Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with \'2c\' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome\' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down\' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps..
When you look at different stages of mitosis, why are the chromosome.pdfkennithdase
When you look at different stages of mitosis, why are the chromosomes in so many different
positions, rather than in one distinct position?
Solution
Answer:
Mitosis refers to the process of cell division where a mother cell divides into two daughter cells
with identical genetic makeup. The different stages of mitosis are G1, S, G2 and M (prophase,
metaphase, anaphase and telophase) phases.
During mitosis, the chromosomes in cells are in different positions. The significance of these
different positions relates to the preparatory and transitional phases for equational DNA division
in the two daughter cells.
The DNA undergoes duplication during the S phase. In G2 phase, the DNA are not distinct but
are intertwined. Condensation of chromosomal material marks the start of prophase. Completion
of prophase is characterized by compact mitotic chromosomes having two chromatids and are
scattered in the cytoplasm. This is a preparatory phase for aligning the chromosomes at the
center for separation in the two daughter cells.
During metaphase, the chromosomes attach to the spindle fibers and are moved to spindle
equator. At anaphase, each chromosome gets separated at the kinetochore for migration towards
the opposite poles. Telophase marks the final stage of chromosome separation with the
chromosomes reaching their respective poles..
1. Describe how variation in meiosis happens and why it is beneficia.pdffashioncollection2
1. Describe how variation in meiosis happens and why it is beneficial? Steps and functions of
mitosis and meiosis? ( I am having a hard time with defining each step in mitosis and meiosis.
There is meiosis I and II and I get confused on the steps with functions. Any ideas how I can
remember?)
TIA :)
Solution
Explanation:-
Variation in meiosis:-
During meiosis in humans, 1 diploid cell (with 46 chromosomes or 23 pairs) undergoes 2 cycles
of cell division but only 1 round of DNA replication. The result is 4 haploidHaving one copy of
each chromosome, or having a single set of chromosomes. Gametes (egg and sperm cells) are
haploid. daughter cells known as gametes or egg and sperm cells (each with 23 chromosomes – 1
from each pair in the diploid cell).
At conception, an egg cell and a sperm cell combine to form a zygote (46 chromosomes or 23
pairs). This is the 1st cell of a new individual. The halving of the number of chromosomes in
gametes ensures that zygotes have the same number of chromosomes from one generation to the
next. This is critical for stable sexual reproduction through successive generations.
Benefits of Meiosis:-
1. New Cell Generation-
The chromosomes created during meiosis are composed of 50% copies of the parent cell and
50% new cells. ‘Â These new cells are produced during the cross-over stages of the cell division
process. ‘Â During this stage half of the genetic material from the parent cell is copied into the
new cells, with the other half having distinct properties and characteristics.
2. DNA Replication
The process of meiosis involves copying or replication of genetic material from the parent cell
into the new cells. ‘Â As much as half of the genetic properties of the parent cell are copied into
the newly-created cells. ‘Â When applied to humans for example, DNA from both parents will
partly be copied onto the cells of their offspring. ‘Â When DNA is copied or replicated, the
offspring will also have similar qualities with either or both of his/her parents.
3. Genetic Variation
With meiosis, only half of the genetic material is replicated into the new cells. ‘Â This simply
means that the remaining half will be composed of unique genetic properties making each cell
different from the other. ‘Â Through this process, humans for example will all have different
genetic material and structure.
With the process involved in meiosis, humans are able to reproduce similar yet distinct offspring.
‘Â The whole process basically explains the fact that babies may share some genetic traits from
their parents but they will also have unique sets of personalities because of their unique genetic
composition.
Meiosis Stages:-
Prophase I
Chromosomes condense, Crossing over occurs
Metaphase I
Homologous chromosomes pair up and align in middle of cell
Anaphase I
Homologous chromosomes pulled apart
Telophase I
Nuclear Envelope reforms
Cytokinesis I
Cell splits into two
Prophase II
Centrioles divide and move to opposite poles
Metaphase II
Chromoso.
When you look at different stages of mitosis, why are the chromosome.pdfkennithdase
When you look at different stages of mitosis, why are the chromosomes in so many different
positions, rather than in one distinct position?
Solution
Answer:
Mitosis refers to the process of cell division where a mother cell divides into two daughter cells
with identical genetic makeup. The different stages of mitosis are G1, S, G2 and M (prophase,
metaphase, anaphase and telophase) phases.
During mitosis, the chromosomes in cells are in different positions. The significance of these
different positions relates to the preparatory and transitional phases for equational DNA division
in the two daughter cells.
The DNA undergoes duplication during the S phase. In G2 phase, the DNA are not distinct but
are intertwined. Condensation of chromosomal material marks the start of prophase. Completion
of prophase is characterized by compact mitotic chromosomes having two chromatids and are
scattered in the cytoplasm. This is a preparatory phase for aligning the chromosomes at the
center for separation in the two daughter cells.
During metaphase, the chromosomes attach to the spindle fibers and are moved to spindle
equator. At anaphase, each chromosome gets separated at the kinetochore for migration towards
the opposite poles. Telophase marks the final stage of chromosome separation with the
chromosomes reaching their respective poles..
1. Describe how variation in meiosis happens and why it is beneficia.pdffashioncollection2
1. Describe how variation in meiosis happens and why it is beneficial? Steps and functions of
mitosis and meiosis? ( I am having a hard time with defining each step in mitosis and meiosis.
There is meiosis I and II and I get confused on the steps with functions. Any ideas how I can
remember?)
TIA :)
Solution
Explanation:-
Variation in meiosis:-
During meiosis in humans, 1 diploid cell (with 46 chromosomes or 23 pairs) undergoes 2 cycles
of cell division but only 1 round of DNA replication. The result is 4 haploidHaving one copy of
each chromosome, or having a single set of chromosomes. Gametes (egg and sperm cells) are
haploid. daughter cells known as gametes or egg and sperm cells (each with 23 chromosomes – 1
from each pair in the diploid cell).
At conception, an egg cell and a sperm cell combine to form a zygote (46 chromosomes or 23
pairs). This is the 1st cell of a new individual. The halving of the number of chromosomes in
gametes ensures that zygotes have the same number of chromosomes from one generation to the
next. This is critical for stable sexual reproduction through successive generations.
Benefits of Meiosis:-
1. New Cell Generation-
The chromosomes created during meiosis are composed of 50% copies of the parent cell and
50% new cells. ‘Â These new cells are produced during the cross-over stages of the cell division
process. ‘Â During this stage half of the genetic material from the parent cell is copied into the
new cells, with the other half having distinct properties and characteristics.
2. DNA Replication
The process of meiosis involves copying or replication of genetic material from the parent cell
into the new cells. ‘Â As much as half of the genetic properties of the parent cell are copied into
the newly-created cells. ‘Â When applied to humans for example, DNA from both parents will
partly be copied onto the cells of their offspring. ‘Â When DNA is copied or replicated, the
offspring will also have similar qualities with either or both of his/her parents.
3. Genetic Variation
With meiosis, only half of the genetic material is replicated into the new cells. ‘Â This simply
means that the remaining half will be composed of unique genetic properties making each cell
different from the other. ‘Â Through this process, humans for example will all have different
genetic material and structure.
With the process involved in meiosis, humans are able to reproduce similar yet distinct offspring.
‘Â The whole process basically explains the fact that babies may share some genetic traits from
their parents but they will also have unique sets of personalities because of their unique genetic
composition.
Meiosis Stages:-
Prophase I
Chromosomes condense, Crossing over occurs
Metaphase I
Homologous chromosomes pair up and align in middle of cell
Anaphase I
Homologous chromosomes pulled apart
Telophase I
Nuclear Envelope reforms
Cytokinesis I
Cell splits into two
Prophase II
Centrioles divide and move to opposite poles
Metaphase II
Chromoso.
The G_1, S, and G_2 phases of the cell cycle are collectively known a.pdfnamarta88
The G_1, S, and G_2 phases of the cell cycle are collectively known as which part of the cell
cycle? Considering mitosis, cytokinesis, and interphase, in which one is nuclear DNA
duplicated? When in the cell cycle do chromosomes first split apart? The division of the
cytoplasm occurs in what part of the cell cycle? Describe the four phases of nuclear (mitotic)
division and what occurs during those phases.
Solution
10) Interphase is the phase with G1 , S and G2 cycles. The cell spends most of its time in
interphase. The phases are -
11) Nuclear DNA is duplicated in interphase S cycle.
12) Chromosomes split apart in mitosis in metaphase and anaphase.
13) The 4 phases of mitosis are Prophase, metaphase , anaphase and telophase.
In prophase cells start break down some structures and build others.
In metaphase, the chromosomes with the spindle apparatus line the chromosomes in the middle
and form metaphase plate.
In anaphase, the sister chromatids pull the chromatids towards opposite poles.
In telophase, is the reversal of prophase..
– Male and female gametes fuse together during fertilization to form a zygote. The chromosome number is halved during the formation of gametes by the process of meiosis. This maintains the chromosome number generations after generations. Meiosis leads to genetic diversity which is very essential for evolution.
Which of the following are true when comparing TCPIP to the OSI Ref.pdfleventhalbrad49439
Which of the following are true when comparing TCP/IP to the OSI Reference Model? (Choose
two.)
A. The TCP/IP model has seven layers, while the OSI model has only four layers.
B. The TCP/IP model has four layers, while the OSI model has seven layers.
C. The TCP/IP Application layer maps to the Application, Session, and Presentation layers of the
OSI Reference Model.
D. The TCP/IP Application layer is virtually identical to the OSI Application layer.
Solution
B and C are correct. The TCP/IP model has four layers. The Application layer maps to the top
three layers of the OSI Reference Model.
A and D are incorrect. A and D are incorrect because they are contrary to B and C..
What is the reason for having most of the blood volume distributed w.pdfleventhalbrad49439
What is the reason for having most of the blood volume distributed within the venous vessels at
any given time?
Solution
The thin, non-elastic wall of veins (as compared to the thick, elastic, muscular walls of arteries)
allows greater volumes of blood to be accommodated within the lumen of the veins. This
accounts for most of the blood volume distributed within the venous vessels at any given time..
More Related Content
Similar to Given the information regarding mitosis and meiosis, clearly and con.pdf
The G_1, S, and G_2 phases of the cell cycle are collectively known a.pdfnamarta88
The G_1, S, and G_2 phases of the cell cycle are collectively known as which part of the cell
cycle? Considering mitosis, cytokinesis, and interphase, in which one is nuclear DNA
duplicated? When in the cell cycle do chromosomes first split apart? The division of the
cytoplasm occurs in what part of the cell cycle? Describe the four phases of nuclear (mitotic)
division and what occurs during those phases.
Solution
10) Interphase is the phase with G1 , S and G2 cycles. The cell spends most of its time in
interphase. The phases are -
11) Nuclear DNA is duplicated in interphase S cycle.
12) Chromosomes split apart in mitosis in metaphase and anaphase.
13) The 4 phases of mitosis are Prophase, metaphase , anaphase and telophase.
In prophase cells start break down some structures and build others.
In metaphase, the chromosomes with the spindle apparatus line the chromosomes in the middle
and form metaphase plate.
In anaphase, the sister chromatids pull the chromatids towards opposite poles.
In telophase, is the reversal of prophase..
– Male and female gametes fuse together during fertilization to form a zygote. The chromosome number is halved during the formation of gametes by the process of meiosis. This maintains the chromosome number generations after generations. Meiosis leads to genetic diversity which is very essential for evolution.
Similar to Given the information regarding mitosis and meiosis, clearly and con.pdf (20)
Which of the following are true when comparing TCPIP to the OSI Ref.pdfleventhalbrad49439
Which of the following are true when comparing TCP/IP to the OSI Reference Model? (Choose
two.)
A. The TCP/IP model has seven layers, while the OSI model has only four layers.
B. The TCP/IP model has four layers, while the OSI model has seven layers.
C. The TCP/IP Application layer maps to the Application, Session, and Presentation layers of the
OSI Reference Model.
D. The TCP/IP Application layer is virtually identical to the OSI Application layer.
Solution
B and C are correct. The TCP/IP model has four layers. The Application layer maps to the top
three layers of the OSI Reference Model.
A and D are incorrect. A and D are incorrect because they are contrary to B and C..
What is the reason for having most of the blood volume distributed w.pdfleventhalbrad49439
What is the reason for having most of the blood volume distributed within the venous vessels at
any given time?
Solution
The thin, non-elastic wall of veins (as compared to the thick, elastic, muscular walls of arteries)
allows greater volumes of blood to be accommodated within the lumen of the veins. This
accounts for most of the blood volume distributed within the venous vessels at any given time..
What is eutectic reaction What is eutectoid reaction What is perife.pdfleventhalbrad49439
What is eutectic reaction? What is eutectoid reaction? What is perifectic reaction? Are these
transformations reversible? in the following phase diagrams (figure QB) these points from points
A, B, C, D, E, F, G, H, K, and write down the reactions at these points. For example point A is
perifectic reaction at this point can be written as Lipid and delta-iron transform to gamma-iron
during cooling.
Solution
Eutectic reaction: A eutectic reaction is a reaction when one a liquid cools down into a mixture
of two solids. Elements have different miscibility at different temperatures. So, when a liquid
containing two elements soluble in one another cools down to a level where the solutbility is
limited, then the excess element is sandwiched between the zones containing the mixture. This
reaction is Eutectic reaction. In the Iron-Carbide diagram above, it is represented by the point D.
Eutectoid reaction: A eutectoid reaction is a reaction where one solid divides into a mixture of
two solids. It has the same concept as eutectic reaction but here, instead of a liquid splitting up
into two solids, a solid breaks down into two solids. For example, B is an eutectoid reaction
Peritectic reaction: A peritectic reaction is where the mixture of a solid and liquid combines into
a single solid. Here, the miscibility is low when there is a phase-equilibrium between the liquid
and the solid but as the temperature decreases, the solubility increases. Hence, it is completely
miscible in the solid. For example, A shows the peritectic reaction happening.
Yes, since all the solutions are in a state of pahse equilibrium, they can be interchanged and the
reaction can go either ways. It is completely reversible..
What are the possible types of interfund transactions What are.pdfleventhalbrad49439
What are the possible types of interfund transactions?
What are the possible types of interfund transactions?
Solution
Two types of interfund transactions are Quasi-External Transactions and Reimbursements
QUASI-EXTERNAL TRANSACTIONS
These are transactions that would be treated as revenues, expenditures or expenses if they
involved organizations external to the State. They are accounted for as ordinary revenues,
expenditures or expenses of the fund involved. Examples are payments by a State agency to the
Office of General Services and Centralized Services Fund billings.
REIMBURSEMENTS
These are transactions that constitute reimbursements of a fund for expenditures or expenses
initially made from it which are properly applicable to another fund. They are recorded as
expenditures or expenses of the reimbursing fund and as reductions of the expenditure or
expense originally charged in the fund that is reimbursed. However, interfund reimbursements do
not include situations whereby interfund receivables or payables have been set up..
What are Sociocultural issues when it come to body imageSolutio.pdfleventhalbrad49439
What are Sociocultural issues when it come to body image?
Solution
Answer:
Sociocultural issues such as dissatisfaction in women and men each they are obese due to eating
disorders & media aspects and cultural aspects often have meticulous influence on individual
body image finally may lead to “development of depressive disorders”. Therefore, unrealistic
images from media & family pressure and society pressure do generate consciousness on
physical beauty via body image. Therefore, “perfect bodies” often maintained by various people
via performing gymnastics, swimming etc. Individual mental health in association with
sociocultural factors such as westernization hypothesis & abnormal eating attitudes can exert
influence on inappropriate body image.
Water can absorb and store a large amount of heat while increasing on.pdfleventhalbrad49439
Water can absorb and store a large amount of heat while increasing only a few degrees in
temperature. Why? The heat must first be used to break the hydrogen bonds rather than raise the
temperature. The heat must first be used to break the ionic bonds rather than raise the
temperature. The heat must first be used to break the covalent bonds rather than raise the
temperature. An increase in temperature causes an increase in adhesion of the water. An
increase in temperature causes an increase in cohesion of the water. A chemical compound that
releases H^+ into a solution is called: a proton. A base. an acid. a hydroxide ion. a hydrogen
ion. Which of the following foods is not a significant source of complex carbohydrates? fresh
fruit rice pasta oatmeal All of the above are significant sources of complex carbohydrates
Solution
Thanks for the question
5.A The hydrogen bonds are reponsible for the high boiling points of water.
6.C an acid is a substance which donates H.
7.E All of the above mentioned foods are good source of complex carbohydrates.
True or false1 Anabolism results in increased production of biomas.pdfleventhalbrad49439
True or false
1 Anabolism results in increased production of biomass.
2 Oxygenic phototrophs can use hydrogen sulfide as an electron donor.
3 Oxidation is the process of accepting electrons.
4 Water appeared on the earth as a result of bombardment by asteroids.
5 The highest abundance of microbial life in soils appears in the O horizon.
6 Typically, increased oxygenic photosynthetic activity results in a decrease in pH values during
the day and an increase in pH values at night.
7 Freshwater represents about 10% of the total water on Earth.
8 Oligotrophic organisms thrive in low nutrient conditions.
9 Establishment of a proton motive force (PMF) is required to generate ATP via substrate level
phosphorylation.
10 Siderphores are used to increase availability of manganese in aerobic environments.
Solution
1. Anabolism: The process of formation of biomolecules during metabolism such as proteins,
polysaccharides, lipids, and nucleic acids etc is known as anabolism, this results in an increase in
the production of biomass.
Therefore, option is: True
2.Oxygenic phototrophs use oxygen as electron donar, while as anoxygenic phototrophs uses
hydrogen sulfide as electron donor.
Therefore, option is: False
3. Oxidation is a process in which a substance looses one or more electrons, while as reduction is
a process of accepting electrons.
Therefore, option is: False
4. The earth got most of the water about 4.6 billion years ago from the asteroid impacts.
Therefore, option is: True.
The diploid number of the hypothetical animal Geneticus introductus .pdfleventhalbrad49439
The diploid number of the hypothetical animal Geneticus introductus is 2n= 44. Each diploid
nucleus contains 5 ng of DNA in G1.
Part A : What amount of DNA is contained in each nucleus at the end of S phase? What amount
of DNA is contained in each nucleus at the end of phase?
a)There are 22 chromosomes containing 2.5 ng of DNA.
b)There are 44 chromosomes containing 5 ng of DNA.
c)There are 44 chromosomes containing 10 ng of DNA.
d)None of the above.
Part B: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase I of meiosis (after cytokinesis). Express your answer as two numbers separated by a
comma (example: 12, 4).
Part C: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Anaphase of mitosis.
Part D: Write the number of chromosomes and amount of DNA (ng) present per cell at the end of
Telophase II of meiosis (after cytokinesis).
Solution
The cell cycle of actively dividing eukaryotic cell consists of gap phases (G 1 and G2), S phase
and M phase. In G 1 phase the cell prepares for cell division .In S phase DNA replication occurs
and the DNA content doubles. In G 2 phase, the cell gets ready for mitosis.
In mitosis, two daughter cells with same genetic content as parent are produced. Mitosis occurs
through five stages: prophase, prometaphase, metaphase, anaphase and telophase.
Prophase: This stage extends over half of mitosis. The nuclear membrane disintegrates. The
centrosome of chromosome duplicates to form two daughter centrosomes and migrates to
opposite poles of the cell. The centrosomes organise the production of microtubules, from which
the spindle fibres arise. The chromosomes condense and consist of two identical chromatids (or
sister chromatids) held together at the centromere.
Prometaphase: The chromosomes, move to the middle of the cell. The spindle fibres bind to
centromere of each chromosome at kinetochore.
Metaphase: The chromosomes align along the metaphase plate of the spindle apparatus.
Anaphase: This is the shortest stage of mitosis. The centromeres divide, and the sister chromatids
of each chromosome are separated and move to the opposite poles of the cell. The separated
sister chromatids ordaughter chromosomesreceives a copy of each parental chromosome.
Telophase: This is the final stage of mitosis. The nuclear membrane reforms, the chromosomes
uncoil, and the spindle fibres disappear.
Cytokinesis: the membrane appears to divide the two daughter cells with same chromosome
number and the DNA content as the parent cell.
The cell then enters interphase - the interval between mitotic divisions.
Meiosis is the cell division occuring in sex cells or gametes (which contain a single copy of each
chromosome). The process involves one DNA replication followed by two successive nuclear
and cellular divisions (Meiosis I and Meiosis II). Similar to mitosis, meiosis is preceded by a
process of DNA replication that converts each chromosome into two sister chrom.
The Antoine equation often is used to describe vapor pressures lnP .pdfleventhalbrad49439
The Antoine equation often is used to describe vapor pressures: lnP = A B /(T + C) . For
temperatures near the normal boiling point:
a. Derive an expression that gives the behavior of Hvap as a function of temperature which is
thermodynamically consistent with the Antoine equation.
Solution
The Antoine equation is a mathematical expression of the relation between the vapour pressure
and the temperturer of pure substances. The equation was first proposed by Ch. Antoine, a french
researcher, in 1888. The basic form of the equation is:
lnP = A - B / ( T + C )
and it can be transformed into this temperature-explicit form:
T = B ( A - lnP ) - C
where: P is the absolute vapor pressure of a substance
T is the temperature of the substance
A, B and C are substance-specific coefficients (i.e., constants or parameters
log is typically either log10 or loge
A simpler form of the equation with only two coefficients is sometimes used:
lnP = A - ( B \\ T )
which can be transformed to:
T = B ( A - lnP )
The Antoine equation is a semi-empirical equation which expresses vapour pressure as a
function of temperature. A new, rapid and highly accurate method for obtaining its three
constants from experimental data is presented and applied to ethanol, water and 14 anaesthetic
substances where p is the vapour pressure, T is temperature and A, B and C are component-
specific constants.
Usually, the Antoine equation cannot be used to describe the entire saturated vapour pressure
curve from the triple point to the critical point, because it is not flexible enough. Therefore,
multiple parameter sets for a single component are commonly used. A low-pressure parameter
set is used to describe the vapour pressure curve up to the normal boiling point and the second
set of parameters is used for the range from the normal boiling point to the critical point.
The Antoine Equation is a vapor pressure equation and describes the relation between vapor
pressure and temperature for pure water between 0°C and 373.946°C.
For T in Kelvins and P in kPa:
273.150 < T 373.150: A = 16.5699, B = 3984.92, C = 39.724
373.150 < T 647.096: A = 16.7285, B = 4169.84, C = 28.665
Antoine(T) = exp(A B / (T + C))
For temperatures above the critical temperature (373.946°C / 647.096K), where water is a
superfluid, the increase in vapor pressure appears essentially linear† with the same slope (i.e.
dP/dT 235.88 kPa/°C). This behavior appears consistent with the Ideal Gas Law and statements
that supercritical water behaves like an ideal gas. Now if we define Tc = 647.096 K, we can
write
P(T Tc) = Antoine(T)
P(T > Tc) = 235.88 (T Tc) + Antoine(Tc)
Hence, the behaviour of delta hvap as a function of temperature which is termodynamically
consistent with the the antoine equation is derived ..
The converson of the English to Christianity began a rich period of .pdfleventhalbrad49439
The converson of the English to Christianity began a rich period of artistic production in
England, especially in the creation of a. secular mocaics, b. illuminated manuscripts, c.
ecclesiastical treatises, d. gothic cathedrals
Solution
b. illuminated manuscripts
Elaborately decorated manuscripts were prepared after the conversion. Thus b is the correct
option..
Table 19. Highest Level of Type of organization germ layers .pdfleventhalbrad49439
Table 19. Highest Level of Type of organization germ layers \"asymmetry (sponges) Cell Noa
Cnidaria Pha An tea cnidarians-- jellyfish, hyvha. Radar 4 nerve Platyhelminthes ema toda
(flatworm) Hoda Nematoda worms) bila uran 3 and have germ Germ associated with
gastrulation. Note that deuterostomes product of developmental in which leads threetissue layers
Solution
3. Bilateral symmetry
5.Value of body cavity:
a) Pseudocoelomic fluid is rich in protein and other dissolved substances which is responsible for
turgescent state of nematode body.
b) Coelom makes the digestive system independent of the body wall. It can act as hydrostatic
skeleton. It transports food, waste and other materials..
Question 12 (5 points)The reason that mitosis can result in the pr.pdfleventhalbrad49439
Question 12 (5 points)
The reason that mitosis can result in the production of haploid gametes in alternation of
generations is because
Question 12 options:
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Save
Question 13 (5 points)
In the fern life cycle, how does fertilization occur?
Question 13 options:
The sporangia on the undersides of the leaves produce gametes, which are flung our of the leaf
and find each other on the ground to fuse
The sporophyte produces sperm and eggs in gametangia. The sperm reach the eggs by swimming
through water
The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
There is no sexual reproduction in ferns
Save
Question 14 (5 points)
The earliest land plants were most likely to have lived in which of the following habitats?
Question 14 options:
arid desert environments
Salt water marshes
Tropical rainforests
fresh water wetlands
Save
The sporophyte and gametophyte are both haploid
Eggs and sperm are produced directly by the diploid sporophyte via meiosis
Eggs and sperm are produced by the diploid gametophyte via mitosis
Eggs and sperm are produced by the haploid gametophyte via mitosis
Solution
Q12 Eggs and sperms are produced by haploid gametophyte by mitosis.
The large, leafy fern is the diploid organism. On the undersurface of its fronds or leaves, its cells
undergo meiosis to create haploid cells. However, these cells do not immediately unite with
others to recreate the diploid state. Instead, they are shed as spores and germinate into small
haploid organisms. Because the diploid organism creates spores, it is called the sporophyte
generation of the life cycle. Upon reaching maturity, the haploid organism creates haploid egg
and sperm cells (gametes) by mitosis . Because the haploid organism creates gametes , it is called
the gametophyte generation of the life cycle. The male gametes (sperm) are then released and
swim to the female egg. Fusion of the gametes creates the new diploid sporophyte, completing
the life cycle.
Q13 The gametophyte produces sperm and eggs in gametangia (antheridia and archegonia). The
sperm reach the eggs by swimming through water
The male organs or antheridia produce and shed numerous motile sperm or antherozoids. These
antherozoids are armed with hair-like or whip-like cilia or flagellae and are able to swim through
water; they do not travel great distances and are only released when free water is available. Only
a continuous film of water is needed. They antherozoids are chemically attracted to the
developing flask-like female organs or archegonia where fertilization of the single egg cell
embedded in each archegonium takes place.
The sporophyte has a double set of chromosomes (i.e. diploid) and i.
peripheral circuit functions built-in to the HC12 microcontroller, t.pdfleventhalbrad49439
peripheral circuit functions built-in to the HC12 microcontroller, that are not part of the CPU
itself or the memories. Access to these is via external pins on the Microcontroller IC package.
What are some of these built-in circuit functions called, and what do they do?
Solution
The peripheral circuit communicate with the Microcontroller using built in cirucit called BUS.
The bus is a set of shared communication lines that serves as the central nervous system of the
computer. Data, address, and control signals are shared by all system components via the bus.
Each component connected to the bus communicates information to and from the bus via its own
bus controller. The data lines,address lines, and control lines allow a specific component to
access data addressed to that component. The data lines are used to communicate words to and
from data.
registers in the various system components such as memory, CPU, and input/output
(I/O) peripherals. The address lines are used to select devices on the bus or specific
data locations within memory. Devices usually have a combinational logic address
decoder circuit that identifies the address code and activates the device..
Mendez Corporation has 10,000 shares of its $100 par value, 7 percen.pdfleventhalbrad49439
Mendez Corporation has 10,000 shares of its $100 par value, 7 percent cumulative
preferred stock outstanding and 50,000 shares of its $1 par value common stock outstanding.
In Mendez’s first four years of operation, its board of directors paid the following
cash dividends: 2011, none; 2012, $120,000; 2013, $140,000; 2014, $140,000.
Determine the dividends per share and total cash dividends paid to the preferred and
common stockholders during each of the four years.
Solution
The dividends per year on preferred stock = 10,000 x $ 100 x 7 %
= $ 10,000 x $ 100 x 0.07 = $ 70,000
Dividend distribution:
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000
Explanation:
First dividend is paid to Preferred stock and then excess amount goes to common share holder.
For cumulative preferred stock, any deficiency in dividend, not paid carries over to subsequent
years and is added to that year\'s dividend.
Preferred Dividends unpaid in 2011 = $70,000
Preferred Dividends on 2012 = $70,000 + $ 70,000 = $ 140,000
Preferred Dividends paid in 2012 = $ 120,000
Preferred Dividends shortage on 2012 = $ 20,000
As the Dividend was insufficient for preferred stocks, common stocks holder was not paid any
dividend during 2011 & 2012.
Preferred Dividends paid in 2013 = $ 70,000 + $ 20,000 = $ 90,000
Common Dividends paid in 2013 = $ 140,000 - $ 90,000 = $ 50,000
Preferred Dividends paid in 2014 = $ 70,000
Common Dividends paid in 2014 = $ 140,000 - $ 70,000 = $ 70,000
Computation of Dividends per share:
2011
2012
2013
2014
Preferred Stock (Dividends per share)
$0
$12.00
$9.00
$7.00
Common Stock (Dividends per share)
$0
$0
$1.00
$1.40
Explanation:
Dividends per share = Dividend paid/No. of shares
Year 2012:
Dividends per share for Preferred Stock = $120,000/10,000 = $ 12
Dividend per share for Common Stock = $ 0
Year 2013:
Dividends per share for Preferred Stock = $90,000/10,000 = $ 9
Dividends per share for Common Stock = $ 50,000/50,000 = $ 1
Year 2014:
Dividends per share for Preferred Stock = $70,000 /10,000 = $ 7
Dividends per share for Common Stock = $ 70,000/50,000 = $ 1.40
Year
Total Dividend Amount
Preferred
Common
2011
-
-
-
2012
$120,000
$120,000
$0
2013
$140,000
$90,000
$50,000
2014
$140,000
$70,000
$70,000.
If the dpy-11 and unc-31 genes are linked on the same chromosome, how.pdfleventhalbrad49439
If the dpy-11 and unc-31 genes are linked on the same chromosome, how many phenotypes
would the F_2 progeny show? If the two genes are on different chromosomes and therefore
independently assorting, how many phenotypes would you see in the F_2 worms? List the
phenotype(s). Strawberry plants have varieties that flower when the days are still short, and
other varieties that flower only when the days get longer. The long-day allele is dominant to the
short-day allele. You\'d like to test whether or not this gene is inherited in a Mendelian fashion.
When you cross between heterozygous plants and count the offspring in the F_1 generation, this
is what you find: 85 plants with no flowers on short days and 35 plants with flowers on short
days What would you expect the ratio of phenotypes be? What is your Chi squared value?
Show your work What does this mean?
Solution
2. Linked genes tend to inherit together for the next-generations. No crossing over ororoccurs
independent assortment occurs. No new characters appear. Where as gegenes are on different
chromosomes, they are non linked. Crossing over or independent assortment takes place. New
characters appear along with parental characters. So dpy-11 & unc-31 present on same
chromosome, they are linked and give only parental characters. Only 2 phenotypes appear. If the
said genes present on different chromosomes, they give 4 different phenotypes
3. a. Heterozygous long day representing as Ll. When crossing or hybridization occur between
these genes
Ll x Ll ----------> 3:1 is the phenotypes ratio. Three are long and one is short.
Chi square value can be calculated by using the formula.
(O-E)^2/E
O = observed phenotypes; E = Expected phenotypes.
How does a wave formation influence the resultant waveSolution.pdfleventhalbrad49439
How does a wave formation influence the resultant wave?
Solution
Waves are usually formed by distant wind.Generally larger waves are more powerful than small
waves . Waves can pile up at shore lines and create a rip current.The resultant wave heights Hi
and the wave period Ti are related to the factors that influence the energy transfer between wind
and the ocean..
How are health services paid for Provide a definition for the term .pdfleventhalbrad49439
How are health services paid for? Provide a definition for the term \"third-party payer\", discuss
the different payers currently operating in the market, and assess the importance of each. Please
use a table format to list the information.
Solution
There are two main payers types: single payer and third-party payer. Single payer is when the
person that obtains the health service is the same which pay for it, here relationship is direct taker
and provider of health services.
However, it is more common the presence of a third party payer, which is an external agent that
covers the other health expenses or part of them, they are insurance companies or public social
protection systems.
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage
Single Payer
Third-Party Payer
Type of relation with provider
Direct
Indirect
Capability to compare Prices/Features
Relatively High
Almost Null
Payment proportion
Full
Low percentage.
First, look to determine what is the earliest era that fossils have .pdfleventhalbrad49439
First, look to determine what is the earliest era that fossils have been found here in Georgia.
Next, see if you can find what the oldest fossil that has been found in Georgia. Give the name of
the fossil, when it was found, and where the fossil was found. Give a brief description of the
organism that was fossilized.
Solution
Trilobites are the oldest fossils found in georgia.They have been found in Georgia’s Appalachian
Plateau which is present the Coosa River in Floyd County.
These fossils are considered to be arthropods,Moreover they had to shed their shells to grow.As a
result the fossils found in these places usually shed shells as bits of individuals or as complete
individuals or as imprints of individuals..
Figure out and write down a definition of a translation that does no.pdfleventhalbrad49439
Figure out and write down a definition of a translation that does not use (x,y) coordinates
Solution
There are many kinds of oprations in mathematics like translation, rotation, reflection, resized
etc. Translation is one of them.
Translation is an operation which is uesd in geometry to move an object in a fixed direction and
for same distance..
explain why right and why wrong Indentify configuration (below) whic.pdfleventhalbrad49439
explain why right and why wrong Indentify configuration (below) which I s not a correct
representation of H and I.
Solution
option C is not correct one.
Explanation:
If we stretch all right hand fingers in the direction of current direction given then the thumb gives
us into the board (into symbol) or out of the board (dot symbol).
In the option C the field should be out of the board, so it is not correct one..
Instructions for Submissions thorugh G- Classroom.pptxJheel Barad
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.
Honest Reviews of Tim Han LMA Course Program.pptxtimhan337
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Embracing GenAI - A Strategic ImperativePeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Francesca Gottschalk - How can education support child empowerment.pptxEduSkills OECD
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
A Strategic Approach: GenAI in EducationPeter Windle
Artificial Intelligence (AI) technologies such as Generative AI, Image Generators and Large Language Models have had a dramatic impact on teaching, learning and assessment over the past 18 months. The most immediate threat AI posed was to Academic Integrity with Higher Education Institutes (HEIs) focusing their efforts on combating the use of GenAI in assessment. Guidelines were developed for staff and students, policies put in place too. Innovative educators have forged paths in the use of Generative AI for teaching, learning and assessments leading to pockets of transformation springing up across HEIs, often with little or no top-down guidance, support or direction.
This Gasta posits a strategic approach to integrating AI into HEIs to prepare staff, students and the curriculum for an evolving world and workplace. We will highlight the advantages of working with these technologies beyond the realm of teaching, learning and assessment by considering prompt engineering skills, industry impact, curriculum changes, and the need for staff upskilling. In contrast, not engaging strategically with Generative AI poses risks, including falling behind peers, missed opportunities and failing to ensure our graduates remain employable. The rapid evolution of AI technologies necessitates a proactive and strategic approach if we are to remain relevant.
Synthetic Fiber Construction in lab .pptxPavel ( NSTU)
Synthetic fiber production is a fascinating and complex field that blends chemistry, engineering, and environmental science. By understanding these aspects, students can gain a comprehensive view of synthetic fiber production, its impact on society and the environment, and the potential for future innovations. Synthetic fibers play a crucial role in modern society, impacting various aspects of daily life, industry, and the environment. ynthetic fibers are integral to modern life, offering a range of benefits from cost-effectiveness and versatility to innovative applications and performance characteristics. While they pose environmental challenges, ongoing research and development aim to create more sustainable and eco-friendly alternatives. Understanding the importance of synthetic fibers helps in appreciating their role in the economy, industry, and daily life, while also emphasizing the need for sustainable practices and innovation.
Given the information regarding mitosis and meiosis, clearly and con.pdf
1. Given the information regarding mitosis and meiosis, clearly and convincingly explain in your
own words how an error in MITOSIS could possibly lead to a gamete containing new genetic
information and thereby giving rise to a new phenotype. Use detail be comprehensive. (7 points)
Solution
The process of mitosis is divided into two phases: Interphase and M-phase.
Interphase is further divided into G1, S and G2 phase. In G1, protein synthesis occurs for S-
phase, in S phase, DNA replicates and in G2 phase, protein synthesis occurs for M-phase to start
properly. Now, during S phase, the ‘2n’ nucleus converts into 4n.
So, the cell entering M-phase is already ‘4n’.
M-phase is the actual mitotic phase, in which the cell divides into two. It contains four phases,
prophase, metaphase, anaphase and telophase. Your question is related with these four phases. In
prophase, the entire DNA converts into chromosomes; the process is known as condensation of
DNA. In metaphase, these chromosomes align on the equator. Now, if alignment at equator is
not correct, then anaphase cannot be started due to metaphase check point.
In anaphase, the aligned chromosomes move towards the opposite poles. This phase is the main
phase, related with formation of gametes. The cell is already ‘4n’. ‘2n’ chromosomes move
towards one end, and the rest of ‘2n’ move towards the other end. Now, suppose an error occurs
in mitosis, and ‘3n’ chromosomes move towards one end, and 1n goes towards the other cell;
then two gametes, ‘3n’ and n will be formed.
This ‘3n’ and ‘n’ gametes are gametes having different genetic information in comparison to
their ‘2n’ counterparts.
Interestingly, Meiosis II will result in gamete formation is mitotic division (equational division).
Haploid cells with '2c' amount of DNA leave Meiosis I. Now, in anaphase, suppose one
chromosome' 2c DNA migrates to one end (due to error) and no DNA migrates to the other end;
then two gametes will be formed; one having all normal chromosomes + one double
chromosome with 2c DNA, and the other having no DNA.
Please note: normally gametes are 1n (haploid) and 1c.
Suppose this gamete with double DNA (2c) fertilizes with a gamete with half DNA (1c), then the
resulting cell will have “all normal chromosomes + one chromosome with ‘3c’ amount of DNA”.
This condition if occurring with chromosome 21 is known as Down' syndrome.
It is important to note that errors in Mitosis result in non-disjunction; which further results in
many Mendelian disorders; as well as ploidy defects.
** Hope it helps.
