Please explain Suppose that I roll a fair die (meaning that each num.pdf
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Please explain Suppose that I roll a fair die (meaning that each number from 1 to 6 is equally likely) until I get a 6. What is the probability (in terms of k) that the first 6 appears the k^th time I roll the die? Suppose Alice and Bob play a game where the alternate rolling the die: First Alice rolls, then Bob, then Alice again, and so on. The first person to roll a 6 wins. What is the probability that Alice wins this game? Solution.
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I need help writing the methods for the song and playList class in J.pdf
I need help writing the methods for the song and playList class in Java. I already have the
interface and the getters and setter. Can someone help with the methods?
Song Class
public class Song
private String artist // the artist performing the song
private String title // the title of the song
private int minutes // number of min in length
private int seconds // number of seconds of length of the song (always less than 60)
// Add Getters / Setters for all fields
// Three constructors (remember to initialize ALL fields in each of the constructors - regardless
of the number of parameters!)
public Song(String artist, String title)
public Song(String artist, String title, int minutes, int seconds)
public Song(Song s)
public boolean equals(Object o) // a song is equal if all four fields are equal
public String toString() { // Use this code for toString EXACTLY
return \"{Song: title = \" + title + \" artist = \" + artist + \"}\";
}
public void play() { // Use this code for play EXACTLY
System.out.printf(\"Playing Song: artist = %-20s title = %s\ \", artist, title);
}
Make the Song class implement the interface Comparable, and make the ordering criteria as
follows: Songs will be ordered by artist in ascending order. If the artists are the same, then by
title in ascending order. If both artist and title are the same, then any order is acceptable.
Don’t forget to include any methods needed to implement the Playable interface.
PlayList Class
public class PlayList
Methods that modify the list should return a boolean return true if they successfully change the
list, otherwise they return false. Note that a PlayList can contain other PlayLists!
Methods that return a Playable element return null if they fail to find the song in question.
private String name // contains the name of the playlist
private ArrayList playableList // ArrayList of Playable elements that make up the play list
// Add Getters / Setters for name and songList
// Add a toString() method that returns the name of the playlist followed by its contents (by
calling toString() on each item it contains). You should only have one name/title per line
// Two constructors (remember to initialize ALL fields in each of the constructors - regardless of
the number of parameters!)
public PlayList() // empty play list named \"Untitled\"
public PlayList(String newName) // empty play list
public boolean loadSongs(String fileName) // loads songs from file, more on this below
public boolean clear() // removes all playable elements in the play list
public boolean addSong(Song s) // adds Song s to the end of the play list
public Playable removePlayable(int index) // removes Playable element at index from the list and
returns it
public Playable removePlayable(Playable p) // removes every occurrence of Playable p from the
list and returns p
public Playable getPlayable(int index) // returns the Playable element at the appropriate index
public boolean addPlayList(PlayList pl) //adds the PlayList that is being pa.
How do autopolyploidy and allopolyploidy differ Only autopolyploidy.pdf
How do autopolyploidy and allopolyploidy differ? Only autopolyploidy involves an increase in
chromosome sets. The chromosomes of an autopolyploid individual come from the same
species, while an allopolyploid has sets of chromosomes from different species. Autopolyploidy
is a mechanism of speciation in plants, while allopolyploid is a mechanism of speciation animals.
Autopolyploidy is sympatric, while allopolyploidy is allopatric.
Solution
Answer:
b). The chromosomes of an autopolyploid individual come from the same species, while an
allopolyploid has sets of chromsomes from different species
Autopolyploidy arises from fusion of gametes of the same species (cell division errors, etc.). In
this process, polyploid individuals develop as a result of cell division errors that double the DNA
content of gametes in individuals from the same species or a self-fertilizing individual. When
these double-dose gametes fuse, the result is an organism with double the chromosome number
of its parent species.
Allopolyploidy occurs when the gametes from two species fuse to form a polyploid generation
and hybrid offspring..
How do you find the order of growth for the following recurrenceT.pdf
How do you find the order of growth for the following recurrence?
T(n) = 9T(n/3) + n2
Solution
The Master Theorem applies to recurrences of the following form:
T (n) = aT (n/b) + f (n)
where a 1 and b > 1 are constants and f (n) is an asymptotically positive function.
There are 3 cases:
Case First:-- f(n) is O(nlogba ). Since the leaves grow faster than f, all work is done at the
leaves, so T(n) is (nlogb a).
Case Second:-- f(n) is (nlogba). The leaves grow at same rate as f, so the same order of work is
done at every level of tree. Therfore tree has O(log n) levels times the work done on one level,
yielding T(n) is (nlogb a log n).
Case Third:-- f(n) is (nlogba + ). Here f grows faster than the number of leaves, which means that
asymptotically the total amount of work is dominated by the work done at the root node. For the
upper bound, we also need an extra smoothness condition on f in this case i.e af(n/b) cf(n) for
some constant c < 1 and large n. In this case T(n) is (f(n)).
Therefore, using Master Method on your above Question
a=9, b=3, =log39 = 2, f(n) = n2, so Case 2, i.e. T(n) = (nlogn) = (n2logn)..
Explain SEAD, SAR and SPAAR routing protocols in more details.So.pdf
Explain SEAD, SAR and SPAAR routing protocols in more details.
Solution
SEAD Protocols: SEAD is a proactive secure routing protocol based on DSDV-SQ-protocol. It
does not rely on asymmetric encryption primitive but instead it relies on one-way hash chain for
security. The algorithm expects that there is some authenticated and secure way to deliver the
initial key KN. This could be done by key delivery in advance or by using public key encryption
and signatures for key delivery. The basic idea of SEAD is, to authenticate the sequence number
and metric of a routing table update message using hash chains elements. In addition, the
receiver also authenticates the sender ensuring that the routing information originates from the
correct node. The source of each routing update message in SEAD must also be authenticated
since an attacker may be able to create routing loops through the impersonation attack. There are
two different approaches proposed such as broadcast authentication mechanism, TESLA and
Message Authentication Codes
SAR routing protocols: SAR takes an approach to routing that incorporates security level of
nodes into traditional routing metrics. In most protocols the length of the route is the only metric
used. It does not target any protocol, but it rather provides security at a more generalized level of
security. The goal is exposing security to the application and to the routing protocol. SAR uses
AODV or DSR as a base protocol; it embeds the security metric into RREQ packet itself and
changes the forwarding behavior of the protocol. When intermediate nodes receive an RREQ
packet with a particular security metric or trust level, the node can only process the packet or
forward it if it can provide the required security or trust level.
SPAAR routing protocols: The SPAAR protocol was developed with the classical managed-
hostile environment in mind, thus meant to provide a very high level of security, and sometimes
at the cost of performance. Among other things, SPAAR also requires that each device to use a
GPS locator to determine its position, although some leeway is given to nodes using a so-called
“locator-proxy” if absolute security is not required. In SPAAR packets are only accepted
between neighboring nodes one hop away from each other, this is to avoid the “invisible node-
attack”. The basic transmission procedure is quite similar to ARAN, although the group
neighborhood key is used for encryption in order to ensure one-hop communication only. Since
all nodes also have information on their location they only forward RREQs if their position is
closer to the destination position.
Dynamic routing protocols are used toa) establish paths real ti.pdf
Dynamic routing protocols are used to:
a) establish path\'s real time between routers
b) define overhead bits within the packet
c) encapsulate the IP packet into a Layer 2 frame
d) reroute traffic during circuit failures and network congestion
e) a and b
f) a, c, and d
g) all of the above
Solution
a,c,d.
Evaluate expression without a calculator. Write answer in radians..pdf
Evaluate expression without a calculator. Write answer in radians.
Please show step by step so I can check my work. My answer -pi/6
Solution
sin^-1(-1/2) =y
range of sin^-1(x) : -pi/2<=y<=pi/2
So, value of y would be less than pi/2and its is -ve so it should be between -pi/2 and 0
y = sin^-1(-1/2)
siny = -1/2
y = -pi/6 radians.
Epithelial tissues that are only one cell thick are called _____ Epi.pdf
Epithelial tissues that are only one cell thick are called _____ Epithelial tissues that are more
than one cell thick are called _____ Of these two basic types of epithelial tissues, which is most
likely better at transport of substances? What structural aspect of this tissue makes it good at
doing this? What is the major function of the other basic type of epithelial tissue? What
structural aspect of this tissue makes It good at doing this? If the cells in an epithelial tissue are
wider than they are tall, they are called _____ cells. If the cells in an epithelial tissue are taller
than they are wide, they are called _____ cells. If the cells In an epithelial tissue are as wide as
they are tall, they are called _____ cells. Of these three types of cell shapes found In epithelial
tissues, which one is best at allowing for simple diffusion? What structural aspect of this tissue
makes it good at doing this? Of these three types of cell shapes found in epithelial tissues, which
ate best at allowing for active transport processes? What structural aspect of this tissue makes it
good at doing this? What type of cell-surface characteristic helps an epithelial tissue reduce
friction? What type of cell-surface characteristic helps an epithelial tissue in absorption and/or
secretion? What type of cell-surface characteristic helps an epithelial tissue allows it to move
particles across its surface?
Solution
- epithelium cells that are only one cell thick are known as simple epithelial tissues
- epithelium cells that are more than one cell thickness are known as stratified epithelial tissue
- simple epithelial tissue is good at transport of substances
- epithelial tissue consists of basement membrane on which it rests. this membrane acts as a
scaffold. through this membrane ther is nerve supply as this membrane is permeable hence it
allows the transport of substances
-the major function of stratified epithelial tisue is protection of the cells
-epithelial tissues acts as a lining surface to the body organs
-.
Describe in five sentences how the TLR system is used to fight HSV-1 .pdf
Describe in five sentences how the TLR system is used to fight HSV-1 encephalitis.
Solution
Toll like receptors are one type of PRRs, which are associate in the recognition of pathogen,
associated molecular patterns. These receptors are initially identified in Drosophila, they are also
found in macrophages, mast cells, dendritic cells in mammals. Toll-like receptors trigger the
production and release of cytokines and also activate other host defense responses associated
with innate and adaptive immune response. In response to HSV -1 (herpes simplex virus 1), the
TLR-2 releases cytokines that cause severe inflammatory response (innate immune response),
which leads to encephalitis. Antagonism of these receptors is likely to provide protection against
encephalitis caused by HSV-1. The TLR-2 are mainly located on the surface of microglia and
astrocytes..
Recommended
I need help writing the methods for the song and playList class in J.pdf
I need help writing the methods for the song and playList class in Java. I already have the
interface and the getters and setter. Can someone help with the methods?
Song Class
public class Song
private String artist // the artist performing the song
private String title // the title of the song
private int minutes // number of min in length
private int seconds // number of seconds of length of the song (always less than 60)
// Add Getters / Setters for all fields
// Three constructors (remember to initialize ALL fields in each of the constructors - regardless
of the number of parameters!)
public Song(String artist, String title)
public Song(String artist, String title, int minutes, int seconds)
public Song(Song s)
public boolean equals(Object o) // a song is equal if all four fields are equal
public String toString() { // Use this code for toString EXACTLY
return \"{Song: title = \" + title + \" artist = \" + artist + \"}\";
}
public void play() { // Use this code for play EXACTLY
System.out.printf(\"Playing Song: artist = %-20s title = %s\ \", artist, title);
}
Make the Song class implement the interface Comparable, and make the ordering criteria as
follows: Songs will be ordered by artist in ascending order. If the artists are the same, then by
title in ascending order. If both artist and title are the same, then any order is acceptable.
Don’t forget to include any methods needed to implement the Playable interface.
PlayList Class
public class PlayList
Methods that modify the list should return a boolean return true if they successfully change the
list, otherwise they return false. Note that a PlayList can contain other PlayLists!
Methods that return a Playable element return null if they fail to find the song in question.
private String name // contains the name of the playlist
private ArrayList playableList // ArrayList of Playable elements that make up the play list
// Add Getters / Setters for name and songList
// Add a toString() method that returns the name of the playlist followed by its contents (by
calling toString() on each item it contains). You should only have one name/title per line
// Two constructors (remember to initialize ALL fields in each of the constructors - regardless of
the number of parameters!)
public PlayList() // empty play list named \"Untitled\"
public PlayList(String newName) // empty play list
public boolean loadSongs(String fileName) // loads songs from file, more on this below
public boolean clear() // removes all playable elements in the play list
public boolean addSong(Song s) // adds Song s to the end of the play list
public Playable removePlayable(int index) // removes Playable element at index from the list and
returns it
public Playable removePlayable(Playable p) // removes every occurrence of Playable p from the
list and returns p
public Playable getPlayable(int index) // returns the Playable element at the appropriate index
public boolean addPlayList(PlayList pl) //adds the PlayList that is being pa.
How do autopolyploidy and allopolyploidy differ Only autopolyploidy.pdf
How do autopolyploidy and allopolyploidy differ? Only autopolyploidy involves an increase in
chromosome sets. The chromosomes of an autopolyploid individual come from the same
species, while an allopolyploid has sets of chromosomes from different species. Autopolyploidy
is a mechanism of speciation in plants, while allopolyploid is a mechanism of speciation animals.
Autopolyploidy is sympatric, while allopolyploidy is allopatric.
Solution
Answer:
b). The chromosomes of an autopolyploid individual come from the same species, while an
allopolyploid has sets of chromsomes from different species
Autopolyploidy arises from fusion of gametes of the same species (cell division errors, etc.). In
this process, polyploid individuals develop as a result of cell division errors that double the DNA
content of gametes in individuals from the same species or a self-fertilizing individual. When
these double-dose gametes fuse, the result is an organism with double the chromosome number
of its parent species.
Allopolyploidy occurs when the gametes from two species fuse to form a polyploid generation
and hybrid offspring..
How do you find the order of growth for the following recurrenceT.pdf
How do you find the order of growth for the following recurrence?
T(n) = 9T(n/3) + n2
Solution
The Master Theorem applies to recurrences of the following form:
T (n) = aT (n/b) + f (n)
where a 1 and b > 1 are constants and f (n) is an asymptotically positive function.
There are 3 cases:
Case First:-- f(n) is O(nlogba ). Since the leaves grow faster than f, all work is done at the
leaves, so T(n) is (nlogb a).
Case Second:-- f(n) is (nlogba). The leaves grow at same rate as f, so the same order of work is
done at every level of tree. Therfore tree has O(log n) levels times the work done on one level,
yielding T(n) is (nlogb a log n).
Case Third:-- f(n) is (nlogba + ). Here f grows faster than the number of leaves, which means that
asymptotically the total amount of work is dominated by the work done at the root node. For the
upper bound, we also need an extra smoothness condition on f in this case i.e af(n/b) cf(n) for
some constant c < 1 and large n. In this case T(n) is (f(n)).
Therefore, using Master Method on your above Question
a=9, b=3, =log39 = 2, f(n) = n2, so Case 2, i.e. T(n) = (nlogn) = (n2logn)..
Explain SEAD, SAR and SPAAR routing protocols in more details.So.pdf
Explain SEAD, SAR and SPAAR routing protocols in more details.
Solution
SEAD Protocols: SEAD is a proactive secure routing protocol based on DSDV-SQ-protocol. It
does not rely on asymmetric encryption primitive but instead it relies on one-way hash chain for
security. The algorithm expects that there is some authenticated and secure way to deliver the
initial key KN. This could be done by key delivery in advance or by using public key encryption
and signatures for key delivery. The basic idea of SEAD is, to authenticate the sequence number
and metric of a routing table update message using hash chains elements. In addition, the
receiver also authenticates the sender ensuring that the routing information originates from the
correct node. The source of each routing update message in SEAD must also be authenticated
since an attacker may be able to create routing loops through the impersonation attack. There are
two different approaches proposed such as broadcast authentication mechanism, TESLA and
Message Authentication Codes
SAR routing protocols: SAR takes an approach to routing that incorporates security level of
nodes into traditional routing metrics. In most protocols the length of the route is the only metric
used. It does not target any protocol, but it rather provides security at a more generalized level of
security. The goal is exposing security to the application and to the routing protocol. SAR uses
AODV or DSR as a base protocol; it embeds the security metric into RREQ packet itself and
changes the forwarding behavior of the protocol. When intermediate nodes receive an RREQ
packet with a particular security metric or trust level, the node can only process the packet or
forward it if it can provide the required security or trust level.
SPAAR routing protocols: The SPAAR protocol was developed with the classical managed-
hostile environment in mind, thus meant to provide a very high level of security, and sometimes
at the cost of performance. Among other things, SPAAR also requires that each device to use a
GPS locator to determine its position, although some leeway is given to nodes using a so-called
“locator-proxy” if absolute security is not required. In SPAAR packets are only accepted
between neighboring nodes one hop away from each other, this is to avoid the “invisible node-
attack”. The basic transmission procedure is quite similar to ARAN, although the group
neighborhood key is used for encryption in order to ensure one-hop communication only. Since
all nodes also have information on their location they only forward RREQs if their position is
closer to the destination position.
Dynamic routing protocols are used toa) establish paths real ti.pdf
Dynamic routing protocols are used to:
a) establish path\'s real time between routers
b) define overhead bits within the packet
c) encapsulate the IP packet into a Layer 2 frame
d) reroute traffic during circuit failures and network congestion
e) a and b
f) a, c, and d
g) all of the above
Solution
a,c,d.
Evaluate expression without a calculator. Write answer in radians..pdf
Evaluate expression without a calculator. Write answer in radians.
Please show step by step so I can check my work. My answer -pi/6
Solution
sin^-1(-1/2) =y
range of sin^-1(x) : -pi/2<=y<=pi/2
So, value of y would be less than pi/2and its is -ve so it should be between -pi/2 and 0
y = sin^-1(-1/2)
siny = -1/2
y = -pi/6 radians.
Epithelial tissues that are only one cell thick are called _____ Epi.pdf
Epithelial tissues that are only one cell thick are called _____ Epithelial tissues that are more
than one cell thick are called _____ Of these two basic types of epithelial tissues, which is most
likely better at transport of substances? What structural aspect of this tissue makes it good at
doing this? What is the major function of the other basic type of epithelial tissue? What
structural aspect of this tissue makes It good at doing this? If the cells in an epithelial tissue are
wider than they are tall, they are called _____ cells. If the cells in an epithelial tissue are taller
than they are wide, they are called _____ cells. If the cells In an epithelial tissue are as wide as
they are tall, they are called _____ cells. Of these three types of cell shapes found In epithelial
tissues, which one is best at allowing for simple diffusion? What structural aspect of this tissue
makes it good at doing this? Of these three types of cell shapes found in epithelial tissues, which
ate best at allowing for active transport processes? What structural aspect of this tissue makes it
good at doing this? What type of cell-surface characteristic helps an epithelial tissue reduce
friction? What type of cell-surface characteristic helps an epithelial tissue in absorption and/or
secretion? What type of cell-surface characteristic helps an epithelial tissue allows it to move
particles across its surface?
Solution
- epithelium cells that are only one cell thick are known as simple epithelial tissues
- epithelium cells that are more than one cell thickness are known as stratified epithelial tissue
- simple epithelial tissue is good at transport of substances
- epithelial tissue consists of basement membrane on which it rests. this membrane acts as a
scaffold. through this membrane ther is nerve supply as this membrane is permeable hence it
allows the transport of substances
-the major function of stratified epithelial tisue is protection of the cells
-epithelial tissues acts as a lining surface to the body organs
-.
Describe in five sentences how the TLR system is used to fight HSV-1 .pdf
Describe in five sentences how the TLR system is used to fight HSV-1 encephalitis.
Solution
Toll like receptors are one type of PRRs, which are associate in the recognition of pathogen,
associated molecular patterns. These receptors are initially identified in Drosophila, they are also
found in macrophages, mast cells, dendritic cells in mammals. Toll-like receptors trigger the
production and release of cytokines and also activate other host defense responses associated
with innate and adaptive immune response. In response to HSV -1 (herpes simplex virus 1), the
TLR-2 releases cytokines that cause severe inflammatory response (innate immune response),
which leads to encephalitis. Antagonism of these receptors is likely to provide protection against
encephalitis caused by HSV-1. The TLR-2 are mainly located on the surface of microglia and
astrocytes..
Daunte is frustrated about managements recent decision to merge hi.pdf
Daunte is frustrated about management\'s recent decision to merge his two teams into one. He
discusses this issue with Claire, a so-worker. Claire responds by saying, \"Its sounds like
management\'s decision really bothers you.\" In the context of listening skills, Clarire\'s
statement best exemplifies a (n) sarcastic tone that blocks responses. (b) active response that
mirrors feelings (c) active response that advises the speaker. (d) blocking response that criticies
the speaker (e) negative response that magnifies the problem.
Solution
(b) active response that mirrors feelings.
Concern over the weather associated with El Nino has increased intere.pdf
Concern over the weather associated with El Nino has increased interest in the possibility that
the climate on earth is getting warmer. The most common theory relates an increase in
atmospheric levels of carbon dioxide (CO_2), a greenhouse gas, to increases in temperature. A
regression analysis of the mean annual CO_2 concentration (in parts per million) in the
atmosphere at the top of Mauna Loa in Hawaii and the mean annual air temperature (in degrees
Celsius) over both land and sea across the globe for 37 years was performed. Assume all
conditions are met for simple linear regression inference. A hypothesis test is to be performed to
determine if CO_2 levels is a significant predictor of air temperature. What is the correct
conclusion based on the results of the hypothesis test? Choose the correct answer below.
Solution
Here, the t statistic is
t = B1 / sB1
here, B1 = 0.004, sB1 = 0.009.
Thus,
t = 4.444444444
For a 0.001 level, the critical t value is, as df = n - 1 = 36,
tcrit= 3.582149701
Thus, as t > 3.5821, then it is OPTION D. [ANSWER].
Clostridium difficile is a bacteria currently making headlines. It h.pdf
Clostridium difficile is a bacteria currently making headlines. It has long been the considered the
most common health-care associated infection, causing death to a large number of patients when
they contract it during hospital stays. The CDC is now reporting an increase in C. difficile cases
even outside of hospitals, which is raising concern. Below is an electron micrograph of C.
difficile. If you were able to do a Gram stain on a sample of the bacteria, you would find that the
cells stain a dark purple.
Given this information, how would you classify C. difficile?
A.
Gram negative, rod shaped
B.
Gram positive, coccus shaped
C.
Gram negative, coccus shaped
D.
Gram positive, rod shaped
Many protists have evolved complex life cycles that include sexual reproduction. This means
that:
A.
haploid cells must undergo meiotic divisions to form gametes.
B.
complex mating behavior likely evolved at the same time as sexual reproduction.
C.
genetic material from two haploid cells is combined at some point in the life cycle.
D.
diploid cells must undergo mitotic cell division in order to reproduce.
Protists:
A.
are all eukaryotes.
B.
are all heterotrophs.
C.
are all photosynthetic.
D.
are all single-celled.
E.
All of the above are correct.
All of the following are examples of the evidence that supports the endosymbiotic theory
EXCEPT:
A.
Mitochondria and chloroplasts have their own genomes.
B.
Mitochondria and chloroplasts divide in a process resembling binary fission.
C.
Mitochondria and chloroplasts are similar in size to bacteria.
D.
Mitochondria and chloroplasts have linear chromosomes.
Solution
Clostridium difficle is a Gram positive, rod shaped bacteria.(option-D) Many protists have
evolved complex sexual reproductive life cycles which includes haploid cells undergo meiotic
division to form gametes. Protists are all eukaryotes, not an animal, plant or fungus. The point
\"mitochondria and chloroplaats have linear chromosomes\" is not an evidence that supports for
endosymbiotic theory..
You cloned an EcoRI digested DNA fragment into a unique EcoRI site o.pdf
You cloned an EcoRI digested DNA fragment into a unique EcoRI site of a vector. You identify
a recombinant vector that you believe has the DNA of interest. To generate a restriction map of
the recombinant plasmid, you take three individual samples of the plasmid and digest them with
EcoRI, HindIII, and with both EcoRI and HindIII. You then run the digested DNA on an
agarose
gel to see the fragments. The gel is then subjected to a Southern hybridization using the EcoRI
insert as a probe. Assume that the insert fragment is smaller than the vector.
15. Which fragment (s) on the gel would hybridize to the probe which represents the inert?
16. Which fragment (s) on the gel would hybridize to a probe representing the 5.3 kb band
observed in the EcoRI lane?
17. Which fragment (s) in the EcoRI lane would hybridize to a probe representing the 4.1 kb
band observed in the HindIII lane?
18. The cloning strategy used being non-directional, the recombinant plasmid may represent one
of two possible orientations of the insert. You perform a hybridization with the 1.5 kb band
observed in the EcoRI lane. Which fragment (s) would not be observed in the opposite
orientation of that represented on the gel?
PLEASE EXPLAIN THOUGHT PROCESS 5.3 4.1 2.7 1.5 EcoRI HindIII EcoRI Hind 3.0 2.3
1.1 0.4
Solution
The band of HINDIII (2.7) and the band of EcoRI+HINDIII (2.3) represent the fragments that
will hybridize to 5.3kb band probe.
The band of EcoRI+HINDIII (3.0) and the band of EcoRI (1.5) represent the fragments that will
hybridize to 4.1kb band probe.
The fragments which will mot be observed in opposite direction are the ones which are dual
enzyme digested. The reason behind this is that when dual enzyme digestion occurs it causes an
incompatible recognition sequence in the opposite direction. Therefore such fragments are found
in the middle of two individual enzyme digestions..
You have constructed a new reporter gene to help you find novel zebr.pdf
You have constructed a new reporter gene to help you find novel zebrafish mutations. The
reporter gene consists of several LEG/TCF binding domains in the enhancer fused to the coding
region for green fluorescent protein (GFP). Wild-type embryos that contain this transgene
normally express it in the brain and eye primordia and have no developmental defects. You then
mutagenize this reporter strain to look for new mutants that affect GFP expression.
Which developmental pathway are these mutants likely to be in and why?
You find a mutant (which you call GLO-1) that ubiquitously expresses your reporter gene
throughout the entire embryo. Which molecule in the above pathway is most likely to be mutated
and why?
Solution
Ans.) Nacre, a zebrafish homolog of MITF is required for pigment cell differentiation. We
isolated a promoter region of shell that contains Tcf/Lef binding sites which will mediate Wnt
responsiveness. This promoter binds to zebrafish Lef1 protein in vitro, and a nacre
communicator construct is powerfully inhibited by dominant-negative Tcf in malignant
melanoma cells. Mutation of Tcf/Lef sites abolishes Lef1 binding and reporter operates in vivo.
Wnt signaling so directly activates shell, which in flip ends up in pigment cell differentiation.
It was determined that the human melanocyte-specific MITF promoter contains many purported
Tcf/Lef binding sites, hallmarks of Wnt/-catenin-mediated transcriptional control. To test the
hypothesis that nacre could be a target of the Wnt/-catenin pathway in zebrafish, a
promoter/enhancer region is isolated for this sequence from a genomic bacteriophage library.
This genomic fragment contains the translation initiation site for nacre which is associated with a
836-bp upstream region with a possible transcription begin site, a cAMP response element
(CRE) and several agreement Tcf/Lef binding sites. These CRE and Tcf/Lef locations are not
fully preserved between the human and zebrafish promoters in either their sequences or their
positions from the transcription begin site. To examine whether the purported putative nacre
promoter/enhancer will direct correct abstraction management of organic phenomenon, injected
one -cell zebrafish embryos with the 836-bp deoxyribonucleic acid fragment driving a green
fluorescent macromolecule (GFP) communicator sequence. Mosaic gene expression is typical for
transient transgenics in zebrafish and because the neural crest springs from only a few precursor
cells, reporter potency is minimized. Despite these, it was ready to analyze the fate of the few
reporter-expressing cells in injected animals..
Write a Java program that will incorporate either a sequential searc.pdf
Write a Java program that will incorporate either a sequential search method or a sort method
Solution
class Search
{
public static void main(String args[])
{ int array[],n;
System.out.println(\"Please enter the sixe of array\");
Scanner s = new Scanner(System.in);
int n = s.nextInt();
System.out.println(\"Please enter\"+ n +\"elements of array\");
for(int i= 0;i.
Answer only with excel instructions.Assume that 96 of ticket hold.pdf
Answer only with excel instructions.
Assume that 96% of ticket holders show up for an airline flight. If a plane seats 200 people, how
many tickets should be sold to make the chance of an overbooked flight (more ticket holders
show up for a flight than there are seats available) is at most 10%.
Solution
Say no. of tickets booked = x
Now only 0.96x will turn up
No. of seats = 200
Now these 0.96x people should get a seat with 90% probability
200 / (0.96x) >=0.9
200/0.864 >= x
x <=231.48
x = 231
Airline will make 231 reservations.
A new species of sea horse has been discovered and is being sequenced.pdf
A new species of sea horse has been discovered and is being sequenced The genome of the new
species is being aligned using the reference sequence from a already-known species of sea horse
Since the two species are evolutionary diverged, there may be some differences between their
sequences Reference sequence: ACCTACATTCACAGAATGAT Fragments a) Align the set
of mini-fragments above. b) How are the sequences of the two sea horse species different? c)
Would this alignment have been easier or more difficult if you didn\'t have any reference
sequence at all (alignment without a reference is called \"de novo\")? Explain d) If there had
been sequencing errors, would this alignment have been easier or more difficult? Explain.
Solution
A) TAGATT TGAATG TCACTGA
B) There is a form of polymorphism occurring in the new sequence of the genes for the new
species of sea horses. There is single nucleotide polymorphism in the new species as C is
replaced by T and A instead of G which may probably occurred due to some sort of mutation
during genetic sequencing.
C) The given alignment would be difficult without reference DNA sequence as reference
sequence gives basic DNA information and a reference database techniques is much easier than
performing a de novo technique though de novo methods are well established using mass spectra
nowadays but still having a reference database helps and gives the better coding information of
protein sequence.
D) if there had been a sequence error there would be difficulty in alignment and in that case we
have to perform de novo sequencing using mass spectra analysis. To know the sequence of the
species and other information regarding the DNA characters..
Which of the following is the mechanism of action for Herceptin Her.pdf
Which of the following is the mechanism of action for Herceptin? Herceptin binds to the HER2
promoter to prevent transcription, thus lowering amounts of receptor. Herceptin binds to
intracellular HER2 proteins to prevent cancer cells from dividing. Herceptin binds to HER3 and
HER4 cell surface receptors to activate the immune system. Herceptin binds to extracellular
domain of HER2 and prevents internalization and subsequent cancer cell division. Herceptin
targets TNF alpha.
Solution
B) herceptin binds to intracellular ER2 proteins to prevent cancer cells from dividing
Trastuzumab (herceptin) activates p27 factor by inhibiting PI3K/Akt, Mirk and hKIS pathways.
HER2 protein can be expressed up to 100 times more in cancer cells. PI3K/Akt, Mirk and hKIS
are intracellular proteins in HER2 pathways are activated which promote cell proliferation.
Signals present in these pathways promote growth of blood vessels in order to supplement for
tumour formation or angiogenesis and therefore resulting in cell proliferation. Thus, when
herceptin inhibits PI3K/Akt, Mirk and hKIS pathways, the cell proliferation is inhibited..
Which of these statements best describes an element An element is c.pdf
Which of these statements best describes an element? An element is composed of three
subatomic particles. An element is formed when two atoms of the same kind are bonded
together. An element is formed when two or more different atoms combine chemically. An
element is composed of identical atoms.
Solution
An element is composed of identical atoms.
Molecules can be made of elements, but elements arent made from molecules. 2 or more atoms
of the same kind make up an element.For example H is hydrogen which is an element formed
from H atoms only whereas H2O is a compount formed from 2 different atoms H and O..
What is the general goal for descriptive statistics How is the goal.pdf
What is the general goal for descriptive statistics? How is the goal served by putting scores in a
frequency distribution? How is the goal served by computing a measure of central tendency?
How is the goal served by computing a measure of variability?
Solution
What is the general goal of descriptive statistics ?
Answer : To simplify , organize, and summarize data so that it is easier for researchers to see
patterns.
How is the goal served by putting scores in a frequency distribution?
A frequncy distribution provides an organized summary of the complete set of scores.
How is the goal served by computing a measure of central tendency?
Summarizes an entires set of scores with a single value that is representative of the whole set.
How is the goal served by computing a measure of variability?
Provides a single number that describes the differences that exist from complete score to another..
What effects are shown on this graph (assume that all effects where.pdf
What effects are shown on this graph? (assume that all effects where the marginal means differ
are significant)
Choose as many as applicable.
A. Main effect of A where scores are higher at A1 than A2
B. Main effect of A where scores are higher at A2 than A1
C, Main effect of B where scores are higher at B1 than B2
D, Main effect of B where scores are higher at B2 than B1
E. Interaction where the effect of A is larger at B1 than B2
F. Interaction where the effect of A is larger at B2 than B1
Solution.
What are adaptive peaks How do organisms move from one adaptive pea.pdf
What are adaptive peaks? How do organisms move from one adaptive peak to another? 100
words or more please
Solution
An equilibrium state in a population when the gene pool has allele frequencies that maximize the
average fitness of a population\'s members. In other words, A high point (perhaps one of several)
on an adaptive landscape; selection tends to drive the genotype composition of the population
toward a combination corresponding to an adaptive peak.
Most extant organisms are thought to sit on the “peaks” of such adaptive landscapes. However,
how does a phenotype shift from one peak to another is not clear. Common hypotheses include
(1) There are “ridges” connecting the adaptive peaks, such that phenotypes along the ridge are
more or less equally fit under a specific environment;
(2) There are no ridges connecting the peaks but when effective population size is small enough,
phenotypes may be able to traverse the less-adaptive region (i.e. adaptive valley) by genetic drift..
We are dealing with Verilog language.......What is a major design .pdf
We are dealing with Verilog language.......
What is a major design advantage when using a ripple counter?
Design simplicity.
Glitch free.
High speed.
Outputs are all synchronized.a.
Design simplicity.b.
Glitch free.c.
High speed.d.
Outputs are all synchronized.
Solution
advantages of design of asynchronous(ripple) counter in verilog code are
1.Removes reset from the data path.
2.multi clock systems needs only one reset.
3.Faster in speed.
So answer is (c).
The text defines three service layers when describing Cloud Computin.pdf
The text defines three service layers when describing Cloud Computing. Identify and describe
these three service layers. Which Service Layer is most likely to be consumed via Public Cloud
service and why? You need to defend your answer.
Solution
First we’ll go high level. Cloud is made feasible through the deployment and interoperability of
three platform types. These three layers are:
IaaS - Infrastructure as a Service
PaaS - Platform as a Service
SaaS - Software as a Service
Now this stack is easily broken down as follows: Think of the “Infrastructure-as-a-Service” as
the road. It’s the basis for communication. It’s the bottom layer that you build your platform on.
The platform are the cars traveling on the infrastructure. PaaS rides on IaaS. But on the top of
that, the goods and passengers inside the cars are the SaaS. It’s the end user experience. It’s the
end result. Let’s take that a step further.
1) Infrastructure-as-a-Service (IaaS)
Cloud Providers offering Infrastructure as a Service tout data-center space, and servers; as well
as network equipment such as routers/switches and software for businesses. These data-centers
are fully outsourced, you need not lift a finger, upgrade an IOS or re-route data. Although this is
the base layer, it allows for scalability and reliability; as well as better security than an
organization may have in a local co-lo or local datacenter. In addition, these services are charged
as utilities, so you pay for what you use, like your water, electric and gas. Depending on your
capacity or usage, your payment is a variable.
Because the IaaS vendors purchase equipment in such bulk, you, Mr. Customer, get the best gear
for the lowest price. Hence, the financial benefits of IaaS are cheaper access to infrastructure.
With the pay-as-you-go model, instead of investing in a fixed capacity infrastructure, which will
either fall short or exceed the organizational need, customers are able to save quite a bit of coin.
Buying hardware that’s barely used is a waste of hardware, air conditioning, space and power.
Operational expenses versus Capital expenses: Cloud is better. Because these computing
resources are basically used and paid for like a utility they can be paid via the operating
expenditures budget versus being paid for via capital investments. In other words, instead of
depreciating the gear over three years, you’re able to expense the monthly charge this year. And
the next year. And the year after that. It’s an elastic service.
1.1)
Application-level security
This level is primarily about identity and access management. Your best investment here isn’t
financial; it’s a policy that limits the ability of users to deploy cloud applications without IT’s
knowledge.
After ensuring policies are in place that offer IT visibility, the next step is to invest in multifactor
authentication and identity management. The first approach uses two or more devices or
applications to permit access. For example, a verification cod.
Social support plays a critical role in successful aging. Is this tr.pdf
Social support plays a critical role in successful aging. Is this true or false? Why?
Solution
Social support plays a critical role in successful aging. The statement is true.
Aging refers to the process of becoming older over a period of time.This includes the changes in
physical , psychological and social changes.At any stage of life social support play a crucial role
in the success of a person this is even more true in case of older adults.with the advancement of
age there are several changes that occur in the life which include changes in the job status, a
person may retire and yet may have family financial commitments to be fulfilled,the children
would have grown up and for their future they might have left the parents for their life
settlements , deterioration of physical health, financial status ,empty nest syndrome, felling
isolated and left out all these may also lead to psychological depression.
A positive approach to life is a very important factor in the process of aging. in older adults the
body immune system becomes fragile , resources for support become depleted and such other
causes will have an effect in the process of aging.Psychological well being is an indication of
successful adaptation to old age .
Acceptance to change as a part of life provides strength in overcoming several psychological
problems and studies have indicated that adults who are active are more likely to be successful in
their old age than those who expect a leisurely and lazy dependent life as this tend to negative
interactions with the family mainly.
As people age remaining socially active provides a positive health benefits to the aged people.
Successful connectedness,isolation, mobility, spirituality all have a say in dealing with the
geriatric problems.one can adapt themselves to the age by being socially active .Voulenteering is
a notable factor in relation to social support for adults.It helps in building companionship and
interact with more people of the community.They can engage themselves in spiritual activities ,
develop more realistic goals ,opportunities for self discovery.
Physical health plays a very important role in successful aging,with the advancement of age
.social support in terms of maintaining day to day affairs also plays a crucial role and studies
have suggested that it also helps in enhanced well being , less depression and often provides a
senses of meaning to life.
To conclude aging is a general process and acceptance with grace provides the main solution for
successful aging and the social support either monetary, physical or psychological all play a very
important role in helping the elderly people in the late years of their life..
sch the Basic Principles of Science 1erms with their prop.pdf
sch the Basic Principles of Science 1erms with their proper defrition. Testable
undeseindiaadammtal processes natural lawaby questions suce, the wok taday are the iame
ihaped the wodd the part, and they will continar Proof is elusive prefmible This nale ilse afire the
English philosophe dange as new Parsimony Repeatability then the eedeseu cannot be
Solution
1. Testable question - 7
Uncertainity - 4
Proof is elusive - 6
parsimony - 3
Repeatability - 5
Empiricism - 1
Uniformitarianism - 2
2. Emergent properties
3. Scientific theory.
A very thin slice is removed from a peanut and treated with Sudan II.pdf
A very thin slice is removed from a peanut and treated with Sudan III stain. Then a drop of
Biuret reagent is added to the peanut slice. When you examine the peanut slice under a
microscope, patches of red and blue-violet are visible. What conclusions can you draw from your
examination?
Solution
Answer:
When Biuret reagent is added to peanut slices, it turns into Blue-Violet colour indicating that
presence of the proteins in peanut.
When peanut is treated with Sudan III stain, slice turns into red colour indicating the presence of
the Lipids..
Question 1. CAM plants have adapted to very dry climates by opening .pdf
Question 1. CAM plants have adapted to very dry climates by opening their stomata only during
the day.
True
False
2 points
QUESTION 2
If the haploid number of chromosomes in an organism is 16, then the diploid number is
24
4
8
32
2 points
QUESTION
In human blood groups, the A, B, and O alleles are codominant.
True
False
2 points
QUESTION
If two plants that show incomplete dominance are crossed, one having red flowers and the other
pink flowers, what will be the phenotypic ratio of their offspring?
1 pink:2 white:1 red
1 red:2 pink:1 white
1 white:2 red:1 pink
2 red:2 pink
2 points
QUESTION
DNA transcription takes place in the nucleus.
True
False
2 points
QUESTION
Which of the following is not haploid?
sperm
gamete
zygote
egg
24
4
8
32
Solution
Ans. 1. False
CAM plants are adapted to very dry climates by opening their stomata during the night. Night
temperature is relatively much lower than that of the day time. So, opening stomata during night
time cause less evaporation (low loss of water) loss of water during transpiration than day time.
Ans. 2. True.
Alleles are said to be co-dominant when both alleles are expressed simultaneously in
heterozygous condition and both the traits retain their identity. For example, in AB blood group,
both the antigens A (gene product of allele IA) and B (gene product of allele IB) are
simultaneously expressed and independently present on RBC cell surface.
In a complete dominance scenario, only the dominant allele is expressed while the recessive is
NOT expressed in heterozygous condition.
Ans. 3. Incomplete dominance shows simultaneously expression of two alleles in the
heterozygous condition, but the two traits blend together to give a new phenotype.
Consider the difference of crosses between homozygous red and homozygous white flower as
RR x rr = RR, Rr, Rr, rr. -
I. Complete dominance: Rr progeny = red flower as that of dominant allele.
II. Incomplete dominance: Rr progeny = Pink flower as blending of red and white color because
R and r are expressed simultaneously.
III. Co-dominance: Rr progeny = A flower with white and red patches of color- both R and r are
expressed simultaneously and both the phenotypes retain their identity.
Now, coming to the question-
Genotype of homozygous red flower: RR
Genotype of homozygous red flower: Rr
Now,
RR x Rr = RR, Rr ; [RR = red ; Rr = pink]
Phenotype ratio = 1: 1 or 2: 2 or 3:3 ; [Note: multiplying a ratio with any factor does
not affect the ration. SO, 1:1 = 2: 2 = 3: 3, and so on].
Ans. 4. True
DNA transcription i.e. formation of RMA from DNA is called transcription. Because DNA is
present in nucleus and does not come out in cytoplasm at any point, the transcription must occur
in nucleus.
Ans. 5. Correct option C. zygote.
Sperm (n) + egg (n) -------> Zygote (2n). Fusion of egg with sperm gives diploid zygote.
Sperm and egg are gametes- all being haploid (n)..
Programming in C.Create a program which turns your Arduino into a .pdf
Programming in C.
Create a program which turns your Arduino into a light based electronic instrument. The
program should play music based on the amount of light hitting the optical sensor. There should
be a way to capture particularly pleasing tones and play those tones back in the order they are
captured.
The program must utilize the following :
• An array to hold the stored tones
• A function : void read_buttons( int *sw1, int *sw2, int *sw3, int *sw4) to read and store the
state of the four buttons on the board
• A function playback( notes[] ) which plays back the notes stored
• A function clear_playlist( *notes) which clears the stored tones
• A function tone_value() which generates a tone based on the current state of the optical sensor
• The possible tones should span the range of human hearing
Solution
#include
#include
using namespace std;
// Class Declaration
class prime
{
//Member Varibale Declaration
int a,k,i;
public:
prime(int x)
{
a=x;
}
// Object Creation For Class
void calculate()
{
k=1;
{
for(i=2;i<=a/2;i++)
if(a%i==0)
{
k=0;
break;
}
else
{
k=1;
}
}
}
void show()
{
if(k==1)
cout<<\"\ \"<>a;
// Object Creation For Class
prime obj(a);
// Call Member Functions
obj.calculate();
obj.show();
getch();
return 0;
}.
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
June 3, 2024 Anti-Semitism Letter Sent to MIT President Kornbluth and MIT Cor...
Letter from the Congress of the United States regarding Anti-Semitism sent June 3rd to MIT President Sally Kornbluth, MIT Corp Chair, Mark Gorenberg
Dear Dr. Kornbluth and Mr. Gorenberg,
The US House of Representatives is deeply concerned by ongoing and pervasive acts of antisemitic
harassment and intimidation at the Massachusetts Institute of Technology (MIT). Failing to act decisively to ensure a safe learning environment for all students would be a grave dereliction of your responsibilities as President of MIT and Chair of the MIT Corporation.
This Congress will not stand idly by and allow an environment hostile to Jewish students to persist. The House believes that your institution is in violation of Title VI of the Civil Rights Act, and the inability or
unwillingness to rectify this violation through action requires accountability.
Postsecondary education is a unique opportunity for students to learn and have their ideas and beliefs challenged. However, universities receiving hundreds of millions of federal funds annually have denied
students that opportunity and have been hijacked to become venues for the promotion of terrorism, antisemitic harassment and intimidation, unlawful encampments, and in some cases, assaults and riots.
The House of Representatives will not countenance the use of federal funds to indoctrinate students into hateful, antisemitic, anti-American supporters of terrorism. Investigations into campus antisemitism by the Committee on Education and the Workforce and the Committee on Ways and Means have been expanded into a Congress-wide probe across all relevant jurisdictions to address this national crisis. The undersigned Committees will conduct oversight into the use of federal funds at MIT and its learning environment under authorities granted to each Committee.
• The Committee on Education and the Workforce has been investigating your institution since December 7, 2023. The Committee has broad jurisdiction over postsecondary education, including its compliance with Title VI of the Civil Rights Act, campus safety concerns over disruptions to the learning environment, and the awarding of federal student aid under the Higher Education Act.
• The Committee on Oversight and Accountability is investigating the sources of funding and other support flowing to groups espousing pro-Hamas propaganda and engaged in antisemitic harassment and intimidation of students. The Committee on Oversight and Accountability is the principal oversight committee of the US House of Representatives and has broad authority to investigate “any matter” at “any time” under House Rule X.
• The Committee on Ways and Means has been investigating several universities since November 15, 2023, when the Committee held a hearing entitled From Ivory Towers to Dark Corners: Investigating the Nexus Between Antisemitism, Tax-Exempt Universities, and Terror Financing. The Committee followed the hearing with letters to those institutions on January 10, 202
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Daunte is frustrated about managements recent decision to merge hi.pdf
Daunte is frustrated about management\'s recent decision to merge his two teams into one. He
discusses this issue with Claire, a so-worker. Claire responds by saying, \"Its sounds like
management\'s decision really bothers you.\" In the context of listening skills, Clarire\'s
statement best exemplifies a (n) sarcastic tone that blocks responses. (b) active response that
mirrors feelings (c) active response that advises the speaker. (d) blocking response that criticies
the speaker (e) negative response that magnifies the problem.
Solution
(b) active response that mirrors feelings.
Concern over the weather associated with El Nino has increased intere.pdf
Concern over the weather associated with El Nino has increased interest in the possibility that
the climate on earth is getting warmer. The most common theory relates an increase in
atmospheric levels of carbon dioxide (CO_2), a greenhouse gas, to increases in temperature. A
regression analysis of the mean annual CO_2 concentration (in parts per million) in the
atmosphere at the top of Mauna Loa in Hawaii and the mean annual air temperature (in degrees
Celsius) over both land and sea across the globe for 37 years was performed. Assume all
conditions are met for simple linear regression inference. A hypothesis test is to be performed to
determine if CO_2 levels is a significant predictor of air temperature. What is the correct
conclusion based on the results of the hypothesis test? Choose the correct answer below.
Solution
Here, the t statistic is
t = B1 / sB1
here, B1 = 0.004, sB1 = 0.009.
Thus,
t = 4.444444444
For a 0.001 level, the critical t value is, as df = n - 1 = 36,
tcrit= 3.582149701
Thus, as t > 3.5821, then it is OPTION D. [ANSWER].
Clostridium difficile is a bacteria currently making headlines. It h.pdf
Clostridium difficile is a bacteria currently making headlines. It has long been the considered the
most common health-care associated infection, causing death to a large number of patients when
they contract it during hospital stays. The CDC is now reporting an increase in C. difficile cases
even outside of hospitals, which is raising concern. Below is an electron micrograph of C.
difficile. If you were able to do a Gram stain on a sample of the bacteria, you would find that the
cells stain a dark purple.
Given this information, how would you classify C. difficile?
A.
Gram negative, rod shaped
B.
Gram positive, coccus shaped
C.
Gram negative, coccus shaped
D.
Gram positive, rod shaped
Many protists have evolved complex life cycles that include sexual reproduction. This means
that:
A.
haploid cells must undergo meiotic divisions to form gametes.
B.
complex mating behavior likely evolved at the same time as sexual reproduction.
C.
genetic material from two haploid cells is combined at some point in the life cycle.
D.
diploid cells must undergo mitotic cell division in order to reproduce.
Protists:
A.
are all eukaryotes.
B.
are all heterotrophs.
C.
are all photosynthetic.
D.
are all single-celled.
E.
All of the above are correct.
All of the following are examples of the evidence that supports the endosymbiotic theory
EXCEPT:
A.
Mitochondria and chloroplasts have their own genomes.
B.
Mitochondria and chloroplasts divide in a process resembling binary fission.
C.
Mitochondria and chloroplasts are similar in size to bacteria.
D.
Mitochondria and chloroplasts have linear chromosomes.
Solution
Clostridium difficle is a Gram positive, rod shaped bacteria.(option-D) Many protists have
evolved complex sexual reproductive life cycles which includes haploid cells undergo meiotic
division to form gametes. Protists are all eukaryotes, not an animal, plant or fungus. The point
\"mitochondria and chloroplaats have linear chromosomes\" is not an evidence that supports for
endosymbiotic theory..
You cloned an EcoRI digested DNA fragment into a unique EcoRI site o.pdf
You cloned an EcoRI digested DNA fragment into a unique EcoRI site of a vector. You identify
a recombinant vector that you believe has the DNA of interest. To generate a restriction map of
the recombinant plasmid, you take three individual samples of the plasmid and digest them with
EcoRI, HindIII, and with both EcoRI and HindIII. You then run the digested DNA on an
agarose
gel to see the fragments. The gel is then subjected to a Southern hybridization using the EcoRI
insert as a probe. Assume that the insert fragment is smaller than the vector.
15. Which fragment (s) on the gel would hybridize to the probe which represents the inert?
16. Which fragment (s) on the gel would hybridize to a probe representing the 5.3 kb band
observed in the EcoRI lane?
17. Which fragment (s) in the EcoRI lane would hybridize to a probe representing the 4.1 kb
band observed in the HindIII lane?
18. The cloning strategy used being non-directional, the recombinant plasmid may represent one
of two possible orientations of the insert. You perform a hybridization with the 1.5 kb band
observed in the EcoRI lane. Which fragment (s) would not be observed in the opposite
orientation of that represented on the gel?
PLEASE EXPLAIN THOUGHT PROCESS 5.3 4.1 2.7 1.5 EcoRI HindIII EcoRI Hind 3.0 2.3
1.1 0.4
Solution
The band of HINDIII (2.7) and the band of EcoRI+HINDIII (2.3) represent the fragments that
will hybridize to 5.3kb band probe.
The band of EcoRI+HINDIII (3.0) and the band of EcoRI (1.5) represent the fragments that will
hybridize to 4.1kb band probe.
The fragments which will mot be observed in opposite direction are the ones which are dual
enzyme digested. The reason behind this is that when dual enzyme digestion occurs it causes an
incompatible recognition sequence in the opposite direction. Therefore such fragments are found
in the middle of two individual enzyme digestions..
You have constructed a new reporter gene to help you find novel zebr.pdf
You have constructed a new reporter gene to help you find novel zebrafish mutations. The
reporter gene consists of several LEG/TCF binding domains in the enhancer fused to the coding
region for green fluorescent protein (GFP). Wild-type embryos that contain this transgene
normally express it in the brain and eye primordia and have no developmental defects. You then
mutagenize this reporter strain to look for new mutants that affect GFP expression.
Which developmental pathway are these mutants likely to be in and why?
You find a mutant (which you call GLO-1) that ubiquitously expresses your reporter gene
throughout the entire embryo. Which molecule in the above pathway is most likely to be mutated
and why?
Solution
Ans.) Nacre, a zebrafish homolog of MITF is required for pigment cell differentiation. We
isolated a promoter region of shell that contains Tcf/Lef binding sites which will mediate Wnt
responsiveness. This promoter binds to zebrafish Lef1 protein in vitro, and a nacre
communicator construct is powerfully inhibited by dominant-negative Tcf in malignant
melanoma cells. Mutation of Tcf/Lef sites abolishes Lef1 binding and reporter operates in vivo.
Wnt signaling so directly activates shell, which in flip ends up in pigment cell differentiation.
It was determined that the human melanocyte-specific MITF promoter contains many purported
Tcf/Lef binding sites, hallmarks of Wnt/-catenin-mediated transcriptional control. To test the
hypothesis that nacre could be a target of the Wnt/-catenin pathway in zebrafish, a
promoter/enhancer region is isolated for this sequence from a genomic bacteriophage library.
This genomic fragment contains the translation initiation site for nacre which is associated with a
836-bp upstream region with a possible transcription begin site, a cAMP response element
(CRE) and several agreement Tcf/Lef binding sites. These CRE and Tcf/Lef locations are not
fully preserved between the human and zebrafish promoters in either their sequences or their
positions from the transcription begin site. To examine whether the purported putative nacre
promoter/enhancer will direct correct abstraction management of organic phenomenon, injected
one -cell zebrafish embryos with the 836-bp deoxyribonucleic acid fragment driving a green
fluorescent macromolecule (GFP) communicator sequence. Mosaic gene expression is typical for
transient transgenics in zebrafish and because the neural crest springs from only a few precursor
cells, reporter potency is minimized. Despite these, it was ready to analyze the fate of the few
reporter-expressing cells in injected animals..
Write a Java program that will incorporate either a sequential searc.pdf
Write a Java program that will incorporate either a sequential search method or a sort method
Solution
class Search
{
public static void main(String args[])
{ int array[],n;
System.out.println(\"Please enter the sixe of array\");
Scanner s = new Scanner(System.in);
int n = s.nextInt();
System.out.println(\"Please enter\"+ n +\"elements of array\");
for(int i= 0;i.
Answer only with excel instructions.Assume that 96 of ticket hold.pdf
Answer only with excel instructions.
Assume that 96% of ticket holders show up for an airline flight. If a plane seats 200 people, how
many tickets should be sold to make the chance of an overbooked flight (more ticket holders
show up for a flight than there are seats available) is at most 10%.
Solution
Say no. of tickets booked = x
Now only 0.96x will turn up
No. of seats = 200
Now these 0.96x people should get a seat with 90% probability
200 / (0.96x) >=0.9
200/0.864 >= x
x <=231.48
x = 231
Airline will make 231 reservations.
A new species of sea horse has been discovered and is being sequenced.pdf
A new species of sea horse has been discovered and is being sequenced The genome of the new
species is being aligned using the reference sequence from a already-known species of sea horse
Since the two species are evolutionary diverged, there may be some differences between their
sequences Reference sequence: ACCTACATTCACAGAATGAT Fragments a) Align the set
of mini-fragments above. b) How are the sequences of the two sea horse species different? c)
Would this alignment have been easier or more difficult if you didn\'t have any reference
sequence at all (alignment without a reference is called \"de novo\")? Explain d) If there had
been sequencing errors, would this alignment have been easier or more difficult? Explain.
Solution
A) TAGATT TGAATG TCACTGA
B) There is a form of polymorphism occurring in the new sequence of the genes for the new
species of sea horses. There is single nucleotide polymorphism in the new species as C is
replaced by T and A instead of G which may probably occurred due to some sort of mutation
during genetic sequencing.
C) The given alignment would be difficult without reference DNA sequence as reference
sequence gives basic DNA information and a reference database techniques is much easier than
performing a de novo technique though de novo methods are well established using mass spectra
nowadays but still having a reference database helps and gives the better coding information of
protein sequence.
D) if there had been a sequence error there would be difficulty in alignment and in that case we
have to perform de novo sequencing using mass spectra analysis. To know the sequence of the
species and other information regarding the DNA characters..
Which of the following is the mechanism of action for Herceptin Her.pdf
Which of the following is the mechanism of action for Herceptin? Herceptin binds to the HER2
promoter to prevent transcription, thus lowering amounts of receptor. Herceptin binds to
intracellular HER2 proteins to prevent cancer cells from dividing. Herceptin binds to HER3 and
HER4 cell surface receptors to activate the immune system. Herceptin binds to extracellular
domain of HER2 and prevents internalization and subsequent cancer cell division. Herceptin
targets TNF alpha.
Solution
B) herceptin binds to intracellular ER2 proteins to prevent cancer cells from dividing
Trastuzumab (herceptin) activates p27 factor by inhibiting PI3K/Akt, Mirk and hKIS pathways.
HER2 protein can be expressed up to 100 times more in cancer cells. PI3K/Akt, Mirk and hKIS
are intracellular proteins in HER2 pathways are activated which promote cell proliferation.
Signals present in these pathways promote growth of blood vessels in order to supplement for
tumour formation or angiogenesis and therefore resulting in cell proliferation. Thus, when
herceptin inhibits PI3K/Akt, Mirk and hKIS pathways, the cell proliferation is inhibited..
Which of these statements best describes an element An element is c.pdf
Which of these statements best describes an element? An element is composed of three
subatomic particles. An element is formed when two atoms of the same kind are bonded
together. An element is formed when two or more different atoms combine chemically. An
element is composed of identical atoms.
Solution
An element is composed of identical atoms.
Molecules can be made of elements, but elements arent made from molecules. 2 or more atoms
of the same kind make up an element.For example H is hydrogen which is an element formed
from H atoms only whereas H2O is a compount formed from 2 different atoms H and O..
What is the general goal for descriptive statistics How is the goal.pdf
What is the general goal for descriptive statistics? How is the goal served by putting scores in a
frequency distribution? How is the goal served by computing a measure of central tendency?
How is the goal served by computing a measure of variability?
Solution
What is the general goal of descriptive statistics ?
Answer : To simplify , organize, and summarize data so that it is easier for researchers to see
patterns.
How is the goal served by putting scores in a frequency distribution?
A frequncy distribution provides an organized summary of the complete set of scores.
How is the goal served by computing a measure of central tendency?
Summarizes an entires set of scores with a single value that is representative of the whole set.
How is the goal served by computing a measure of variability?
Provides a single number that describes the differences that exist from complete score to another..
What effects are shown on this graph (assume that all effects where.pdf
What effects are shown on this graph? (assume that all effects where the marginal means differ
are significant)
Choose as many as applicable.
A. Main effect of A where scores are higher at A1 than A2
B. Main effect of A where scores are higher at A2 than A1
C, Main effect of B where scores are higher at B1 than B2
D, Main effect of B where scores are higher at B2 than B1
E. Interaction where the effect of A is larger at B1 than B2
F. Interaction where the effect of A is larger at B2 than B1
Solution.
What are adaptive peaks How do organisms move from one adaptive pea.pdf
What are adaptive peaks? How do organisms move from one adaptive peak to another? 100
words or more please
Solution
An equilibrium state in a population when the gene pool has allele frequencies that maximize the
average fitness of a population\'s members. In other words, A high point (perhaps one of several)
on an adaptive landscape; selection tends to drive the genotype composition of the population
toward a combination corresponding to an adaptive peak.
Most extant organisms are thought to sit on the “peaks” of such adaptive landscapes. However,
how does a phenotype shift from one peak to another is not clear. Common hypotheses include
(1) There are “ridges” connecting the adaptive peaks, such that phenotypes along the ridge are
more or less equally fit under a specific environment;
(2) There are no ridges connecting the peaks but when effective population size is small enough,
phenotypes may be able to traverse the less-adaptive region (i.e. adaptive valley) by genetic drift..
We are dealing with Verilog language.......What is a major design .pdf
We are dealing with Verilog language.......
What is a major design advantage when using a ripple counter?
Design simplicity.
Glitch free.
High speed.
Outputs are all synchronized.a.
Design simplicity.b.
Glitch free.c.
High speed.d.
Outputs are all synchronized.
Solution
advantages of design of asynchronous(ripple) counter in verilog code are
1.Removes reset from the data path.
2.multi clock systems needs only one reset.
3.Faster in speed.
So answer is (c).
The text defines three service layers when describing Cloud Computin.pdf
The text defines three service layers when describing Cloud Computing. Identify and describe
these three service layers. Which Service Layer is most likely to be consumed via Public Cloud
service and why? You need to defend your answer.
Solution
First we’ll go high level. Cloud is made feasible through the deployment and interoperability of
three platform types. These three layers are:
IaaS - Infrastructure as a Service
PaaS - Platform as a Service
SaaS - Software as a Service
Now this stack is easily broken down as follows: Think of the “Infrastructure-as-a-Service” as
the road. It’s the basis for communication. It’s the bottom layer that you build your platform on.
The platform are the cars traveling on the infrastructure. PaaS rides on IaaS. But on the top of
that, the goods and passengers inside the cars are the SaaS. It’s the end user experience. It’s the
end result. Let’s take that a step further.
1) Infrastructure-as-a-Service (IaaS)
Cloud Providers offering Infrastructure as a Service tout data-center space, and servers; as well
as network equipment such as routers/switches and software for businesses. These data-centers
are fully outsourced, you need not lift a finger, upgrade an IOS or re-route data. Although this is
the base layer, it allows for scalability and reliability; as well as better security than an
organization may have in a local co-lo or local datacenter. In addition, these services are charged
as utilities, so you pay for what you use, like your water, electric and gas. Depending on your
capacity or usage, your payment is a variable.
Because the IaaS vendors purchase equipment in such bulk, you, Mr. Customer, get the best gear
for the lowest price. Hence, the financial benefits of IaaS are cheaper access to infrastructure.
With the pay-as-you-go model, instead of investing in a fixed capacity infrastructure, which will
either fall short or exceed the organizational need, customers are able to save quite a bit of coin.
Buying hardware that’s barely used is a waste of hardware, air conditioning, space and power.
Operational expenses versus Capital expenses: Cloud is better. Because these computing
resources are basically used and paid for like a utility they can be paid via the operating
expenditures budget versus being paid for via capital investments. In other words, instead of
depreciating the gear over three years, you’re able to expense the monthly charge this year. And
the next year. And the year after that. It’s an elastic service.
1.1)
Application-level security
This level is primarily about identity and access management. Your best investment here isn’t
financial; it’s a policy that limits the ability of users to deploy cloud applications without IT’s
knowledge.
After ensuring policies are in place that offer IT visibility, the next step is to invest in multifactor
authentication and identity management. The first approach uses two or more devices or
applications to permit access. For example, a verification cod.
Social support plays a critical role in successful aging. Is this tr.pdf
Social support plays a critical role in successful aging. Is this true or false? Why?
Solution
Social support plays a critical role in successful aging. The statement is true.
Aging refers to the process of becoming older over a period of time.This includes the changes in
physical , psychological and social changes.At any stage of life social support play a crucial role
in the success of a person this is even more true in case of older adults.with the advancement of
age there are several changes that occur in the life which include changes in the job status, a
person may retire and yet may have family financial commitments to be fulfilled,the children
would have grown up and for their future they might have left the parents for their life
settlements , deterioration of physical health, financial status ,empty nest syndrome, felling
isolated and left out all these may also lead to psychological depression.
A positive approach to life is a very important factor in the process of aging. in older adults the
body immune system becomes fragile , resources for support become depleted and such other
causes will have an effect in the process of aging.Psychological well being is an indication of
successful adaptation to old age .
Acceptance to change as a part of life provides strength in overcoming several psychological
problems and studies have indicated that adults who are active are more likely to be successful in
their old age than those who expect a leisurely and lazy dependent life as this tend to negative
interactions with the family mainly.
As people age remaining socially active provides a positive health benefits to the aged people.
Successful connectedness,isolation, mobility, spirituality all have a say in dealing with the
geriatric problems.one can adapt themselves to the age by being socially active .Voulenteering is
a notable factor in relation to social support for adults.It helps in building companionship and
interact with more people of the community.They can engage themselves in spiritual activities ,
develop more realistic goals ,opportunities for self discovery.
Physical health plays a very important role in successful aging,with the advancement of age
.social support in terms of maintaining day to day affairs also plays a crucial role and studies
have suggested that it also helps in enhanced well being , less depression and often provides a
senses of meaning to life.
To conclude aging is a general process and acceptance with grace provides the main solution for
successful aging and the social support either monetary, physical or psychological all play a very
important role in helping the elderly people in the late years of their life..
sch the Basic Principles of Science 1erms with their prop.pdf
sch the Basic Principles of Science 1erms with their proper defrition. Testable
undeseindiaadammtal processes natural lawaby questions suce, the wok taday are the iame
ihaped the wodd the part, and they will continar Proof is elusive prefmible This nale ilse afire the
English philosophe dange as new Parsimony Repeatability then the eedeseu cannot be
Solution
1. Testable question - 7
Uncertainity - 4
Proof is elusive - 6
parsimony - 3
Repeatability - 5
Empiricism - 1
Uniformitarianism - 2
2. Emergent properties
3. Scientific theory.
A very thin slice is removed from a peanut and treated with Sudan II.pdf
A very thin slice is removed from a peanut and treated with Sudan III stain. Then a drop of
Biuret reagent is added to the peanut slice. When you examine the peanut slice under a
microscope, patches of red and blue-violet are visible. What conclusions can you draw from your
examination?
Solution
Answer:
When Biuret reagent is added to peanut slices, it turns into Blue-Violet colour indicating that
presence of the proteins in peanut.
When peanut is treated with Sudan III stain, slice turns into red colour indicating the presence of
the Lipids..
Question 1. CAM plants have adapted to very dry climates by opening .pdf
Question 1. CAM plants have adapted to very dry climates by opening their stomata only during
the day.
True
False
2 points
QUESTION 2
If the haploid number of chromosomes in an organism is 16, then the diploid number is
24
4
8
32
2 points
QUESTION
In human blood groups, the A, B, and O alleles are codominant.
True
False
2 points
QUESTION
If two plants that show incomplete dominance are crossed, one having red flowers and the other
pink flowers, what will be the phenotypic ratio of their offspring?
1 pink:2 white:1 red
1 red:2 pink:1 white
1 white:2 red:1 pink
2 red:2 pink
2 points
QUESTION
DNA transcription takes place in the nucleus.
True
False
2 points
QUESTION
Which of the following is not haploid?
sperm
gamete
zygote
egg
24
4
8
32
Solution
Ans. 1. False
CAM plants are adapted to very dry climates by opening their stomata during the night. Night
temperature is relatively much lower than that of the day time. So, opening stomata during night
time cause less evaporation (low loss of water) loss of water during transpiration than day time.
Ans. 2. True.
Alleles are said to be co-dominant when both alleles are expressed simultaneously in
heterozygous condition and both the traits retain their identity. For example, in AB blood group,
both the antigens A (gene product of allele IA) and B (gene product of allele IB) are
simultaneously expressed and independently present on RBC cell surface.
In a complete dominance scenario, only the dominant allele is expressed while the recessive is
NOT expressed in heterozygous condition.
Ans. 3. Incomplete dominance shows simultaneously expression of two alleles in the
heterozygous condition, but the two traits blend together to give a new phenotype.
Consider the difference of crosses between homozygous red and homozygous white flower as
RR x rr = RR, Rr, Rr, rr. -
I. Complete dominance: Rr progeny = red flower as that of dominant allele.
II. Incomplete dominance: Rr progeny = Pink flower as blending of red and white color because
R and r are expressed simultaneously.
III. Co-dominance: Rr progeny = A flower with white and red patches of color- both R and r are
expressed simultaneously and both the phenotypes retain their identity.
Now, coming to the question-
Genotype of homozygous red flower: RR
Genotype of homozygous red flower: Rr
Now,
RR x Rr = RR, Rr ; [RR = red ; Rr = pink]
Phenotype ratio = 1: 1 or 2: 2 or 3:3 ; [Note: multiplying a ratio with any factor does
not affect the ration. SO, 1:1 = 2: 2 = 3: 3, and so on].
Ans. 4. True
DNA transcription i.e. formation of RMA from DNA is called transcription. Because DNA is
present in nucleus and does not come out in cytoplasm at any point, the transcription must occur
in nucleus.
Ans. 5. Correct option C. zygote.
Sperm (n) + egg (n) -------> Zygote (2n). Fusion of egg with sperm gives diploid zygote.
Sperm and egg are gametes- all being haploid (n)..
Programming in C.Create a program which turns your Arduino into a .pdf
Programming in C.
Create a program which turns your Arduino into a light based electronic instrument. The
program should play music based on the amount of light hitting the optical sensor. There should
be a way to capture particularly pleasing tones and play those tones back in the order they are
captured.
The program must utilize the following :
• An array to hold the stored tones
• A function : void read_buttons( int *sw1, int *sw2, int *sw3, int *sw4) to read and store the
state of the four buttons on the board
• A function playback( notes[] ) which plays back the notes stored
• A function clear_playlist( *notes) which clears the stored tones
• A function tone_value() which generates a tone based on the current state of the optical sensor
• The possible tones should span the range of human hearing
Solution
#include
#include
using namespace std;
// Class Declaration
class prime
{
//Member Varibale Declaration
int a,k,i;
public:
prime(int x)
{
a=x;
}
// Object Creation For Class
void calculate()
{
k=1;
{
for(i=2;i<=a/2;i++)
if(a%i==0)
{
k=0;
break;
}
else
{
k=1;
}
}
}
void show()
{
if(k==1)
cout<<\"\ \"<>a;
// Object Creation For Class
prime obj(a);
// Call Member Functions
obj.calculate();
obj.show();
getch();
return 0;
}.
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