The DNA content of a diploid cell in the G1 phase of the cell cycle .pdf
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The DNA content of a diploid cell in the G1 phase of the cell cycle is measured. If this DNA content is X, then the DNA content of the same cell at metaphase of meiosis II would be: Question 22 options: 0.25X 0.5X X 2X 4X 0.25X 0.5X X 2X 4X Solution During meiosis, the chromosome number is halved. This usually occurs for the production of gametes. After fusion of gametes, the original chromosome number as well as total DNA content are restored. Before, meiosis starts, the DNA content is doubled during the S-phase of Interphaseand then the G2 phase comes. During the Interphase, G1 occurs first, then S phase occurs and G2 occurs. DNA replication occurs only during the S-phase. So, if the DNA content was X during the G1 Phase, it will become 2X during S-phase. At the end of Meiosis-2 it will become X. However, at Metaphase2, the chromosomes have aligned themselves at the centre of the cell but not segregated yet. So, DNA content will be the same at Metaphase-2 but it will be halved at the end of Meiosis2. After fusion the original DNA content will remain the same. Therefore, DNA Content will be X at Metaphase2..
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Given the information regarding mitosis and meiosis, clearly and con.pdf
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.pdf
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..
Answer1. Chromatin is unwound DNA found in the nucleus during Int.pdf
Answer:
1. Chromatin is unwound DNA found in the nucleus during Interphase.
2. Replicated chromosome is a chromosome that has doubled and is made of 2 chromatids.
3. Chromatid is a half of a replicated chromosome
4. Chromosome is a “package” of 1-5 thousand genes. Humans have 46 of these.
5. Mitosis is a process that divides the nucleus of a cell.
6. Interphase is the stage most cells are in during the cell cycle when growth, replication of
DNA, and preparation for division occurs.
7. Prophase is the phase in mitosis where the chromosomes become visible as the nuclear
membrane breaks down
8. Metaphase is the phase of mitosis where replicated chromosomes line up at the “equator” of
the cell.
9. Anaphase is the phase of mitosis where sister chromatids of the replicated chromosomes split
and head for the “pole”
10. Telophase is the last stage of mitosis when the nucleus reforms and cytokinesis occurs.
11. Cytokinesis is the last part of the Telophase when the division of the cell cytoplasm into 2
cells occurs.
12. Cell cycle is all events that occur in a cell from the time it divides to the time it divides again.
13. G1 phse is the Interphase time of cell growth and replication of organelles.
14. S phase is the Interphase time of DNA replication
15. G2 phase is the Interphase time of preparation for cell division by checking for DNA
mistakes
16. Homologous chromosomes are chromosomes with the same kinds of genes on them
17. Diploid is twice the number of chromosomes (2n=46 in humans).
18. Haploid is half the diploid number of chromosomes (n=23 in humans)
19. Meiosis is the cell division process in gametes in 2 haploid cells.
20. Meiosis I is the 1st division of meiosis resulting in 2 haploid cells
21. Meiosis II is the 2nd division of meiosis resulting in 4 haploid cells.
22. Independent assortment is the principle that Mendel discovered that the occurenc of 1 trait
has nothing to do with another tait if on different chromosomes
23. Gene map is a diagram of where genes occur on chromosomes.
Solution
Answer:
1. Chromatin is unwound DNA found in the nucleus during Interphase.
2. Replicated chromosome is a chromosome that has doubled and is made of 2 chromatids.
3. Chromatid is a half of a replicated chromosome
4. Chromosome is a “package” of 1-5 thousand genes. Humans have 46 of these.
5. Mitosis is a process that divides the nucleus of a cell.
6. Interphase is the stage most cells are in during the cell cycle when growth, replication of
DNA, and preparation for division occurs.
7. Prophase is the phase in mitosis where the chromosomes become visible as the nuclear
membrane breaks down
8. Metaphase is the phase of mitosis where replicated chromosomes line up at the “equator” of
the cell.
9. Anaphase is the phase of mitosis where sister chromatids of the replicated chromosomes split
and head for the “pole”
10. Telophase is the last stage of mitosis when the nucleus reforms and cytokinesis occurs.
11. Cytokinesis is the last part of the Telophase wh.
Recommended
Given the information regarding mitosis and meiosis, clearly and con.pdf
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.pdf
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..
Answer1. Chromatin is unwound DNA found in the nucleus during Int.pdf
Answer:
1. Chromatin is unwound DNA found in the nucleus during Interphase.
2. Replicated chromosome is a chromosome that has doubled and is made of 2 chromatids.
3. Chromatid is a half of a replicated chromosome
4. Chromosome is a “package” of 1-5 thousand genes. Humans have 46 of these.
5. Mitosis is a process that divides the nucleus of a cell.
6. Interphase is the stage most cells are in during the cell cycle when growth, replication of
DNA, and preparation for division occurs.
7. Prophase is the phase in mitosis where the chromosomes become visible as the nuclear
membrane breaks down
8. Metaphase is the phase of mitosis where replicated chromosomes line up at the “equator” of
the cell.
9. Anaphase is the phase of mitosis where sister chromatids of the replicated chromosomes split
and head for the “pole”
10. Telophase is the last stage of mitosis when the nucleus reforms and cytokinesis occurs.
11. Cytokinesis is the last part of the Telophase when the division of the cell cytoplasm into 2
cells occurs.
12. Cell cycle is all events that occur in a cell from the time it divides to the time it divides again.
13. G1 phse is the Interphase time of cell growth and replication of organelles.
14. S phase is the Interphase time of DNA replication
15. G2 phase is the Interphase time of preparation for cell division by checking for DNA
mistakes
16. Homologous chromosomes are chromosomes with the same kinds of genes on them
17. Diploid is twice the number of chromosomes (2n=46 in humans).
18. Haploid is half the diploid number of chromosomes (n=23 in humans)
19. Meiosis is the cell division process in gametes in 2 haploid cells.
20. Meiosis I is the 1st division of meiosis resulting in 2 haploid cells
21. Meiosis II is the 2nd division of meiosis resulting in 4 haploid cells.
22. Independent assortment is the principle that Mendel discovered that the occurenc of 1 trait
has nothing to do with another tait if on different chromosomes
23. Gene map is a diagram of where genes occur on chromosomes.
Solution
Answer:
1. Chromatin is unwound DNA found in the nucleus during Interphase.
2. Replicated chromosome is a chromosome that has doubled and is made of 2 chromatids.
3. Chromatid is a half of a replicated chromosome
4. Chromosome is a “package” of 1-5 thousand genes. Humans have 46 of these.
5. Mitosis is a process that divides the nucleus of a cell.
6. Interphase is the stage most cells are in during the cell cycle when growth, replication of
DNA, and preparation for division occurs.
7. Prophase is the phase in mitosis where the chromosomes become visible as the nuclear
membrane breaks down
8. Metaphase is the phase of mitosis where replicated chromosomes line up at the “equator” of
the cell.
9. Anaphase is the phase of mitosis where sister chromatids of the replicated chromosomes split
and head for the “pole”
10. Telophase is the last stage of mitosis when the nucleus reforms and cytokinesis occurs.
11. Cytokinesis is the last part of the Telophase wh.
7. Which statements about financial forecasting is most likely TRUE.pdf
7. Which statements about financial forecasting is most likely TRUE:
a. Net income flows through to cash flows from operations on the balance sheet
b. Net income flows to shareholders equity on the cash flow statement
c. Marketable securities is the same as cash on the balance sheet
d. Marketable securities is the same as short term debt on the balance sheet
e. Marketable securities is good to use as a plug figure when forecasting financial statements
Solution
e. Marketable securities is good to use as a plug figure when forecasting financial statements.
Thanks.
Inputs in the light reaction of photosynthesis Inputs in the li.pdf
Inputs in the light reaction of photosynthesis
Inputs in the light reaction of photosynthesis
Solution
Inputs in the light reaction of photosynthesis are
Light energy, ADP, Pi, NADP and H2O
2H2O + 2NADP+ + 3ADP + 3Pi O2 + 2NADPH + 3ATP
These reactions begin with the absorption of light energy by pigments, and end with the
production of stored chemical energy in the form of NADPH and ATP.
How do I obtain the solution Also, how would I convert the limit .pdf
How do I obtain the solution: ? Also, how would I convert the limit into a limit that I could apply
the L\'hopital rule to? Step-by-step solution would be appreciated.
Solution
e^x-1-x/(x*(e^x-1) apply the rule e^x-1/((e^x-1)+xe^x) apply the rule again
e^x/(e^x+e^x+xe^x) not putx=0 answer=1/2.
If both of Johns parents are heterozygous for sickle cell anemia a.pdf
If both of John\'s parents are heterozygous for sickle cell anemia and Sarah\'s parents are AA and
Aa what are the chances that John and Sarah\'s child will have Sickle Cell Anemia? (Sickle Cell
Anemia is an autosomal recessive disease)
Solution
Sickle cell disease is an autosomal recessive condition. Assume that the gene coding for normal
phenotype is A and the gene coding for sickle cell phenotype is a; A is dominant over a.
Given that, John\'s parents are heterozygous for sickle cell anemia.
Aa* Aa ----> AA (1/4), Aa (1/2), aa (1/4)
The probability of John to be a heterozygous carrier is 2/3. The probability of Sarah to be a
carrier is 1/2.
The probability that the Jogn and Sarah will have Sickle cell anemia is = 1/2*2/3*1/2*1/2 = 1/12.
I need help with ALL of the following bacteria -trypanosoma gam.pdf
I need help with ALL of the following bacteria:
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
I need help with ALL of the following bacteria:
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
Solution
Trypanosoma gambiense: it is a protozoan parasite.
Movement- with the help of flagellum.
Reproduction: By longitudinal binary fission.
Who is affected- the disease caused by trypanosoma is known as sleeping sickness. The parasites
are transmitted to humans by tsetse fly.
How they live- eating habits the blood, lymph and the intercellular spaces of different tissues and
organs of man Hogs,goats, cattle and sheep are the potential Reservoir hosts.
Class mastigophora and genus trypanosoma.
2. Radiolaria: also called radiozoa are protozoans
Movement: amoeboid movement
Reproduction: the reproduce asexually and probably sexually also Asexual fission, binary fission
and Multiple Fission and budding.
Whom they effect: they are found as you plant on throughout the ocean and their skeletal
remains make up a large part of the cover of the ocean floor as siliceous ooze.
The skeletons of ancient radiolarians are used in geological dating including Oil exploration and
determination of ancient climates.
Classes: polycystinea, acantharea, sticholonchea.
3. Euglena: it is single celled flagellate it is the best known and most widely studied member of
the class Euglenoidea. They are autotrophic in nature due to the presence of chloroplast. All
euglenoids how two flagella rooted in basal bodies. The longer flagellum helps the organism to
swim. Euglena reproduces asexually through binary fission. It belongs to the class euglenoidea
and genus euglena.
4. Paramoecium: it belongs to the class oligohymenophorea and genus paramoecium it is
commonly studied as a representative of the ciliate group there are widespread in freshwater
brackish and Marine environments they prepare themselves by whiplash moments of their cilia
which are arranged around the outside of their body. They reproduce asexually by binary fission,
they also follow conjugation which is a sexual phenomenon.
5. Vorticella: it belongs to the class oligihymenophorea and genus vorticella. It is a genus of
ciliates with 16 known species they are stalked inverted bell shaped ciliates, live mainly in
freshwater Ponds and streams they reproduce by budding where cell undergoes longitudinal
fission. Each cell has a separate sta.
I having trouble with this problem. Can someone out there help me J.pdf
I having trouble with this problem. Can someone out there help me? John Thildore, a retired
army officer, opened Thildore\'s Catering Service. As his accountant, analyze the transactions
listed and present them in proper form. a. The analysis of the transactions by using the expanded
accounting equation. b. A balance sheet showing the financial position of the firm before
opening on October 31, 201X. C. An income statement for the month of November. d. A
statement of owner\'s equity for November. e. A balance sheet as of November 30, 201X. 201X
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transaction (You should now prepare your balance sheet as of October 31, 201X.) Nov. 1
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Solution
a.
b.
c.
d.EventAssetsLiabilitiesOwners equityCash+Accounts receivable+Equipment=Accounts
payable+John capital-John withdrawl+Revenue-Salaries expense-Telephone bill-Rent expense-
Supplies expenseOct 25 25,000 25,000Oct 27 -1,200 1,200Oct 28
800 800Oct 29 -400 -400Total 23,400
- 2,000 400 25,000Nov 1
2,200 2,200Nov 5 -550 550Nov 8 250
250Nov 10 80 -80Nov 15 -50 50Nov
17 -120 120Nov 20 2,500 2,500Nov 25 1,300
1,300Nov 28 650 650Nov 30 -550
550Total 26,910 170 3,300 2,350
25,000 120 4,950 550 50
650 550.
ExplainAnswer these genetic questions please. A pure-breeding line .pdf
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trait is dominant? You self the pink F1 progeny and observe a 15:1 ratio of pink:white flowering
F2 progeny. Draw a potential biochemical pathway for production of pink pigment, and indicate
how the gene(s) responsible for producing pink pigment fit into this pathway.
Solution
(a) Let us denote the genotype of the pure breeding white flowering tulips as WW and the
genotype of the pure breeding pink flowering tulips as PP.
The F1 generation cross will give all progeny as: WWPP
It is given that all the F1 generation tulips are pink in color. Therefore, according to Mendel\'s
laws of inheritance, the dominant phenotype is physically expressed in Mendelian crosses, when
present in homozygous or heterozygous condition.
Therefore n the heterozygous condition of WP, pink (PP) is the dominant trait.
(b) Now that it is known that pink is the dominant trait, let us rename the genotypes as:
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white: pp
So F1 generation offsprings will have genotype: Pp
When F1 offsprings are self-crossed, the offsprings are:
Pp x Pp
F2 offsprings: PP, Pp, Pp, pp
According to Mendel\'s law, the F2 offspring phenotype ratio should have been 3:1. But, the
observed ratio is 15:1 (Pink flower: white flower). This anomaly is due to duplication of genes in
the case of tetraploid genetics..
Early version of UNIX were available only with a command-driven inte.pdf
Early version of UNIX were available only with a command-driven interface. In more recent
years, graphical user interfaces have become popular. Explain in your own words why these GUI
made an impact on the popularity of this operating system.
Solution
Graphical user interface is important for the novice users. Linux is mostly used by developers or
guys from computer science background. It is easier to interact with the computer with the help
of GUI. Most of the user\'s are not familer with the command line interface. Also, to interact
efficiently wee need to remember many commands in the unix. GUI is visually intuitivie hence
most of the new users prefer GUI or windows based operating system over linux operating
system. User can viewe, control, manipulate and toggle through multiple programs. Using mouse
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Environmental sciencePlease answer it ASAP 1. Why do water man.pdf
Environmental science
Please answer it ASAP 1. Why do water managers care about turbidity? How do water managers
estimate the bacterial disease risk associated with drinking water? 3. What is eutrophication? 4.
In a stratified lake (has fayers), which layer has the best drinking water? Bonus: What facility is
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Solution
Q1).The measurement of turbidity is a key test ofwater quality.
Fluids can contain suspended solid matter consisting of particles of many different sizes. While
some suspended material will be large enough and heavy enough to settle rapidly to the bottom
of the container if a liquid sample is left to stand (the settable solids), very small particles will
settle only very slowly or not at all if the sample is regularly agitated or the particles arecolloidal.
These small solid particles cause the liquid to appear turbid.
Q2.Water, the environment and people’s health • Part 4
53
MANAGEMENT ISSUES FOR DOMESTIC WATER SUPPLY.
What factors are important in the management of the microbial quality of domestic
water supplies?
The important factors are (i) disinfected drinking water, (ii) sound disposal of faecal wastes, (iii)
application of good personal, food and water use hygiene, (iv) use only of safe water for crops
eaten raw.
In the management of water quality for domestic purposes it is important to bear in mind that the
achievement of the goal of providing safe water for drinking and food preparation rests on the
management of the safety of the water supply along the whole supply chain. The water
supply chain consists of the water source, treatment works, distribution system, the supply tap,
the
drinking water container, jug or cup. (See vol., 5, Management Guide, DWAF, DOH & WRC,
2002).
Dirt that has gained access to the water supply line or tank (e.g. high-rise buildings) often causes
problems in the management of water for domestic purposes,
• In the case of the repair of a burst pipe where dirt may gain access to the supply line, this may
cause a temporary elevation of the microbiological total plate count in the supply line, but the
consumer is usually alerted to the problem by brown water coming out of the tap. The situation
is usually remedied by flushing the tap and allowing the plug of contaminated water to flow
down the drain.
• Rodents may gain access to gravity feed tanks on top of high-rise buildings which may not be
covered properly. This typically results in foul smelling and tasting water, and the consumer
is alerted to the problem. Such tanks will need flushing and proper screening of any access
points.
The other situation which needs attention is to ensure that containers that are used in the home
for storing drinking water are clean and that they are properly covered to ensure that flies and
dirt
do not settle into the water in the container. One of the quickest ways to contaminate water in a
bucket, for example, is to dip an unwas.
Discuss a specific public health concern for each of the following g.pdf
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b. African Americans
c. Asian Americans
d. Pacific Islanders
e. American Indians & Alaskan Natives
Solution
a). The prevalence of diabetes is found to be different in different groups of people. In United
states, the prevalence is more in case of Hispanic Americans that that of the White Americans or
black population. Also, African Americans have higher incidence than the white Americans.
b). African Americans are at higher risk of developing heart diseases than any other population,
about 3/4th of the population develop high blood pressure after 40 years of age. Their health
access is very less and do not visit a physician for regular checkup. Means they are less likely to
take any preventive medicines, which increases the risk of developing high blood pressure and
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Due to your interference with organisms on the planet, one species of flying insectivorous
organism that formerly lived near Lake Agassiz on Wallace have moved to a location just south
of Lake Lyell. Unfortunately, there is already a different species of flying insectivorous organism
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Solution
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competition could be observed between these two species. The interspecific competition if it is
drastic might lead to complete change in the population size of unfitted organism through natural
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competent. The adaptations such as flexibility of the newer species to change itself to newer
climatic conditions, acclimatization and learning helpful for the survival of a new species which
gets migrated to a newer habitat..
Complete the following chart, and then circle the layer that is found.pdf
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the cells of stratum corneum and stratum basale differ? ____ State the specific location of
melanin observed in dark skin ____ Which is darker, the anterior forearm or the posterior
forearm ____ Explain why ____ What special qualities, due to the presence of fibers, does the
connective tissue of the dermis have? ___How is the structure of the thick skin of the palms of
the hands and soles of the feet properly suited for its function? ____
Solution
1. ANS:
A. Epidermis
Location: outermost layer of the skin
Tissues: Epithelial tissue
Characteristics: It is a protective waterproof barrier that also keeps pathogens at bay and
regulates body temperature
B. Dermis:
Location: Middle layer present in beneath the epidermis of the skin
Tissues: Connective tissue
Characteristics: It is a thick layer of fibrous and elastic tissue (made mostly of collagen, elastin,
and fibrillin) that gives the skin its flexibility and strength.
C. Hypo dermis:
Location: Deeper subcutaneous tissue layer of the skin
Tissues: Fat and connective tissue
Characteristics: It maintains the body temperature.
2. ANS:
Stratum corneum and stratum basale difference:
The cells of stratum basale are cuboidal to columnar and the stratum corneum cells are made
with keratin.
3. ANS:
In the location of stratum basale, melanin content specifically observed in dark (heavily
pigmented) skin.
4. ANS:
The dorsal side (posterior) of the forearm is the \"top\" side-- the side that usually faces upward
and is generally darker than anterioir side. It is mainly due to melanin and hair follicles.
5. ANS:
The special qualities of dermis connective tissue can contains with Collagenous fibers, elastic
fibers, and a gel-like substance.
6. ANS:
The sole of feet and the palms of hands are structured to perform day to day functions. The
hardness on the skin on these locations is due to the layers of epidermis that contain dead
keratinocytes are the Stratum corneum and stratum lucidum
Stratum lucidum: Thin, clear layer present in the thick skin of the palms and soles; no remaining
living cells.
Stratum corneum: Several layers of keratinized corneum; a collagenous matrix composed of the
products of dead cells. Keratin and the thickened plasma membrane protect the skin against
abrasion and penetration, and the glycolipid between its cells nearly waterproofs this layer..
1a. What kinds of deductions are prohibited as a matter of public po.pdf
1a. What kinds of deductions are prohibited as a matter of public policy? bribes, kickbacks, and
\"other\" illegal payments as nondeductibleT.
1b.Why might Congress deem it important to disallow deductions for expenditures against public
policy?
Solution
The IRC lists bribes, kickbacks, and \"other\" illegal payments as nondeductibleCongress didn\'t
want the tax benefits associated with deductions to benefit or subsidize wrongdoing of course,
this rationale doesn\'t really explain the prohibition against deducting political contributions
which is probably better explained by the potential perception that political efforts are being
subsidized by taxpayers..
3. IQ tests are given to fifteen pairs of siblings. In each pair, one.pdf
3. IQ tests are given to fifteen pairs of siblings. In each pair, one sibling has phenylketonuria
(PKU, a metabolism disorder); the other doesn\'t. Is there any evidence that, in general, the PKU
cases have higher or lower IQs from their siblings? Perform a nonparametric test with sufficient
power, and justify your choice of test
Solution.
A rectangular board is 9 by 17 units. A string is connected to the f.pdf
A rectangular board is 9 by 17 units. A string is connected to the foci and pulled taut by a pencil
in order to draw the ellipse. Find the length of the string
Solution
length of major axis = 2a = 17
a = 17/2 = 8.5
length of minor axis = 2b = 9
b = 4.5
foci c^2 = a^2 - b^2
c^2 = 8.5^2 - 4.5^2 = 52
c = 7.21
length of the string = two times foci = 2*7.21 = 14.42 units.
You have 20 marbles and 20 rocks. Place them into two boxes to incre.pdf
You have 20 marbles and 20 rocks. Place them into two boxes to increase chances of getting the
marbles. Boxes are mixed up. Marbles and rocks have to be seperated for best chance to win the
marbles.
Solution
SOl)
20 rocks in 1st box
20 marbles on 2nd box
Total boxes=2
Probability to choose 20 marbles box = 1/2.
Write a method named convertInfixToPostfix that accepts an Infix exp.pdf
Write a method named convertInfixToPostfix that accepts an Infix expression as a parameter and
converts it to its Postfix equivalent by implementing the algorithm given in Figure 6.10 of the
required text book. Note that the Infix expression does not need to be a fully parenthesized
expression.
Your method should have the following signature and should be included in a class called
ExpressionConversion.
public String convertInfixToPostfix(String infixExp) { /* your implementation here*/ }
Solution
import java.util.Scanner; //for user input
public class ExpressionConversion
{
public static void main(String[] args) //main function
{
String infixExp,postfixExp;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter Infix Expression : \");
infixExp = in.next();
postfixExp = convertInfixToPostfix(infixExp);
System.out.println(\"Postfix Expression : \" + postfixExp);
}
public static String convertInfixToPostfix(String infixExp) //function to convert infix to
postfix
{
int len = infixExp.length();
Stack s = new Stack(len);
int i=0;
char op;
String postfixExp = \"\";
for(i=0; i < len; i++)
{
if(infixExp.charAt(i) == \'+\' || infixExp.charAt(i) == \'-\' || infixExp.charAt(i) == \'*\' ||
infixExp.charAt(i) == \'/\' || infixExp.charAt(i) == \'(\' || infixExp.charAt(i) == \')\')
{
switch(infixExp.charAt(i))
{
case \'(\' :
s.push(infixExp.charAt(i));
break;
case \')\' :
op = s.pop();
while(op != \'(\')
{
postfixExp = postfixExp + op;
op = s.pop();
}
break;
default:
while(preceed(s.peek()) >= preceed(infixExp.charAt(i)))
postfixExp = postfixExp + s.pop();
s.push(infixExp.charAt(i));
}
}
else
postfixExp = postfixExp + infixExp.charAt(i);
}
while(!s.isEmpty())
postfixExp = postfixExp + s.pop();
return postfixExp;
}
static int preceed(char st) //function to check the precedenceof operators
{
switch(st)
{
case \'*\' :
case \'/\' :
return 3;
case \'+\' :
case \'-\' :
return 2;
default :
return 1;
}
}
}
class Stack //creating stack to implement infix to postfix conversion
{
int maxSize;
char[] stackArray;
int top;
public Stack(int max)
{
maxSize = max;
stackArray = new char[maxSize];
top = -1;
}
public void push(char item)
{
stackArray[++top] = item;
}
public char pop()
{
return stackArray[top--];
}
public char peek()
{
if(!isEmpty())
return stackArray[top];
else
return 0;
}
public boolean isEmpty()
{
return (top == -1);
}
}.
X and Y are continuous random variables with joint p.d.f. f(x,y) = c.pdf
X and Y are continuous random variables with joint p.d.f. f(x,y) = c(1 + x + y)
A) find E(X + Y)
B) find Cov(X,Y)
I could use some helping finding the answer here. I managed to get one however my classmates
found a different answer
I forgot to mention but the steps would be extremelly helpful. Thanks
Solution.
What is meant by cultural emergent and how does this differ from oth.pdf
What is meant by cultural emergent and how does this differ from other approaches to culture?
Explain how culture can be both persistent and ever-changing.
Solution
Cultural emergent and how does this differ from other approaches to culture discussion of the
scope and limitations of current diagnostic practice criticisms from different quarters and the role
and relevance of culture in the diagnostic encounter precede the examination of advantages and
disadvantages of the approach the practical aspects include the acceptance of cultural
discordances as a component of an updated definition of mental disorder and the use of a
refurbished cultural formulation.
This is why the recognition of cultural diversity can help to ensure that ownership of
development and peace initiatives is vested in the populations concerned the message of
sustainable development is that the planet is essentially finite and that the resources humanity
hoped to discover in its environment must now be found within itself in its very diversity cultural
diversity invites us to think in terms of a plural humanity embodying a creative potential that
precludes any prescribed model of development.
The acknowledgement of cultural diversity and intercultural dialogue help to defuse the tensions
that can arise in multicultural societies when a majority and minorities confront each other over
recognition of their rights although outcome research is limited culturally responsive behavioral
health counseling results in greater counselor credibility.
An archaeological eye to an examination of human response to unpredictable environmental
conditions informed by an understanding of contemporary traditional peoples sustainable
Lifeways enhances our understanding of both the forces of conservatism and innovation which
may have been in play in major transitions in the past..
What are two main causes of recombination Errors in DNA replication.pdf
What are two main causes of recombination? Errors in DNA replication and damage to DNA
induced by chemicals or radiation Independent assortment of chromosomes and crossing over
between nonhomologous chromosomes during meiosis Independent assortment of chromosomes
and crossing over between homologous chromosomes during meiosis Fusion of male and female
gametes bearing different alleles and crossing over between homologous chromosomes during
mitosis Crossing over between sister chromatids during mitosis and crossing over between
homologous chromosomes during meiosis
Solution
The two main causes of recombination are:
3. Independent assortment of chromosomes and crossing over between homologous
chromosomes during meiosis
5. Crossing over between sister chromatids during mitosis and crossing over between
homologous chromosomes during meiosis.
Note: Mitotic recombination is primarily a result of DNA repair processes responding to
spontaneous or induced damages..
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7. Which statements about financial forecasting is most likely TRUE.pdf
7. Which statements about financial forecasting is most likely TRUE:
a. Net income flows through to cash flows from operations on the balance sheet
b. Net income flows to shareholders equity on the cash flow statement
c. Marketable securities is the same as cash on the balance sheet
d. Marketable securities is the same as short term debt on the balance sheet
e. Marketable securities is good to use as a plug figure when forecasting financial statements
Solution
e. Marketable securities is good to use as a plug figure when forecasting financial statements.
Thanks.
Inputs in the light reaction of photosynthesis Inputs in the li.pdf
Inputs in the light reaction of photosynthesis
Inputs in the light reaction of photosynthesis
Solution
Inputs in the light reaction of photosynthesis are
Light energy, ADP, Pi, NADP and H2O
2H2O + 2NADP+ + 3ADP + 3Pi O2 + 2NADPH + 3ATP
These reactions begin with the absorption of light energy by pigments, and end with the
production of stored chemical energy in the form of NADPH and ATP.
How do I obtain the solution Also, how would I convert the limit .pdf
How do I obtain the solution: ? Also, how would I convert the limit into a limit that I could apply
the L\'hopital rule to? Step-by-step solution would be appreciated.
Solution
e^x-1-x/(x*(e^x-1) apply the rule e^x-1/((e^x-1)+xe^x) apply the rule again
e^x/(e^x+e^x+xe^x) not putx=0 answer=1/2.
If both of Johns parents are heterozygous for sickle cell anemia a.pdf
If both of John\'s parents are heterozygous for sickle cell anemia and Sarah\'s parents are AA and
Aa what are the chances that John and Sarah\'s child will have Sickle Cell Anemia? (Sickle Cell
Anemia is an autosomal recessive disease)
Solution
Sickle cell disease is an autosomal recessive condition. Assume that the gene coding for normal
phenotype is A and the gene coding for sickle cell phenotype is a; A is dominant over a.
Given that, John\'s parents are heterozygous for sickle cell anemia.
Aa* Aa ----> AA (1/4), Aa (1/2), aa (1/4)
The probability of John to be a heterozygous carrier is 2/3. The probability of Sarah to be a
carrier is 1/2.
The probability that the Jogn and Sarah will have Sickle cell anemia is = 1/2*2/3*1/2*1/2 = 1/12.
I need help with ALL of the following bacteria -trypanosoma gam.pdf
I need help with ALL of the following bacteria:
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
I need help with ALL of the following bacteria:
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
-trypanosoma gambiense
-Radiolaria
-Euglena
-Paramecium
-Vorticella
I need to know about EACH BACTERIA :
-how they move
-how they reproduce
-what/who they effect
-how they live
-class/genus
**ALL general information **
Thank you!!
Solution
Trypanosoma gambiense: it is a protozoan parasite.
Movement- with the help of flagellum.
Reproduction: By longitudinal binary fission.
Who is affected- the disease caused by trypanosoma is known as sleeping sickness. The parasites
are transmitted to humans by tsetse fly.
How they live- eating habits the blood, lymph and the intercellular spaces of different tissues and
organs of man Hogs,goats, cattle and sheep are the potential Reservoir hosts.
Class mastigophora and genus trypanosoma.
2. Radiolaria: also called radiozoa are protozoans
Movement: amoeboid movement
Reproduction: the reproduce asexually and probably sexually also Asexual fission, binary fission
and Multiple Fission and budding.
Whom they effect: they are found as you plant on throughout the ocean and their skeletal
remains make up a large part of the cover of the ocean floor as siliceous ooze.
The skeletons of ancient radiolarians are used in geological dating including Oil exploration and
determination of ancient climates.
Classes: polycystinea, acantharea, sticholonchea.
3. Euglena: it is single celled flagellate it is the best known and most widely studied member of
the class Euglenoidea. They are autotrophic in nature due to the presence of chloroplast. All
euglenoids how two flagella rooted in basal bodies. The longer flagellum helps the organism to
swim. Euglena reproduces asexually through binary fission. It belongs to the class euglenoidea
and genus euglena.
4. Paramoecium: it belongs to the class oligohymenophorea and genus paramoecium it is
commonly studied as a representative of the ciliate group there are widespread in freshwater
brackish and Marine environments they prepare themselves by whiplash moments of their cilia
which are arranged around the outside of their body. They reproduce asexually by binary fission,
they also follow conjugation which is a sexual phenomenon.
5. Vorticella: it belongs to the class oligihymenophorea and genus vorticella. It is a genus of
ciliates with 16 known species they are stalked inverted bell shaped ciliates, live mainly in
freshwater Ponds and streams they reproduce by budding where cell undergoes longitudinal
fission. Each cell has a separate sta.
I having trouble with this problem. Can someone out there help me J.pdf
I having trouble with this problem. Can someone out there help me? John Thildore, a retired
army officer, opened Thildore\'s Catering Service. As his accountant, analyze the transactions
listed and present them in proper form. a. The analysis of the transactions by using the expanded
accounting equation. b. A balance sheet showing the financial position of the firm before
opening on October 31, 201X. C. An income statement for the month of November. d. A
statement of owner\'s equity for November. e. A balance sheet as of November 30, 201X. 201X
John Thildore invested $25,000 in the catering business from his personal savings account. Oct.
25 27 Bought equipment for cash from Small Co., $1,200. 28 Bought additional equipment on
account from Ryan Co., $800. 29 Paid $400 to Ryan Co. as partial payment of the Oct. 28
transaction (You should now prepare your balance sheet as of October 31, 201X.) Nov. 1
Catered a graduation and collected cash, $2,200. 5 Paid salaries of employees, $550. 8 Prepared
desserts for customers on account, $250. 10 Received $80 cash as partial payment of Nov. 8
transaction. 15 Paid telephone bill, $50 17 Paid his home electric bill from the company\'s
checkbook, $120. 20 Catered a wedding and received cash, $2,500 25 Bought additional
equipment on account, $1,300 28 Rent expense due but unpaid, $650. 30 Paid supplies expense,
$550.
Solution
a.
b.
c.
d.EventAssetsLiabilitiesOwners equityCash+Accounts receivable+Equipment=Accounts
payable+John capital-John withdrawl+Revenue-Salaries expense-Telephone bill-Rent expense-
Supplies expenseOct 25 25,000 25,000Oct 27 -1,200 1,200Oct 28
800 800Oct 29 -400 -400Total 23,400
- 2,000 400 25,000Nov 1
2,200 2,200Nov 5 -550 550Nov 8 250
250Nov 10 80 -80Nov 15 -50 50Nov
17 -120 120Nov 20 2,500 2,500Nov 25 1,300
1,300Nov 28 650 650Nov 30 -550
550Total 26,910 170 3,300 2,350
25,000 120 4,950 550 50
650 550.
ExplainAnswer these genetic questions please. A pure-breeding line .pdf
Explain/Answer these genetic questions please. A pure-breeding line of white flowering tulips is
crossed with a pure-breeding line of pink tulips. The F1 progeny all have pink flowers, Which
trait is dominant? You self the pink F1 progeny and observe a 15:1 ratio of pink:white flowering
F2 progeny. Draw a potential biochemical pathway for production of pink pigment, and indicate
how the gene(s) responsible for producing pink pigment fit into this pathway.
Solution
(a) Let us denote the genotype of the pure breeding white flowering tulips as WW and the
genotype of the pure breeding pink flowering tulips as PP.
The F1 generation cross will give all progeny as: WWPP
It is given that all the F1 generation tulips are pink in color. Therefore, according to Mendel\'s
laws of inheritance, the dominant phenotype is physically expressed in Mendelian crosses, when
present in homozygous or heterozygous condition.
Therefore n the heterozygous condition of WP, pink (PP) is the dominant trait.
(b) Now that it is known that pink is the dominant trait, let us rename the genotypes as:
Pink: PP
white: pp
So F1 generation offsprings will have genotype: Pp
When F1 offsprings are self-crossed, the offsprings are:
Pp x Pp
F2 offsprings: PP, Pp, Pp, pp
According to Mendel\'s law, the F2 offspring phenotype ratio should have been 3:1. But, the
observed ratio is 15:1 (Pink flower: white flower). This anomaly is due to duplication of genes in
the case of tetraploid genetics..
Early version of UNIX were available only with a command-driven inte.pdf
Early version of UNIX were available only with a command-driven interface. In more recent
years, graphical user interfaces have become popular. Explain in your own words why these GUI
made an impact on the popularity of this operating system.
Solution
Graphical user interface is important for the novice users. Linux is mostly used by developers or
guys from computer science background. It is easier to interact with the computer with the help
of GUI. Most of the user\'s are not familer with the command line interface. Also, to interact
efficiently wee need to remember many commands in the unix. GUI is visually intuitivie hence
most of the new users prefer GUI or windows based operating system over linux operating
system. User can viewe, control, manipulate and toggle through multiple programs. Using mouse
and keyboard it is much easier for novice user to interact with the opearting system..
Environmental sciencePlease answer it ASAP 1. Why do water man.pdf
Environmental science
Please answer it ASAP 1. Why do water managers care about turbidity? How do water managers
estimate the bacterial disease risk associated with drinking water? 3. What is eutrophication? 4.
In a stratified lake (has fayers), which layer has the best drinking water? Bonus: What facility is
the largest nonmilitary user of explosives in the world? It is located in Florida.
Solution
Q1).The measurement of turbidity is a key test ofwater quality.
Fluids can contain suspended solid matter consisting of particles of many different sizes. While
some suspended material will be large enough and heavy enough to settle rapidly to the bottom
of the container if a liquid sample is left to stand (the settable solids), very small particles will
settle only very slowly or not at all if the sample is regularly agitated or the particles arecolloidal.
These small solid particles cause the liquid to appear turbid.
Q2.Water, the environment and people’s health • Part 4
53
MANAGEMENT ISSUES FOR DOMESTIC WATER SUPPLY.
What factors are important in the management of the microbial quality of domestic
water supplies?
The important factors are (i) disinfected drinking water, (ii) sound disposal of faecal wastes, (iii)
application of good personal, food and water use hygiene, (iv) use only of safe water for crops
eaten raw.
In the management of water quality for domestic purposes it is important to bear in mind that the
achievement of the goal of providing safe water for drinking and food preparation rests on the
management of the safety of the water supply along the whole supply chain. The water
supply chain consists of the water source, treatment works, distribution system, the supply tap,
the
drinking water container, jug or cup. (See vol., 5, Management Guide, DWAF, DOH & WRC,
2002).
Dirt that has gained access to the water supply line or tank (e.g. high-rise buildings) often causes
problems in the management of water for domestic purposes,
• In the case of the repair of a burst pipe where dirt may gain access to the supply line, this may
cause a temporary elevation of the microbiological total plate count in the supply line, but the
consumer is usually alerted to the problem by brown water coming out of the tap. The situation
is usually remedied by flushing the tap and allowing the plug of contaminated water to flow
down the drain.
• Rodents may gain access to gravity feed tanks on top of high-rise buildings which may not be
covered properly. This typically results in foul smelling and tasting water, and the consumer
is alerted to the problem. Such tanks will need flushing and proper screening of any access
points.
The other situation which needs attention is to ensure that containers that are used in the home
for storing drinking water are clean and that they are properly covered to ensure that flies and
dirt
do not settle into the water in the container. One of the quickest ways to contaminate water in a
bucket, for example, is to dip an unwas.
Discuss a specific public health concern for each of the following g.pdf
Discuss a specific public health concern for each of the following groups:
a. Americans of Hispanic Origin
b. African Americans
c. Asian Americans
d. Pacific Islanders
e. American Indians & Alaskan Natives
Solution
a). The prevalence of diabetes is found to be different in different groups of people. In United
states, the prevalence is more in case of Hispanic Americans that that of the White Americans or
black population. Also, African Americans have higher incidence than the white Americans.
b). African Americans are at higher risk of developing heart diseases than any other population,
about 3/4th of the population develop high blood pressure after 40 years of age. Their health
access is very less and do not visit a physician for regular checkup. Means they are less likely to
take any preventive medicines, which increases the risk of developing high blood pressure and
lethal heart problems.
c). As per the health statistics, Asian Americans are less prone to diabetes and hypertension.
d). Pacific Islanders are at higher risk of developing type 1 diabetes..
Due to your interference with organisms on the planet, one species o.pdf
Due to your interference with organisms on the planet, one species of flying insectivorous
organism that formerly lived near Lake Agassiz on Wallace have moved to a location just south
of Lake Lyell. Unfortunately, there is already a different species of flying insectivorous organism
very similar to the first organism living near Lake Lyell. Describe in detail what things (multiple
things)might happen as these two species interact. and what methods they can adopt to avoid
extinction with example.
Solution
The migration of flying insectivorous organism from Lake Agassiz on Wallace to Lake Lyell.
This leads to a competition between two organisms for their sustainability. Interspecific
competition could be observed between these two species. The interspecific competition if it is
drastic might lead to complete change in the population size of unfitted organism through natural
selection. The species which has been gets suppressed due to over - aggressive species gets
completely excluded or the unfitted species gets change itself its natural characteristics if it is
competent. The adaptations such as flexibility of the newer species to change itself to newer
climatic conditions, acclimatization and learning helpful for the survival of a new species which
gets migrated to a newer habitat..
Complete the following chart, and then circle the layer that is found.pdf
Complete the following chart, and then circle the layer that is found beneath the skin. How do
the cells of stratum corneum and stratum basale differ? ____ State the specific location of
melanin observed in dark skin ____ Which is darker, the anterior forearm or the posterior
forearm ____ Explain why ____ What special qualities, due to the presence of fibers, does the
connective tissue of the dermis have? ___How is the structure of the thick skin of the palms of
the hands and soles of the feet properly suited for its function? ____
Solution
1. ANS:
A. Epidermis
Location: outermost layer of the skin
Tissues: Epithelial tissue
Characteristics: It is a protective waterproof barrier that also keeps pathogens at bay and
regulates body temperature
B. Dermis:
Location: Middle layer present in beneath the epidermis of the skin
Tissues: Connective tissue
Characteristics: It is a thick layer of fibrous and elastic tissue (made mostly of collagen, elastin,
and fibrillin) that gives the skin its flexibility and strength.
C. Hypo dermis:
Location: Deeper subcutaneous tissue layer of the skin
Tissues: Fat and connective tissue
Characteristics: It maintains the body temperature.
2. ANS:
Stratum corneum and stratum basale difference:
The cells of stratum basale are cuboidal to columnar and the stratum corneum cells are made
with keratin.
3. ANS:
In the location of stratum basale, melanin content specifically observed in dark (heavily
pigmented) skin.
4. ANS:
The dorsal side (posterior) of the forearm is the \"top\" side-- the side that usually faces upward
and is generally darker than anterioir side. It is mainly due to melanin and hair follicles.
5. ANS:
The special qualities of dermis connective tissue can contains with Collagenous fibers, elastic
fibers, and a gel-like substance.
6. ANS:
The sole of feet and the palms of hands are structured to perform day to day functions. The
hardness on the skin on these locations is due to the layers of epidermis that contain dead
keratinocytes are the Stratum corneum and stratum lucidum
Stratum lucidum: Thin, clear layer present in the thick skin of the palms and soles; no remaining
living cells.
Stratum corneum: Several layers of keratinized corneum; a collagenous matrix composed of the
products of dead cells. Keratin and the thickened plasma membrane protect the skin against
abrasion and penetration, and the glycolipid between its cells nearly waterproofs this layer..
1a. What kinds of deductions are prohibited as a matter of public po.pdf
1a. What kinds of deductions are prohibited as a matter of public policy? bribes, kickbacks, and
\"other\" illegal payments as nondeductibleT.
1b.Why might Congress deem it important to disallow deductions for expenditures against public
policy?
Solution
The IRC lists bribes, kickbacks, and \"other\" illegal payments as nondeductibleCongress didn\'t
want the tax benefits associated with deductions to benefit or subsidize wrongdoing of course,
this rationale doesn\'t really explain the prohibition against deducting political contributions
which is probably better explained by the potential perception that political efforts are being
subsidized by taxpayers..
3. IQ tests are given to fifteen pairs of siblings. In each pair, one.pdf
3. IQ tests are given to fifteen pairs of siblings. In each pair, one sibling has phenylketonuria
(PKU, a metabolism disorder); the other doesn\'t. Is there any evidence that, in general, the PKU
cases have higher or lower IQs from their siblings? Perform a nonparametric test with sufficient
power, and justify your choice of test
Solution.
A rectangular board is 9 by 17 units. A string is connected to the f.pdf
A rectangular board is 9 by 17 units. A string is connected to the foci and pulled taut by a pencil
in order to draw the ellipse. Find the length of the string
Solution
length of major axis = 2a = 17
a = 17/2 = 8.5
length of minor axis = 2b = 9
b = 4.5
foci c^2 = a^2 - b^2
c^2 = 8.5^2 - 4.5^2 = 52
c = 7.21
length of the string = two times foci = 2*7.21 = 14.42 units.
You have 20 marbles and 20 rocks. Place them into two boxes to incre.pdf
You have 20 marbles and 20 rocks. Place them into two boxes to increase chances of getting the
marbles. Boxes are mixed up. Marbles and rocks have to be seperated for best chance to win the
marbles.
Solution
SOl)
20 rocks in 1st box
20 marbles on 2nd box
Total boxes=2
Probability to choose 20 marbles box = 1/2.
Write a method named convertInfixToPostfix that accepts an Infix exp.pdf
Write a method named convertInfixToPostfix that accepts an Infix expression as a parameter and
converts it to its Postfix equivalent by implementing the algorithm given in Figure 6.10 of the
required text book. Note that the Infix expression does not need to be a fully parenthesized
expression.
Your method should have the following signature and should be included in a class called
ExpressionConversion.
public String convertInfixToPostfix(String infixExp) { /* your implementation here*/ }
Solution
import java.util.Scanner; //for user input
public class ExpressionConversion
{
public static void main(String[] args) //main function
{
String infixExp,postfixExp;
Scanner in = new Scanner(System.in);
System.out.println(\"Enter Infix Expression : \");
infixExp = in.next();
postfixExp = convertInfixToPostfix(infixExp);
System.out.println(\"Postfix Expression : \" + postfixExp);
}
public static String convertInfixToPostfix(String infixExp) //function to convert infix to
postfix
{
int len = infixExp.length();
Stack s = new Stack(len);
int i=0;
char op;
String postfixExp = \"\";
for(i=0; i < len; i++)
{
if(infixExp.charAt(i) == \'+\' || infixExp.charAt(i) == \'-\' || infixExp.charAt(i) == \'*\' ||
infixExp.charAt(i) == \'/\' || infixExp.charAt(i) == \'(\' || infixExp.charAt(i) == \')\')
{
switch(infixExp.charAt(i))
{
case \'(\' :
s.push(infixExp.charAt(i));
break;
case \')\' :
op = s.pop();
while(op != \'(\')
{
postfixExp = postfixExp + op;
op = s.pop();
}
break;
default:
while(preceed(s.peek()) >= preceed(infixExp.charAt(i)))
postfixExp = postfixExp + s.pop();
s.push(infixExp.charAt(i));
}
}
else
postfixExp = postfixExp + infixExp.charAt(i);
}
while(!s.isEmpty())
postfixExp = postfixExp + s.pop();
return postfixExp;
}
static int preceed(char st) //function to check the precedenceof operators
{
switch(st)
{
case \'*\' :
case \'/\' :
return 3;
case \'+\' :
case \'-\' :
return 2;
default :
return 1;
}
}
}
class Stack //creating stack to implement infix to postfix conversion
{
int maxSize;
char[] stackArray;
int top;
public Stack(int max)
{
maxSize = max;
stackArray = new char[maxSize];
top = -1;
}
public void push(char item)
{
stackArray[++top] = item;
}
public char pop()
{
return stackArray[top--];
}
public char peek()
{
if(!isEmpty())
return stackArray[top];
else
return 0;
}
public boolean isEmpty()
{
return (top == -1);
}
}.
X and Y are continuous random variables with joint p.d.f. f(x,y) = c.pdf
X and Y are continuous random variables with joint p.d.f. f(x,y) = c(1 + x + y)
A) find E(X + Y)
B) find Cov(X,Y)
I could use some helping finding the answer here. I managed to get one however my classmates
found a different answer
I forgot to mention but the steps would be extremelly helpful. Thanks
Solution.
What is meant by cultural emergent and how does this differ from oth.pdf
What is meant by cultural emergent and how does this differ from other approaches to culture?
Explain how culture can be both persistent and ever-changing.
Solution
Cultural emergent and how does this differ from other approaches to culture discussion of the
scope and limitations of current diagnostic practice criticisms from different quarters and the role
and relevance of culture in the diagnostic encounter precede the examination of advantages and
disadvantages of the approach the practical aspects include the acceptance of cultural
discordances as a component of an updated definition of mental disorder and the use of a
refurbished cultural formulation.
This is why the recognition of cultural diversity can help to ensure that ownership of
development and peace initiatives is vested in the populations concerned the message of
sustainable development is that the planet is essentially finite and that the resources humanity
hoped to discover in its environment must now be found within itself in its very diversity cultural
diversity invites us to think in terms of a plural humanity embodying a creative potential that
precludes any prescribed model of development.
The acknowledgement of cultural diversity and intercultural dialogue help to defuse the tensions
that can arise in multicultural societies when a majority and minorities confront each other over
recognition of their rights although outcome research is limited culturally responsive behavioral
health counseling results in greater counselor credibility.
An archaeological eye to an examination of human response to unpredictable environmental
conditions informed by an understanding of contemporary traditional peoples sustainable
Lifeways enhances our understanding of both the forces of conservatism and innovation which
may have been in play in major transitions in the past..
What are two main causes of recombination Errors in DNA replication.pdf
What are two main causes of recombination? Errors in DNA replication and damage to DNA
induced by chemicals or radiation Independent assortment of chromosomes and crossing over
between nonhomologous chromosomes during meiosis Independent assortment of chromosomes
and crossing over between homologous chromosomes during meiosis Fusion of male and female
gametes bearing different alleles and crossing over between homologous chromosomes during
mitosis Crossing over between sister chromatids during mitosis and crossing over between
homologous chromosomes during meiosis
Solution
The two main causes of recombination are:
3. Independent assortment of chromosomes and crossing over between homologous
chromosomes during meiosis
5. Crossing over between sister chromatids during mitosis and crossing over between
homologous chromosomes during meiosis.
Note: Mitotic recombination is primarily a result of DNA repair processes responding to
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