It was noted that excess water can kill plants. If water is not toxi.pdf
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It was noted that excess water can kill plants. If water is not toxic, and if all plants require water, then how can excess water kill plants? What is actually killing the plants? Solution When excess water is present, the roots will be deprived of oxygen because the water fills the pores in the soil. This mostly occurs in clay soils. Also, excess water drain out the nutrients and promote root rot disease, which gradually kills the plant..
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In biology, the roles of transition metal ions can be broadly groupe.pdf
In biology, the roles of transition metal ions can be broadly grouped into two classifications:
redox chemistry (catalysis, electron transfer, etc.) and acid-base chemistry (catalysis, structural,
etc.).
A) List three metal ions that commonly participate in biological redox chemistry, and give
examples of how they participate.
B) Compare this list to metal ions commonly participating in acid-base chemistry. Discuss the
overlap of these two groups of metal ions. (e.g. Why are some used for one of these functions
and not for the other?)
C) List some common biological ligands of these metal ions. Can you see patterns that would
allow you to group these ligands into sets with common properties, or that bind common metals?
D) Why are trace metals used for these biological functions, and not other more-abundant
elements like C,N,O, etc?
Solution
(a) iron, copper, cobalt, manganese etc are used in biology redox reactions like molecular di
oxygen, superoxide, hydrogen peroxide thus inhibiting the effect or formation of these
compound by use of the certain other enzymes..
(c) the ions and the Ligands for them are Ike gluthaione peroxide which can bind to chromium,
arsenic etc.. Iron binds to ferrodoxin which is associated with the photosynthesis..
(d) transition metals are used because they have empty d-orbitals in with them due to which they
can easily transfer or accept the electrons or exchange the electrons among the metals and
enzymes.. Which is not possible in other trace metals like carbon, nitrogen etc...
Image that a researcher examined how the people’s heights are associ.pdf
Image that a researcher examined how the people’s heights are associated with their salaries. She
found that the correlation between heights (measured in inches) and salary range (measured in
dollars) was .36. Had she measured the height in cm instead, what would the correlation have
been?
Solution
Correlation would not change because it is unaffected by the units..
I need help with this one Question 18 of 18 Sapling Learning macmill.pdf
I need help with this one Question 18 of 18 Sapling Learning macmillan earning Integral
membrane proteins have different characteristics and functions than peripheral membrane
proteins. Sort the statements into those that are properties of integral membrane proteins and
those that are properties of peripheral membrane proteins. Integral membrane proteins Peripheral
membrane proteins can move laterally (sideways) in the hydrophobic part of the membrane
found at the surface of the cell membrane reach through all or part of the membrane pump
substances across the cell membrane
Solution
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These proteins
constitute about 20-30 % of total membrane proteins.
Reach through all or part of the membrane (Most of the integral proteins span the entire
phospholipid bilayer -from the inner surface to the outer surface. These transmembrane proteins
contain one or more membrane-spanning domains as well as domains, from four to several
hundred residues long, extending into the aqueous medium on each side of the bilayer
Pump substances across the cell membrane (Peripheral proteins at the cytosolic face of the
plasma membrane include the cytoskeletal proteins spectrin and actin in erythrocytes and the
enzyme protein kinase C. This enzyme shuttles between the cytosol and the cytosolic face of the
plasma membrane and plays a role in signal transduction)
Can move laterally (sideways) in the hydrophobic part of the membrane. Integral proteins are
tightly bound and are directly interacting with the hydrophobic core of the plasma membrane
Do not interact with the hydrophobic core in between two layers of phospholipids.
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These pro.
Explain how recombination increases the amount of genetic variation i.pdf
Explain how recombination increases the amount of genetic variation in offspring: Explain why
it is not possible to have a recombination frequency of greater than 50% (half recombinant
progeny): A second pair of Drosophila are mated. The female is Cucu YY (straight wing, gray
body), while the male is Cucu yy (straight wing, yellow body). Assuming recombination,
perform the cross and list the offspring genotypes and phenotypes.
Solution
Que-3:
The recombination frequency is the measure of genetic linkage at different loci. The value of
recombination frequency is 50% if the chromosomal genes are located at different regions on the
different chromosome because of independent assortment. If the genes are located closely
together on the same chromosome, then those genes considered as genetically linked; because
they have not assorted independently due to \"crossing over\", then the recombination frequency
value is less than 50%. Linked genes are due to linkage often do not obey Mendel\'s laws of
inheritance such as independent assortment because of \"crossing over with reciprocal exchange
of hereditary genetic material meticulously in between non-sister chromatids\" at the time of
\"synapsis formation\" (propahse -I) in reduction division or meiosis 1. These events are leading
to formation of gamete cells with complete different genotypically from diploid parent
reproductive cell via meiosis finally induce \"genetic variation\" in phenotype due to
\"fertilization\" events between sperm cells & ovum.
The recombination of chromosomes enables meiosis process of cell division with the generation
of novel recombinant nucleotide -sequences by crossing over during the cell division in both
unicellular and multicellular species. Sometimes, double strand breaks in DNA during
homologous chromosomal recombination may be produced due to exposure any harmful
radiation finally may cause higher genetic variation via duplication followed by fertilization to
form a phenotype with higher genetic variation
Que-4:
Crossing over is somewhat randomly distributed over the length of the chromosome because if
the two genes are close together then it is very change to get exchange of chromosomal breaks
therefore, if the two genes are located farther apart then there will be higher room for the
exchange of genetic material with chromosomal breaks. Therefore, when two genetic are far
apart on the same chromosome are more likely to have a crossover between them than two loci
that are close together. This recombination frequency is considered as a measure of genetic
linkage during the crossover of the homologous chromosomes. Linkage disequilibrium defined
as the existence of alleles at different loci with the absence of genetic linkage between them even
though there is no equilibrium with allelic frequencies independently. The recombination
frequency is the measure of genetic linkage at different loci. The value of recombination
frequency is 50% \"if the chromosomal genes are .
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col pox.pdf
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col poxio pon 2 the stol sle shiage
type head eschane
Solution
since there are two different fluids takes place that is hot body and cold body initially
condensation takes place in the condensation process the hot body temperature will maintain low
temperature than that of surroundings and afterwards evaporation takes palces to maintain the
temperature inside the system so that the pressure variation does not takes place and gets
maintained to the differntlevel of fluids..
DNA is a macromolecule that is critical for life. Much of the functi.pdf
DNA is a macromolecule that is critical for life. Much of the functionality of DNA lies in
itscomplex structure. What are the 2 main forces responsible for holding together the structureof
DNA and what does each force hold together?
Solution
The double helical structure of DNA (deoxyribonucleic acid) is formed by the coiling of two
polynucleotide chains, similar to a spiral staircase.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds and hydrostatic forces.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds..
Chapter 7 , book Health in the Later Years, 5th EditionGive a his.pdf
Chapter 7 , book: Health in the Later Years, 5th Edition
Give a historical perspective on advocacy efforts for the mentally ill in the U.S.
Solution
Answer:
Though it is not clear what was the student/client really asked for. As per the studies concern
making the fact easily understandable is the goal of our company. So, here is the brief of the fact
asked for-
Mental health services throughout the world are in the midst of unprecedented change. New
thinking is emerging about what constitutes \"best practice\" in mental health services.
Community mental health treatment and support services are now the primary locus of care for
people with serious mental illness in many areas. Fundamental components of mental health
service systems such as methods of designing and delivering services, mechanisms for
establishing and funding service priorities, methods of ensuring system accountability, and
definitions of staff role, needed competencies, and training approaches, and so forth, are
undergoing scrutiny and restructuring. Historic distinctions between clinical treatment services
and disability support or rehabilitation services are no longer simple and examples of integrated
service systems are becoming more common. In this context, traditional psychiatric hospital
services are undergoing a parallel process which is re-shaping their role, services, and position as
partners with community treatment and rehabilitation services, disability services, and self-
help/peer support services in an integrated system. Mental health consumers and their family
members are exercising their \"voice\" and are assuming increasingly vital roles in governance,
policy formulation, and service design and delivery which is helping to shape how these changes
will occur in our thinking and practice.\'.
Discuss the benefits of event-driven programming.Contrast event-dr.pdf
Discuss the benefits of event-driven programming.
Contrast event-driven programming with procedural programming.
Discuss how messages from other programs can be viewed as events.
Give examples. Give examples of tasks that require the program to guide the user.
Should programs be purely event driven, procedural, or a mixture of both? Why?
Solution
The benefit of event-driven programming is that it\'s very intuitive and naturally well-suited to
applications whose control flow are based, not on its structure, but rather on internal or external
events. This is why event-driven programming is so popular with GUI programming. Likweise,
SOA and client-server architectures are naturally event-driven as well.
We if we find contrast between Procedural and event-based then applications written in
procedural languages execute by proceeding logically through the program code, one line at a
time. Logic flow can be temporarily transferred to other parts of the program through the GoTo,
GoSub, and Call statements, directing the program from beginning to end.
In contrast, program statements in an event-driven application execute only when a specific
event calls a section of code assigned to that event. Events can be triggered by keyboard input,
mouse actions, the operating system, or code in the application. For example, consider what
happens when the user clicks a command button named Command1 on a form. The mouse click
is an event. When the Click event occurs, Visual Basic executes the code in the
Sub procedure named Command1_Click. When the code has finished running, Visual Basic
waits for the next event.
Messages from other programs/threads can be seen as events, as they trigger some instance of
code to execute..
Recommended
In biology, the roles of transition metal ions can be broadly groupe.pdf
In biology, the roles of transition metal ions can be broadly grouped into two classifications:
redox chemistry (catalysis, electron transfer, etc.) and acid-base chemistry (catalysis, structural,
etc.).
A) List three metal ions that commonly participate in biological redox chemistry, and give
examples of how they participate.
B) Compare this list to metal ions commonly participating in acid-base chemistry. Discuss the
overlap of these two groups of metal ions. (e.g. Why are some used for one of these functions
and not for the other?)
C) List some common biological ligands of these metal ions. Can you see patterns that would
allow you to group these ligands into sets with common properties, or that bind common metals?
D) Why are trace metals used for these biological functions, and not other more-abundant
elements like C,N,O, etc?
Solution
(a) iron, copper, cobalt, manganese etc are used in biology redox reactions like molecular di
oxygen, superoxide, hydrogen peroxide thus inhibiting the effect or formation of these
compound by use of the certain other enzymes..
(c) the ions and the Ligands for them are Ike gluthaione peroxide which can bind to chromium,
arsenic etc.. Iron binds to ferrodoxin which is associated with the photosynthesis..
(d) transition metals are used because they have empty d-orbitals in with them due to which they
can easily transfer or accept the electrons or exchange the electrons among the metals and
enzymes.. Which is not possible in other trace metals like carbon, nitrogen etc...
Image that a researcher examined how the people’s heights are associ.pdf
Image that a researcher examined how the people’s heights are associated with their salaries. She
found that the correlation between heights (measured in inches) and salary range (measured in
dollars) was .36. Had she measured the height in cm instead, what would the correlation have
been?
Solution
Correlation would not change because it is unaffected by the units..
I need help with this one Question 18 of 18 Sapling Learning macmill.pdf
I need help with this one Question 18 of 18 Sapling Learning macmillan earning Integral
membrane proteins have different characteristics and functions than peripheral membrane
proteins. Sort the statements into those that are properties of integral membrane proteins and
those that are properties of peripheral membrane proteins. Integral membrane proteins Peripheral
membrane proteins can move laterally (sideways) in the hydrophobic part of the membrane
found at the surface of the cell membrane reach through all or part of the membrane pump
substances across the cell membrane
Solution
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These proteins
constitute about 20-30 % of total membrane proteins.
Reach through all or part of the membrane (Most of the integral proteins span the entire
phospholipid bilayer -from the inner surface to the outer surface. These transmembrane proteins
contain one or more membrane-spanning domains as well as domains, from four to several
hundred residues long, extending into the aqueous medium on each side of the bilayer
Pump substances across the cell membrane (Peripheral proteins at the cytosolic face of the
plasma membrane include the cytoskeletal proteins spectrin and actin in erythrocytes and the
enzyme protein kinase C. This enzyme shuttles between the cytosol and the cytosolic face of the
plasma membrane and plays a role in signal transduction)
Can move laterally (sideways) in the hydrophobic part of the membrane. Integral proteins are
tightly bound and are directly interacting with the hydrophobic core of the plasma membrane
Do not interact with the hydrophobic core in between two layers of phospholipids.
Integral membrane proteins ( intrinsic proteins)
Peripheral membrane proteins, or extrinsic proteins
Mainly found either fully or partially submerged in the phospholipids bilayer of the plasma
membrane. They have hydrophobic side chains that interact with fatty acyl groups of the
membrane phospholipids, thus anchoring the protein to the membrane.
Integral proteins represent around 70% of plasma membrane proteins.
Present at the surface ie on the innermost and outermost of phospholipids bilayer. These proteins
are loosely bound to the plasma membrane either directly by interactions with polar heads of
phospholipids bilayer or indirectly by interactions with integral proteins. These pro.
Explain how recombination increases the amount of genetic variation i.pdf
Explain how recombination increases the amount of genetic variation in offspring: Explain why
it is not possible to have a recombination frequency of greater than 50% (half recombinant
progeny): A second pair of Drosophila are mated. The female is Cucu YY (straight wing, gray
body), while the male is Cucu yy (straight wing, yellow body). Assuming recombination,
perform the cross and list the offspring genotypes and phenotypes.
Solution
Que-3:
The recombination frequency is the measure of genetic linkage at different loci. The value of
recombination frequency is 50% if the chromosomal genes are located at different regions on the
different chromosome because of independent assortment. If the genes are located closely
together on the same chromosome, then those genes considered as genetically linked; because
they have not assorted independently due to \"crossing over\", then the recombination frequency
value is less than 50%. Linked genes are due to linkage often do not obey Mendel\'s laws of
inheritance such as independent assortment because of \"crossing over with reciprocal exchange
of hereditary genetic material meticulously in between non-sister chromatids\" at the time of
\"synapsis formation\" (propahse -I) in reduction division or meiosis 1. These events are leading
to formation of gamete cells with complete different genotypically from diploid parent
reproductive cell via meiosis finally induce \"genetic variation\" in phenotype due to
\"fertilization\" events between sperm cells & ovum.
The recombination of chromosomes enables meiosis process of cell division with the generation
of novel recombinant nucleotide -sequences by crossing over during the cell division in both
unicellular and multicellular species. Sometimes, double strand breaks in DNA during
homologous chromosomal recombination may be produced due to exposure any harmful
radiation finally may cause higher genetic variation via duplication followed by fertilization to
form a phenotype with higher genetic variation
Que-4:
Crossing over is somewhat randomly distributed over the length of the chromosome because if
the two genes are close together then it is very change to get exchange of chromosomal breaks
therefore, if the two genes are located farther apart then there will be higher room for the
exchange of genetic material with chromosomal breaks. Therefore, when two genetic are far
apart on the same chromosome are more likely to have a crossover between them than two loci
that are close together. This recombination frequency is considered as a measure of genetic
linkage during the crossover of the homologous chromosomes. Linkage disequilibrium defined
as the existence of alleles at different loci with the absence of genetic linkage between them even
though there is no equilibrium with allelic frequencies independently. The recombination
frequency is the measure of genetic linkage at different loci. The value of recombination
frequency is 50% \"if the chromosomal genes are .
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col pox.pdf
exclange I ren peretr) ol and eleaced by the maxtit wadr Ho t Col poxio pon 2 the stol sle shiage
type head eschane
Solution
since there are two different fluids takes place that is hot body and cold body initially
condensation takes place in the condensation process the hot body temperature will maintain low
temperature than that of surroundings and afterwards evaporation takes palces to maintain the
temperature inside the system so that the pressure variation does not takes place and gets
maintained to the differntlevel of fluids..
DNA is a macromolecule that is critical for life. Much of the functi.pdf
DNA is a macromolecule that is critical for life. Much of the functionality of DNA lies in
itscomplex structure. What are the 2 main forces responsible for holding together the structureof
DNA and what does each force hold together?
Solution
The double helical structure of DNA (deoxyribonucleic acid) is formed by the coiling of two
polynucleotide chains, similar to a spiral staircase.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds and hydrostatic forces.
Of the two polynucleotide chains, the sugar-phosphate backbones are exposed to the outside of
the spiral staircase. While the base pairs (side chains) extend towards inside of the helix. The two
strands are connected together by hydrogen bonds..
Chapter 7 , book Health in the Later Years, 5th EditionGive a his.pdf
Chapter 7 , book: Health in the Later Years, 5th Edition
Give a historical perspective on advocacy efforts for the mentally ill in the U.S.
Solution
Answer:
Though it is not clear what was the student/client really asked for. As per the studies concern
making the fact easily understandable is the goal of our company. So, here is the brief of the fact
asked for-
Mental health services throughout the world are in the midst of unprecedented change. New
thinking is emerging about what constitutes \"best practice\" in mental health services.
Community mental health treatment and support services are now the primary locus of care for
people with serious mental illness in many areas. Fundamental components of mental health
service systems such as methods of designing and delivering services, mechanisms for
establishing and funding service priorities, methods of ensuring system accountability, and
definitions of staff role, needed competencies, and training approaches, and so forth, are
undergoing scrutiny and restructuring. Historic distinctions between clinical treatment services
and disability support or rehabilitation services are no longer simple and examples of integrated
service systems are becoming more common. In this context, traditional psychiatric hospital
services are undergoing a parallel process which is re-shaping their role, services, and position as
partners with community treatment and rehabilitation services, disability services, and self-
help/peer support services in an integrated system. Mental health consumers and their family
members are exercising their \"voice\" and are assuming increasingly vital roles in governance,
policy formulation, and service design and delivery which is helping to shape how these changes
will occur in our thinking and practice.\'.
Discuss the benefits of event-driven programming.Contrast event-dr.pdf
Discuss the benefits of event-driven programming.
Contrast event-driven programming with procedural programming.
Discuss how messages from other programs can be viewed as events.
Give examples. Give examples of tasks that require the program to guide the user.
Should programs be purely event driven, procedural, or a mixture of both? Why?
Solution
The benefit of event-driven programming is that it\'s very intuitive and naturally well-suited to
applications whose control flow are based, not on its structure, but rather on internal or external
events. This is why event-driven programming is so popular with GUI programming. Likweise,
SOA and client-server architectures are naturally event-driven as well.
We if we find contrast between Procedural and event-based then applications written in
procedural languages execute by proceeding logically through the program code, one line at a
time. Logic flow can be temporarily transferred to other parts of the program through the GoTo,
GoSub, and Call statements, directing the program from beginning to end.
In contrast, program statements in an event-driven application execute only when a specific
event calls a section of code assigned to that event. Events can be triggered by keyboard input,
mouse actions, the operating system, or code in the application. For example, consider what
happens when the user clicks a command button named Command1 on a form. The mouse click
is an event. When the Click event occurs, Visual Basic executes the code in the
Sub procedure named Command1_Click. When the code has finished running, Visual Basic
waits for the next event.
Messages from other programs/threads can be seen as events, as they trigger some instance of
code to execute..
Before 1900, despite its weaknesses in effective management of worke.pdf
Before 1900, despite its weaknesses in effective management of workers, manufacturing
leadership was well provided by top management. They were technological entrepreneurs,
archictects of productive systems, veritable lions of industry. But when they delegated their
production responsibilities to a second-level department, the factory institution never recovered
its vitality. The lion was tamed. It\'s management systems became protective and generally were
neither enterpreneual nor strategic. Production managers since then have typically had little to do
with initiating substantially new process technology-in contrast to their predecessors before 1900
(skinner 1985).
D) how is Japan (or Germany) different from (or the same as) America with regards to this trend
in manufacturing leadership?
E) taking the structural charestaristics of manufacturing enterprises (e.g., scale, complexity, pace
of technological change) as given, what can be done to revitalize manufacturing leadership?
Solution
Strategic Windows: their nature.
The nature and purpose of strategy and how it is formulated. The nature of marketing strategy
and how this should take account of the interests of various stakeholders when involving such
things as, product/service development and delivery, promotional mix, support services,
manufacturing and production processes, R&D, and material purchasing affect the stakeholders.
Other factors in the business environment that influence marketing strategy: political, economic,
socio-cultural and technological (PEST).
Marketing and competitors: how a firm must be able to position itself competitively in the minds
of its customers so that its products and services stand out very favourably in important respects
in relationship to competitors.
Matching the firm’s products / services with opportunities and threats in the market place. The
limited periods during which the fit between the key requirements of a market and the particular
competencies of a firm competing in that market are at an optimum. Investment in a product line
or a market area should be timed to coincide with periods during which a strategic window is
open. Correspondingly, withdrawal should be considered where something which was a good fit,
is no longer a good fit. Ways in which a market can evolve and how firms might develop a
competitive strategy to take advantage of Strategic Windows.
Portfolio Analysis
How organisations create their own environments rather than simply adapt to existing ones. How
they select the strategic windows of opportunities and threats through which they want to look
out into the world and develop and market product and services to meet the needs of what they
observe to be required in the face of environmental turbulence.
How well the fit between an organization’s products/services meet the needs presented by the
windows of opportunities and threats is a fitting start for exploring the subject of strategic
marketing. It introduces the many factors t.
Alpha particle radiation sensor. A silicon diode radiation sensor is .pdf
Alpha particle radiation sensor. A silicon diode radiation sensor is used to check the radioactive
source in a smoke detector. According to the label, the source is given as 1 music. The source is
Ain-241 with an energy of 486 times 10^6 e V and an activity of 195 times 10^10 B q. The
energy needed to generate an electron-hole pair is 61 e V. The diode is connected to an inverse
potential of 12 V, has a thickness of 0.5 mm, and the mobility\'s are 450 cm^2/V s for holes and
1350 cm^2/V s for electrons. The radiation from Ain-241 is mostly a particles that penetrate very
little into the silicon and hence may be assumed to be absorbed at the surface of the diode For
this reason the junction is exposed directly so radiation does not pass through the electrodes.
What is the current through the diode if all particles emitted are trapped by the diode?
Solution
Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their
production in standard alpharadioactive decay, alpha particles generally have a kinetic energy of
about 5 MeV, and a velocity in the vicinity of 5% the speed of light. (See discussion below for
the limits of these figures in alpha decay.) They are a highly ionizing form of particle radiation,
and (when resulting from radioactive alpha decay) have low penetration depth. They are able to
be stopped by a few centimeters of air, or by the skin. However, so-called long range alpha
particles from ternary fission are three times as energetic, and penetrate three times as far. As
noted, the helium nuclei that form 10–12% of cosmic rays are also usually of much higher
energy than those produced by nuclear decay processes, and are thus capable of being highly
penetrating and able to traverse the human body and also many meters of dense solid shielding,
depending on their energy. To a lesser extent, this is also true of very high-energy helium nuclei
produced by particle accelerators..
A ship was caught in a storm and driven to a small island in the mid.pdf
A ship was caught in a storm and driven to a small island in the middle of the South Pacific. For
generations, no one encountered this small colony, who despite the odds made a thriving
settlement on their island. After many years, what characteristics would be unique about this
population?
A: The people on the island would no longer be considered humans after evolving independently
for many generations.
B: The people on the island would all look very different from each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
C: The people on the island would all look very different from each other, and they may have
evolved some odd characteristics that aren\'t seen in other humans.
D: The people on the island would all look very similar to each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
Solution
There is a small group of population who lived seperately from the human for many many years .
In this case, the small population would have bred amongst themselves are continued their
colony.
This small group say after a thousand years look more like each other than any other human on
the earth as they had been isolated from the rest of the world.
So the answer should be D)The people on the island would all look very similar to each other,
and they may exhibit abnormally high occurrences of traits that are rare in the general human
population.
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its rea.pdf
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its reactions?
1. 2. 3.
4.
10. What is the BASIC difference between substrate-level phosphorylation and oxidative
phosphorylation? (do not tell me which pathways are involved).
11. Exactly….and I mean exactly…..which chemical reaction in the aerobic metabolic pathways,
actually uses oxygen, and what does it do to the oxygen that it uses.
Solution
8).
Enzymes are the proteins, which decrease the activation energy of reaction and catalyse the
reaction process. All enzymes are catalysts because,
Cells decrease the rate of reaction by
1). Altering pH
2). Accumulation of products
3). Altered temperature
4). Inactivation of catalyst.
As leaf temperature increases from 20°C to 35°C, the quantum yield o.pdf
As leaf temperature increases from 20°C to 35°C, the quantum yield of C3 photosynthesis is
directly affected by
The decreasing ratio CO2 to O2 in the gas phase
The increased reaction rates of photorespiration
The decreasing ratio CO2 to O2 in the aqueous phase
Degradation of the thylakoid membrane and proteins
RubisCO’s improved affinity for O2.
Solution
Answer is :
The decreasing ratio CO2 to O2 in the aqueous phase.
Reason: When the leaves are hot, CO2 becomes less soluble in the aqueous phase faster than
O2, and the CO2/O2 ratio drops.
Among 10 people traveling in a group, 2 have outdated passports. It .pdf
Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will
check the passports of 20% of the people in any group passing their desks. The group can go as
a
whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How
should
members of the group arrange themselves to maximize the probability of getting by the
inspectors
without having the outdated passports detected? I do not know how to figure out each
probability... The section is on counting rules - i am guessing it is without replacement and order
does not matter. Thank you for your help
Solution
Inspectors would check 20% of people in the group
Hence if 10 people, 2 would be checked or if 5 one person would be checked
If 10 people go to one desk, p for having outdated passport = 0.2
X - no of people having correct passports in 10 is binomial with p = 0.8
They would be caught only if Inspector selects any one person from the two.
Y - no of people having outdated passports and p for Y =0.2
P(Y>=1) = 1-P(Y=0)
= 1-(0.8)10
= 0.8926
---------------------------------------------
If goes into two groups then either each group can have one outdated or one group no outdated
and other groups 2 outdated.
For each group one outdated prob of being caught
= P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group)
as these two events are independent
= {1-(0.8)5}+{1-(0.8)5}-{1-(0.8)5}{1-(0.8)5}
=2(0,67232)-0.672322
= 0.8926 (same as the first case)
----------------------------------------------
If one group has both outdated and other has none then prob of being not caught
= 1(Prob of atleast one in the group having outdated 2)
Here p for good passport =0.6
= P(Y>=1)
= 1-P(y=0)
= 1-(0.6)5
=0.92224
Hence this is the best method to go.
Divide into two groups with 2 outdated in one group.
1.) The normal eye color of Drosophila is red, but strains in which .pdf
1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are
available. Similarly, wings are normally long, but there are strains with short wings. A female
from a pure line with brown eyes and short wings is crossed with a male from a normal pure line.
The F1 consists of normal females and short-winged males. An F2 is then produced by
intercrossing the F1. Both sexes of F2 flies show phenotypes as follows:
3/8 red eyes, long wings
3/8 red eyes, short wings
1/8 brown eyes, long wings
1/8 brown eyes, short wings
Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own
invention. State the genotypes of all three generations and the genotypic proportions of the F1
and F2.
2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs
divides mitotically, what are the genotypes of the daughter cells?
3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ;
b = brown body; F = forked bristles; f = unforked bristles.) The results are
black, forked 230
brown, forked 240
black, unforked 210
brown, unforked 250
Use the 2 test to determine if these results fit the results expected from testcrossing the
hypothesized dihybrid.
4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an
autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is
crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in
each sex? (Invent and define your own gene symbols.)
Solution
Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the
daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will
give rise to daughter cells with the genotype A/a; B/b; C/c.
However, if the cell was to divide meiotically (reductional division), the genetic constitution
would be different as only one half of the parental chromosomes goes to the daughter cell. So the
genotypes would be as follows:
ABC or AbC or ABc or Abc or aBC or abC or aBc or abc
Answer 3. For the given cross
BbFf X bbff
For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf).
For dihybrid cross, the degree of freedom will be 4 – 1 = 3
Chi square test is as follows
BF
Bf
bF
bf
Totals
Observed phenotype
230
210
240
250
930
Expected phenotype
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930
Observed – expected
-2.5
-22.5
7.5
17.5
(Obs – Exp)2
Expected
6.25/232.5= 0.0268
506.25/232.5= 2.1774
56.25/232.5= 0.2419
306.25/232.5= 1.3172
3.7633
Looking at the probability value table for degree of freedom = 3,
df
.90
.70
.50
.30
.20
.10
.05
.01
3
0.58
1.42
2.37
3.67
4.64
6.25
7.82
11.35
Hence we can deduce that there is more than 20% probability for the occurrence of this event.
Answer 4. Let XD = Normal phenotype (tall)
Xd = Dwarf phenotype
P = Pink coat
p = Br.
29. Code an application program that keeps track of student informat.pdf
29. Code an application program that keeps track of student information at your college. Include
their names, identification numbers, and grade point averages in a fully encapsulated,
homogeneous sorted singly linked list structure. When launched, the user will be asked to input
the initial number of students and the initial data set. Once this is complete, the user will be
presented with the following menu: Enter: 1 to insert a new student\'s information, 2 to fetch and
output a student\'s information, 3 to delete a student\'s information, 4 to update a student\'s
information, 5 to output all the student information in sorted order, and 6 to exit the program.
The program should perform an unlimited number of operations until the user enters a 6 to exit
the program. If the user requests an operation on a node not in the structure, the program output
should be “node not in structure.” Otherwise, the message “operation complete” should be
output. in java
Solution
Dear Asker,
Following 3 classes consist the colution to this problem. This will give you an idea to approach
this problem:
//Student Class
package studentinfo;
public class Student implements Comparable {
private String name;
private String id;
private double gpa;
public Student(String id, String name, double gpa) {
this.name = name;
this.id = id;
this.gpa = gpa;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the id
*/
public String getId() {
return id;
}
/**
* @param id the id to set
*/
public void setId(String id) {
this.id = id;
}
/**
* @return the gpa
*/
public double getGpa() {
return gpa;
}
/**
* @param gpa the gpa to set
*/
public void setGpa(double gpa) {
this.gpa = gpa;
}
@Override
public int compareTo(Student o) {
if (this.id.compareTo(o.id) < 0) {
return -1;
} else if (this.id.compareTo(o.id) > 0) {
return 1;
} else if (this.gpa < o.gpa) {
return -1;
} else if (this.gpa > o.gpa) {
return 1;
}
return 0;
}
public boolean equals(Student o) {
if (this.name == o.name && this.id == o.id && this.gpa == o.gpa) {
return true;
}
return false;
}
}
//Student container class
package studentinfo;
import java.util.Collections;
import java.util.LinkedList;
public class StudentContainer {
LinkedList studentList;
public StudentContainer() {
this.studentList = new LinkedList();
}
public boolean hasStudent(String id, String name, double gpa) {
return this.studentList.contains(new Student(id, name, gpa));
}
public boolean insertStudent(String id, String name, double gpa) {
if (!this.isValidString(id) || !this.isValidString(name) || !this.isValidGpa(gpa)) {
return false;
}
this.studentList.add(new Student(id, name, gpa));
return true;
}
public boolean deleteStudentById(String id) {
Student toRemove = null;
for (Student e : studentList) {
if (e.getId() == id) {
toRemove = e;
break;
}
}
if (toRemove != null) {
this.studentList.remove(toRemove);
return true;
}
return false;.
Write the code for a small function called myStack, which creates a .pdf
Write the code for a small function called myStack, which creates a Stack object of String
elements, named cities. Add \"Sydney\", then \"London\", then \"New York\" and then
\"Beijing\" to cities. Then print each element on the screen, one per line. Print each element in a
way that causes cities to be empty (i.e. have 0 elements) afterwards. (6 mks)
Solution
PROGRAM CODE:
package stack;
import java.util.Stack;
public class MyStackClass {
//stack function
public static void myStack()
{
//creating a stack object called cities
Stack cities = new Stack();
//Adding cities into the stack
cities.push(\"Sydney\");
cities.push(\"London\");
cities.push(\"New York\");
cities.push(\"Beijing\");
//using while loop to loop through the stack till its empty
while(!cities.isEmpty())
{
// pop function is used to remove elements from the stack
// pop() returns the elemets which is printed on the csreen
System.out.println(cities.pop());
}
}
// Main function for the class
public static void main(String args[])
{
myStack(); // function call to the stack method
}
}
OUTPUT:
Beijing
New York
London
Sydney.
Why does the neutralization of an acid by a base often produce water.pdf
Neutralization often produces water because acids release H+ ions and bases release OH- ions. When an acid and base react, the H+ and OH- ions combine to form water. For example, when HCl and NaOH react, they produce NaCl and water as H+ from HCl combines with OH- from NaOH.
Which statement about phloem transport is falseIt takes place in .pdf
Which statement about phloem transport is false?
It takes place in sieve tubes
It requires metabolic energy
A high pressure potential is maintained in the sieve tubes
At sinks, solutes are actively transported into sieve tube elements
The mycorrhizal relationship allows a plant’s roots to ??
increase the amount of soil than can be accessed for nutrients.
grow in nutrient-poor soils.
avoid direct interaction with the soil.
None of the above
Solution
1. At sinks, solutes are actively transported into sieve tube elements. (False)
Phloem is a vascular tissue in plants that actively transports synthesized food from the source
tissue ( photosynthetic) to sink tissue (non-photosynthetic). The sieve tube elements (STEs) are
the phloem tissue compartments which help in the transportation of food from source to sink.
Phloem transport requires metabolic energy and a high-pressure potential is maintained at the
sieve tubes.
2. Increase the amount of soil that can be accessed for nutrients
Mycorrhiza is the symbiotic relationship between fungi and the plant roots on which they
extensively grow. The network of fungal growth or mycorrhiza helps in increasing the surface
area of the roots so that they can access more amount of soil and take up nutrients essential for
plant growth..
Which of the following is not a primitive data type Which of th.pdf
Which of the following is not a primitive data type
Which of the following is not a primitive data type
Solution
Hi, PLease find primitive data type list in C:
The primitive data types in c language are the inbuilt data types provided by the c language
itself. Thus, all c compilers provide support for these data types.
The following primitive data types in c are available:
Integer Data Type
int , short, long
Float data Type
float
Double Data Type
double
Character Data Type
char
Void Data Type
void.
What are the three types of column chromatography and the principles.pdf
What are the three types of column chromatography and the principles behind how they work?
For the most specificity/highest purification which type would you expect to achieve this.
Solution
Three type of column chromatography are as follows:
1.Gel permeation chromatograhy
2.Ion exchange chromatography
3.Adsorption Chromatography
GEL PERMEATION CHROMATOGRAPHY
his chromatographic technique for the separation of molecules on the basis of their
molecular size and shape exploits the
molecular sieve
properties of a variety of porous
materials. The terms
exclusion
or
permeation chromatography
or
gel filtration
describe
all molecular separation processes using molecular sieves. The general principle of
exclusion chromatography is quite simple. A column of microparticulate cross-linked
copolymersgenerally of either styreneor divinylbenzene andwith a narrow range of pore
sizes is in equilibrium with a suitable mobile phase for the analytes to be separated. Large
analytes that are completely excluded from the pores will pass through the interstitial
spaces between the particles and will appear first in the eluate. Smaller analytes will be
distributed between the mobile phase inside and outside the particles and will therefore
pass through the column at a slower rate, hence appearing last in the eluate
ION EXCHANGE CHROMATOGRAPHY
Separation of molecules that contain electrically charged groups is often accomplished best by
ion
exchange chromatography.The technique depends upon interactions between the charged groups
of the molecules being separated and fixed ionic groups on an immobile matrix. Separation
depends upon small differences in pK a values and net charges and upon varying interactions of
nonpolar parts of the molecules being separated with the matrix. Since changes in pH can affect
both the charges on the molecules being separated and those of the ion exchange material, the
affinities of the molecules being separated are strongly dependent on pH. For example, proteins
and most amino acids are held tightly by cation exchangers at low pH but not at all at high
pH.Aqueous solutions are usually employed and the columns are packed with beads of ion
exchange resins, porous materials containing bound ionic groups such as – SO 3 – , – COO – , –
NH 3+ , or quaternary nitrogen atoms. Synthetic resins based on a cross- linked polystyrene are
usually employed for separation of small molecules. For larger molecules chemical derivatives
of cellulose or of crosslinked dextrans (Sephadex), agarose, or polyacrylamide are more
appropriate. Positively charged ions, such as amino acids in a low pH solution, are placed on a
cation exchange resin such as Dowex 50, which contains dissociated sulfonic acid groups as well
as counter ions such as Na + , K + , or H + . The adsorbed amino
acids are usually eluted with buffers of increasing pH containing sodium or lithium ions.Ion
exchange chromatography of proteins and peptides is often done with such ion exchange
materi.
What are the benefits of endocrine signaling in comparison to neuron.pdf
What are the benefits of endocrine signaling in comparison to neuronal signaling?
Solution
The endocrine signalling is long lasting,broad impact and is difficult to reverse.The endocrine
system relies on hormones to elicit responses from target cells.
The nervous system conveys high-speed electrical signals along specialized cells called
neurons,these signals regulate other cells.The response to stimuli by the nervous system is near
instantaneous,although the effects are often short lived..
What does this statement mean...genes code for gene products W.pdf
What does this statement mean...genes code for gene products?
What does this statement mean...genes code for gene products?
Solution
Gene codes for gene product.The statement means gene expression which is a process by which
a gene\'s information is used for the synthesis of a gene product that is functional.The gene
products are proteins in case of protein coding genes while in case of non protein coding genes
i,e t-RNA or snRNA , the gene product is RNA that is functional..
What a program to toggle PD3, PH7, and PC5 continuously without, dist.pdf
What a program to toggle PD3, PH7, and PC5 continuously without, disturbing the rest of the
bits. Write a program to monitor bit PC3. When it is HIGH, send 0 times 55 to PORTD.
Solution
//Accessing Ports as SFRs using sfr data type
sfr PD3=0x80;
sfr PD7=0x90;
sfr PC5=0xA0;
void MSDelay(unsigned int);
void main(void)
{
while (1)
{
PD3=0x55;
PD7=0x55;
PC5=0x55;
MSDelay(0);
PD3=0xAA;
PD7=0xAA;
PC5=0xAA;
MSDelay(0); } }.
True of False results that are practically significant will always .pdf
True of False? results that are practically significant will always be statistically significant.
Solution
False, some times sample mean or sample mean difference may seem very high, (practically
significant), but are not statistically significant (p-value is not less than alpha), implying that they
could occur by chance..
The waiting time T between successive occurrences of an event E in a .pdf
The waiting time T between successive occurrences of an event E in a discrete-time renewal
process has the probability distribution P(T = 1) = 0.7, P(T = 2)- 0.3. The waiting time to the
sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10). 0.0595 0.0494
0.0393
Solution
Let the waiting time to 1st occurence be T1 , between occurence 1 and 2 be T2 , and.... between
occurence 5 and 6 be T6.
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) :
For the 6 time intervals to add up to 10, as the only possible interval values are 1 and 2, 4 of the
intervals have to have the value 2 and remaining 2 have to have the value 1
2+2+2+2+1+1 = 10
These values can be assigned to 6 intervals in : 6C4 * 2C2 ways = 15 ways
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) = 15 * [P(T=1)]2 * [P(T=2)]4 = 15*0.72 *0.34 = 0.0595
Answer: 0.0595.
The half-life of Carbon-14 is about 5700 years. You have found a fos.pdf
The half-life of Carbon-14 is about 5700 years. You have found a fossil in sedimentary rocks
that have 1/128 the normal (modern) expected ratio of Carbon-14C. Based on this information,
you estimate that it is about _______ years old.
44
5700
39,900
729,600
Solution
The fossil in 39,900 years old. At 1/128 the carbon-14 of living organisms, it has gone through
seven half-lives of carbon-14 (5700 years x 7 = 39,900 years).
The (circle one meanmedian) would be a better measure of central te.pdf
The (circle one: mean/median) would be a better measure of central tendency if you are
describing the annual incomes of all people in Minnesota.
Solution
Income is a measure that is typically skewed towards right and has extreme outliers towards
right. Median is a better measure with data with outliers..
clinical examination of hip joint (1).pdf
described clinical examination all orthopeadic conditions .
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Before 1900, despite its weaknesses in effective management of worke.pdf
Before 1900, despite its weaknesses in effective management of workers, manufacturing
leadership was well provided by top management. They were technological entrepreneurs,
archictects of productive systems, veritable lions of industry. But when they delegated their
production responsibilities to a second-level department, the factory institution never recovered
its vitality. The lion was tamed. It\'s management systems became protective and generally were
neither enterpreneual nor strategic. Production managers since then have typically had little to do
with initiating substantially new process technology-in contrast to their predecessors before 1900
(skinner 1985).
D) how is Japan (or Germany) different from (or the same as) America with regards to this trend
in manufacturing leadership?
E) taking the structural charestaristics of manufacturing enterprises (e.g., scale, complexity, pace
of technological change) as given, what can be done to revitalize manufacturing leadership?
Solution
Strategic Windows: their nature.
The nature and purpose of strategy and how it is formulated. The nature of marketing strategy
and how this should take account of the interests of various stakeholders when involving such
things as, product/service development and delivery, promotional mix, support services,
manufacturing and production processes, R&D, and material purchasing affect the stakeholders.
Other factors in the business environment that influence marketing strategy: political, economic,
socio-cultural and technological (PEST).
Marketing and competitors: how a firm must be able to position itself competitively in the minds
of its customers so that its products and services stand out very favourably in important respects
in relationship to competitors.
Matching the firm’s products / services with opportunities and threats in the market place. The
limited periods during which the fit between the key requirements of a market and the particular
competencies of a firm competing in that market are at an optimum. Investment in a product line
or a market area should be timed to coincide with periods during which a strategic window is
open. Correspondingly, withdrawal should be considered where something which was a good fit,
is no longer a good fit. Ways in which a market can evolve and how firms might develop a
competitive strategy to take advantage of Strategic Windows.
Portfolio Analysis
How organisations create their own environments rather than simply adapt to existing ones. How
they select the strategic windows of opportunities and threats through which they want to look
out into the world and develop and market product and services to meet the needs of what they
observe to be required in the face of environmental turbulence.
How well the fit between an organization’s products/services meet the needs presented by the
windows of opportunities and threats is a fitting start for exploring the subject of strategic
marketing. It introduces the many factors t.
Alpha particle radiation sensor. A silicon diode radiation sensor is .pdf
Alpha particle radiation sensor. A silicon diode radiation sensor is used to check the radioactive
source in a smoke detector. According to the label, the source is given as 1 music. The source is
Ain-241 with an energy of 486 times 10^6 e V and an activity of 195 times 10^10 B q. The
energy needed to generate an electron-hole pair is 61 e V. The diode is connected to an inverse
potential of 12 V, has a thickness of 0.5 mm, and the mobility\'s are 450 cm^2/V s for holes and
1350 cm^2/V s for electrons. The radiation from Ain-241 is mostly a particles that penetrate very
little into the silicon and hence may be assumed to be absorbed at the surface of the diode For
this reason the junction is exposed directly so radiation does not pass through the electrodes.
What is the current through the diode if all particles emitted are trapped by the diode?
Solution
Alpha particles, like helium nuclei, have a net spin of zero. Due to the mechanism of their
production in standard alpharadioactive decay, alpha particles generally have a kinetic energy of
about 5 MeV, and a velocity in the vicinity of 5% the speed of light. (See discussion below for
the limits of these figures in alpha decay.) They are a highly ionizing form of particle radiation,
and (when resulting from radioactive alpha decay) have low penetration depth. They are able to
be stopped by a few centimeters of air, or by the skin. However, so-called long range alpha
particles from ternary fission are three times as energetic, and penetrate three times as far. As
noted, the helium nuclei that form 10–12% of cosmic rays are also usually of much higher
energy than those produced by nuclear decay processes, and are thus capable of being highly
penetrating and able to traverse the human body and also many meters of dense solid shielding,
depending on their energy. To a lesser extent, this is also true of very high-energy helium nuclei
produced by particle accelerators..
A ship was caught in a storm and driven to a small island in the mid.pdf
A ship was caught in a storm and driven to a small island in the middle of the South Pacific. For
generations, no one encountered this small colony, who despite the odds made a thriving
settlement on their island. After many years, what characteristics would be unique about this
population?
A: The people on the island would no longer be considered humans after evolving independently
for many generations.
B: The people on the island would all look very different from each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
C: The people on the island would all look very different from each other, and they may have
evolved some odd characteristics that aren\'t seen in other humans.
D: The people on the island would all look very similar to each other, and they may exhibit
abnormally high occurrences of traits that are rare in the general human population.
Solution
There is a small group of population who lived seperately from the human for many many years .
In this case, the small population would have bred amongst themselves are continued their
colony.
This small group say after a thousand years look more like each other than any other human on
the earth as they had been isolated from the rest of the world.
So the answer should be D)The people on the island would all look very similar to each other,
and they may exhibit abnormally high occurrences of traits that are rare in the general human
population.
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its rea.pdf
8. List 4 GENERAL ways that a CELL, can DECREASE the rate of its reactions?
1. 2. 3.
4.
10. What is the BASIC difference between substrate-level phosphorylation and oxidative
phosphorylation? (do not tell me which pathways are involved).
11. Exactly….and I mean exactly…..which chemical reaction in the aerobic metabolic pathways,
actually uses oxygen, and what does it do to the oxygen that it uses.
Solution
8).
Enzymes are the proteins, which decrease the activation energy of reaction and catalyse the
reaction process. All enzymes are catalysts because,
Cells decrease the rate of reaction by
1). Altering pH
2). Accumulation of products
3). Altered temperature
4). Inactivation of catalyst.
As leaf temperature increases from 20°C to 35°C, the quantum yield o.pdf
As leaf temperature increases from 20°C to 35°C, the quantum yield of C3 photosynthesis is
directly affected by
The decreasing ratio CO2 to O2 in the gas phase
The increased reaction rates of photorespiration
The decreasing ratio CO2 to O2 in the aqueous phase
Degradation of the thylakoid membrane and proteins
RubisCO’s improved affinity for O2.
Solution
Answer is :
The decreasing ratio CO2 to O2 in the aqueous phase.
Reason: When the leaves are hot, CO2 becomes less soluble in the aqueous phase faster than
O2, and the CO2/O2 ratio drops.
Among 10 people traveling in a group, 2 have outdated passports. It .pdf
Among 10 people traveling in a group, 2 have outdated passports. It is known that inspectors will
check the passports of 20% of the people in any group passing their desks. The group can go as
a
whole (all 10) to one desk or can split into two groups of 5 and use two different desks. How
should
members of the group arrange themselves to maximize the probability of getting by the
inspectors
without having the outdated passports detected? I do not know how to figure out each
probability... The section is on counting rules - i am guessing it is without replacement and order
does not matter. Thank you for your help
Solution
Inspectors would check 20% of people in the group
Hence if 10 people, 2 would be checked or if 5 one person would be checked
If 10 people go to one desk, p for having outdated passport = 0.2
X - no of people having correct passports in 10 is binomial with p = 0.8
They would be caught only if Inspector selects any one person from the two.
Y - no of people having outdated passports and p for Y =0.2
P(Y>=1) = 1-P(Y=0)
= 1-(0.8)10
= 0.8926
---------------------------------------------
If goes into two groups then either each group can have one outdated or one group no outdated
and other groups 2 outdated.
For each group one outdated prob of being caught
= P(y=1) in I group + P(y=1 in second group)-P( P(y=1) in I group *P(y=1 in second group)
as these two events are independent
= {1-(0.8)5}+{1-(0.8)5}-{1-(0.8)5}{1-(0.8)5}
=2(0,67232)-0.672322
= 0.8926 (same as the first case)
----------------------------------------------
If one group has both outdated and other has none then prob of being not caught
= 1(Prob of atleast one in the group having outdated 2)
Here p for good passport =0.6
= P(Y>=1)
= 1-P(y=0)
= 1-(0.6)5
=0.92224
Hence this is the best method to go.
Divide into two groups with 2 outdated in one group.
1.) The normal eye color of Drosophila is red, but strains in which .pdf
1.) The normal eye color of Drosophila is red, but strains in which all flies have brown eyes are
available. Similarly, wings are normally long, but there are strains with short wings. A female
from a pure line with brown eyes and short wings is crossed with a male from a normal pure line.
The F1 consists of normal females and short-winged males. An F2 is then produced by
intercrossing the F1. Both sexes of F2 flies show phenotypes as follows:
3/8 red eyes, long wings
3/8 red eyes, short wings
1/8 brown eyes, long wings
1/8 brown eyes, short wings
Deduce the inheritance of these phenotypes; use clearly defined genetic symbols of your own
invention. State the genotypes of all three generations and the genotypic proportions of the F1
and F2.
2.) When a cell of genotype A/a ; B/b ; C/c having all the genes on separate chromosome pairs
divides mitotically, what are the genotypes of the daughter cells?
3.) A presumed dihybrid in Drosophila, B/b ; F/f, is test-crossed with b/b ; f/f. (B = black body ;
b = brown body; F = forked bristles; f = unforked bristles.) The results are
black, forked 230
brown, forked 240
black, unforked 210
brown, unforked 250
Use the 2 test to determine if these results fit the results expected from testcrossing the
hypothesized dihybrid.
4.) In mice, dwarfism is caused by an X-linked recessive allele, and pink coat is caused by an
autosomal dominant allele (coats are normally brownish). If a dwarf female from a pure line is
crossed with a pink male from a pure line, what will be the phenotypic ratios in the F1 and F2 in
each sex? (Invent and define your own gene symbols.)
Solution
Answer 2. If a cell divides mitotically (equational division), the genetic constitution of the
daughter cells remains the same as the parents. So a parent with the genotype A/a; B/b; C/c will
give rise to daughter cells with the genotype A/a; B/b; C/c.
However, if the cell was to divide meiotically (reductional division), the genetic constitution
would be different as only one half of the parental chromosomes goes to the daughter cell. So the
genotypes would be as follows:
ABC or AbC or ABc or Abc or aBC or abC or aBc or abc
Answer 3. For the given cross
BbFf X bbff
For such a test cross the resultant phenotypic ration obtained is 1 (BF):1 (Bf):1 (bF):1 (bf).
For dihybrid cross, the degree of freedom will be 4 – 1 = 3
Chi square test is as follows
BF
Bf
bF
bf
Totals
Observed phenotype
230
210
240
250
930
Expected phenotype
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930x ¼= 232.5
930
Observed – expected
-2.5
-22.5
7.5
17.5
(Obs – Exp)2
Expected
6.25/232.5= 0.0268
506.25/232.5= 2.1774
56.25/232.5= 0.2419
306.25/232.5= 1.3172
3.7633
Looking at the probability value table for degree of freedom = 3,
df
.90
.70
.50
.30
.20
.10
.05
.01
3
0.58
1.42
2.37
3.67
4.64
6.25
7.82
11.35
Hence we can deduce that there is more than 20% probability for the occurrence of this event.
Answer 4. Let XD = Normal phenotype (tall)
Xd = Dwarf phenotype
P = Pink coat
p = Br.
29. Code an application program that keeps track of student informat.pdf
29. Code an application program that keeps track of student information at your college. Include
their names, identification numbers, and grade point averages in a fully encapsulated,
homogeneous sorted singly linked list structure. When launched, the user will be asked to input
the initial number of students and the initial data set. Once this is complete, the user will be
presented with the following menu: Enter: 1 to insert a new student\'s information, 2 to fetch and
output a student\'s information, 3 to delete a student\'s information, 4 to update a student\'s
information, 5 to output all the student information in sorted order, and 6 to exit the program.
The program should perform an unlimited number of operations until the user enters a 6 to exit
the program. If the user requests an operation on a node not in the structure, the program output
should be “node not in structure.” Otherwise, the message “operation complete” should be
output. in java
Solution
Dear Asker,
Following 3 classes consist the colution to this problem. This will give you an idea to approach
this problem:
//Student Class
package studentinfo;
public class Student implements Comparable {
private String name;
private String id;
private double gpa;
public Student(String id, String name, double gpa) {
this.name = name;
this.id = id;
this.gpa = gpa;
}
/**
* @return the name
*/
public String getName() {
return name;
}
/**
* @param name the name to set
*/
public void setName(String name) {
this.name = name;
}
/**
* @return the id
*/
public String getId() {
return id;
}
/**
* @param id the id to set
*/
public void setId(String id) {
this.id = id;
}
/**
* @return the gpa
*/
public double getGpa() {
return gpa;
}
/**
* @param gpa the gpa to set
*/
public void setGpa(double gpa) {
this.gpa = gpa;
}
@Override
public int compareTo(Student o) {
if (this.id.compareTo(o.id) < 0) {
return -1;
} else if (this.id.compareTo(o.id) > 0) {
return 1;
} else if (this.gpa < o.gpa) {
return -1;
} else if (this.gpa > o.gpa) {
return 1;
}
return 0;
}
public boolean equals(Student o) {
if (this.name == o.name && this.id == o.id && this.gpa == o.gpa) {
return true;
}
return false;
}
}
//Student container class
package studentinfo;
import java.util.Collections;
import java.util.LinkedList;
public class StudentContainer {
LinkedList studentList;
public StudentContainer() {
this.studentList = new LinkedList();
}
public boolean hasStudent(String id, String name, double gpa) {
return this.studentList.contains(new Student(id, name, gpa));
}
public boolean insertStudent(String id, String name, double gpa) {
if (!this.isValidString(id) || !this.isValidString(name) || !this.isValidGpa(gpa)) {
return false;
}
this.studentList.add(new Student(id, name, gpa));
return true;
}
public boolean deleteStudentById(String id) {
Student toRemove = null;
for (Student e : studentList) {
if (e.getId() == id) {
toRemove = e;
break;
}
}
if (toRemove != null) {
this.studentList.remove(toRemove);
return true;
}
return false;.
Write the code for a small function called myStack, which creates a .pdf
Write the code for a small function called myStack, which creates a Stack object of String
elements, named cities. Add \"Sydney\", then \"London\", then \"New York\" and then
\"Beijing\" to cities. Then print each element on the screen, one per line. Print each element in a
way that causes cities to be empty (i.e. have 0 elements) afterwards. (6 mks)
Solution
PROGRAM CODE:
package stack;
import java.util.Stack;
public class MyStackClass {
//stack function
public static void myStack()
{
//creating a stack object called cities
Stack cities = new Stack();
//Adding cities into the stack
cities.push(\"Sydney\");
cities.push(\"London\");
cities.push(\"New York\");
cities.push(\"Beijing\");
//using while loop to loop through the stack till its empty
while(!cities.isEmpty())
{
// pop function is used to remove elements from the stack
// pop() returns the elemets which is printed on the csreen
System.out.println(cities.pop());
}
}
// Main function for the class
public static void main(String args[])
{
myStack(); // function call to the stack method
}
}
OUTPUT:
Beijing
New York
London
Sydney.
Why does the neutralization of an acid by a base often produce water.pdf
Neutralization often produces water because acids release H+ ions and bases release OH- ions. When an acid and base react, the H+ and OH- ions combine to form water. For example, when HCl and NaOH react, they produce NaCl and water as H+ from HCl combines with OH- from NaOH.
Which statement about phloem transport is falseIt takes place in .pdf
Which statement about phloem transport is false?
It takes place in sieve tubes
It requires metabolic energy
A high pressure potential is maintained in the sieve tubes
At sinks, solutes are actively transported into sieve tube elements
The mycorrhizal relationship allows a plant’s roots to ??
increase the amount of soil than can be accessed for nutrients.
grow in nutrient-poor soils.
avoid direct interaction with the soil.
None of the above
Solution
1. At sinks, solutes are actively transported into sieve tube elements. (False)
Phloem is a vascular tissue in plants that actively transports synthesized food from the source
tissue ( photosynthetic) to sink tissue (non-photosynthetic). The sieve tube elements (STEs) are
the phloem tissue compartments which help in the transportation of food from source to sink.
Phloem transport requires metabolic energy and a high-pressure potential is maintained at the
sieve tubes.
2. Increase the amount of soil that can be accessed for nutrients
Mycorrhiza is the symbiotic relationship between fungi and the plant roots on which they
extensively grow. The network of fungal growth or mycorrhiza helps in increasing the surface
area of the roots so that they can access more amount of soil and take up nutrients essential for
plant growth..
Which of the following is not a primitive data type Which of th.pdf
Which of the following is not a primitive data type
Which of the following is not a primitive data type
Solution
Hi, PLease find primitive data type list in C:
The primitive data types in c language are the inbuilt data types provided by the c language
itself. Thus, all c compilers provide support for these data types.
The following primitive data types in c are available:
Integer Data Type
int , short, long
Float data Type
float
Double Data Type
double
Character Data Type
char
Void Data Type
void.
What are the three types of column chromatography and the principles.pdf
What are the three types of column chromatography and the principles behind how they work?
For the most specificity/highest purification which type would you expect to achieve this.
Solution
Three type of column chromatography are as follows:
1.Gel permeation chromatograhy
2.Ion exchange chromatography
3.Adsorption Chromatography
GEL PERMEATION CHROMATOGRAPHY
his chromatographic technique for the separation of molecules on the basis of their
molecular size and shape exploits the
molecular sieve
properties of a variety of porous
materials. The terms
exclusion
or
permeation chromatography
or
gel filtration
describe
all molecular separation processes using molecular sieves. The general principle of
exclusion chromatography is quite simple. A column of microparticulate cross-linked
copolymersgenerally of either styreneor divinylbenzene andwith a narrow range of pore
sizes is in equilibrium with a suitable mobile phase for the analytes to be separated. Large
analytes that are completely excluded from the pores will pass through the interstitial
spaces between the particles and will appear first in the eluate. Smaller analytes will be
distributed between the mobile phase inside and outside the particles and will therefore
pass through the column at a slower rate, hence appearing last in the eluate
ION EXCHANGE CHROMATOGRAPHY
Separation of molecules that contain electrically charged groups is often accomplished best by
ion
exchange chromatography.The technique depends upon interactions between the charged groups
of the molecules being separated and fixed ionic groups on an immobile matrix. Separation
depends upon small differences in pK a values and net charges and upon varying interactions of
nonpolar parts of the molecules being separated with the matrix. Since changes in pH can affect
both the charges on the molecules being separated and those of the ion exchange material, the
affinities of the molecules being separated are strongly dependent on pH. For example, proteins
and most amino acids are held tightly by cation exchangers at low pH but not at all at high
pH.Aqueous solutions are usually employed and the columns are packed with beads of ion
exchange resins, porous materials containing bound ionic groups such as – SO 3 – , – COO – , –
NH 3+ , or quaternary nitrogen atoms. Synthetic resins based on a cross- linked polystyrene are
usually employed for separation of small molecules. For larger molecules chemical derivatives
of cellulose or of crosslinked dextrans (Sephadex), agarose, or polyacrylamide are more
appropriate. Positively charged ions, such as amino acids in a low pH solution, are placed on a
cation exchange resin such as Dowex 50, which contains dissociated sulfonic acid groups as well
as counter ions such as Na + , K + , or H + . The adsorbed amino
acids are usually eluted with buffers of increasing pH containing sodium or lithium ions.Ion
exchange chromatography of proteins and peptides is often done with such ion exchange
materi.
What are the benefits of endocrine signaling in comparison to neuron.pdf
What are the benefits of endocrine signaling in comparison to neuronal signaling?
Solution
The endocrine signalling is long lasting,broad impact and is difficult to reverse.The endocrine
system relies on hormones to elicit responses from target cells.
The nervous system conveys high-speed electrical signals along specialized cells called
neurons,these signals regulate other cells.The response to stimuli by the nervous system is near
instantaneous,although the effects are often short lived..
What does this statement mean...genes code for gene products W.pdf
What does this statement mean...genes code for gene products?
What does this statement mean...genes code for gene products?
Solution
Gene codes for gene product.The statement means gene expression which is a process by which
a gene\'s information is used for the synthesis of a gene product that is functional.The gene
products are proteins in case of protein coding genes while in case of non protein coding genes
i,e t-RNA or snRNA , the gene product is RNA that is functional..
What a program to toggle PD3, PH7, and PC5 continuously without, dist.pdf
What a program to toggle PD3, PH7, and PC5 continuously without, disturbing the rest of the
bits. Write a program to monitor bit PC3. When it is HIGH, send 0 times 55 to PORTD.
Solution
//Accessing Ports as SFRs using sfr data type
sfr PD3=0x80;
sfr PD7=0x90;
sfr PC5=0xA0;
void MSDelay(unsigned int);
void main(void)
{
while (1)
{
PD3=0x55;
PD7=0x55;
PC5=0x55;
MSDelay(0);
PD3=0xAA;
PD7=0xAA;
PC5=0xAA;
MSDelay(0); } }.
True of False results that are practically significant will always .pdf
True of False? results that are practically significant will always be statistically significant.
Solution
False, some times sample mean or sample mean difference may seem very high, (practically
significant), but are not statistically significant (p-value is not less than alpha), implying that they
could occur by chance..
The waiting time T between successive occurrences of an event E in a .pdf
The waiting time T between successive occurrences of an event E in a discrete-time renewal
process has the probability distribution P(T = 1) = 0.7, P(T = 2)- 0.3. The waiting time to the
sixth occurrence of E is denoted by W6. Find the probability P(W6 = 10). 0.0595 0.0494
0.0393
Solution
Let the waiting time to 1st occurence be T1 , between occurence 1 and 2 be T2 , and.... between
occurence 5 and 6 be T6.
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) :
For the 6 time intervals to add up to 10, as the only possible interval values are 1 and 2, 4 of the
intervals have to have the value 2 and remaining 2 have to have the value 1
2+2+2+2+1+1 = 10
These values can be assigned to 6 intervals in : 6C4 * 2C2 ways = 15 ways
P(T1 + T2 + T3 + T4 + T5 + T6 = 10) = 15 * [P(T=1)]2 * [P(T=2)]4 = 15*0.72 *0.34 = 0.0595
Answer: 0.0595.
The half-life of Carbon-14 is about 5700 years. You have found a fos.pdf
The half-life of Carbon-14 is about 5700 years. You have found a fossil in sedimentary rocks
that have 1/128 the normal (modern) expected ratio of Carbon-14C. Based on this information,
you estimate that it is about _______ years old.
44
5700
39,900
729,600
Solution
The fossil in 39,900 years old. At 1/128 the carbon-14 of living organisms, it has gone through
seven half-lives of carbon-14 (5700 years x 7 = 39,900 years).
The (circle one meanmedian) would be a better measure of central te.pdf
The (circle one: mean/median) would be a better measure of central tendency if you are
describing the annual incomes of all people in Minnesota.
Solution
Income is a measure that is typically skewed towards right and has extreme outliers towards
right. Median is a better measure with data with outliers..
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LAND USE LAND COVER AND NDVI OF MIRZAPUR DISTRICT, UP
This Dissertation explores the particular circumstances of Mirzapur, a region located in the
core of India. Mirzapur, with its varied terrains and abundant biodiversity, offers an optimal
environment for investigating the changes in vegetation cover dynamics. Our study utilizes
advanced technologies such as GIS (Geographic Information Systems) and Remote sensing to
analyze the transformations that have taken place over the course of a decade.
The complex relationship between human activities and the environment has been the focus
of extensive research and worry. As the global community grapples with swift urbanization,
population expansion, and economic progress, the effects on natural ecosystems are becoming
more evident. A crucial element of this impact is the alteration of vegetation cover, which plays a
significant role in maintaining the ecological equilibrium of our planet.Land serves as the foundation for all human activities and provides the necessary materials for
these activities. As the most crucial natural resource, its utilization by humans results in different
'Land uses,' which are determined by both human activities and the physical characteristics of the
land.
The utilization of land is impacted by human needs and environmental factors. In countries
like India, rapid population growth and the emphasis on extensive resource exploitation can lead
to significant land degradation, adversely affecting the region's land cover.
Therefore, human intervention has significantly influenced land use patterns over many
centuries, evolving its structure over time and space. In the present era, these changes have
accelerated due to factors such as agriculture and urbanization. Information regarding land use and
cover is essential for various planning and management tasks related to the Earth's surface,
providing crucial environmental data for scientific, resource management, policy purposes, and
diverse human activities.
Accurate understanding of land use and cover is imperative for the development planning
of any area. Consequently, a wide range of professionals, including earth system scientists, land
and water managers, and urban planners, are interested in obtaining data on land use and cover
changes, conversion trends, and other related patterns. The spatial dimensions of land use and
cover support policymakers and scientists in making well-informed decisions, as alterations in
these patterns indicate shifts in economic and social conditions. Monitoring such changes with the
help of Advanced technologies like Remote Sensing and Geographic Information Systems is
crucial for coordinated efforts across different administrative levels. Advanced technologies like
Remote Sensing and Geographic Information Systems
9
Changes in vegetation cover refer to variations in the distribution, composition, and overall
structure of plant communities across different temporal and spatial scales. These changes can
occur natural.
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