the conducting tissue found in vascular plants that functions in tra.pdf
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the conducting tissue found in vascular plants that functions in transportation of materials throughout the plant, but does not lend structure support to the stem is A. xylem B. phloem C.sporopollenin D.endosperm e. LYCOPHYLL Solution Answer : B. Phloem Phloem is responsible for the transporting sugars,proteins, and other organic molecules in plants. xylem is responsible for the transport of water and water soluble nutrients,taken up by the roots and also primary component of wood. Vascular plants are able to higher than other plants due to the rigidity of xylem cells, which support the plant..
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Are hydrophobic interactions weak or strongSolutionHydrophobi.pdf
Are hydrophobic interactions weak or strong?
Solution
Hydrophobic interactions describe the relations between water and hydrophobes (Low-water
soluble molecule). Hydrophobes are non-polar molecules and usually have a long chain of
carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
Vander Waals forces and hydrogen bonds). The strength of hydrophobic interactions depends on
several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
increases also. However, at extreme temperatures, hydrophobic interactions denature.
2. Number of carbons on hydrophobes: Molecules with greatest number of carbon atoms will
have the strongest hydrophobic interactions.
3. The shape of hydrophobes: Aliphatic organic molecules will have stronger interactions than
aromatic organic molecules..
What is another word for account based forecast What is anothe.pdf
What is another word for account based forecast?
What is another word for account based forecast?
Solution
The other word for account based forecasting is Budgeting.
A Budget is a quantitative statement in a functional area forecasting the expected values. For
example, a cash budget estimates various cash inflows and cash outflows that can arise during
the controllable period. Thus accounting based forecast uses budgets as the means for estimating
the variables, so it is also called as budgeting..
What function can be used to return multiple numeric data such as in.pdf
What function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program?
A) Input
B)Eval
C)Output
D)return
Solution
Answer: Input
Input function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program. Input function will get the inputs from keyboard and seent it to
program for execution..
Answer the following questions in reference to Drosophila melanogast.pdf
Answer the following questions in reference to Drosophila melanogaster (fruit fly) that has 6
chromosomes.
How many combinations are possible when chromosomes assort independently into gametes
(ignoring crossing over)?
6 raised to the 2nd power
2 raised to the 6th power
6 x 2 = 12
2 raised to the 3rd power
3 raised to the 2nd power
How many tetrads align at the metaphase plate of meiosis I?
0
12
6
3
Solution
Ans.1 No. of possible combination during independent assortment =
2n
Where, 2 indicates ploidy of the somatic cells undergoing meiosis,
n = number of haploid chromosomes.
Given, 2n = 6
Hence, n = 3.
Thus,
No. of possible combination during independent assortment = 2n = 23
Correct option: 2 raised to 3rd power.
Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n)
To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has
two copies of each chromosome- one inherited from mother and the other from father. So,
diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal
chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each
other, randomly scattered in nucleus. We will proceed with one chromosome pair, for simplicity.
The following events occur when a diploid cell enters meiosis-
A1 and A2 are called homologues pair. During S-phase, DNA is duplicated.
The resultant-
I. A1A1 remain joined to each other. Are called sister chromatids.
The same happens to A2A2, and rest other chromosomes.
II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are
duplicated, they are (A1A1) and (A2A2) respectively.
III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together.
Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together
as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and
A2 from father remain together as one single homologous chromosome).
Thus, number of treads in drosophila = n = 3
Correct option: D (3).
There are five feasible solutions to a three-objective optimization p.pdf
There are five feasible solutions to a three-objective optimization problem (solns A - E shown
below with their objective function values). Indicate whether each solution is dominated or non-
dominated. If a solution is dominated, indicate which other solution(s) dominate(s) it.
Solution
soln Dom or non-dom Dominated by
A Dominated E
B Non-dominated
C Non-dominated
D Dominated B
E Non-dominated.
An air-filled parallel-plate capacitor has capacitance Co. If two ide.pdf
An air-filled parallel-plate capacitor has capacitance Co. If two identically sized dielectric slabs
of dielectric constants K_1 and K_2 are inserted as shown in the following figure, what is the
new capacitance? (Hnt: treat this as two capacitor in combinations)
Solution
it would act as capacitors in parallel.
C= K0A/d.
Non C1= A/(d/2)*K1=2C/k1
Simlarly C2= 2C/K2
Since it can be treated as capacitor is parallel so Cfinal= 2C/K1 + 2c/k2 = 2C(K1+K2/K1*K2).
The incandescent lightbulb was actually invented by Humphry Davy in 1.pdf
The incandescent lightbulb was actually invented by Humphry Davy in 1802. It was not
commercializable in that form. Joseph Swan was successful in making a commercializable unit
in the late 1870s. He may have been granted the first light bulb patent (British patent 4933,
issued in 1880). Thomas Edison was also developing electric lighting at that time. He applied for
US patents as early as 1878 So why is Edison credited as the inventor? Partly because he was the
first to design and manufacture a robust lighting system having a bulb with an effective
incandescent material, higher vacuum inside the bulb, and high resistance so power distribution
from a central source was possible. Perhaps more importantly was the development ofsupply
sources and materials for distribution networks. A modem incandescent lightbulb is shown
below. Overall efficiency, defined as light power out divided by electrical power in, is about
2.5% for a 100 W unit. 1 1. outline of glass bulb 2. Low pressure inert gas (argon, nitrogen,
krypton, xenon) 3. Tun ten filament 5) 4. Contact wire (goes out of stem) 5. Contact wire (goes
into stem) 6. Support wires (one end embedded in stem; conduct no current) 7. Stem (glass
mount) 8. Contact wire (goes out of Identify five different systems that could be used for an
energy analysis (some combination Draw arrows for each systems\' energy flow process(es)-W
Qooeduction, Qconvection, radiation.
Solution
To conduction and convection of heat must have the medium.
in vaccum only radiation is there.
given problem the heat coming out from the tungsten filament(3).
the work supplied by the electrical current through the wires.
first humphry davy invented incandescent bulb having vaccum inside so there is no possibilty of
heat transfer through conduction or convection.
next edison filled with the low pressure inert gases. beacuse of that from hot Tungsten filamet to
inert gases heat transfer occurs through convection.
next inertgasses to glass or stem heated by the convection (fluid and solid interaction)
again glass to sourroundings is convection(solid and fluid interaction)
radiation is always there beacuse it possible with medium and with out medium
contact wires are heated up through the conduction..
The fossil record provides little information about ancient mosses. D.pdf
The fossil record provides little information about ancient mosses. Do you think nonvascular
plants could have ever been as large as trees or shrubs? Provide evidence based on your
knowledge of nonvascular plant anatomy.
Solution
2). No, the nonvascular plants could have never been as large as trees or shrubs because they
cannot provide the water and nutrients to the plant parts that are away from the source of water.
The nonvascular plants cannot control their water supply internally, it is solely depends upon
their external environment that provides water to the plant. So, they most likely grown over a
lake or rocks to get the water and propagated vegetative (clonal moss)..
Recommended
Are hydrophobic interactions weak or strongSolutionHydrophobi.pdf
Are hydrophobic interactions weak or strong?
Solution
Hydrophobic interactions describe the relations between water and hydrophobes (Low-water
soluble molecule). Hydrophobes are non-polar molecules and usually have a long chain of
carbon that do not interact with water molecules.
Strength of hydrophobic bonds are relatively stronger than other weak intermolecular forces (i.e.,
Vander Waals forces and hydrogen bonds). The strength of hydrophobic interactions depends on
several factors like:
1. Temperature: As the temperature increases, the strength of hydrophobic interaction also
increases also. However, at extreme temperatures, hydrophobic interactions denature.
2. Number of carbons on hydrophobes: Molecules with greatest number of carbon atoms will
have the strongest hydrophobic interactions.
3. The shape of hydrophobes: Aliphatic organic molecules will have stronger interactions than
aromatic organic molecules..
What is another word for account based forecast What is anothe.pdf
What is another word for account based forecast?
What is another word for account based forecast?
Solution
The other word for account based forecasting is Budgeting.
A Budget is a quantitative statement in a functional area forecasting the expected values. For
example, a cash budget estimates various cash inflows and cash outflows that can arise during
the controllable period. Thus accounting based forecast uses budgets as the means for estimating
the variables, so it is also called as budgeting..
What function can be used to return multiple numeric data such as in.pdf
What function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program?
A) Input
B)Eval
C)Output
D)return
Solution
Answer: Input
Input function can be used to return multiple numeric data such as integer and/or float entered at
the keyboard to the program. Input function will get the inputs from keyboard and seent it to
program for execution..
Answer the following questions in reference to Drosophila melanogast.pdf
Answer the following questions in reference to Drosophila melanogaster (fruit fly) that has 6
chromosomes.
How many combinations are possible when chromosomes assort independently into gametes
(ignoring crossing over)?
6 raised to the 2nd power
2 raised to the 6th power
6 x 2 = 12
2 raised to the 3rd power
3 raised to the 2nd power
How many tetrads align at the metaphase plate of meiosis I?
0
12
6
3
Solution
Ans.1 No. of possible combination during independent assortment =
2n
Where, 2 indicates ploidy of the somatic cells undergoing meiosis,
n = number of haploid chromosomes.
Given, 2n = 6
Hence, n = 3.
Thus,
No. of possible combination during independent assortment = 2n = 23
Correct option: 2 raised to 3rd power.
Ans. 2. Number of tetrad during meiosis = Number of haploid chromosomes (n)
To understand it, let’s number the three haploid chromosomes as A, B and C. A diploid cell has
two copies of each chromosome- one inherited from mother and the other from father. So,
diploid cell has following chromosomes- A1, A2, B1, B2, C1 and C3, where 1= maternal
chromosomes and 2 = fraternal chromosomes. All the chromosomes remain independent of each
other, randomly scattered in nucleus. We will proceed with one chromosome pair, for simplicity.
The following events occur when a diploid cell enters meiosis-
A1 and A2 are called homologues pair. During S-phase, DNA is duplicated.
The resultant-
I. A1A1 remain joined to each other. Are called sister chromatids.
The same happens to A2A2, and rest other chromosomes.
II. Homologous chromosome A1 and A2 pair together prophase I. Since each of A1 and A2 are
duplicated, they are (A1A1) and (A2A2) respectively.
III. Homologous pairing forms a tetrad – A1A1/A2A2 i.e. four chromosomes remaining together.
Note: Both the maternal and fraternal chromosomes and their duplicated copies remain together
as a tetrad. That is, the number of tetrad in half of diploid number (because A1 from mother and
A2 from father remain together as one single homologous chromosome).
Thus, number of treads in drosophila = n = 3
Correct option: D (3).
There are five feasible solutions to a three-objective optimization p.pdf
There are five feasible solutions to a three-objective optimization problem (solns A - E shown
below with their objective function values). Indicate whether each solution is dominated or non-
dominated. If a solution is dominated, indicate which other solution(s) dominate(s) it.
Solution
soln Dom or non-dom Dominated by
A Dominated E
B Non-dominated
C Non-dominated
D Dominated B
E Non-dominated.
An air-filled parallel-plate capacitor has capacitance Co. If two ide.pdf
An air-filled parallel-plate capacitor has capacitance Co. If two identically sized dielectric slabs
of dielectric constants K_1 and K_2 are inserted as shown in the following figure, what is the
new capacitance? (Hnt: treat this as two capacitor in combinations)
Solution
it would act as capacitors in parallel.
C= K0A/d.
Non C1= A/(d/2)*K1=2C/k1
Simlarly C2= 2C/K2
Since it can be treated as capacitor is parallel so Cfinal= 2C/K1 + 2c/k2 = 2C(K1+K2/K1*K2).
The incandescent lightbulb was actually invented by Humphry Davy in 1.pdf
The incandescent lightbulb was actually invented by Humphry Davy in 1802. It was not
commercializable in that form. Joseph Swan was successful in making a commercializable unit
in the late 1870s. He may have been granted the first light bulb patent (British patent 4933,
issued in 1880). Thomas Edison was also developing electric lighting at that time. He applied for
US patents as early as 1878 So why is Edison credited as the inventor? Partly because he was the
first to design and manufacture a robust lighting system having a bulb with an effective
incandescent material, higher vacuum inside the bulb, and high resistance so power distribution
from a central source was possible. Perhaps more importantly was the development ofsupply
sources and materials for distribution networks. A modem incandescent lightbulb is shown
below. Overall efficiency, defined as light power out divided by electrical power in, is about
2.5% for a 100 W unit. 1 1. outline of glass bulb 2. Low pressure inert gas (argon, nitrogen,
krypton, xenon) 3. Tun ten filament 5) 4. Contact wire (goes out of stem) 5. Contact wire (goes
into stem) 6. Support wires (one end embedded in stem; conduct no current) 7. Stem (glass
mount) 8. Contact wire (goes out of Identify five different systems that could be used for an
energy analysis (some combination Draw arrows for each systems\' energy flow process(es)-W
Qooeduction, Qconvection, radiation.
Solution
To conduction and convection of heat must have the medium.
in vaccum only radiation is there.
given problem the heat coming out from the tungsten filament(3).
the work supplied by the electrical current through the wires.
first humphry davy invented incandescent bulb having vaccum inside so there is no possibilty of
heat transfer through conduction or convection.
next edison filled with the low pressure inert gases. beacuse of that from hot Tungsten filamet to
inert gases heat transfer occurs through convection.
next inertgasses to glass or stem heated by the convection (fluid and solid interaction)
again glass to sourroundings is convection(solid and fluid interaction)
radiation is always there beacuse it possible with medium and with out medium
contact wires are heated up through the conduction..
The fossil record provides little information about ancient mosses. D.pdf
The fossil record provides little information about ancient mosses. Do you think nonvascular
plants could have ever been as large as trees or shrubs? Provide evidence based on your
knowledge of nonvascular plant anatomy.
Solution
2). No, the nonvascular plants could have never been as large as trees or shrubs because they
cannot provide the water and nutrients to the plant parts that are away from the source of water.
The nonvascular plants cannot control their water supply internally, it is solely depends upon
their external environment that provides water to the plant. So, they most likely grown over a
lake or rocks to get the water and propagated vegetative (clonal moss)..
The DNA sequence below is a shortened version of the sequence on the .pdf
The DNA sequence below is a shortened version of the sequence on the previous page. This
DNA sequence corresponds to the mRNA sequence only; two introns are not included. The
numbers indicate the position of nucleotides within the sequence and indicates intervening
nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are
very long DNA molecules with millions of nucleotides, so this sequence is preceded and
followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the
sequence of the first 10 nucleotides of the complementary strand of DNA.
Solution
ans. DNA is a double stranded which consist of two strands and one strand is complemetary to
the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two
hydrogen bonds guanine to cytosine by three hydrogen bonds
The seque of complementary strand of the DNA molecule is
5\' TGTAAACGAA.
Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf
Show that f(x)=5/7^x+1 and f^-1(x)=7x-7/5 are inverses Choose the appropriate equality that
will show that f(x) and f^-1(x) are inverses (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
(f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
Solution
Replace x by f^-1(x) in f(x)
=> 5/7(7x-7)/5+1
=> (7x - 7)/7 + 1 = > x - 1 + 1 => x
=> f(f^-1(x)) = x.
Read the short article Horizontal Gene Transfer in E Coli and answer.pdf
Read the short article Horizontal Gene Transfer in E Coli and answer the following
questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to
use your own words; plagiarism of any kind will not be tolerated (see syllabus for further
information on what constitutes plagiarism).
(1) What is horizontal gene transfer (HGT)?
(2) In the posted article, how many gene transfer elements were found for E. coli O104:H4?
(3) What is the source of these gene transfer elements?
(4) Should we be concerned about HGT in prokaryotes? If so, why?
Solution
(1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of
genetic material takes place between unicellular or multicellular organisms across species. This
gene transfer is different from the vertical gene transfer in which the transmission of DNA
occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same
as well as different species, through this means by aquiring one of the processes; transformation,
transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that
transfer genes.
(2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu,
identified 38 gene transfer elements for E. coli O104:H4.
(3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They
showed that the these elements act as genetic weapons protecting the bacteria from antibiotics.
The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the
process bacteria aquired these elements from viruses.
(4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through
HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT,
genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to
another species of bacteria, and is the primary reason for the spread of antibiotic resistance.
Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds
(such as different insecticides and pesticides). Thus the bacterial strains after HGT are much
more dangerous as they are multi drug resistant and more virulent. These transfer elements when
transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to
antibiotics available with us and thus cannot be trated easily..
Please select the best answer and click submit. The owners of fou.pdf
Please select the best answer and click \"submit.\" The owners of four companies competing for
a contract are shown in the table below. If a report is released that advocates company A. which
of the people having funded the report should result in the most skepticism?
Solution
Option d.
Debby Smith would result in the most skepticism because his company is in the second place in
the list..
One function of the spliceosome is to (Select all that apply.)A. .pdf
One function of the spliceosome is to: (Select all that apply.)
A. recognize the beginning and end of the introns.
B. recognize the 5\' end of the primary transcript.
C recognize the beginning and end of the exons.
D recognize the 3\' end of the primary transcript.
Solution
The correct answer is :A. recognize the beginning and end of the introns.
Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This
complex binds to the beginning and end of the introns and forms a loop which is later excised
and the ends of exons are sealed..
Match the following organelles with their key functions in the cell .pdf
Match the following organelles with their key functions in the cell nucleus rough endoplasmic
reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon
fixation site of protein synthesis ready for cellular distribution recycling of intracellular
materials modification and packing of proteins for secreation site of protein synthesis that
contains the majority of metabolic pathways containment of toxic reactions the site of the core
genome oxidative phosphorylation
Solution
1: Nucleus: The site of core genome;g
2: Rough ER: site of protein synthesis ready for cellular distribution; b
3: Golgi apparatus: Modification and packing of proteins for secretion; d
4: Lysosome: recylcing of intracellular material; c
5: Mitochondria: Oxidative phosphorylation;h
6: Chloroplast: Carbon fixation; a
7: Peroxisomes: Containment of toxic reaction; f
8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.
please choose the correct answerIn the ABO blood-type phenotype, w.pdf
please choose the correct answer
In the ABO blood-type phenotype, which gene is the most epistatic?
a.) A
b.) B
c.) O
d.)I
e.) H
Solution
Answer is e) H- Epistasis is a product of gene- gene interaction. when one genes expression is
controlled by other gene, it is called epistasis. H antigen is the precursor of A and B antigens.
mo Press Submit until all questions are colored green and you have do.pdf
mo Press Submit until all questions are colored green and you have double c Question 1 Convert
the following degree measure to radians: 210 14 21 20 12 7 d) f ONone of the above.
Solution
We know that,
Radians = Degree*(pi/180)
Given Degree measures = 210o
Using the formunla
Radian Measures = 210(pi/180) = 7(pi/6) Radians
Hence (d) is the correct answer..
matching chose the stage of the light reactions which the statement.pdf
matching: chose the stage of the light reactions which the statement occurs in.
1. _Produces ATP.
2. _Passes energized electrons down an electron transport chain to create ATP.
3. _Produces NADPH.
4. _Splits water to get electrons to replace energized electrons.
5. _Replaces energized electrons from electrons that have completed the 1st electron transport
chain.
a. water-splitting photosystem.
b. NADPH-Producing photosystem.
Solution
During water splitting photosystem ,the electrons in water are removed and passed to the
chlorophyll. The hydrogen ions extracted from the water molecule, and after two water
molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ).
In photosynthesis, water splitting donates electrons to the electron transport chain in
photosystem II.
So it splits water to get electrons to replace energized electrons.
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel via the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH.
So it produces NADPH.
Let E be the set of all even integers, and let O be the set of all o.pdf
Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O.
A. Every integer is both even and odd.
B. Some integers are both even and odd.
C. Even integers and odd integers are integers.
D. Every integer is either even or odd.
E. No integer is both even and odd.
Solution
ans : D
take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.
JavaIm not sure how to implement this code can someone help me .pdf
Java
I\'m not sure how to implement this code can someone help me / guide me? thank you
(Updated with PrefixMap interface included at the bottom of the code).
import java.util.List;
import java.util.Map;
public class Storage implements PrefixMap {
private Map keys;
private Map> prefixes;
public Storage() {
// TODO Initialise your maps
}
/*
* How many keys are stored in the map
*/
@Override
public int size() {
// TODO Implement this, then remove this comment
return 0;
}
@Override
public boolean isEmpty() {
// TODO Implement this, then remove this comment
return false;
}
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String get(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {
// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boole.
Indexes prevent the database engine from having to scan every record .pdf
Indexes prevent the database engine from having to scan every record in the database. True
False
Solution
Answer: True
Indexes prevent the database engine from having to scan every record in the database.
Indexes provides the improvement of query performance by providing the database engine with
fast way of locating a particular record based on the a column or columns provided in where
condition..
Is this a probability Distribution If so, find the Mean (mu) and the.pdf
Is this a probability Distribution? If so, find the Mean (mu) and the Standard deviation (a). If
this is not a Probability Distribution, let the summation of P(x): (LP(x))= A, then find: A and let
that he your missing row, Then find the Usual Values.!
Solution
The average population during 1994 was (204500 + 215000)/2 = 419500/2 = 209750. Incidence
rate refers to the rate of occurrence of new cases of disease in a population over a specified
period of time which also means the number of new cases per unit of population.
Incidence rate of lyme disease in 1994 = Number of new cases of lyme disease in 1994/ average
population during 1994 = 25/ 209750 = 0.011918951 % or, 0.12 % ( on rounding off to 2
decimal places).
in Java (ignore the last line thats hidden) Create a doubly linked l.pdf
in Java (ignore the last line thats hidden) Create a doubly linked list whose nodes contain Strings.
Your list should include the following methods: Insert a node in the list in alphabetical order
Find a node that matches a String Traverse the list forwards and print Traverse the list backwards
and print Delete a node from the list Delete/destroy the list In addition to creating the class(es)
for your linked list, you\'ll need to create a main method that thoroughly tests each function you
build to show that it works. You should do these things in
Solution
/ Java program to create , insert , traverse and delete a node from doubly linked list
class LinkedList {
static Node head = null;
class Node {
int data;
Node next, prev;
Node(int d) {
data = d;
next = prev = null;
}
}
/* Let us create the doubly linked list 10<->8<->4<->2 */
list.push(head, 2);
list.push(head, 4);
list.push(head, 8);
list.push(head, 10);
System.out.println(\"Original Linked list \");
list.printList(head);
/*Function to delete a node in a Doubly Linked List.
head_ref --> pointer to head node pointer.
del --> pointer to node to be deleted. */
void deleteNode(Node head_ref, Node del) {
/* base case */
if (head == null || del == null) {
return;
}
/* If node to be deleted is head node */
if (head == del) {
head = del.next;
}
/* Change next only if node to be deleted is NOT the last node */
if (del.next != null) {
del.next.prev = del.prev;
}
/* Change prev only if node to be deleted is NOT the first node */
if (del.prev != null) {
del.prev.next = del.next;
}
/* Finally, free the memory occupied by del*/
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(Node head_ref, int new_data) {
/* allocate node */
Node new_node = new Node(new_data);
/* since we are adding at the begining,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = (head);
/* change prev of head node to new node */
if ((head) != null) {
(head).prev = new_node;
}
/* move the head to point to the new node */
(head) = new_node;
}
/*Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(Node node) {
while (node != null) {
System.out.print(node.data + \" \");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
/* delete nodes from the doubly linked list */
list.deleteNode(head, head); /*delete first node*/
list.deleteNode(head, head.next); /*delete middle node*/
list.deleteNode(head, head.next); /*delete last node*/
System.out.println(\"\");
/* Modified linked list will be NULL<-8->NULL */
System.out.println(\"Modified Linked List\");
list.printList(head);
}
}.
Imagine that you are looking at the amino acid sequences of the same .pdf
Imagine that you are looking at the amino acid sequences of the same protein from many
different organisms. All of the organisms have a Cys at the same place in the protein sequence.
What does that tell you about the role of that Cys in the protein? All of the organisms have a
stretch of 3 amino acids that are Arg or Lys-Ser or Thr-Arg or Lys. What does that tell you about
this part of the protein? Another part of the protein seems to have all of the possible amino acid
sequences in the different organisms. What does that tell you about that part of the protein?
Solution
Answer:
1) Cysteine (Cys) and Methionine are the sulphur containng aminaocids. Cysteine presence in
the teritary structure of a protein is a common thing due to formation of the disulphide linkage.
Disulphide likages are formed by cysteine only but not methionine. The thiol group is oxidised to
form disulphide linkage.
2)
Arg or Lys-Ser, Thr-Arg or Lys
From above sequence, it can be concluded that Basic group containing aminoacids such as
Arginine (Arg) and Lysine (Lys) are forming bond with hydroxyl group containg aminoacids
such as Serine (Ser) and Threonine(Thr). As basic aminoacids are insoluble in water, so they
participate in protein structure by association with polar aminoacids (water loving) in the protein.
3)
There are 20 aminacids with 64 codons which code for each aminoacid. Proteins seems to have
all possible aminoacid sequences in different organisms. The aminoacids sequence of different
protein vary with there functional properties, each protein function is based on the teritary
arrangement of the polypeptide. Every gene has specific property and it codes for respective
protein which is essential biological system, similarly, every enzyme has its owns aminoacid
sequence to exhibit its catalytic activity..
I need help implementing a Stack with this java programming assignme.pdf
I need help implementing a Stack with this java programming assignment:
Given a doubly linked list (code provided below) implement a Stack. You must use the doubly
link list class and utilize it as an object in your implementation of the stack.
A Stack is a Last in First Out (LiFO) structure which translates to the processes of outputting the
last element which was inputted in the collection. The Stack is based on the process of putting
things on top of one another and taking the item on top (think of a stack of papers).
Operations to implement:
push (E): Add an element to the start of the sequence
pop: Remove an element from the start of the sequence
Peek: Return the first element of the sequence without removing it
atIndex(x): Return the element at the given index (x) Or throw an exception if it is out of bound
(if you can control the user input then do that instead)
Size: Return the size of the Stack
isEmpty: Boolean, returns true if the Stack is empty
Empty: Empty the Stack
Code Part 1:
public class DoublyLinkedList{
private DNode head;
private DNode tail;
private int size;
public DoublyLinkedList(){ // construct an empty list
this.tail = new DNode (null, null, this.head);
this.head = new DNode (null, this.tail, null);
this.size = 0;
}
public DoublyLinkedList(DNode next){ // constructs a list
// out of a single node
this.tail = new DNode (null, null, next);
this.head = new DNode (null, next, null);
next.changeNext(this.tail);
next.changePrev(this.head);
this.size = 1;
}
public DoublyLinkedList(Object [] objectArray){ // construct a list out of
// an array
this.tail = new DNode (null, null, this.head);
this.head = new DNode (null, this.tail, null);
DNode temp = this.head;
for (Object e : objectArray)
{
//Anonomus function
new DNode (e, temp.getNext(),temp).insertBetweenNodes(temp, temp.getNext());
temp = temp.getNext();
this.size += 1;
}
}
public void addToFrontofList(Object toAdd){ // Appends the begining
// of the list
DNode temp = new DNode (toAdd, this.head.getNext(), this.head);
this.head.getNext().changePrev(temp);
this.head.changeNext(temp);
this.size += 1;
}
public void addToendofList (Object toAdd) // appends the end of the list
// with a node
{
DNode temp = new DNode (toAdd, this.tail, this.tail.getPrev());
this.tail.getPrev().changeNext(temp);
this.tail.changePrev(temp);
this.size += 1;
}
public void insertAfterNode(DNode current, Object input)// Inserts a new
// a new node after
// current node
{
current.insertAfterNode(input);
this.size += 1;
}
public int getSize() // returns the size of the list
{
return this.size;
}
public String toString()
{
String result = \"\";
for (DNode temp = this.head.getNext();
temp.hasNext(); temp = temp.getNext())
{
result += temp.getValue();
result += \" -> \";
}
result += \"End of list\";
return result;
}
}
and code part2:
public class DNode{
private Object e; // place holder for the type of object the
//collection is storing
private DNode next; // reference to the next node in the sequence
// of.
If a trait is sexually selected, the trait could appear the same in .pdf
If a trait is sexually selected, the trait could appear the same in both males and females.
Agree/Disagree? Explain:
Solution
No, all the sexually selected traits could not appear the same in both males and females. For
example, in sex-linked diseases, the defective allele is present in X-chromosome. X- linked or
sex-linked diseases could be developed in females only when they inherit two X-linked recessive
alleles, one from the mother and another from the father. Whereas in males, only one defective
allele will result in expression of disease phenotype. Similarly, Y-linked genes are not expressed
in females as females do not contain the Y-chromosome..
html js node Create a function that reads from a file and returns a .pdf
html js node Create a function that reads from a file and returns a string with the contents of that
file. Create a Node server that serves the above HTML document in its response. Have it
viewable at localhost:8000/.
Solution
var fs = require(\'fs\');
fs.readFile(\'DATA\', \'utf8\', function(err, contents) {
var data = contents;
});.
How does phospholipids structure relate to the selective permeability.pdf
How does phospholipids structure relate to the selective permeability of the plasma membrane?
critical feature of the plasma membrane is that it is selectively permeable This a lows the plasma
membrane to regulate transport across cellular boundaries a function essential Drag the labels to
fill in the table. Use only white targets, Pink tables for pink targets, and blue labels for blue
targets.
Solution
1. Hydrophobic can cross easily No transport protein required
In lipid bilayer it has polar head and nonpolar tail. So, Non polar molecule can easily cross the
membrane. It does not require any special mechanism for transport this molecules.
2. Polar molecules - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein
require for efficient transport.
Hydrophobic molecules hates the hydrophilic so it does not allow the polar molecules easily
cross in the lipid bilayers.
3. Ions - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein require for
efficient transport..
Geneticists use a shorthand notation to refer to the genotype of the.pdf
Geneticists use a shorthand notation to refer to the genotype of the different individuals with
which they are working. For instance, the symbol denoting the recessive allele for the ability to
roll your tongue might be written TRR. An individual that is heterozygous for this trait and also
carries the wild-type allele might be denoted as ______. Note:
Select all correct answers.
+/TRR
TR+/TRR
TRR/TRR
+/+
TR+/TR+
Solution
In +/TRR the two alleles are \'+\' which denotes wild type, and TRR which is the recessive allele
and since both the alleles are different it is heterozygous.
In TR+/TRR the two alleles are \'TR+\' which has one wild type gene (+) and the other recessive
(TR). The single wild type gene may confer the wild type trait depending on other factors such as
dominance. An individual with these alleles are also heterozygous.
TRR/TRR is homozygous recessive.
+/+ is homozygous wild type.
TR+/TR+ is homozygous..
The Challenger.pdf DNHS Official Publication
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!
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The DNA sequence below is a shortened version of the sequence on the .pdf
The DNA sequence below is a shortened version of the sequence on the previous page. This
DNA sequence corresponds to the mRNA sequence only; two introns are not included. The
numbers indicate the position of nucleotides within the sequence and indicates intervening
nucleotides that are not shown. The gene is 626 nucleotides long. Recall that chromosomes are
very long DNA molecules with millions of nucleotides, so this sequence is preceded and
followed by seemingly endless sequences of nucleotides. IDNA is double-stranded. Write the
sequence of the first 10 nucleotides of the complementary strand of DNA.
Solution
ans. DNA is a double stranded which consist of two strands and one strand is complemetary to
the other.DNA contains complemetary base pairs in which adenine is linke d to tymine by two
hydrogen bonds guanine to cytosine by three hydrogen bonds
The seque of complementary strand of the DNA molecule is
5\' TGTAAACGAA.
Show that f(x)=57^x+1 and f^-1(x)=7x-75 are inverses Choose the ap.pdf
Show that f(x)=5/7^x+1 and f^-1(x)=7x-7/5 are inverses Choose the appropriate equality that
will show that f(x) and f^-1(x) are inverses (f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
(f.f^-1)(x)= (f^-1.f)(x)=1 (f.f^-1)(x)= (f^-1.f)(x)=x
Solution
Replace x by f^-1(x) in f(x)
=> 5/7(7x-7)/5+1
=> (7x - 7)/7 + 1 = > x - 1 + 1 => x
=> f(f^-1(x)) = x.
Read the short article Horizontal Gene Transfer in E Coli and answer.pdf
Read the short article Horizontal Gene Transfer in E Coli and answer the following
questions.(https://www.sciencedaily.com/releases/2015/05/150519105900.htm) Remember to
use your own words; plagiarism of any kind will not be tolerated (see syllabus for further
information on what constitutes plagiarism).
(1) What is horizontal gene transfer (HGT)?
(2) In the posted article, how many gene transfer elements were found for E. coli O104:H4?
(3) What is the source of these gene transfer elements?
(4) Should we be concerned about HGT in prokaryotes? If so, why?
Solution
(1) Horizontal gene transfer (HGT) is the lateral gene movement, wherein the movement of
genetic material takes place between unicellular or multicellular organisms across species. This
gene transfer is different from the vertical gene transfer in which the transmission of DNA
occurs from parent to offspring. Often genetic material swaping occurs between bacteria of same
as well as different species, through this means by aquiring one of the processes; transformation,
transduction or conjugation. During HGT, plasmids, transposons or integrons act as vectors that
transfer genes.
(2) In the given article, the team from Madurai Kamaraj University in Madurai, Tamil Nadu,
identified 38 gene transfer elements for E. coli O104:H4.
(3) The research team identified 38 gene transfer prophage elements for E.coli O104:H4. They
showed that the these elements act as genetic weapons protecting the bacteria from antibiotics.
The source of these elements were viruses (the bacteriophages), that infect bacteria, and in the
process bacteria aquired these elements from viruses.
(4) In prokaryotes, horizontal gene transfer is a major concern. Aquisition of new genes through
HGT is ability of Bacteria and Archaea to adapt to new changing environments. Druing HGT,
genes that are responsible for antibiotic resistance in one species of bacteria can be transferred to
another species of bacteria, and is the primary reason for the spread of antibiotic resistance.
Besides antibiotic resistance, HGT also gives bacteria an ability to degrade novel compounds
(such as different insecticides and pesticides). Thus the bacterial strains after HGT are much
more dangerous as they are multi drug resistant and more virulent. These transfer elements when
transferred to a pathogenic bacteia, can become a serious health hazard as they are resistant to
antibiotics available with us and thus cannot be trated easily..
Please select the best answer and click submit. The owners of fou.pdf
Please select the best answer and click \"submit.\" The owners of four companies competing for
a contract are shown in the table below. If a report is released that advocates company A. which
of the people having funded the report should result in the most skepticism?
Solution
Option d.
Debby Smith would result in the most skepticism because his company is in the second place in
the list..
One function of the spliceosome is to (Select all that apply.)A. .pdf
One function of the spliceosome is to: (Select all that apply.)
A. recognize the beginning and end of the introns.
B. recognize the 5\' end of the primary transcript.
C recognize the beginning and end of the exons.
D recognize the 3\' end of the primary transcript.
Solution
The correct answer is :A. recognize the beginning and end of the introns.
Reason: Spliceosome consists of snRNAs and snRNPs which together make a complex.This
complex binds to the beginning and end of the introns and forms a loop which is later excised
and the ends of exons are sealed..
Match the following organelles with their key functions in the cell .pdf
Match the following organelles with their key functions in the cell nucleus rough endoplasmic
reticulum Golgi apparatus lysosomes mitochondria chloroplasts peroxisomes cytosol carbon
fixation site of protein synthesis ready for cellular distribution recycling of intracellular
materials modification and packing of proteins for secreation site of protein synthesis that
contains the majority of metabolic pathways containment of toxic reactions the site of the core
genome oxidative phosphorylation
Solution
1: Nucleus: The site of core genome;g
2: Rough ER: site of protein synthesis ready for cellular distribution; b
3: Golgi apparatus: Modification and packing of proteins for secretion; d
4: Lysosome: recylcing of intracellular material; c
5: Mitochondria: Oxidative phosphorylation;h
6: Chloroplast: Carbon fixation; a
7: Peroxisomes: Containment of toxic reaction; f
8: Cytosol: Site of protein synthesis that contains majority of metabolic pathways; e.
please choose the correct answerIn the ABO blood-type phenotype, w.pdf
please choose the correct answer
In the ABO blood-type phenotype, which gene is the most epistatic?
a.) A
b.) B
c.) O
d.)I
e.) H
Solution
Answer is e) H- Epistasis is a product of gene- gene interaction. when one genes expression is
controlled by other gene, it is called epistasis. H antigen is the precursor of A and B antigens.
mo Press Submit until all questions are colored green and you have do.pdf
mo Press Submit until all questions are colored green and you have double c Question 1 Convert
the following degree measure to radians: 210 14 21 20 12 7 d) f ONone of the above.
Solution
We know that,
Radians = Degree*(pi/180)
Given Degree measures = 210o
Using the formunla
Radian Measures = 210(pi/180) = 7(pi/6) Radians
Hence (d) is the correct answer..
matching chose the stage of the light reactions which the statement.pdf
matching: chose the stage of the light reactions which the statement occurs in.
1. _Produces ATP.
2. _Passes energized electrons down an electron transport chain to create ATP.
3. _Produces NADPH.
4. _Splits water to get electrons to replace energized electrons.
5. _Replaces energized electrons from electrons that have completed the 1st electron transport
chain.
a. water-splitting photosystem.
b. NADPH-Producing photosystem.
Solution
During water splitting photosystem ,the electrons in water are removed and passed to the
chlorophyll. The hydrogen ions extracted from the water molecule, and after two water
molecules are splitted,the oxygen atoms takes electron and form molecular oxygen (O 2 ).
In photosynthesis, water splitting donates electrons to the electron transport chain in
photosystem II.
So it splits water to get electrons to replace energized electrons.
In the photosystem II (PSII) reaction center, energy from sunlight is used to extract electrons
from water. The electrons travel via the chloroplast electron transport chain to photosystem I
(PSI), which reduces NADP+ to NADPH.
So it produces NADPH.
Let E be the set of all even integers, and let O be the set of all o.pdf
Let E be the set of all even integers, and let O be the set of all odd integers. Explain why Z E O.
A. Every integer is both even and odd.
B. Some integers are both even and odd.
C. Even integers and odd integers are integers.
D. Every integer is either even or odd.
E. No integer is both even and odd.
Solution
ans : D
take any integer z, z is either odd or even so, for all z , Z is a subset of EUO.
JavaIm not sure how to implement this code can someone help me .pdf
Java
I\'m not sure how to implement this code can someone help me / guide me? thank you
(Updated with PrefixMap interface included at the bottom of the code).
import java.util.List;
import java.util.Map;
public class Storage implements PrefixMap {
private Map keys;
private Map> prefixes;
public Storage() {
// TODO Initialise your maps
}
/*
* How many keys are stored in the map
*/
@Override
public int size() {
// TODO Implement this, then remove this comment
return 0;
}
@Override
public boolean isEmpty() {
// TODO Implement this, then remove this comment
return false;
}
/*
* Get the value corresponding to the key (or null if the key is not found)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String get(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Insert the value into the data structure, using the given key. If the key
* already existed, replace and return the old value (otherwise return null)
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key or value is null, throw IllegalArgumentException
*/
@Override
public String put(String key, String value) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Remove the value corresponding to the given key from the data structure,
* if it exists. Return the old value, or null if no value was found.
* if the key contains any character other than A,C,G,T, throw MalformedKeyException
* if the key is null, throw IllegalArgumentException
*/
@Override
public String remove(String key) {
// TODO Implement this, then remove this comment
return null;
}
/*
* return the number of keys which start with the given prefix if the prefix
* contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public int countKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return 0;
}
/*
* return the collection of keys which start with the given prefix if the
* prefix contains any character other than A,C,G,T, throw MalformedKeyException
* if the prefix is null, throw IllegalArgumentException
*/
@Override
public List getKeysMatchingPrefix(String prefix) {
// TODO Implement this, then remove this comment
return null;
}
/*
* Return the number of unique prefixes
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 8
* because the possible prefixes are G, GA, GAT, GATT, GATTC, GATTA, GATTAC,
GATTACA
*/
@Override
public int countPrefixes() {
// TODO Implement this, then remove this comment
return 0;
}
/*
* Return the sum of the lengths of all keys
* e.g. if the tree stores keys GAT, GATTC, GATTACA, this method will return 15
*/
@Override
public int sumKeyLengths() {
// TODO Implement this, then remove this comment
return 0;
}
}
import java.util.List;
public interface PrefixMap {
public boole.
Indexes prevent the database engine from having to scan every record .pdf
Indexes prevent the database engine from having to scan every record in the database. True
False
Solution
Answer: True
Indexes prevent the database engine from having to scan every record in the database.
Indexes provides the improvement of query performance by providing the database engine with
fast way of locating a particular record based on the a column or columns provided in where
condition..
Is this a probability Distribution If so, find the Mean (mu) and the.pdf
Is this a probability Distribution? If so, find the Mean (mu) and the Standard deviation (a). If
this is not a Probability Distribution, let the summation of P(x): (LP(x))= A, then find: A and let
that he your missing row, Then find the Usual Values.!
Solution
The average population during 1994 was (204500 + 215000)/2 = 419500/2 = 209750. Incidence
rate refers to the rate of occurrence of new cases of disease in a population over a specified
period of time which also means the number of new cases per unit of population.
Incidence rate of lyme disease in 1994 = Number of new cases of lyme disease in 1994/ average
population during 1994 = 25/ 209750 = 0.011918951 % or, 0.12 % ( on rounding off to 2
decimal places).
in Java (ignore the last line thats hidden) Create a doubly linked l.pdf
in Java (ignore the last line thats hidden) Create a doubly linked list whose nodes contain Strings.
Your list should include the following methods: Insert a node in the list in alphabetical order
Find a node that matches a String Traverse the list forwards and print Traverse the list backwards
and print Delete a node from the list Delete/destroy the list In addition to creating the class(es)
for your linked list, you\'ll need to create a main method that thoroughly tests each function you
build to show that it works. You should do these things in
Solution
/ Java program to create , insert , traverse and delete a node from doubly linked list
class LinkedList {
static Node head = null;
class Node {
int data;
Node next, prev;
Node(int d) {
data = d;
next = prev = null;
}
}
/* Let us create the doubly linked list 10<->8<->4<->2 */
list.push(head, 2);
list.push(head, 4);
list.push(head, 8);
list.push(head, 10);
System.out.println(\"Original Linked list \");
list.printList(head);
/*Function to delete a node in a Doubly Linked List.
head_ref --> pointer to head node pointer.
del --> pointer to node to be deleted. */
void deleteNode(Node head_ref, Node del) {
/* base case */
if (head == null || del == null) {
return;
}
/* If node to be deleted is head node */
if (head == del) {
head = del.next;
}
/* Change next only if node to be deleted is NOT the last node */
if (del.next != null) {
del.next.prev = del.prev;
}
/* Change prev only if node to be deleted is NOT the first node */
if (del.prev != null) {
del.prev.next = del.next;
}
/* Finally, free the memory occupied by del*/
return;
}
/* UTILITY FUNCTIONS */
/* Function to insert a node at the beginning of the Doubly Linked List */
void push(Node head_ref, int new_data) {
/* allocate node */
Node new_node = new Node(new_data);
/* since we are adding at the begining,
prev is always NULL */
new_node.prev = null;
/* link the old list off the new node */
new_node.next = (head);
/* change prev of head node to new node */
if ((head) != null) {
(head).prev = new_node;
}
/* move the head to point to the new node */
(head) = new_node;
}
/*Function to print nodes in a given doubly linked list
This function is same as printList() of singly linked lsit */
void printList(Node node) {
while (node != null) {
System.out.print(node.data + \" \");
node = node.next;
}
}
public static void main(String[] args) {
LinkedList list = new LinkedList();
/* delete nodes from the doubly linked list */
list.deleteNode(head, head); /*delete first node*/
list.deleteNode(head, head.next); /*delete middle node*/
list.deleteNode(head, head.next); /*delete last node*/
System.out.println(\"\");
/* Modified linked list will be NULL<-8->NULL */
System.out.println(\"Modified Linked List\");
list.printList(head);
}
}.
Imagine that you are looking at the amino acid sequences of the same .pdf
Imagine that you are looking at the amino acid sequences of the same protein from many
different organisms. All of the organisms have a Cys at the same place in the protein sequence.
What does that tell you about the role of that Cys in the protein? All of the organisms have a
stretch of 3 amino acids that are Arg or Lys-Ser or Thr-Arg or Lys. What does that tell you about
this part of the protein? Another part of the protein seems to have all of the possible amino acid
sequences in the different organisms. What does that tell you about that part of the protein?
Solution
Answer:
1) Cysteine (Cys) and Methionine are the sulphur containng aminaocids. Cysteine presence in
the teritary structure of a protein is a common thing due to formation of the disulphide linkage.
Disulphide likages are formed by cysteine only but not methionine. The thiol group is oxidised to
form disulphide linkage.
2)
Arg or Lys-Ser, Thr-Arg or Lys
From above sequence, it can be concluded that Basic group containing aminoacids such as
Arginine (Arg) and Lysine (Lys) are forming bond with hydroxyl group containg aminoacids
such as Serine (Ser) and Threonine(Thr). As basic aminoacids are insoluble in water, so they
participate in protein structure by association with polar aminoacids (water loving) in the protein.
3)
There are 20 aminacids with 64 codons which code for each aminoacid. Proteins seems to have
all possible aminoacid sequences in different organisms. The aminoacids sequence of different
protein vary with there functional properties, each protein function is based on the teritary
arrangement of the polypeptide. Every gene has specific property and it codes for respective
protein which is essential biological system, similarly, every enzyme has its owns aminoacid
sequence to exhibit its catalytic activity..
I need help implementing a Stack with this java programming assignme.pdf
I need help implementing a Stack with this java programming assignment:
Given a doubly linked list (code provided below) implement a Stack. You must use the doubly
link list class and utilize it as an object in your implementation of the stack.
A Stack is a Last in First Out (LiFO) structure which translates to the processes of outputting the
last element which was inputted in the collection. The Stack is based on the process of putting
things on top of one another and taking the item on top (think of a stack of papers).
Operations to implement:
push (E): Add an element to the start of the sequence
pop: Remove an element from the start of the sequence
Peek: Return the first element of the sequence without removing it
atIndex(x): Return the element at the given index (x) Or throw an exception if it is out of bound
(if you can control the user input then do that instead)
Size: Return the size of the Stack
isEmpty: Boolean, returns true if the Stack is empty
Empty: Empty the Stack
Code Part 1:
public class DoublyLinkedList{
private DNode head;
private DNode tail;
private int size;
public DoublyLinkedList(){ // construct an empty list
this.tail = new DNode (null, null, this.head);
this.head = new DNode (null, this.tail, null);
this.size = 0;
}
public DoublyLinkedList(DNode next){ // constructs a list
// out of a single node
this.tail = new DNode (null, null, next);
this.head = new DNode (null, next, null);
next.changeNext(this.tail);
next.changePrev(this.head);
this.size = 1;
}
public DoublyLinkedList(Object [] objectArray){ // construct a list out of
// an array
this.tail = new DNode (null, null, this.head);
this.head = new DNode (null, this.tail, null);
DNode temp = this.head;
for (Object e : objectArray)
{
//Anonomus function
new DNode (e, temp.getNext(),temp).insertBetweenNodes(temp, temp.getNext());
temp = temp.getNext();
this.size += 1;
}
}
public void addToFrontofList(Object toAdd){ // Appends the begining
// of the list
DNode temp = new DNode (toAdd, this.head.getNext(), this.head);
this.head.getNext().changePrev(temp);
this.head.changeNext(temp);
this.size += 1;
}
public void addToendofList (Object toAdd) // appends the end of the list
// with a node
{
DNode temp = new DNode (toAdd, this.tail, this.tail.getPrev());
this.tail.getPrev().changeNext(temp);
this.tail.changePrev(temp);
this.size += 1;
}
public void insertAfterNode(DNode current, Object input)// Inserts a new
// a new node after
// current node
{
current.insertAfterNode(input);
this.size += 1;
}
public int getSize() // returns the size of the list
{
return this.size;
}
public String toString()
{
String result = \"\";
for (DNode temp = this.head.getNext();
temp.hasNext(); temp = temp.getNext())
{
result += temp.getValue();
result += \" -> \";
}
result += \"End of list\";
return result;
}
}
and code part2:
public class DNode{
private Object e; // place holder for the type of object the
//collection is storing
private DNode next; // reference to the next node in the sequence
// of.
If a trait is sexually selected, the trait could appear the same in .pdf
If a trait is sexually selected, the trait could appear the same in both males and females.
Agree/Disagree? Explain:
Solution
No, all the sexually selected traits could not appear the same in both males and females. For
example, in sex-linked diseases, the defective allele is present in X-chromosome. X- linked or
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html js node Create a function that reads from a file and returns a string with the contents of that
file. Create a Node server that serves the above HTML document in its response. Have it
viewable at localhost:8000/.
Solution
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fs.readFile(\'DATA\', \'utf8\', function(err, contents) {
var data = contents;
});.
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How does phospholipids structure relate to the selective permeability of the plasma membrane?
critical feature of the plasma membrane is that it is selectively permeable This a lows the plasma
membrane to regulate transport across cellular boundaries a function essential Drag the labels to
fill in the table. Use only white targets, Pink tables for pink targets, and blue labels for blue
targets.
Solution
1. Hydrophobic can cross easily No transport protein required
In lipid bilayer it has polar head and nonpolar tail. So, Non polar molecule can easily cross the
membrane. It does not require any special mechanism for transport this molecules.
2. Polar molecules - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein
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Hydrophobic molecules hates the hydrophilic so it does not allow the polar molecules easily
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3. Ions - Hydrophilic - Have difficulty to cross hydrophobic part - transport protein require for
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which they are working. For instance, the symbol denoting the recessive allele for the ability to
roll your tongue might be written TRR. An individual that is heterozygous for this trait and also
carries the wild-type allele might be denoted as ______. Note:
Select all correct answers.
+/TRR
TR+/TRR
TRR/TRR
+/+
TR+/TR+
Solution
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(TR). The single wild type gene may confer the wild type trait depending on other factors such as
dominance. An individual with these alleles are also heterozygous.
TRR/TRR is homozygous recessive.
+/+ is homozygous wild type.
TR+/TR+ is homozygous..
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