondary active
transport. The glucose transporters (GLUT) present in the basolateral membrane now allow the
glucose transport into the peritubular capillaries. Inhibition of the Na+/K+/ATPase fails to
generate energy required for the SGLT secondary transporters, which inhibits glucose uptake.
Sugars such as glucose, galactose and free amino acid (alanine) uptake into the microvilli
epithelial cells is based on sodium mediated cotransport whereas fructose cannot be transported
by sodium mediated transport during absorption after digestion. H+ mediated oligopeptide
transport is present in gut epithelium.
Scientist PV Roder has analyzed the function of SGLTs and GLUTs in the apical membrane
using “radiotracer glucose gavage“, in which mice was taken of two different types such as
“Sglt1 expressing wild type” and ” mice with sglt1-knockout” finally C14 radioactive tracing
performed for the moveme.
Describe how rock fossils are formed.What are two different ways t.pdf
Describe how rock fossils are formed.
What are two different ways that geologists can use to age fossils?
Solution
1). Permineralization or petrification is the most typical form of fossil formation. The soft tissues
of the organism decays within the sediment, leaving behind the hard structures such as skeleton.
Skeleton contains minerals, when water seeps into the skeleton, the minerals are dissolved in the
water and form crystals within the spaces of remains. This crystallization process turns the
skeleton remains into hard structures while encasing the sedimentary rock..
You are working on an older PC and need to replace a faulty CPU. As .pdf
You are working on an older PC and need to replace a faulty CPU. As you remove the CPU, you
notice that the processor unit is a rectangular package with two uniformly spaced parallel rows of
pins pointing downward. What general socket type takes this sort of chip?
A. DIP
B. LIF
C. PGA
D. ZIF
Solution
A. The DIP, or dual in-line package, is characterized by a chip with two rows of pins arranged in
parallel, and this design can be found on a number of different electronic components, such as
CPUs and other integrated circuits..
write a short essay report to describe a topic of your interest rela.pdf
write a short essay report to describe a topic of your interest related to the past, or current, or
future technologies of cellular communications, bandwidths, antennas or their impact to the
human society
Solution
past, or current, or future technologies of cellular communications:
Mobile computing is to describe technologies thatenable people to access network services
anyplace, anytime, and anywhere.
past,present technologies of cellular communications:
1896 Guglielmo Marconi,First demonstration of wireless telegraphy.Based on long wave,
requiring very large transmitters
1907 Commercial Trans-Atlantic Wireless Service.Huge ground stations: 30 x 100m antenna
masts
1920 Discovery of short waves by Marconi.Cheaper, smaller, better quality transmitters by
vacuum tube
1982 Start of GSM in Europe (1G analog)
1983 Start of AMPS in America (1G analog)
1992 Start of GSM (2G digital)
1997 Wireless LAN - IEEE802.11
1998 Iridium satellite system.
1999 Standardization of additional wireless LANs
2000 GSM with higher data rates (2.5G digital).HSCSD offers up to 57,6kbit/s.First GPRS trials
with up to 50 kbit/s
2001 Start of 3G systems.IMT - 2000, several “members” of a “family”, CDMA2000 in Korea,
UMTS tests in Europe.
GPRS: General Packet Radio Service
It is a mobile data service available to users of GSM mobile phones. It is often described as
\"2.5G“.
GPRS is packet-switched which means that multiple users share the same transmission channel,
only transmitting when they have data to send.
GPRS provides moderate speed data transfer, by allocating unused cell bandwidth to transmit
data.Poor bit rate in busy cells.Usually, GPRS data is billed per kilobytes of information
transceived.
future technologies of cellular communications:
Improved radio technology and antennas.smart antennas, beam forming, multiple-input multiple-
output (MIMO) 802.11N.dynamic spectrum allocation.Core network convergence.IP-based,
quality of service, mobile IP.Ad-hoc technologies.spontaneous communication, power saving,
redundancy.Simple and open service platform.intelligence at the edge, not in the network (as
with IN).more service providers, not network operators only.
Impact on Society:
Since the introduction of mobile phones, concerns (both scientific and public) have been raised
about the potential health impacts from regular use.But by 2008, American mobile phones
transmitted and received more text messages than phone calls.Numerous studies have reported
no significant relationship between mobile phone use and health, but the effect of mobile phone
usage on health continues to be an area of public concern.
For example, at the request of some of their customers, Verizon created usage controls that meter
service and can switch phones off, so that children could get some sleep.There have also been
attempts to limit use by persons operating moving trains or automobiles, coaches when writing to
potential players on their teams, and movie theater audiences. By one measure, near.
Write C language sequence to output the upper byte of the unsigned in.pdf
Write C language sequence to output the upper byte of the unsigned integer variable x to port PI
and lower byte to port P0.
Solution
#define lowbyte(i ) ((unsigned char) (i)) #define highbyte(i ) ((unsigned char) (((unsigned int)
(i)) >> 8)) unsigned int a; unsigned char p, q; void shift (void) { p = lowbyte(a); q =
highbyte(a); }.
Capital Budgeting Select a government agency (localstatefederal) a.pdf
Capital Budgeting Select a government agency (local/state/federal) and develop a four- to five-
page APA-style paper summarizing the following: Explain how the debt capacity of the
governmental entity is determined. Evaluate the effect of refunding or reorganizing existing debt
obligations. Analyze various funding alternatives that can be used to support debt obligations.
Your paper must be four to five pages (not including title and reference pages) and must be
formatted according to APA style as outlined in the approved APA style guide. You must cite at
least three scholarly sources in addition to the textbook.
Solution
The ability of any government entity to generate revenue from its economy underlie the
government\'s debt capacity.
GFOA (Government Finance Officers Association) determine its debt capacity through five steps
methodology which includes the following:
(1) defining the scope of analysis,
(2) peer group development,
(3)analysis of current position,
(4) scenario analysis, and
(5) financial policy development.
Along with debts various other sources that could be used for funding includesales and use tax,
property taxes, reserves etc..
can anybody think of any world events that changed information syste.pdf
can anybody think of any world events that changed information systems?
Solution
Systems range from very simple to very complex. There are numerous types of systems. For
example, there are biological systems, mechanical systems, human/mechanical systems,
ecological systems, and social systems.
Following are some technologies that will change technology:-
- 3D printers,
- sensor networks,
- virtual humans
technologies under development now will drastically change our world in the decade to come.
Internet Things:-
We have passed the threshold where more things are connected to the Internet than people. The
transition to IPV6 also supports seemingly limitless connectivity.
Sensor Network:-
Nowadays this technology grows very fast by using this we can track song pollution, air
pollution and we also can track temperature. This technology is growing day by day and many
companies are coming with new ideas related to sensor networks.
Not just Big data, but zettabyte :-
About 5 exabytes of unique information were created in 2008. That\'s 1 billion DVDs. Fast
forward three years and we are creating 1.2 zettabytes, with one zettabyte equal to 1,024
exabytes. \"This is the same as every person on Earth tweeting for 100 years or 125 million years
of your favorite one-hour TV show,\" says Evans. Our love of high-definition video accounts for
much of the increase.
Yes, there are many other things too but I tried to tell you things I know..
Recommended
Effects of protease inhibitors in the treatment of secondary infecti.pdf
Effects of protease inhibitors in the treatment of secondary infections in HIV-positive patients
Find the average complication rate in the two groups of HIV-positive patients. Protease
inhibitors (PIs) arc used in the treatment of infection by human immunodeficiency virus (HIV).
A military hospital provides unlimited access to a full spectrum of HIV medications. Has this
access reduced admissions caused by secondary infections? In particular, has unlimited access to
PIs reduced the annual admission rate of patients with pulmonary complications? The number of
admissions for HIV. separated into patients with and those without pulmonary complications,
was obtained for the 4 years (1992-1995) before access to PIs and for the 2 years (1996-1997)
after PIs became available.
Solution
The complication rate before PIs are available is: number of complications in the four years/total
number of patients in those four years
=194/485=0.4
The complication rate before PIs are available is: number of complications in the two years/total
number of patients in those two years
=25/92=0.27
Thus average complication rate is: (0.4+0.27)/2=0.335.
Describe the process (step by step) how a glucose molecule is absorb.pdf
Describe the process (step by step) how a glucose molecule is absorbed from the lumen of the
intestine to the blood. Include what energy drives this process. Then explain how researchers like
Pia Roder were able to use radioactivity to determine where and how the transporters functioned.
(The answer must be in a long paragraph form)
Solution
Carbohydrates initially undergo chemical digestion by salivary amylases result in formation of
simple monomer sugars of \"glucose\" etc finally absorbed directly into the sublingual
circulation. Later mixed food material enters into the pharynx from mouth due to the peristaltic
movement of the visceral smooth muscles finally enters into the stomach
Disaccharidases are the enzymes secreted from the brush border cells of intestine. These
enzymes are sucrase, maltase, and lactase. These enzymes acts on disachraides such as maltose,
sucrose and lactose and digests them into monomer sugar units of glucose & galactose. Simple,
soluble and absorbable monomer glucose molecules \"diffused\" (diffusion) into the blood
portals finally food chyme enters into the large intestine for reabsorption.
It has clearly observed the \"sodium -glucose symport\" enable the transport of glucose
monomers after digestion into the intestinal epithelial cells via microvilli (improve absorption)
finally into the interstitial space followed by movement of glucose into the blood down the
concentration gradient (crossing basilar membrane). Primary active transport using Na+-K+-
ATPase improves movement sodium inside and potassium outside using energy followed by
building up of interstitial sodium concentration. Sodium –glucose symport has considered as
secondary active transport as it is a co-transporter.
P V Roder analysis of \"glucose transporter functions\" via radioactivity principle:
The sodium-glucose linked transporter (SGLTs, a co-transporter) uses the energy generated by
the ATPase pump through the downhill sodium ion gradient for the glucose transportation across
the apical membrane (against the glucose gradient). SGLTs are an example of secondary active
transport. The glucose transporters (GLUT) present in the basolateral membrane now allow the
glucose transport into the peritubular capillaries. Inhibition of the Na+/K+/ATPase fails to
generate energy required for the SGLT secondary transporters, which inhibits glucose uptake.
Sugars such as glucose, galactose and free amino acid (alanine) uptake into the microvilli
epithelial cells is based on sodium mediated cotransport whereas fructose cannot be transported
by sodium mediated transport during absorption after digestion. H+ mediated oligopeptide
transport is present in gut epithelium.
Scientist PV Roder has analyzed the function of SGLTs and GLUTs in the apical membrane
using “radiotracer glucose gavage“, in which mice was taken of two different types such as
“Sglt1 expressing wild type” and ” mice with sglt1-knockout” finally C14 radioactive tracing
performed for the moveme.
Describe how rock fossils are formed.What are two different ways t.pdf
Describe how rock fossils are formed.
What are two different ways that geologists can use to age fossils?
Solution
1). Permineralization or petrification is the most typical form of fossil formation. The soft tissues
of the organism decays within the sediment, leaving behind the hard structures such as skeleton.
Skeleton contains minerals, when water seeps into the skeleton, the minerals are dissolved in the
water and form crystals within the spaces of remains. This crystallization process turns the
skeleton remains into hard structures while encasing the sedimentary rock..
You are working on an older PC and need to replace a faulty CPU. As .pdf
You are working on an older PC and need to replace a faulty CPU. As you remove the CPU, you
notice that the processor unit is a rectangular package with two uniformly spaced parallel rows of
pins pointing downward. What general socket type takes this sort of chip?
A. DIP
B. LIF
C. PGA
D. ZIF
Solution
A. The DIP, or dual in-line package, is characterized by a chip with two rows of pins arranged in
parallel, and this design can be found on a number of different electronic components, such as
CPUs and other integrated circuits..
write a short essay report to describe a topic of your interest rela.pdf
write a short essay report to describe a topic of your interest related to the past, or current, or
future technologies of cellular communications, bandwidths, antennas or their impact to the
human society
Solution
past, or current, or future technologies of cellular communications:
Mobile computing is to describe technologies thatenable people to access network services
anyplace, anytime, and anywhere.
past,present technologies of cellular communications:
1896 Guglielmo Marconi,First demonstration of wireless telegraphy.Based on long wave,
requiring very large transmitters
1907 Commercial Trans-Atlantic Wireless Service.Huge ground stations: 30 x 100m antenna
masts
1920 Discovery of short waves by Marconi.Cheaper, smaller, better quality transmitters by
vacuum tube
1982 Start of GSM in Europe (1G analog)
1983 Start of AMPS in America (1G analog)
1992 Start of GSM (2G digital)
1997 Wireless LAN - IEEE802.11
1998 Iridium satellite system.
1999 Standardization of additional wireless LANs
2000 GSM with higher data rates (2.5G digital).HSCSD offers up to 57,6kbit/s.First GPRS trials
with up to 50 kbit/s
2001 Start of 3G systems.IMT - 2000, several “members” of a “family”, CDMA2000 in Korea,
UMTS tests in Europe.
GPRS: General Packet Radio Service
It is a mobile data service available to users of GSM mobile phones. It is often described as
\"2.5G“.
GPRS is packet-switched which means that multiple users share the same transmission channel,
only transmitting when they have data to send.
GPRS provides moderate speed data transfer, by allocating unused cell bandwidth to transmit
data.Poor bit rate in busy cells.Usually, GPRS data is billed per kilobytes of information
transceived.
future technologies of cellular communications:
Improved radio technology and antennas.smart antennas, beam forming, multiple-input multiple-
output (MIMO) 802.11N.dynamic spectrum allocation.Core network convergence.IP-based,
quality of service, mobile IP.Ad-hoc technologies.spontaneous communication, power saving,
redundancy.Simple and open service platform.intelligence at the edge, not in the network (as
with IN).more service providers, not network operators only.
Impact on Society:
Since the introduction of mobile phones, concerns (both scientific and public) have been raised
about the potential health impacts from regular use.But by 2008, American mobile phones
transmitted and received more text messages than phone calls.Numerous studies have reported
no significant relationship between mobile phone use and health, but the effect of mobile phone
usage on health continues to be an area of public concern.
For example, at the request of some of their customers, Verizon created usage controls that meter
service and can switch phones off, so that children could get some sleep.There have also been
attempts to limit use by persons operating moving trains or automobiles, coaches when writing to
potential players on their teams, and movie theater audiences. By one measure, near.
Write C language sequence to output the upper byte of the unsigned in.pdf
Write C language sequence to output the upper byte of the unsigned integer variable x to port PI
and lower byte to port P0.
Solution
#define lowbyte(i ) ((unsigned char) (i)) #define highbyte(i ) ((unsigned char) (((unsigned int)
(i)) >> 8)) unsigned int a; unsigned char p, q; void shift (void) { p = lowbyte(a); q =
highbyte(a); }.
Capital Budgeting Select a government agency (localstatefederal) a.pdf
Capital Budgeting Select a government agency (local/state/federal) and develop a four- to five-
page APA-style paper summarizing the following: Explain how the debt capacity of the
governmental entity is determined. Evaluate the effect of refunding or reorganizing existing debt
obligations. Analyze various funding alternatives that can be used to support debt obligations.
Your paper must be four to five pages (not including title and reference pages) and must be
formatted according to APA style as outlined in the approved APA style guide. You must cite at
least three scholarly sources in addition to the textbook.
Solution
The ability of any government entity to generate revenue from its economy underlie the
government\'s debt capacity.
GFOA (Government Finance Officers Association) determine its debt capacity through five steps
methodology which includes the following:
(1) defining the scope of analysis,
(2) peer group development,
(3)analysis of current position,
(4) scenario analysis, and
(5) financial policy development.
Along with debts various other sources that could be used for funding includesales and use tax,
property taxes, reserves etc..
can anybody think of any world events that changed information syste.pdf
can anybody think of any world events that changed information systems?
Solution
Systems range from very simple to very complex. There are numerous types of systems. For
example, there are biological systems, mechanical systems, human/mechanical systems,
ecological systems, and social systems.
Following are some technologies that will change technology:-
- 3D printers,
- sensor networks,
- virtual humans
technologies under development now will drastically change our world in the decade to come.
Internet Things:-
We have passed the threshold where more things are connected to the Internet than people. The
transition to IPV6 also supports seemingly limitless connectivity.
Sensor Network:-
Nowadays this technology grows very fast by using this we can track song pollution, air
pollution and we also can track temperature. This technology is growing day by day and many
companies are coming with new ideas related to sensor networks.
Not just Big data, but zettabyte :-
About 5 exabytes of unique information were created in 2008. That\'s 1 billion DVDs. Fast
forward three years and we are creating 1.2 zettabytes, with one zettabyte equal to 1,024
exabytes. \"This is the same as every person on Earth tweeting for 100 years or 125 million years
of your favorite one-hour TV show,\" says Evans. Our love of high-definition video accounts for
much of the increase.
Yes, there are many other things too but I tried to tell you things I know..
Biology research I need help in finding out they specific topic .pdf
Biology research: I need help in finding out they specific topic on Genetics or cell biology to
write a research paper. Could you give me some suggestions of specific topics to write about?
Thank you!
Biology research: I need help in finding out they specific topic on Genetics or cell biology to
write a research paper. Could you give me some suggestions of specific topics to write about?
Thank you!
Thank you!
Solution
Specific topics on genetics :
Features of inheritance
Molecular basis for inheritance
Gene expression
Mutations
Specific topics on Cell biology:
Growth and development
Techniques used to study cells
Organelles.
B. Please provide complete correct detailed solution. Thank you. Con.pdf
B. Please provide complete correct detailed solution. Thank you. Consider the distribution of
individual incomes, yi in a population that is divided into two subgroups, indicated by the
random variable D {0,1}. Assume: So mu 0, v 0 are the mean and variance of yi in group 0, and
mu1, v1 are the mean and variance of y, in group 1. Suppose that P(D,i = 1) = p. Find mu =
E[yi]> the overall mean of yi.
Solution.
3.35. Left-handedness. Left-handedness can be an inconvenience for a .pdf
3.35. Left-handedness. Left-handedness can be an inconvenience for a portion of the human
population and is a study of interest to researchers throughout the world. About 6-10% of the
human population exhibits left-handedness. Studies suggest that left-handedness is due to
genetics rather than environmental causes. It is found to be more common in males (about 8-
12%) than females (about 5-9%), and also that the probability of left-handedness increases when
a mother or father has it. The table (McManus, 1991; Table 4, page 262) provides empirical
probabilities of a child being left-handed depending on gender and maternal/paternal handedness.
Using these probabilities, answer parts (a-c) (a) Assume that parents are both right-handed. What
is the probability that their child is right-handed? Probabilities of a son or daughter are assumed
equal. (b) Ann is left-handed and expecting a son. No information about handedness of baby\'s
father Ben is available, but it is known that Ben\'s parents are right-handed. What is the
probability that the baby is going be left-handed. (c) Adam\'s parents are both right-handed.
Adam\'s daughter Bethany and wife Claudia are both left-handed. What is the probability that
Adam is right-handed. Does Claudia\'s handedness matter?
Solution
a) if the parents are both right handed
Probability ther child ( can be son or daughter) is righ handed = (0.8951+0.9137) / 2 = 0.9044
b)
ben could be left or right
if ben is left with ann that is left
probability will be (left) = 0.2940
if ben is right with ann that is left
probabilty that the son will be (left) = 0.2150
PROBABILITY THAT THE BABY (SON) IS GOING TO BE LEFT-HANDED WILL BE =
0.2150+0.2940 = 0.509
c)
probability that adam is right handed will be 0.8951.
Write a generic method named genericListSort that takes a generic li.pdf
Write a generic method named genericListSort that takes a generic linked list with elements of
type E and rearranges the nodes so that the elements stored in the list are sorted into the order of
smallest to largest. Use the comparable to decide if an element of the list is larger, smaller or
equal to another element. To enable the use of comparable, your generic Node class must
implement Java\'s Comparable interface.
The method listSort should be in a Java class called SortedGenericLinkedList as shown in the
following.
public class SortedGenericLinkedList {
public Node genericListSort(Node head) { /* your implementation */}
}
Solution
public class SortedGenericLinkedList {
private static Node head;
private static Node tail;
public Node genericListSort(Node head) {
Node tmp = head;
while(true){
Node tmp1 = tmp.getNextRef();
while(tmp1!=null){
if(tmp.getValue()>tmp1.getValue()){
int temp=tmp.getValue();
tmp.getValue()=tmp1.getValue();
tmp1.getValue()=temp;
}
tmp1= tmp1.getNextRef();
}
tmp = tmp.getNextRef();
}
return head;
}
public void add(T element){
Node nd = new Node();
nd.setValue(element);
System.out.println(\"Adding: \"+element);
/**
* check if the list is empty
*/
if(head == null){
//since there is only one element, both head and
//tail points to the same object.
head = nd;
tail = nd;
} else {
//set current tail next link to new node
tail.setNextRef(nd);
//set tail as newly created node
tail = nd;
}
}
public void display(){
Node tmp = head;
while(true){
if(tmp == null){
break;
}
System.out.println(tmp.getValue());
tmp = tmp.getNextRef();
}
}
public static void main(String a[]){
SortedGenericLinkedList sl = new SortedGenericLinkedList();
sl.add(32);
sl.add(3);
sl.add(54);
sl.add(89);
sl.display();
sl.genericListSort(head);
sl.display();
}
}.
Why is it recommended that you use the broth (regardless of medium) .pdf
Why is it recommended that you use the broth (regardless of medium) with the most growth to
serve as the \"standard\" for scoring a \"3\"? (3 for adundant growth)
Why is it important to use the uninoculated controls for each medium to serve as the \"standard\"
for scorinh a \"0\" in yheir respective tubes? (0 for no growth)
Suppose afte performing this experiment, you examine the growth of two different organisms in
Medium X. Both show growth and Organism 1 has produced less turnidity than organism 2.
Explain how it could be possible that Organism 1\'s needs are actually better supplied by the
medium than organism 2\', yet there is less turbidity?
What is the biochemical basis for the spectrum of fastidousness seen in the mircobiol world?
That is, why are some organism fastidious and others are nonfastidious?
What are some consequences of leabing a stain on a bacterial smear too long? (over staining)
What are some conswquences of not leabing a stain on a smear long enough? (under-staining)
Consider a coccus and a rod of equal volume.
Which is more likely to survive in a dry enviroment? Why?
Which would be better adapeted to a moist enviroment? why?
Solution
1. Broth with the maximum growth is used as a standard for measuring growth because broth
helps in maintaining optimal concentration of bacteria. Broth results in uniform growth on solid
media. Additionally growth can be measured with respect to turbidity either with the help of
spectrophotometer or through visual examination of the broth.
2. Uninoculated medium serves as a negative control. It is used as an indicator for testing the
sterility of the medium. In case any growth is present on uninoculated medium, it is an indicator
of contanimation.
3. There might be various reasons for less turbidity shown by organism 1. Optimum temperature
might not be maintained for Organism 1, which leads to slower growth. The growth rate of
organism 1 may be less than organism 2. It might also be possible that there is production of
inhibitory substances by organism 2 which are interfering with the growth of organism 1.
4. Some microorganisms are fastidious as they lack the metabolic pathways required to
syntheisze certain nutrients. Therefore they rely on particular media for fulfilling their complex
nutritional reqirements.
5. Overstaining a bacterial smear will cause cell wall disruption as a result of which the cell
looses its true morphological characteristics.
6. Understaining will cause false Gram negative results. In case of simple staining understaining
will cause difficulty in observing the cells.
7. Cocci are more likely to survive in a dry environment because they have a lower surface area
to volume ratio which causes them to dehydrate slowly.
8. Rods are more likely to survive in a moist environment because they have a larfer surface area
to volume ratio due to which they can easily exchange materials with their surroundings..
Which of the following groups produce strobili (Mark all that apply.pdf
Which of the following groups produce strobili? (Mark all that apply.)
a.lycophytes
b.mosses
c.liverworts
d.hornworts
e.ferns
a.lycophytes
b.mosses
c.liverworts
d.hornworts
e.ferns
Solution
Answer: a lycophytes
Strobili is a cone shaped structure containing sporangia which is present aggregated at the tip of
the stem.
Ferns, mosses, liverworts and hornworts does not have strobili..
What sources could be used to evaluate the performance of people wor.pdf
What sources could be used to evaluate the performance of people working in the following
jobs?
Sales representative, TV reporter, Director of Nursing in a hospital, HR manager, and an Air
Traffic Controller
Solution
Evaluation can be made of the followin jobs in the below mentioned ways
Sales representative: The representatives should be evaluated on how well they are able to deal
with the consumers and their ability to close the deal. The whole process should be evaluated
under which the sales representative is working, rather than just evaluating the results or the
targets. The representatives should be evaluated on team work, coordination and cooperative
behaviour with the employees and with the customers.
TV Reporter :
The reporter can be evaluated on the basis of initiative taking ability and single handedly
effectively and efficiently dealing with the news without any biasedness towards any section.
The coopration with other employees is also considered as an important factor for evaluating
performance of the journalist. The flexibility to work any time and any any place is also
considered important keeping in mind the nature of work invoved in TV reporting.
Communication also gives the journalist and important profile. The TV reporter should have the
ability to communicate clearly in a concise manner.
Director of Nursing in a hospital: The nursing director will be evaluated on the basis of handling
day to day activities of the nursing care. The ability to motivate nurses towards efficiency in
work by guiding them towards work as a humane job. Nursing director is also evaluated on the
basis of flexibility to work any shift, be it in odd hours. Communication with the nurses and the
patients is also considered important while evaluating the performance.
HR Manager: The performance of HR manager will be on the basis of HR relations with whole
of the employees of the organization. The existent culture in the organization. The overall
performance of the organization. The attrition rate and ability to support and retain employees.
The motivation of the employees of the organization and the performance of the newly hired
employees.
Air Traffic Controller: The communication ability to different language speaking pilots and crew
of varoius airlines. Flexibility is utmost important for evaluating the performance of ATC.
Knowledge of various aircrafts and problem resolving abilities..
What is the term that describes the process in which nutrients pass t.pdf
What is the term that describes the process in which nutrients pass through the wall of the
gastrointestinal tract? Digestion Absorption Elimination Segmentation
Solution
THE TERM THATDESCRIBES THE PROCESS WHERE NUTRIENTS PASS THROUGH
THE WALL OF THE GASTROINTESTINAL TRACT?
ANS-
ABSORPTION
Nutrients paas through the wall of the intestine and reach the blood/lymph stream through the
process absorption..
2. The old human telephone “switches” were quaint but very slow. You.pdf
2. The old human telephone “switches” were quaint but very slow. You would signal the
operator and say you wanted to speak to Jill, but the operator knowing that Jill was visiting Sally,
would connect you there. Other than slowness and inefficiency, name two disadvantages of this
scheme.
Solution
The disadvantage of old telephone switching schene are like the absence of end to end
encryption which discloses the message directly if man in the middle like attacks are done ..
moreover it also causes time delay...there is no confidentiality of user call log because he has to
inform the human operator.
what is testosterones role in the mestrual cycle SolutionHigh.pdf
what is testosterones role in the mestrual cycle ?
Solution
High testosterone levels in women may leads to development of sterility and male secondary
sexual features & there will be no menstrual cycle. Therefore, low level of testosterone in
women is going to promote “menstrual cycle” & this testosterone is going to maintain
equilibrium level of balancing estrogen and progesterone hormones during the menstrual cycle as
explained below.
Estrogens and progesterone have negative feedback regulation on FSH and LH through the
anterior pituitary. The FSH and LH cause the increased secretion of estrogen and progesterone
from the ovaries.
The female sex hormones (estrogens and progesterone) are secreted by several structures such as
ovaries, adrenal cortices and the placenta. The two abundant estrogens include the estrone and
estradiol.
Under the stimulation of estrogens, the reproductive organs including the vagina, uterus, uterine
tubes, ovaries and the other external structures become enlarged. The endometrium becomes
thicken..
What is the present worth of a cash flow that gives you $5 in time p.pdf
What is the present worth of a cash flow that gives you $5 in time period zero and $5 every
period until time period 20?
$5(P/A,i, 21)/(1+i)
$5+$5(P/A,i, 20)
$5(P/A,i, 21)
105/(1+i)^20
$5(P/A,i, 21)/(1+i)
$5+$5(P/A,i, 20)
$5(P/A,i, 21)
105/(1+i)^20
Solution
b) 5 + 5/(i+i) + 5/[(i+i)^2] +...+ 5/[(i+i)^20]
= $5+$5(P/A,i, 20).
This pedigree shows a family with Tay-Sachs disease which is inherite.pdf
This pedigree shows a family with Tay-Sachs disease which is inherited in an autosomal
recessive manner. Members of the family with filled symbols have Tay-Sachs. Which members
of this family are definitely heterozygous (Tt)? 1-1, 1-2, II-2, II-3, and III -1 1-1, 1-2, II-2, and
II-3 only 1-1, 1-2, and II-3 only I-1 and I-2 and only
Solution
Answer:
D). I-1 and I-2 and only
I-1 & I-2 are definitely are heterozygous
II-2, II-3 & III-1 are may or may not heterozygous.
This is for C++7) Parameters areA) The limits of what the progr.pdf
This is for C++
7) Parameters are:
A) The limits of what the program can do
B) Variables that exist in the Global scope that provide information for a function to use
C) A list of local variables that are created as part of the function that take information from
outside the function
D) The values returned from a function which are used to finish calculations where the function
was called
E) The documentation of what a function does
Choose one of the following. I will rate right away! Thank you!
Solution
The answer is C. Parameters are values that are passed to the function. These values are passed
when we call the function. In C++ we provide the parameter data type and name of the paramter
in function declaration..
The sum of internal energy (U) and the product of pressure and volume.pdf
The sum of internal energy (U) and the product of pressure and volume Enthalpy Entropy
Solution
Entropy is a measure of disorder or randomness of a system. An ordered system has low entropy.
A disordered system has high entropy.
Enthalpy is defined as the sum of internal energy of a system and the product of the pressure and
volume of the system. H = U + pv
Therefore answer is enthalpy.
1.Reproductive isolation between geographically separated members of.pdf
1.Reproductive isolation between geographically separated members of a population is an
example of _____ speciation.allopatricsympatric2.Polyploidy in a member of a population is
typically an example of the initial barrier to gene flow in _____
speciation.allopatricsympatric1.Reproductive isolation between geographically separated
members of a population is an example of _____ speciation.allopatricsympatric
Solution
1:-Answer-Allopatric speciation
Explanation:
Reproductive isolation between geographically separated members of a population is an example
of allopatric peciation.
2:-Answer:-Sympatric speciation
Reproductive isolation between geographically separated members of a population is an
example of allopatric peciation.
2:-Answer:-Sympatric speciation.
the classic field studies of Robert Paine that established the Keyst.pdf
the classic field studies of Robert Paine that established the Keystone Species concept. This
concept--the disproportional importance of one species in helping to maintain the biodiversity of
the community--is dependent upon the presence of indirect and diffuse interactions among
organisms. What are other keystone species that have been studied? What are the domino-effects
of indirect/diffuse interactions that causes these species to be keystone? 100 words please
Solution
the role of keystone species in ecosystem is anologus to tthe keystone in arch.while the keystone
is under least presure but without that keystone arch will collapse. in the same way an ecosystem
will show drastic change if the key stone species is removed. even though it was a small part in
the ecosystem in mass and bioproductivity. it became a popular concept in conservation biology.
it has gots the hails of critics for over symplifying complex ecological systems. it also allows
easy communication between ecologists and conservation polocy makers.
a key stone generally means a small predator which stops the herbivore to eliminate pdominant
plant species..
1. List and give an example of each of the four types of market fail.pdf
1. List and give an example of each of the four types of market failure (i.e. the reasons that we
need government intervention.)
Solution
The main reasons because of which we need government intervention are as follows:-
1.) Poverty alleviation:- Poverty is one of major problem in underdeveloped and developing
nations. Government runs various poverty alleviation programmes. For example, Prime
Minister\'s Rozgar Yojna (PMRY) in India. Through these programmes, government accelerates
the economic growth. It helps in population control through awareness regarding various family
planning programmes. It tries to reduces income inequalities in the nation.
2) Human capital formation:- It implies development of abilities and skills among the population
of country. It is the process of acquiring and increasing the number of people, who have skills,
experience and education, which are critical for social, economical and political development
and growth of the nation. The state and union governments have been earmarking substantial
financial outlays for development of education and health sectors in the country.
3) Rural development:- Government plays a very vital role in developing rural life through
providing drinking water facility, proper sanitation and severage facility, Basic amenities like
Housing etc. Government aims at improving social and economic conditions of people living in
villages through various awareness programmes and campaigns.
4) Employment Generation:- Union and state governments played vital role in generating
employment or creating opportunities for employment generation. Government provides \'Direct
Employment\' by employing people in various departments for administrative purposes and in
different industries, hotels, transport companies etc. run by the government.
Therefore these are the Four areas ( Poverty alleviation, Human capital formation, Rural
development and Employment generation etc.) which need government intervention to the large
extent..
Suppose that you have a particular S. cerevisiae mutant strain that .pdf
Suppose that you have a particular S. cerevisiae mutant strain that cannot make the amino acid
leucine. Will the strain grow on a YPD plate? What about a SD plate? How could you modify
the SD plate to allow this strain to grow?
Solution
The leu2 allele in the yeast Saccharomyces cerevisiae is a stable mutation. Because the allele
does not revert, it has been used, for example, to demonstrate the delivery of large bacterial
conjugative plasmids carrying the LE U2 gene during matings between bacteria and yeast).
However, small colonies often do form during selection for Leu’ cells on medium devoid of
leucine, even in controls without bacterial donors. Nevertheless, cells from these small colonies
remain leucine auxotrophs. We considered the possibility that growth was due to variants that
have regained a limited ability to synthesize leucine (by reversion or extragenic suppression of
leu,). Alternatively, phenotypic suppression of the Leu- phenotype could be due to variants that
have acquired the ability to better concentrate leucine from their environment by alterations of
leucine uptake pathways. Growth of Leu- cells might be due to leucine ‘ scavengers ’, variants
within the population that are Leubut grow because they accumulate sufficient quantities of
leucine by some other means, such as transporting leucine released by dying neighbours.
In a comparison with several nutrients, observed that cells respond differently to starvation for a
nutrient made necessary by mutation (leucine in a leucine auxotroph), compared with a “natural”
starvation for phosphate or sulfate. Notably, in the latter cases, nearly all of the cells in the
culture appeared to have arrested in G0/G1 (unbudded), whereas there was no similarly rapid,
coordinated arrest upon starvation for leucine and uracil in the respective auxotrophs. In contrast,
when cultures of exponentially growing cells in nutrient-limited chemostats were compared with
the same limiting nutrients, no difference in the budding index between the natural and
auxotrophic limitations was observed. however, observed that chemostat cultures limited for
uracil or leucine differed from those limited by phosphate, sulfate, or ammonium in another way:
The latter conserved much of the excess glucose available to them, whereas the former
completely fermented the excess glucose to ethanol.
Both the lack of a uniform arrest during starvation of auxotrophs and the wasting of excess
glucose suggest a failure of regulation. Specifically, they raise the possibility that one or more
signals given during starvation for the natural nutrients, leading to cell-cycle arrest and cessation
of glucose fermentation, are lacking when a nutrient made necessary only by a mutation runs out.
It seemed likely that this failure of regulation might have consequences for the viability of cells
that were starved of auxotrophic nutrients..
statisticsSolutionIn the testing of hypothesis, there are two .pdf
statistics
Solution
In the testing of hypothesis, there are two types of error
Type I error is the probability of rejecting of Ho the null hypothesis when H0 is true.
Type II error is the probability of acceptance H0 when the H1 is true.
For a given sample size, when we increase the probability of Type I error, the probability of
Type II error is decrease because as the power increases the significance level decreases..
Snouters are small critters that live on both sides of the Ohio Rive.pdf
Snouters are small critters that live on both sides of the Ohio River. A population on the north
side of the Ohio River is fixed for an allele for blue feet. The population on the south side of the
river is fixed for an allele for pink feet. Forty blue-footed snouters from the north and 60 pink-
footed snouters from the south colonize a newly formed island in the middle of the river.
Please calculate the following.
1.)What are the genotype frequencies on the island at the time of colonization (Parental
Generation, P1)? (Just the colonizers before reproduction)
2.)What are the allele frequencies on the island at the time of colonization (P1)? (Just the
colonizers before reproduction, N = 100)
3.)What are the genotype frequencies of the offspring of the next generation (First generation of
offspring, F1 ), assuming random mating?
4.) What are the allele frequencies of the offspring of the next generation (F1 )? Heterozygotes
have purple feet that are susceptible to snouter purple-foot fungus that sterilizes purple-footed
snouters (They cannot reproduce). Assume that the F1 generation above is the parents for the
next set of questions.
5.)Calculate the allele frequencies of the parents that can breed (F1 with no purple footed
individuals). Calculate the genotype and allele frequencies of their offspring (F2) in the second
generation assuming random mating (Hint: the sum of the allele frequencies represented by the
homozygotes are equal to 1, keep this in mind when you calculate the allele frequency of F1 as
parents. It may help if you assume a population size for F1 prior to reproduction.) (2 pt) Has
evolution occurred? Why? If the purple foot fungus were to continue on this island for several
generations, what color feet would most likely become fixed on the island and why?
Solution
1). Assume that the genotype for the blue feet snouters is BB and for the pink feet snouters is PP.
Thus the genotypic ratio before reproduction would be: Blue: Pink = 40: 60 = 2:3
Genotypic frequencies of BB = 0.4
The genotypic frequency of PP = 0.6.
Polish students' mobility in the Czech Republic
Polish students mobility to the Czech Republic within eTwinning project "Medieval adventures with Marco Polo"
Instructions for Submissions thorugh G- Classroom.pptx
This presentation provides a briefing on how to upload submissions and documents in Google Classroom. It was prepared as part of an orientation for new Sainik School in-service teacher trainees. As a training officer, my goal is to ensure that you are comfortable and proficient with this essential tool for managing assignments and fostering student engagement.
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Biology research I need help in finding out they specific topic .pdf
Biology research: I need help in finding out they specific topic on Genetics or cell biology to
write a research paper. Could you give me some suggestions of specific topics to write about?
Thank you!
Biology research: I need help in finding out they specific topic on Genetics or cell biology to
write a research paper. Could you give me some suggestions of specific topics to write about?
Thank you!
Thank you!
Solution
Specific topics on genetics :
Features of inheritance
Molecular basis for inheritance
Gene expression
Mutations
Specific topics on Cell biology:
Growth and development
Techniques used to study cells
Organelles.
B. Please provide complete correct detailed solution. Thank you. Con.pdf
B. Please provide complete correct detailed solution. Thank you. Consider the distribution of
individual incomes, yi in a population that is divided into two subgroups, indicated by the
random variable D {0,1}. Assume: So mu 0, v 0 are the mean and variance of yi in group 0, and
mu1, v1 are the mean and variance of y, in group 1. Suppose that P(D,i = 1) = p. Find mu =
E[yi]> the overall mean of yi.
Solution.
3.35. Left-handedness. Left-handedness can be an inconvenience for a .pdf
3.35. Left-handedness. Left-handedness can be an inconvenience for a portion of the human
population and is a study of interest to researchers throughout the world. About 6-10% of the
human population exhibits left-handedness. Studies suggest that left-handedness is due to
genetics rather than environmental causes. It is found to be more common in males (about 8-
12%) than females (about 5-9%), and also that the probability of left-handedness increases when
a mother or father has it. The table (McManus, 1991; Table 4, page 262) provides empirical
probabilities of a child being left-handed depending on gender and maternal/paternal handedness.
Using these probabilities, answer parts (a-c) (a) Assume that parents are both right-handed. What
is the probability that their child is right-handed? Probabilities of a son or daughter are assumed
equal. (b) Ann is left-handed and expecting a son. No information about handedness of baby\'s
father Ben is available, but it is known that Ben\'s parents are right-handed. What is the
probability that the baby is going be left-handed. (c) Adam\'s parents are both right-handed.
Adam\'s daughter Bethany and wife Claudia are both left-handed. What is the probability that
Adam is right-handed. Does Claudia\'s handedness matter?
Solution
a) if the parents are both right handed
Probability ther child ( can be son or daughter) is righ handed = (0.8951+0.9137) / 2 = 0.9044
b)
ben could be left or right
if ben is left with ann that is left
probability will be (left) = 0.2940
if ben is right with ann that is left
probabilty that the son will be (left) = 0.2150
PROBABILITY THAT THE BABY (SON) IS GOING TO BE LEFT-HANDED WILL BE =
0.2150+0.2940 = 0.509
c)
probability that adam is right handed will be 0.8951.
Write a generic method named genericListSort that takes a generic li.pdf
Write a generic method named genericListSort that takes a generic linked list with elements of
type E and rearranges the nodes so that the elements stored in the list are sorted into the order of
smallest to largest. Use the comparable to decide if an element of the list is larger, smaller or
equal to another element. To enable the use of comparable, your generic Node class must
implement Java\'s Comparable interface.
The method listSort should be in a Java class called SortedGenericLinkedList as shown in the
following.
public class SortedGenericLinkedList {
public Node genericListSort(Node head) { /* your implementation */}
}
Solution
public class SortedGenericLinkedList {
private static Node head;
private static Node tail;
public Node genericListSort(Node head) {
Node tmp = head;
while(true){
Node tmp1 = tmp.getNextRef();
while(tmp1!=null){
if(tmp.getValue()>tmp1.getValue()){
int temp=tmp.getValue();
tmp.getValue()=tmp1.getValue();
tmp1.getValue()=temp;
}
tmp1= tmp1.getNextRef();
}
tmp = tmp.getNextRef();
}
return head;
}
public void add(T element){
Node nd = new Node();
nd.setValue(element);
System.out.println(\"Adding: \"+element);
/**
* check if the list is empty
*/
if(head == null){
//since there is only one element, both head and
//tail points to the same object.
head = nd;
tail = nd;
} else {
//set current tail next link to new node
tail.setNextRef(nd);
//set tail as newly created node
tail = nd;
}
}
public void display(){
Node tmp = head;
while(true){
if(tmp == null){
break;
}
System.out.println(tmp.getValue());
tmp = tmp.getNextRef();
}
}
public static void main(String a[]){
SortedGenericLinkedList sl = new SortedGenericLinkedList();
sl.add(32);
sl.add(3);
sl.add(54);
sl.add(89);
sl.display();
sl.genericListSort(head);
sl.display();
}
}.
Why is it recommended that you use the broth (regardless of medium) .pdf
Why is it recommended that you use the broth (regardless of medium) with the most growth to
serve as the \"standard\" for scoring a \"3\"? (3 for adundant growth)
Why is it important to use the uninoculated controls for each medium to serve as the \"standard\"
for scorinh a \"0\" in yheir respective tubes? (0 for no growth)
Suppose afte performing this experiment, you examine the growth of two different organisms in
Medium X. Both show growth and Organism 1 has produced less turnidity than organism 2.
Explain how it could be possible that Organism 1\'s needs are actually better supplied by the
medium than organism 2\', yet there is less turbidity?
What is the biochemical basis for the spectrum of fastidousness seen in the mircobiol world?
That is, why are some organism fastidious and others are nonfastidious?
What are some consequences of leabing a stain on a bacterial smear too long? (over staining)
What are some conswquences of not leabing a stain on a smear long enough? (under-staining)
Consider a coccus and a rod of equal volume.
Which is more likely to survive in a dry enviroment? Why?
Which would be better adapeted to a moist enviroment? why?
Solution
1. Broth with the maximum growth is used as a standard for measuring growth because broth
helps in maintaining optimal concentration of bacteria. Broth results in uniform growth on solid
media. Additionally growth can be measured with respect to turbidity either with the help of
spectrophotometer or through visual examination of the broth.
2. Uninoculated medium serves as a negative control. It is used as an indicator for testing the
sterility of the medium. In case any growth is present on uninoculated medium, it is an indicator
of contanimation.
3. There might be various reasons for less turbidity shown by organism 1. Optimum temperature
might not be maintained for Organism 1, which leads to slower growth. The growth rate of
organism 1 may be less than organism 2. It might also be possible that there is production of
inhibitory substances by organism 2 which are interfering with the growth of organism 1.
4. Some microorganisms are fastidious as they lack the metabolic pathways required to
syntheisze certain nutrients. Therefore they rely on particular media for fulfilling their complex
nutritional reqirements.
5. Overstaining a bacterial smear will cause cell wall disruption as a result of which the cell
looses its true morphological characteristics.
6. Understaining will cause false Gram negative results. In case of simple staining understaining
will cause difficulty in observing the cells.
7. Cocci are more likely to survive in a dry environment because they have a lower surface area
to volume ratio which causes them to dehydrate slowly.
8. Rods are more likely to survive in a moist environment because they have a larfer surface area
to volume ratio due to which they can easily exchange materials with their surroundings..
Which of the following groups produce strobili (Mark all that apply.pdf
Which of the following groups produce strobili? (Mark all that apply.)
a.lycophytes
b.mosses
c.liverworts
d.hornworts
e.ferns
a.lycophytes
b.mosses
c.liverworts
d.hornworts
e.ferns
Solution
Answer: a lycophytes
Strobili is a cone shaped structure containing sporangia which is present aggregated at the tip of
the stem.
Ferns, mosses, liverworts and hornworts does not have strobili..
What sources could be used to evaluate the performance of people wor.pdf
What sources could be used to evaluate the performance of people working in the following
jobs?
Sales representative, TV reporter, Director of Nursing in a hospital, HR manager, and an Air
Traffic Controller
Solution
Evaluation can be made of the followin jobs in the below mentioned ways
Sales representative: The representatives should be evaluated on how well they are able to deal
with the consumers and their ability to close the deal. The whole process should be evaluated
under which the sales representative is working, rather than just evaluating the results or the
targets. The representatives should be evaluated on team work, coordination and cooperative
behaviour with the employees and with the customers.
TV Reporter :
The reporter can be evaluated on the basis of initiative taking ability and single handedly
effectively and efficiently dealing with the news without any biasedness towards any section.
The coopration with other employees is also considered as an important factor for evaluating
performance of the journalist. The flexibility to work any time and any any place is also
considered important keeping in mind the nature of work invoved in TV reporting.
Communication also gives the journalist and important profile. The TV reporter should have the
ability to communicate clearly in a concise manner.
Director of Nursing in a hospital: The nursing director will be evaluated on the basis of handling
day to day activities of the nursing care. The ability to motivate nurses towards efficiency in
work by guiding them towards work as a humane job. Nursing director is also evaluated on the
basis of flexibility to work any shift, be it in odd hours. Communication with the nurses and the
patients is also considered important while evaluating the performance.
HR Manager: The performance of HR manager will be on the basis of HR relations with whole
of the employees of the organization. The existent culture in the organization. The overall
performance of the organization. The attrition rate and ability to support and retain employees.
The motivation of the employees of the organization and the performance of the newly hired
employees.
Air Traffic Controller: The communication ability to different language speaking pilots and crew
of varoius airlines. Flexibility is utmost important for evaluating the performance of ATC.
Knowledge of various aircrafts and problem resolving abilities..
What is the term that describes the process in which nutrients pass t.pdf
What is the term that describes the process in which nutrients pass through the wall of the
gastrointestinal tract? Digestion Absorption Elimination Segmentation
Solution
THE TERM THATDESCRIBES THE PROCESS WHERE NUTRIENTS PASS THROUGH
THE WALL OF THE GASTROINTESTINAL TRACT?
ANS-
ABSORPTION
Nutrients paas through the wall of the intestine and reach the blood/lymph stream through the
process absorption..
2. The old human telephone “switches” were quaint but very slow. You.pdf
2. The old human telephone “switches” were quaint but very slow. You would signal the
operator and say you wanted to speak to Jill, but the operator knowing that Jill was visiting Sally,
would connect you there. Other than slowness and inefficiency, name two disadvantages of this
scheme.
Solution
The disadvantage of old telephone switching schene are like the absence of end to end
encryption which discloses the message directly if man in the middle like attacks are done ..
moreover it also causes time delay...there is no confidentiality of user call log because he has to
inform the human operator.
what is testosterones role in the mestrual cycle SolutionHigh.pdf
what is testosterones role in the mestrual cycle ?
Solution
High testosterone levels in women may leads to development of sterility and male secondary
sexual features & there will be no menstrual cycle. Therefore, low level of testosterone in
women is going to promote “menstrual cycle” & this testosterone is going to maintain
equilibrium level of balancing estrogen and progesterone hormones during the menstrual cycle as
explained below.
Estrogens and progesterone have negative feedback regulation on FSH and LH through the
anterior pituitary. The FSH and LH cause the increased secretion of estrogen and progesterone
from the ovaries.
The female sex hormones (estrogens and progesterone) are secreted by several structures such as
ovaries, adrenal cortices and the placenta. The two abundant estrogens include the estrone and
estradiol.
Under the stimulation of estrogens, the reproductive organs including the vagina, uterus, uterine
tubes, ovaries and the other external structures become enlarged. The endometrium becomes
thicken..
What is the present worth of a cash flow that gives you $5 in time p.pdf
What is the present worth of a cash flow that gives you $5 in time period zero and $5 every
period until time period 20?
$5(P/A,i, 21)/(1+i)
$5+$5(P/A,i, 20)
$5(P/A,i, 21)
105/(1+i)^20
$5(P/A,i, 21)/(1+i)
$5+$5(P/A,i, 20)
$5(P/A,i, 21)
105/(1+i)^20
Solution
b) 5 + 5/(i+i) + 5/[(i+i)^2] +...+ 5/[(i+i)^20]
= $5+$5(P/A,i, 20).
This pedigree shows a family with Tay-Sachs disease which is inherite.pdf
This pedigree shows a family with Tay-Sachs disease which is inherited in an autosomal
recessive manner. Members of the family with filled symbols have Tay-Sachs. Which members
of this family are definitely heterozygous (Tt)? 1-1, 1-2, II-2, II-3, and III -1 1-1, 1-2, II-2, and
II-3 only 1-1, 1-2, and II-3 only I-1 and I-2 and only
Solution
Answer:
D). I-1 and I-2 and only
I-1 & I-2 are definitely are heterozygous
II-2, II-3 & III-1 are may or may not heterozygous.
This is for C++7) Parameters areA) The limits of what the progr.pdf
This is for C++
7) Parameters are:
A) The limits of what the program can do
B) Variables that exist in the Global scope that provide information for a function to use
C) A list of local variables that are created as part of the function that take information from
outside the function
D) The values returned from a function which are used to finish calculations where the function
was called
E) The documentation of what a function does
Choose one of the following. I will rate right away! Thank you!
Solution
The answer is C. Parameters are values that are passed to the function. These values are passed
when we call the function. In C++ we provide the parameter data type and name of the paramter
in function declaration..
The sum of internal energy (U) and the product of pressure and volume.pdf
The sum of internal energy (U) and the product of pressure and volume Enthalpy Entropy
Solution
Entropy is a measure of disorder or randomness of a system. An ordered system has low entropy.
A disordered system has high entropy.
Enthalpy is defined as the sum of internal energy of a system and the product of the pressure and
volume of the system. H = U + pv
Therefore answer is enthalpy.
1.Reproductive isolation between geographically separated members of.pdf
1.Reproductive isolation between geographically separated members of a population is an
example of _____ speciation.allopatricsympatric2.Polyploidy in a member of a population is
typically an example of the initial barrier to gene flow in _____
speciation.allopatricsympatric1.Reproductive isolation between geographically separated
members of a population is an example of _____ speciation.allopatricsympatric
Solution
1:-Answer-Allopatric speciation
Explanation:
Reproductive isolation between geographically separated members of a population is an example
of allopatric peciation.
2:-Answer:-Sympatric speciation
Reproductive isolation between geographically separated members of a population is an
example of allopatric peciation.
2:-Answer:-Sympatric speciation.
the classic field studies of Robert Paine that established the Keyst.pdf
the classic field studies of Robert Paine that established the Keystone Species concept. This
concept--the disproportional importance of one species in helping to maintain the biodiversity of
the community--is dependent upon the presence of indirect and diffuse interactions among
organisms. What are other keystone species that have been studied? What are the domino-effects
of indirect/diffuse interactions that causes these species to be keystone? 100 words please
Solution
the role of keystone species in ecosystem is anologus to tthe keystone in arch.while the keystone
is under least presure but without that keystone arch will collapse. in the same way an ecosystem
will show drastic change if the key stone species is removed. even though it was a small part in
the ecosystem in mass and bioproductivity. it became a popular concept in conservation biology.
it has gots the hails of critics for over symplifying complex ecological systems. it also allows
easy communication between ecologists and conservation polocy makers.
a key stone generally means a small predator which stops the herbivore to eliminate pdominant
plant species..
1. List and give an example of each of the four types of market fail.pdf
1. List and give an example of each of the four types of market failure (i.e. the reasons that we
need government intervention.)
Solution
The main reasons because of which we need government intervention are as follows:-
1.) Poverty alleviation:- Poverty is one of major problem in underdeveloped and developing
nations. Government runs various poverty alleviation programmes. For example, Prime
Minister\'s Rozgar Yojna (PMRY) in India. Through these programmes, government accelerates
the economic growth. It helps in population control through awareness regarding various family
planning programmes. It tries to reduces income inequalities in the nation.
2) Human capital formation:- It implies development of abilities and skills among the population
of country. It is the process of acquiring and increasing the number of people, who have skills,
experience and education, which are critical for social, economical and political development
and growth of the nation. The state and union governments have been earmarking substantial
financial outlays for development of education and health sectors in the country.
3) Rural development:- Government plays a very vital role in developing rural life through
providing drinking water facility, proper sanitation and severage facility, Basic amenities like
Housing etc. Government aims at improving social and economic conditions of people living in
villages through various awareness programmes and campaigns.
4) Employment Generation:- Union and state governments played vital role in generating
employment or creating opportunities for employment generation. Government provides \'Direct
Employment\' by employing people in various departments for administrative purposes and in
different industries, hotels, transport companies etc. run by the government.
Therefore these are the Four areas ( Poverty alleviation, Human capital formation, Rural
development and Employment generation etc.) which need government intervention to the large
extent..
Suppose that you have a particular S. cerevisiae mutant strain that .pdf
Suppose that you have a particular S. cerevisiae mutant strain that cannot make the amino acid
leucine. Will the strain grow on a YPD plate? What about a SD plate? How could you modify
the SD plate to allow this strain to grow?
Solution
The leu2 allele in the yeast Saccharomyces cerevisiae is a stable mutation. Because the allele
does not revert, it has been used, for example, to demonstrate the delivery of large bacterial
conjugative plasmids carrying the LE U2 gene during matings between bacteria and yeast).
However, small colonies often do form during selection for Leu’ cells on medium devoid of
leucine, even in controls without bacterial donors. Nevertheless, cells from these small colonies
remain leucine auxotrophs. We considered the possibility that growth was due to variants that
have regained a limited ability to synthesize leucine (by reversion or extragenic suppression of
leu,). Alternatively, phenotypic suppression of the Leu- phenotype could be due to variants that
have acquired the ability to better concentrate leucine from their environment by alterations of
leucine uptake pathways. Growth of Leu- cells might be due to leucine ‘ scavengers ’, variants
within the population that are Leubut grow because they accumulate sufficient quantities of
leucine by some other means, such as transporting leucine released by dying neighbours.
In a comparison with several nutrients, observed that cells respond differently to starvation for a
nutrient made necessary by mutation (leucine in a leucine auxotroph), compared with a “natural”
starvation for phosphate or sulfate. Notably, in the latter cases, nearly all of the cells in the
culture appeared to have arrested in G0/G1 (unbudded), whereas there was no similarly rapid,
coordinated arrest upon starvation for leucine and uracil in the respective auxotrophs. In contrast,
when cultures of exponentially growing cells in nutrient-limited chemostats were compared with
the same limiting nutrients, no difference in the budding index between the natural and
auxotrophic limitations was observed. however, observed that chemostat cultures limited for
uracil or leucine differed from those limited by phosphate, sulfate, or ammonium in another way:
The latter conserved much of the excess glucose available to them, whereas the former
completely fermented the excess glucose to ethanol.
Both the lack of a uniform arrest during starvation of auxotrophs and the wasting of excess
glucose suggest a failure of regulation. Specifically, they raise the possibility that one or more
signals given during starvation for the natural nutrients, leading to cell-cycle arrest and cessation
of glucose fermentation, are lacking when a nutrient made necessary only by a mutation runs out.
It seemed likely that this failure of regulation might have consequences for the viability of cells
that were starved of auxotrophic nutrients..
statisticsSolutionIn the testing of hypothesis, there are two .pdf
statistics
Solution
In the testing of hypothesis, there are two types of error
Type I error is the probability of rejecting of Ho the null hypothesis when H0 is true.
Type II error is the probability of acceptance H0 when the H1 is true.
For a given sample size, when we increase the probability of Type I error, the probability of
Type II error is decrease because as the power increases the significance level decreases..
Snouters are small critters that live on both sides of the Ohio Rive.pdf
Snouters are small critters that live on both sides of the Ohio River. A population on the north
side of the Ohio River is fixed for an allele for blue feet. The population on the south side of the
river is fixed for an allele for pink feet. Forty blue-footed snouters from the north and 60 pink-
footed snouters from the south colonize a newly formed island in the middle of the river.
Please calculate the following.
1.)What are the genotype frequencies on the island at the time of colonization (Parental
Generation, P1)? (Just the colonizers before reproduction)
2.)What are the allele frequencies on the island at the time of colonization (P1)? (Just the
colonizers before reproduction, N = 100)
3.)What are the genotype frequencies of the offspring of the next generation (First generation of
offspring, F1 ), assuming random mating?
4.) What are the allele frequencies of the offspring of the next generation (F1 )? Heterozygotes
have purple feet that are susceptible to snouter purple-foot fungus that sterilizes purple-footed
snouters (They cannot reproduce). Assume that the F1 generation above is the parents for the
next set of questions.
5.)Calculate the allele frequencies of the parents that can breed (F1 with no purple footed
individuals). Calculate the genotype and allele frequencies of their offspring (F2) in the second
generation assuming random mating (Hint: the sum of the allele frequencies represented by the
homozygotes are equal to 1, keep this in mind when you calculate the allele frequency of F1 as
parents. It may help if you assume a population size for F1 prior to reproduction.) (2 pt) Has
evolution occurred? Why? If the purple foot fungus were to continue on this island for several
generations, what color feet would most likely become fixed on the island and why?
Solution
1). Assume that the genotype for the blue feet snouters is BB and for the pink feet snouters is PP.
Thus the genotypic ratio before reproduction would be: Blue: Pink = 40: 60 = 2:3
Genotypic frequencies of BB = 0.4
The genotypic frequency of PP = 0.6.
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