A bacterium that has flagella arranged as a tuft at one end of the b.pdf
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A bacterium that has flagella arranged as a tuft at one end has a lophotrichous flagella arrangement. Lophotrichous means several flagella extend from one end of the cell in a tuft. Flagellum is a thread-like structure that enables bacteria and other microorganisms to swim.
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In the negative feedback control system of the papillary light refle.pdf
In the negative feedback control system of the papillary light reflex, what are the physiological
processes that represent what engineers call the controller, plant, feedback element, controlling,
controlled, feedback variables?
Solution
EdingerWestphal nuclei of brain (controller) lead to change in neural traffic down the efferent
nerves back to the eyes (controller action). It affects the plant retina by feedback sensors
sphincter muscles and radial dilator muscles which relax and contract respectively. The output is
increase in pupil area..
images of a Hindu deity such as Shiva, Vishnu, Kali, Ganesh, or any .pdf
images of a Hindu deity such as Shiva, Vishnu, Kali, Ganesh, or any other Hindu deity that
interests you. What did you notice about these images? Were there common features that you
could identity from image to image?
Solution
Images of the deities portray them as the supreme power. They all fought against Asura or the
Demons. They all have some or other kind of Astra or weapons to fight. Trishula being supreme
and most powerful weapon. Another common thing they all have is they all have a Vahana i.e
vehicle of gods and these vahanas all are animals..
Identify each of the enzymes involved in DNA replication and whether.pdf
Identify each of the enzymes involved in DNA replication and whether they are used during
synthesis of the leading strand, lagging strand, or both
Solution
ANS:
REPLICATION:
Replication is a process of duplication of DNA .the replication had specific time in S
phase. All the cellular molecules had timing programed inside the cells, the time depends on
species. Replication occur during the S phase of eukaryotic cell cycle.it is highly accurate
process and involved the faithful transmission of the genetically information from parental DNA
to two daughter DNA duplexes.
Replication involves the polymerization of precursor of d NTPs (d ATP, d GTP, d CTP, d TTP).
B. ENZYMOLOGY IN REPLICATION:
Replication is a complex process the various enzymes/proteins are involved in this.
1. DNA polymerase:
It can polymerized the d NTP’s directed by template strand of DNA so called DNA
DEPENDENT DNA POLYMARASE.
In eukaryotes 5 different DNA polymerase are there (Pol Alfa, Pol Beta, Pol Gama, Pol Delta
and Pol Epsilon).
DNA Pol Alfa can initiate replication in both leading and lagging strands, it dual activity so often
called as Pol Alfa/ primase enzyme.
DNA pol Delta is a major replication enzyme involved in the replication of both leading and
lagging strands.
2. DNA A protein: also called initiation protein, it involved in recognition & binding to origin.
3. DNA HELICASE: 1st component in replisoma, it possess an ATPase activity, it separates the
2 parental strands of DNA, It always binds to lagging strand.
4. PRIMASE: Involved in synthesis of RNA primer, always binds to helicase results the
formation of primosome.
5. SSB protein: It prevents reannealing of DNA, it removes the secondary structure prevent in
DNA.
6. DNA Topoisomerase 1: Relaxes tension due to un coiling of dna.it binds downstream of
replication fork.
7. DNA Gyrase: It binds up stream of replication fork and introduces Negative supercoiling.
8. DNA LIGASE: It covalently seals the nick between the adjacent okazaki fragments.
9. Termination protein: Also called TUS protein, binds the TER sites results termination.
10. DNA Topoisomerase 4: Involved in decapitation of the 2 circular daughter DNA duplexes
obtained after replication.
Synthesis of the leading strand and lagging strand:
Semi –conservative mechanism:
In this the 2 strands of DNA separated from each other and each strand act as template for
synthesis of anew complimentary strand ,such daughter duplex contain 1 parental and 1 newly
synthesize strand.
Replication always began from fixed point called origin.as unwinding DNA produced a Y
shaped junction called REPLICATIONPORK. That represents origin of replication.
As the replication pork moved in one direction the DNA copies continuously in both strands.
Then the replication missionary should be able to synthesize in both 5\'-3\' and 3\'-5\' direction.
But the DNA polymerase involved in replication can added a new nucleotide only to 3\'end of
growingchain.so DNA always synthesized in 5\'-3\' .
How and what do sea stars eatSolutionSea Star eat clams, oyste.pdf
How and what do sea stars eat?
Solution
Sea Star eat clams, oysters sand dollars and mussels, They all are attached to rocks or move very
slowly. They can eat injured fish and snails also. but not all starfish eat Fishes, however. Some
species of sea star find decomposing plants and fishes or other sea animal in the sea water and
consume them.
Most of Sea Star eat by prying inside their stomach which have following Step.
Sea star also digesting prey from outside their bodies which help to sea star to eat larger animal..
Estimate the time needed for the Point in Polyhedron test for a poly.pdf
Estimate the time needed for the Point in Polyhedron test for a polyhedron with v vertices, e
edges, and f faces? Please explain in details.
THIS IS WHAT I KNOW.
The line L intersects P if and only if there is a face F that has a point q in common with L. On the
plane containing the face F, + defines a ray beginning at q. This ray intersects F in an odd
number of sides, since q lies inside F. It follows that + intersects an odd number of edges
bounding F.
If, however, the line and a face do not intersect, the half-plane can intersect only an even number
of edges of this face. This proves the theorem.
This theorem provides the basis for the following algorithm which solves the intersection of a
line and a polyhedron problem.
Consider an edge of P and test for intersection with +. If the edge and the half-plane have only
one point in common, then the counters for the two faces containing the edge are incremented by
1 (the counters for the faces of P are initially zeroed.) If after all of the edges have been tested at
least one of the counters has an odd value, then L and P intersect.
This algorithm is based on a simple test for the intersection of an edge (line segment) and a half-
plane.
Consider a point p and a polyhedron P. Take an arbitrary line L passing through p and an
arbitrary plane passing through L. Let L+ denote one of the parts of L divided by p, and let + be
one of the half-planes obtained by dividing with L. We say that an edge e of P is marked if +
intersects e and only one of the two planes of faces of P that share e intersects L+. Suppose +
does not pass through vertices of P.
Then p belongs to P if and only if the number of marked edges is odd, and p lies outside P if this
number is even.
Solution
and look at the network we get after performing Step 1 just once. Now, by drawing a diagonal
we added one edge. Our original face has become two faces, so we have added one to the
number of faces. We haven\'t changed the number of vertices. The network now has V vertices,
E + 1 edges and F + 1 faces. So how has V - E + F changed after we performed Step 1 once?
Using what we know about the changes in V, E and F we can see that V - E + F has become V -
(E + 1) + (F + 1). Now we have
V - (E + 1) + (F + 1) = V - E - 1 + F + 1 = V - E + F..
Discuss the binary representation of Char strings. What makes th.pdf
Discuss the binary representation of Char strings. What makes this tricky?
Discuss the binary representation of Char strings. What makes this tricky?
Discuss the binary representation of Char strings. What makes this tricky?
Solution
void strings_to_binary(char* st)
{
// temporary buffer for conversion
char temp[9];
while (*st)
{
// itoa is built in function that converts to bits
itoa(*st, output, 2);
putchar(output);
putchar(\'/n\')
++st;
}
}
output:
strings_to_binary(\'efg\');
01100101
01100110
01100111.
Describe a possible difference in the DNA content of chromosome 1.pdf
Describe a possible difference in the DNA content of chromosome 1 between a chromatid in a
somatic cell and a monad in a sperm cell.
Describe a possible difference in the DNA content of chromosome 1 between a chromatid in a
somatic cell and a monad in a sperm cell.
Solution
Answer:
Chromatid of somatic cell consists of two two DNA molecules whereas monad of a sperm cell
consists of a single DNA..
Consider the following function. Find the inverse function State the.pdf
Consider the following function. Find the inverse function State the domain and range of f.
(Enter your answers using interval notation.)
Solution
Fisrt part is correct ie. F^-1 = sqrt(7x +7)
domain and range of function f (x) = 1/2 x^2 - 1 is all values of x greater than equal to 0
domain ( -inf, +inf)
Range : find Y intercpt by plugging x=0 we get y =-1
[ -1, + inf)
Domain of inverse function: Check where function does not exist: 7x+7 >=0
x> = -1
Domain:: [ -1 , inf)
Range : f(x) exist for >=0
So, [0, inf).
Recommended
In the negative feedback control system of the papillary light refle.pdf
In the negative feedback control system of the papillary light reflex, what are the physiological
processes that represent what engineers call the controller, plant, feedback element, controlling,
controlled, feedback variables?
Solution
EdingerWestphal nuclei of brain (controller) lead to change in neural traffic down the efferent
nerves back to the eyes (controller action). It affects the plant retina by feedback sensors
sphincter muscles and radial dilator muscles which relax and contract respectively. The output is
increase in pupil area..
images of a Hindu deity such as Shiva, Vishnu, Kali, Ganesh, or any .pdf
images of a Hindu deity such as Shiva, Vishnu, Kali, Ganesh, or any other Hindu deity that
interests you. What did you notice about these images? Were there common features that you
could identity from image to image?
Solution
Images of the deities portray them as the supreme power. They all fought against Asura or the
Demons. They all have some or other kind of Astra or weapons to fight. Trishula being supreme
and most powerful weapon. Another common thing they all have is they all have a Vahana i.e
vehicle of gods and these vahanas all are animals..
Identify each of the enzymes involved in DNA replication and whether.pdf
Identify each of the enzymes involved in DNA replication and whether they are used during
synthesis of the leading strand, lagging strand, or both
Solution
ANS:
REPLICATION:
Replication is a process of duplication of DNA .the replication had specific time in S
phase. All the cellular molecules had timing programed inside the cells, the time depends on
species. Replication occur during the S phase of eukaryotic cell cycle.it is highly accurate
process and involved the faithful transmission of the genetically information from parental DNA
to two daughter DNA duplexes.
Replication involves the polymerization of precursor of d NTPs (d ATP, d GTP, d CTP, d TTP).
B. ENZYMOLOGY IN REPLICATION:
Replication is a complex process the various enzymes/proteins are involved in this.
1. DNA polymerase:
It can polymerized the d NTP’s directed by template strand of DNA so called DNA
DEPENDENT DNA POLYMARASE.
In eukaryotes 5 different DNA polymerase are there (Pol Alfa, Pol Beta, Pol Gama, Pol Delta
and Pol Epsilon).
DNA Pol Alfa can initiate replication in both leading and lagging strands, it dual activity so often
called as Pol Alfa/ primase enzyme.
DNA pol Delta is a major replication enzyme involved in the replication of both leading and
lagging strands.
2. DNA A protein: also called initiation protein, it involved in recognition & binding to origin.
3. DNA HELICASE: 1st component in replisoma, it possess an ATPase activity, it separates the
2 parental strands of DNA, It always binds to lagging strand.
4. PRIMASE: Involved in synthesis of RNA primer, always binds to helicase results the
formation of primosome.
5. SSB protein: It prevents reannealing of DNA, it removes the secondary structure prevent in
DNA.
6. DNA Topoisomerase 1: Relaxes tension due to un coiling of dna.it binds downstream of
replication fork.
7. DNA Gyrase: It binds up stream of replication fork and introduces Negative supercoiling.
8. DNA LIGASE: It covalently seals the nick between the adjacent okazaki fragments.
9. Termination protein: Also called TUS protein, binds the TER sites results termination.
10. DNA Topoisomerase 4: Involved in decapitation of the 2 circular daughter DNA duplexes
obtained after replication.
Synthesis of the leading strand and lagging strand:
Semi –conservative mechanism:
In this the 2 strands of DNA separated from each other and each strand act as template for
synthesis of anew complimentary strand ,such daughter duplex contain 1 parental and 1 newly
synthesize strand.
Replication always began from fixed point called origin.as unwinding DNA produced a Y
shaped junction called REPLICATIONPORK. That represents origin of replication.
As the replication pork moved in one direction the DNA copies continuously in both strands.
Then the replication missionary should be able to synthesize in both 5\'-3\' and 3\'-5\' direction.
But the DNA polymerase involved in replication can added a new nucleotide only to 3\'end of
growingchain.so DNA always synthesized in 5\'-3\' .
How and what do sea stars eatSolutionSea Star eat clams, oyste.pdf
How and what do sea stars eat?
Solution
Sea Star eat clams, oysters sand dollars and mussels, They all are attached to rocks or move very
slowly. They can eat injured fish and snails also. but not all starfish eat Fishes, however. Some
species of sea star find decomposing plants and fishes or other sea animal in the sea water and
consume them.
Most of Sea Star eat by prying inside their stomach which have following Step.
Sea star also digesting prey from outside their bodies which help to sea star to eat larger animal..
Estimate the time needed for the Point in Polyhedron test for a poly.pdf
Estimate the time needed for the Point in Polyhedron test for a polyhedron with v vertices, e
edges, and f faces? Please explain in details.
THIS IS WHAT I KNOW.
The line L intersects P if and only if there is a face F that has a point q in common with L. On the
plane containing the face F, + defines a ray beginning at q. This ray intersects F in an odd
number of sides, since q lies inside F. It follows that + intersects an odd number of edges
bounding F.
If, however, the line and a face do not intersect, the half-plane can intersect only an even number
of edges of this face. This proves the theorem.
This theorem provides the basis for the following algorithm which solves the intersection of a
line and a polyhedron problem.
Consider an edge of P and test for intersection with +. If the edge and the half-plane have only
one point in common, then the counters for the two faces containing the edge are incremented by
1 (the counters for the faces of P are initially zeroed.) If after all of the edges have been tested at
least one of the counters has an odd value, then L and P intersect.
This algorithm is based on a simple test for the intersection of an edge (line segment) and a half-
plane.
Consider a point p and a polyhedron P. Take an arbitrary line L passing through p and an
arbitrary plane passing through L. Let L+ denote one of the parts of L divided by p, and let + be
one of the half-planes obtained by dividing with L. We say that an edge e of P is marked if +
intersects e and only one of the two planes of faces of P that share e intersects L+. Suppose +
does not pass through vertices of P.
Then p belongs to P if and only if the number of marked edges is odd, and p lies outside P if this
number is even.
Solution
and look at the network we get after performing Step 1 just once. Now, by drawing a diagonal
we added one edge. Our original face has become two faces, so we have added one to the
number of faces. We haven\'t changed the number of vertices. The network now has V vertices,
E + 1 edges and F + 1 faces. So how has V - E + F changed after we performed Step 1 once?
Using what we know about the changes in V, E and F we can see that V - E + F has become V -
(E + 1) + (F + 1). Now we have
V - (E + 1) + (F + 1) = V - E - 1 + F + 1 = V - E + F..
Discuss the binary representation of Char strings. What makes th.pdf
Discuss the binary representation of Char strings. What makes this tricky?
Discuss the binary representation of Char strings. What makes this tricky?
Discuss the binary representation of Char strings. What makes this tricky?
Solution
void strings_to_binary(char* st)
{
// temporary buffer for conversion
char temp[9];
while (*st)
{
// itoa is built in function that converts to bits
itoa(*st, output, 2);
putchar(output);
putchar(\'/n\')
++st;
}
}
output:
strings_to_binary(\'efg\');
01100101
01100110
01100111.
Describe a possible difference in the DNA content of chromosome 1.pdf
Describe a possible difference in the DNA content of chromosome 1 between a chromatid in a
somatic cell and a monad in a sperm cell.
Describe a possible difference in the DNA content of chromosome 1 between a chromatid in a
somatic cell and a monad in a sperm cell.
Solution
Answer:
Chromatid of somatic cell consists of two two DNA molecules whereas monad of a sperm cell
consists of a single DNA..
Consider the following function. Find the inverse function State the.pdf
Consider the following function. Find the inverse function State the domain and range of f.
(Enter your answers using interval notation.)
Solution
Fisrt part is correct ie. F^-1 = sqrt(7x +7)
domain and range of function f (x) = 1/2 x^2 - 1 is all values of x greater than equal to 0
domain ( -inf, +inf)
Range : find Y intercpt by plugging x=0 we get y =-1
[ -1, + inf)
Domain of inverse function: Check where function does not exist: 7x+7 >=0
x> = -1
Domain:: [ -1 , inf)
Range : f(x) exist for >=0
So, [0, inf).
Compare and contrast the ideas that led to the Bohr model of the ato.pdf
Compare and contrast the ideas that led to the Bohr model of the atom to the ideas that led to the
quantum mecchanical theory that gave rise to the hydrogen atom wave functions.
Solution
The difficulty with the planetary picture provided before the Bohr model is that it is inconsistent
with a well established fact of classical electrodynamics which says that an accelerated electric
charge must continually radiate energy. If electrons actually followed such a trajectory, all atoms
would act as source of radiation. The radiated energy would come from the kinetic energy of the
orbiting electron; as this energy gets radiated away, As a result the electron would quickly fall
into the nucleus. According to classical physics, no atom based on this model could exist for
more than a brief fraction of a second.
In contrary The Bohr model is a mixture of classical physics and quantum physics. The main
postulated of Bohr\'s orbit are as follows:
1.) The electron in the atom moves in a circular orbit centred on its nucleus. Its motion in the
orbit is governed by the Coulomb electric force between the negatively charged electron and the
positively charged proton.The radius of orbit can be given as [r = nh/(2*pi*m*v), where n is
principal quantum number.Emission spectrum of hydrogen is nicely explained by Bohr model.
However, one serious difficulty with the Bohr model was that it was unable to explain the
spectrum of any atom more complicated than hydrogen.
2.) An electron in a Bohr orbit does not continuously radiate electromagnetic radiation. Its
energy is therefore constant. The orbit is referred to as a stationary orbit.
3.) Electromagnetic radiation isonlyemitted when the electron changes from one orbit to another
of a lower total energy. (The electron is said to undergo atransition.) In such a case, the energy
lost, E, is emitted asonequantum of radiation of frequencyfas given by thePlanck–Einstein
formula:
E = hf
Although the Bohr model hypothesis for the quantization of angular momentum can be justified
in terms of electron wave ideas, the Bohr model remains profoundly unsatisfactory as a wave
model for the atom and also was not able to answer that why it doesn\'t fall into the nucleus. The
main failures of Bohr\'s model are:
- a mixture of classical and quantum ideas (electrons move classically on orbits, but their
possible energy states are quantified)
- postulates that on the allowed orbits electrons do not radiate
(conflict with Maxwell’s theory)
- could not account for the maximal electron numbers on one shell
- could not explain splitting of the spectral lines in magnetic fields
- it is a non-relativistic theory although the speed of the electrons is close to c
In the Quantum Mechanical Model, the electron is treated mathematically as a wave. The
electron has properties of both particles and waves. The advantage of Quantum Mechanical
model over Bohr model is that it was successful to give explaination of the fact that electron does
not fall in.
Blackboard Learn Realtor Rite Aid Citizens Bank peoples CPOBIE UI.pdf
Blackboard Learn Realtor Rite Aid Citizen\'s Bank people\'s CPOBIE UI Blackboard Learn KN
Question Completion Status: QUESTION 19 For what values of x does the function shown in
the graph appear to be increasing? all values of no values of X 20
Solution
In the graph first we can see that the function is decreasing through out. So for no value of x the
function is increasing. Option D
In next graph the function is negative for x <0 so Option A.
According to the Shannon-Hartley theorem, the capacity of a communic.pdf
According to the Shannon-Hartley theorem, the capacity of a communications channel in bits per
second is given by C=Blog2 ((s/n)+1) where B is the frequency bandwidth of the channel in
hertz and s/n is its signal-to-noise ratio. It is physically impossible to exceed this limit. Solve the
equation for the signal-to-noise ratio s/n.
Solution
C = B*log2( 1+s/n)
divide both sides by B:
(C/B) = log2(1+s/n)
Use the log property : loga(b) = x ----: b = a^x
So, 1+ s/n = 2^(C/B)
( signal-to-noise ratio) S/N = 2^(C/B) -1.
A fellow student, one that is not as careful as you, finds the follow.pdf
A fellow student, one that is not as careful as you, finds the following: They have produced 2
plates of media, and inoculated each plate with three different bacterial samples. Unfortunately,
they have labeled none of their plates. The plates have the following appearances: Plate 1 The
media is a dark red (burgundy) color, 3 different bacteria were inoculated onto the plate and
bacteria are growing in all 3 of the sections of the plate. Section A: you observe growth of
bacteria with a translucent/clear region around all colonies. Section B: you observe growth with
no change in the color of the media around all colonies. Section C: you observe growth with a
greenish color around all colonies. Plate 2: The media is a pale purple color, 3 different bacteria
were inoculated onto the plate, and organisms are growing in 2 of the sections of the plate.
Section A: you observe no growth of bacteria at all. Section B: you observe bacterial growth
with pink colonies. Section C: you observe bacterial growth with colorless/white colonies.
Using the diagrams on the next page, COMPLETELY interpret these results-this means I want to
know what enzymes are active. First tell me what media are represented for each plate, what the
media\'s properties are (selective, differential, any pH indicators), whether the media is selective
and/or differential. Second fully interpret the type organisms that are present in each of the three
sections of each plate, including enzymes, and whether the bacteria is Gram positive or Gram
negative.
Solution
Answer:
Plate 1: The dark red colour media is Eosine-methylene blue (EMB) agar due to the presence of
methylene blue and eosine dye. It is a selective and differencial culture medium. It is used for
selective growth of gram negative bacteria and differentiate the growth of lactose fermenting and
lactose non-fermenting coliforms. The details is given below:
Lactose fermenter
E. coli
Explaination: In the presence of an acidic environment (low pH) after the fermentation of
lactose, EMB forms a complex that precipitates out onto the coliform colonies, producing dark
centers and greenish sheen around the colonies. This reaction is characterstics of Escherichia
Coli. In addition, other weaker fermentation of lactose results in colonies with a pinkish-purple
color with no change around the colonies as their is low change of pH of growth medium.
However, in absence of an acidic environment the dye is not absorbed. Thus, the colonies of
lactose non-fermentors will be colorless.
Plate 2: The pale purple colour media is MacConkey agar due to the presence of crystal violet
dye. It is a selective and differencial culture medium used for selective growth of gram negative
bacteria and differentiate the growth of lactose fermenting and lactose non-fermenting coliforms.
The details is given below:
Explaination: MacConkey agar is detect the growth of coliforms and enteric pathogens based on
their ability to ferment lactose. Lactose fermenting c.
2. There are special proteins (called transporters) that are able to.pdf
2. There are special proteins (called transporters) that are able to move protons (H+) into or out
of cells. These proteins can be seen to be at work if the pH inside and/or outside the cell changes.
However, the pH is only seen to go down outside the cell when ATP (adenosine triphosphate, an
energy providing molecule) is present. Is the proton movement an example of active or passive
movement? Which way are the protons moving? Is this an example of an endergonic or an
exogonic reaction?
Solution
Movement of proton is an example of primary active transport as it uses redox energy of
mitochondrial electron transport from NADH and makes protons moves across the inner
mitochondrial membrane against their concentration gradient. Movment of proton depends upon
the concentration gradient which way it is low it will move that way. As it uses the energy from
ATP it is considered to be endergonic reaction..
11. Recessive epistasis is described by all of the following EXCEPT.pdf
11. Recessive epistasis is described by all of the following EXCEPT:
A. it is observed in coat color in Labrador retrievers.
B. it has two genes interacting to produce three phenotypes instead of four.
C. it involves one gene controlling the expression of another.
D. it gives the modified Mendelian ratio of 12:3:1.
E. Actually, all of these describe recessive epistasis.
12. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce
a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and
1/16 white. What is the genotype of the white progeny?
A. aa bb
B. A_ B_ and A_ bb
C. A_ B_
D. A_ bb
E. aa B_
13. Traits that have differential expression in the two sexes are known as:
A. sex-linked.
B. sex-influenced.
C. autosomal.
D. sex-determined.
E. None of the above.
14. Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons
possess extra fingers or toes that are fully functional, whereas others possess only a small tag of
extra skin. This is an example of
A. complete dominance.
B. cytoplasmic inheritance.
C. complementation.
D. independent assortment.
E. variable expressivity.
15. A mother with blood type A has a child with blood type A. Give all possible blood types for
the father of this child.
A. A, AB
B. A, B, AB, O
C. O
D. B, AB
E. A, B, O
16. Considering cases of incomplete dominance:
A. the heterozygote has a phenotype intermediate between the homozygotes
B. the homozygote has a phenotype intermediate between the heterozygotes.
C. the heterozygote has a phenotype identical to one of the homozygotes.
D. the homozygote has a phenotype identical to one of the heterozygotes.
E. None of the above.
17. A mother with blood type A has a child with blood type AB. Give all possible blood types
for the father of this child.
A. A, B, AB, O
B. A, B, O
C. O
D. B, AB
E. A, AB
18. Gene interaction produces all of the following phenomena EXCEPT:
A. recessive epistasis.
B. duplicate recessive epistasis.
C. dominant epistasis.
D. duplicate interaction.
E. Actually, all of these are recognized forms of gene interaction
19. In order to determine if mutations from different organisms that exhibit the same phenotype
are allelic, which test would you perform?
A. Allelic series test
B. Biochemical test
C. Epistasis test
D. Complementation test
E. Test cross
20. Interactions among the human ABO blood group alleles involve _______ and ________.
A. complete dominance; incomplete dominance
B. co-dominance; complete dominance
C. epistasis; complementation
D. codominance; incomplete dominance
E. continuous variation; environmental variation
Solution
Answer:
11. (D) It gives the modified Mendelian ratio of 12:3:1
12. (B) A_ B_ and A_ bb
13. (B) Sex-influenced
Characters whose expression in a particular genotype depends on whether the individual is male
or, female, are known as sex influenced traits. In one of the sexes, an allele may be dominant
while it may be reces.
Write complete VHDL codes for the following schematic. Solution.pdf
Write complete VHDL codes for the following schematic.
Solution
for shift reg:
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
-- Uncomment the following library declaration if using
-- arithmetic functions with Signed or Unsigned values
--use IEEE.NUMERIC_STD.ALL;
-- Uncomment the following library declaration if instantiating
-- any Xilinx primitives in this code.
--library UNISIM;
--use UNISIM.VComponents.all;
entity pipo is
Port ( ClrN,load,clk : in STD_LOGIC;
D3,D2,D1,D0 : in STD_LOGIC;
Q3,Q2,Q1,Q0 : out STD_LOGIC);
end pipo;
architecture Behavioral of pipo is
component D_ff is
Port ( clock,ClrN : in STD_LOGIC;
D : in STD_LOGIC;
Q : out STD_LOGIC);
end component;
signal clock:STD_LOGIC;
begin
clock<= load and clk;
u3:D_Ff port map(clock,ClrN,D3,Q3);
u2:D_Ff port map(clock,ClrN,D2,Q2);
u1:D_Ff port map(clock,ClrN,D1,Q1);
U0:D_Ff port map(clock,ClrN,D0,Q0);
end Behavioral;
for D-Flipflop:
entity D_FF is
Port ( clock,ClrN : in STD_LOGIC;
D : in STD_LOGIC;
Q : out STD_LOGIC);
end D_FF;
architecture Behavioral of D_FF is
begin
process(clk,ClrN)
begin
if(ClrN = \'0\') then
if(falling_edge(clk)) then
Q <= D;
end if;
else
Q <= \'0\';
end if;
end process;
end Behavioral;.
Why do carrier-mediated processes become saturated (choose the best .pdf
Why do carrier-mediated processes become saturated (choose the best answer)? Question 8
options:
A:The carriers are very specific.
B:The number of carriers in a membrane is limited.
C:The concentration gradient is too strong.
D:The amount of ATP in the cell is limited.
E:The concentration gradient is too weak
Solution
Answer is E.
It is a passive process in which the solutes move down the concentration gradient..
Why is it that linked lists are better when it comes to needing freq.pdf
Why is it that linked lists are better when it comes to needing frequent add(), remove(), and get()
operations?
Solution
1.Linked list permits the constant time insertions or removals with iterators.
2.The items are able to add or remove from the center of the list.
3.interpretating the starting size of an array is not needed.
4.insert and delete manipulations are simply accomplished in a linkedlist.
5.allocation and deallocation is possible at the time of programm running.
6.the main advantage of using linkedlist is reusing of current iterators to insert and delete
element..
Which of the following isare characteristic of the parasympathetic .pdf
Which of the following is/are characteristic of the parasympathetic nervous system, but not of
the sympathetic nervous system?
A ganglia in or near target tissues
B nicotinic receptors on postganglionic neurons
C muscarinic receptors on some target tissues
D 1 receptors on some target tissues
E 1 receptors on some target tissues
Solution
A. ganglia in or near target tissues. The parasympathetic nervous system is a slower system that
moves along longer pathways. Preganglionic fibres from the medulla or spinal cord project
ganglia close to the target organ. They create a synapse, which eventually creates the desired
response..
What statistic is often interpreted as providing an estimate of the .pdf
What statistic is often interpreted as providing an estimate of the average difference between a
statistic and a parameter?
Solution
The standard deviation, which finds the root of the mean squared-difference. It is interpreted as a
measure of the average difference between the sample mean (statistic) and population mean
(parameter)..
The waiting time X until delivery of a new component for an indus.pdf
The waiting time X until delivery of a new component for an industrial operation is uniformly
distributed over the interval from 1 to 5 days. The cost of this delay is given by probability
density function for Y.
Solution
uniform distribution over the interval from 1 to 5
x = 1 / (5-1 ) = 1/4
y = 2* (1/4)^2 + 3
y = 3.125.
The Latin term a priori describes the origin of knowledge developed .pdf
The Latin term a priori describes the origin of knowledge developed rationally, and the term a
posteriori describes knowledge developed through observation and experience. What is produced
through a priori rational thought that applies to ethics? nalysis systems Principles of ethics
Concern for outcomes Emotions Subjective sense of identity
Solution
please explain the question.
Suppose X and Y are two independent random variables with the follow.pdf
Suppose X and Y are two independent random variables with the following moment generating
functions: Mx(t)= e^(-t+(t^2)), My(t)= e^(2t+1/2(t^2))
Let Z = X
Solution
Given Mx(t) = e-t+t^2)
= 1-t+t^2)+(-t+t^2)^2/2!+....
= 1-t+t^2+t^2/2+t^3+t^4/2+...
Coefficient of t = -1
Mean of x = -1
E(X^2) = coeff of t^2 =3/2
Variance x = 3/2-1 = 1/2
My(t) = e^(2t+t^2/2)
= 1+2t+t^2/2 +4t^2+...
Mean of Y = 2
E(Y^2) = 9/2
Var(Y) = 0.5
Hence Var (z)
= Var(x-5y) = Var(x)+25Var(Y)
= 1/2 + 25(0.5)
=0.5+12.5 = 13..
The (circle one meanmedian) would be a better measure of central ten.pdf
The (circle one mean/median) would be a better measure of central tendency if you are
describing the annual incomes of all people n Minnesota.
Solution
vaiables like Annual incomes have a lower limit (zero) but no upper limit, and thus have huge
outliers on right. Median is a better measure of central tendency for such skewed distributions..
Summary of Ethics and Information Security (one page)Solution.pdf
Summary of Ethics and Information Security (one page)
Solution
Ethics and Information Security
here Ethics is nothing the principles and standards that guide our behavior towards other people
secondly Intellectual property is creative work in physical form
Copyright is nothing but the legal protection afforded an expression of an idea, such as song,
video game, and sometypes of proprietary documents
Pirated software is the unauthorized use, duplication, distribution, or sale of copyrighted
software
Counterfeit software is a software that is manufactured to look like the real thing and sold as
such
here Privacy include
1. Right to be left alone
2. Control your own personal possessions
3. Not to be observedwithout your consent
Epolicies are policies and procedures that address the ethical use of computers and internet usage
in thebusiness environment
and CIO Magazine’s Size Principles for Ethical Information Management
finally Information is a valuable corporate assets and should be managed as such, like cash,
facilities, orany other corporate asset2.
The CIO is steward of corporate information and is responsible for managing it over its life
cycle-from its generation to its appropriate destruction3.
The CIO is responsible for controlling access to and use of information, as determined
bygovernmental regulation and corporate policy4.
The CIO is responsible for preventing the inappropriate destruction of information5.
The CIO is responsible for bringing technological knowledge to the development of
informationmanagement practices and policies6.
The CIO should partner with executive peers to
develop and execute the organization’s
information management policies
Ethical computer use policy
contains general principles to guide computer user behavior. Example : theethical computer use
policy might explicitly state that users should refrain from playing computer gamesduring
working hours
Information privacy policy
contains general principles regarding information privacy
Acceptable use policy (AUP)
is a policy that a user must agree to follow in order to be provided accessto a network or to the
internet.
Nonrepudiation is nothing but a contractual stipulation to ensure that ebusiness participants do
not deny their onlineactions.
suppose f is continuous on the open interval from negative infinity .pdf
suppose f is continuous on the open interval from negative infinity to positive infinity and
f(1)=1. I was given the hint that this problem involves using the concepts of limits, derivative,
integral, L\'Hopital\'s rule, chain rule, and the fundamental theorem of calculus.
Solution
apply l\' hospital -sinx^2 * 2x*f(cosx^2)/4x^3 -sinx^2**f(cosx^2)/2x^2 put the
limit -1/2.
So I already have most of the code and now I have to1. create an .pdf
So I already have most of the code and now I have to:
1. create an index file to keep track of all the inventory IDs and their locations
2. modify my class to be able to display any record with a given ID without searching through
the entire file
// This program displays the contents of the inventory file.
#include
#include
#include
using namespace std;
class DB{
// Declaration of InventoryItem structure
class InventoryItem
{
public:
char Id[5];
char desc[31];
int qty;
float price;
};
string DBname;
fstream inventory;// (DBname, ios::in | ios::binary);
public: DB(string N){ DBname = N; }// constructor
private:void Input(InventoryItem &Inv)
{
cout << \"Please type Id\" << endl;
cin >> Inv.Id;
cout << \"Please type Desc\" << endl;
cin >> Inv.desc;
cout << \"Please type qty\" << endl;
cin >> Inv.qty;
cout << \"Please type price\"<> Inv.price;
}
public: void Create(int Nrec)
{
inventory.open(DBname, ios::out | ios::binary);
// fstream inventory(\"Inventory.dat\", ios::out | ios::binary);
InventoryItem record = { \"\", \" \", 0, 0.0 };
// Write the blank records
for (int count = 0; count < Nrec; count++)
{
Input(record);
cout << \"Now writing record \" << count << endl;
inventory.write(reinterpret_cast(&record), sizeof(record));
}
// Close the file.
inventory.close();
return ;
}
public: void Display()
{
inventory.open(DBname, ios::in | ios::binary);
InventoryItem record = { \"\",\"\", 0, 0.0 };
// Now read and display the records
inventory.read(reinterpret_cast(&record),
sizeof(record));
while (!inventory.eof())
{
cout << \"Inventory Id: \";
cout << record.Id << endl;
cout << \"Description: \";
cout << record.desc << endl;
cout << \"Quantity: \";
cout << record.qty << endl;
cout << \"Price: \";
cout << record.price << endl << endl;
inventory.read(reinterpret_cast(&record), sizeof(record));
}
// Close the file.
inventory.close();
return;
}
};
int main()
{
DB Mydb(\"Inventory.dat\"); // declare a database
Mydb.Create(3); // load data
cout << \"***** display ***\" << endl;
Mydb.Display(); // print entire database
// implement the following function
// Mydb.Show(\"AB001\"); // display record with given ID
char C; cin >> C; return 0;
}
Solution
Programming Code in C++
#include
#include
#include
#include
#include
#include
using namespace std;
//Structure.
struct InvStruct
{
string iName;
int iQuantity;
double iPrice;
};
const int MAXSIZE = 9;
void addToInventory(InvStruct aList[], int& aSize);
void displayInventory(const InvStruct aList[], int aSize);
void saveToFile(const InvStruct aList[], int aSize);
void openAFile(InvStruct aList[], int& aSize);
char menuResponse();
int main(int argc, char *argv[])
{
InvStruct iRecords[MAXSIZE];
int noOfRec = 0;
bool reRun = true;
do
{
cout << \"Inventory has \" << noOfRec << \" items\" << endl;
switch (menuResponse())
{
case \'A\':
addToInventory(iRecords, noOfRec);
break;
case \'D\':
displayInventory(iRecords, noOfRec);
break;
case \'O\':
openAFile(iRecords, noOfRec);
break;
case \'S\':
saveToFile(iRecords, noOfRe.
Question 48 of 75. Cynthia (9) is the stepdaughter of the taxpayer .pdf
Question 48 of 75. Cynthia (9) is the stepdaughter of the taxpayer She is a Brazilan cfizen and
resident The Elane (19) is less than a half-tme student who earns S4000Ste mined te t1payers
daughter and provided less than 50% of her own support she is not eated taxpayer paid over half
her support and lived with her for seven months Cymthia dld not contribute to her own support O
Kenny (21) is the taxpayer\'s nephew and a tus tme student He lived with the taxpayer for thiree
months before moving into another Mana (5) is the daughter of the taxpayer with whom she hed
for seven morts nte arrent yea Maa s anurnamed u steen who ad not provide over hat hei om
upport aunts home The other aunt is willing to give the dependency exomption to the taxpayer
Solution.
1)Why does a fetal pig have anything in its intestine2) Are there.pdf
1)Why does a fetal pig have anything in its intestine?
2) Are there any differences between the thyroid gland, thymus gland, and spleen in a male and
female pig? Briefly explain.
3) What is the significance of the different length:volume ratios of the digestive organs in a male
and female pig? Explain you answer.
4) Is there a reason why a fetal pig\'s intestinal contents are a certain color?
Please help ):
Solution
Ans:
1) Why does a fetal pig have anything in its intestine?
The fetal pig\'s digestive organs are well developed before birth, although it does not ingest food.
These organs include the esophagus, stomach, small and large intestines. Mesenteries serve to
connect the organs of the fetal pig together. In order for digestion to occur, the fetal pig would
have to ingest food. Instead, it gains much needed nutrition from the mother pig via the umbilical
cord. In the adult pig, food will follow the general flow through the esophagus, which can be
located behind the tracheae. From the oral cavity, the esophagus leads to the stomach, small
intestine, and large intestine. Other organs developing during fetal pig development such as the
gallbladder, pancreas and spleen are all critical in contributing to the overall flow of the digestive
system because they contain digestive enzymes that will perform chemical digestion of food.
After food is digested and nutrients are absorbed, the food follows through the large intestine and
solid wastes are excreted through the anus. In the fetal pig however, the metabolic wastes are
sent back to the mother through the umbilical cord where the mother excretes the wastes. Other
remaining wastes remain in the fetal pig until birth.
2) Are there any differences between the thyroid gland, thymus gland, and spleen in a male and
female pig? Briefly explain.
Thyroid gland: Lying atop the trachea, locate the pinkish-brown, V shaped structure called the
thyroid gland. This gland secretes hormones that control growth and metabolism. Difference
between male and femalepigs was detected in the follicular hyperplasia and hypertrophy of the
thyroid gland
Thymus gland: Thymus was a large bilobed structure situated in the superior mediastinum. The
difference between the histological appearance of male and female is in the female the lobules
are highly vascular and Hassall’s corpuscles are either absent or only one is present. In the male
2-3 Hassall’s corpuscles are seen in the lobules. There by showing some sex difference.
Spleen: a flattened organ that lies across the stomach and toward the extreme left side of the pig.
The spleen stores blood and is not part of the digestive system. On the underside of the spleen,
locate the splenic artery. There is no difference in spleen between males and females, but the
stomach and small intestine were heavier in females than in males.
3) What is the significance of the different length:volume ratios of the digestive organs in a male
and female pig? Explain you answer.
A range of .
The geography of Taylor Swift - some ideas
Geographical themes connected with Taylor Swift's ERAS tour - coming to the UK in June 2024
Home assignment II on Spectroscopy 2024 Answers.pdf
Answers to Home assignment on UV-Visible spectroscopy: Calculation of wavelength of UV-Visible absorption
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Compare and contrast the ideas that led to the Bohr model of the ato.pdf
Compare and contrast the ideas that led to the Bohr model of the atom to the ideas that led to the
quantum mecchanical theory that gave rise to the hydrogen atom wave functions.
Solution
The difficulty with the planetary picture provided before the Bohr model is that it is inconsistent
with a well established fact of classical electrodynamics which says that an accelerated electric
charge must continually radiate energy. If electrons actually followed such a trajectory, all atoms
would act as source of radiation. The radiated energy would come from the kinetic energy of the
orbiting electron; as this energy gets radiated away, As a result the electron would quickly fall
into the nucleus. According to classical physics, no atom based on this model could exist for
more than a brief fraction of a second.
In contrary The Bohr model is a mixture of classical physics and quantum physics. The main
postulated of Bohr\'s orbit are as follows:
1.) The electron in the atom moves in a circular orbit centred on its nucleus. Its motion in the
orbit is governed by the Coulomb electric force between the negatively charged electron and the
positively charged proton.The radius of orbit can be given as [r = nh/(2*pi*m*v), where n is
principal quantum number.Emission spectrum of hydrogen is nicely explained by Bohr model.
However, one serious difficulty with the Bohr model was that it was unable to explain the
spectrum of any atom more complicated than hydrogen.
2.) An electron in a Bohr orbit does not continuously radiate electromagnetic radiation. Its
energy is therefore constant. The orbit is referred to as a stationary orbit.
3.) Electromagnetic radiation isonlyemitted when the electron changes from one orbit to another
of a lower total energy. (The electron is said to undergo atransition.) In such a case, the energy
lost, E, is emitted asonequantum of radiation of frequencyfas given by thePlanck–Einstein
formula:
E = hf
Although the Bohr model hypothesis for the quantization of angular momentum can be justified
in terms of electron wave ideas, the Bohr model remains profoundly unsatisfactory as a wave
model for the atom and also was not able to answer that why it doesn\'t fall into the nucleus. The
main failures of Bohr\'s model are:
- a mixture of classical and quantum ideas (electrons move classically on orbits, but their
possible energy states are quantified)
- postulates that on the allowed orbits electrons do not radiate
(conflict with Maxwell’s theory)
- could not account for the maximal electron numbers on one shell
- could not explain splitting of the spectral lines in magnetic fields
- it is a non-relativistic theory although the speed of the electrons is close to c
In the Quantum Mechanical Model, the electron is treated mathematically as a wave. The
electron has properties of both particles and waves. The advantage of Quantum Mechanical
model over Bohr model is that it was successful to give explaination of the fact that electron does
not fall in.
Blackboard Learn Realtor Rite Aid Citizens Bank peoples CPOBIE UI.pdf
Blackboard Learn Realtor Rite Aid Citizen\'s Bank people\'s CPOBIE UI Blackboard Learn KN
Question Completion Status: QUESTION 19 For what values of x does the function shown in
the graph appear to be increasing? all values of no values of X 20
Solution
In the graph first we can see that the function is decreasing through out. So for no value of x the
function is increasing. Option D
In next graph the function is negative for x <0 so Option A.
According to the Shannon-Hartley theorem, the capacity of a communic.pdf
According to the Shannon-Hartley theorem, the capacity of a communications channel in bits per
second is given by C=Blog2 ((s/n)+1) where B is the frequency bandwidth of the channel in
hertz and s/n is its signal-to-noise ratio. It is physically impossible to exceed this limit. Solve the
equation for the signal-to-noise ratio s/n.
Solution
C = B*log2( 1+s/n)
divide both sides by B:
(C/B) = log2(1+s/n)
Use the log property : loga(b) = x ----: b = a^x
So, 1+ s/n = 2^(C/B)
( signal-to-noise ratio) S/N = 2^(C/B) -1.
A fellow student, one that is not as careful as you, finds the follow.pdf
A fellow student, one that is not as careful as you, finds the following: They have produced 2
plates of media, and inoculated each plate with three different bacterial samples. Unfortunately,
they have labeled none of their plates. The plates have the following appearances: Plate 1 The
media is a dark red (burgundy) color, 3 different bacteria were inoculated onto the plate and
bacteria are growing in all 3 of the sections of the plate. Section A: you observe growth of
bacteria with a translucent/clear region around all colonies. Section B: you observe growth with
no change in the color of the media around all colonies. Section C: you observe growth with a
greenish color around all colonies. Plate 2: The media is a pale purple color, 3 different bacteria
were inoculated onto the plate, and organisms are growing in 2 of the sections of the plate.
Section A: you observe no growth of bacteria at all. Section B: you observe bacterial growth
with pink colonies. Section C: you observe bacterial growth with colorless/white colonies.
Using the diagrams on the next page, COMPLETELY interpret these results-this means I want to
know what enzymes are active. First tell me what media are represented for each plate, what the
media\'s properties are (selective, differential, any pH indicators), whether the media is selective
and/or differential. Second fully interpret the type organisms that are present in each of the three
sections of each plate, including enzymes, and whether the bacteria is Gram positive or Gram
negative.
Solution
Answer:
Plate 1: The dark red colour media is Eosine-methylene blue (EMB) agar due to the presence of
methylene blue and eosine dye. It is a selective and differencial culture medium. It is used for
selective growth of gram negative bacteria and differentiate the growth of lactose fermenting and
lactose non-fermenting coliforms. The details is given below:
Lactose fermenter
E. coli
Explaination: In the presence of an acidic environment (low pH) after the fermentation of
lactose, EMB forms a complex that precipitates out onto the coliform colonies, producing dark
centers and greenish sheen around the colonies. This reaction is characterstics of Escherichia
Coli. In addition, other weaker fermentation of lactose results in colonies with a pinkish-purple
color with no change around the colonies as their is low change of pH of growth medium.
However, in absence of an acidic environment the dye is not absorbed. Thus, the colonies of
lactose non-fermentors will be colorless.
Plate 2: The pale purple colour media is MacConkey agar due to the presence of crystal violet
dye. It is a selective and differencial culture medium used for selective growth of gram negative
bacteria and differentiate the growth of lactose fermenting and lactose non-fermenting coliforms.
The details is given below:
Explaination: MacConkey agar is detect the growth of coliforms and enteric pathogens based on
their ability to ferment lactose. Lactose fermenting c.
2. There are special proteins (called transporters) that are able to.pdf
2. There are special proteins (called transporters) that are able to move protons (H+) into or out
of cells. These proteins can be seen to be at work if the pH inside and/or outside the cell changes.
However, the pH is only seen to go down outside the cell when ATP (adenosine triphosphate, an
energy providing molecule) is present. Is the proton movement an example of active or passive
movement? Which way are the protons moving? Is this an example of an endergonic or an
exogonic reaction?
Solution
Movement of proton is an example of primary active transport as it uses redox energy of
mitochondrial electron transport from NADH and makes protons moves across the inner
mitochondrial membrane against their concentration gradient. Movment of proton depends upon
the concentration gradient which way it is low it will move that way. As it uses the energy from
ATP it is considered to be endergonic reaction..
11. Recessive epistasis is described by all of the following EXCEPT.pdf
11. Recessive epistasis is described by all of the following EXCEPT:
A. it is observed in coat color in Labrador retrievers.
B. it has two genes interacting to produce three phenotypes instead of four.
C. it involves one gene controlling the expression of another.
D. it gives the modified Mendelian ratio of 12:3:1.
E. Actually, all of these describe recessive epistasis.
12. Coat color is determined by two loci in large cats. Two pink panthers fall in love and produce
a large litter of baby panthers with the following phenotypic ratios: 12/16 pink; 3/16 black; and
1/16 white. What is the genotype of the white progeny?
A. aa bb
B. A_ B_ and A_ bb
C. A_ B_
D. A_ bb
E. aa B_
13. Traits that have differential expression in the two sexes are known as:
A. sex-linked.
B. sex-influenced.
C. autosomal.
D. sex-determined.
E. None of the above.
14. Polydactyly is the condition of having extra fingers or toes. Some polydactylous persons
possess extra fingers or toes that are fully functional, whereas others possess only a small tag of
extra skin. This is an example of
A. complete dominance.
B. cytoplasmic inheritance.
C. complementation.
D. independent assortment.
E. variable expressivity.
15. A mother with blood type A has a child with blood type A. Give all possible blood types for
the father of this child.
A. A, AB
B. A, B, AB, O
C. O
D. B, AB
E. A, B, O
16. Considering cases of incomplete dominance:
A. the heterozygote has a phenotype intermediate between the homozygotes
B. the homozygote has a phenotype intermediate between the heterozygotes.
C. the heterozygote has a phenotype identical to one of the homozygotes.
D. the homozygote has a phenotype identical to one of the heterozygotes.
E. None of the above.
17. A mother with blood type A has a child with blood type AB. Give all possible blood types
for the father of this child.
A. A, B, AB, O
B. A, B, O
C. O
D. B, AB
E. A, AB
18. Gene interaction produces all of the following phenomena EXCEPT:
A. recessive epistasis.
B. duplicate recessive epistasis.
C. dominant epistasis.
D. duplicate interaction.
E. Actually, all of these are recognized forms of gene interaction
19. In order to determine if mutations from different organisms that exhibit the same phenotype
are allelic, which test would you perform?
A. Allelic series test
B. Biochemical test
C. Epistasis test
D. Complementation test
E. Test cross
20. Interactions among the human ABO blood group alleles involve _______ and ________.
A. complete dominance; incomplete dominance
B. co-dominance; complete dominance
C. epistasis; complementation
D. codominance; incomplete dominance
E. continuous variation; environmental variation
Solution
Answer:
11. (D) It gives the modified Mendelian ratio of 12:3:1
12. (B) A_ B_ and A_ bb
13. (B) Sex-influenced
Characters whose expression in a particular genotype depends on whether the individual is male
or, female, are known as sex influenced traits. In one of the sexes, an allele may be dominant
while it may be reces.
Write complete VHDL codes for the following schematic. Solution.pdf
Write complete VHDL codes for the following schematic.
Solution
for shift reg:
library IEEE;
use IEEE.STD_LOGIC_1164.ALL;
-- Uncomment the following library declaration if using
-- arithmetic functions with Signed or Unsigned values
--use IEEE.NUMERIC_STD.ALL;
-- Uncomment the following library declaration if instantiating
-- any Xilinx primitives in this code.
--library UNISIM;
--use UNISIM.VComponents.all;
entity pipo is
Port ( ClrN,load,clk : in STD_LOGIC;
D3,D2,D1,D0 : in STD_LOGIC;
Q3,Q2,Q1,Q0 : out STD_LOGIC);
end pipo;
architecture Behavioral of pipo is
component D_ff is
Port ( clock,ClrN : in STD_LOGIC;
D : in STD_LOGIC;
Q : out STD_LOGIC);
end component;
signal clock:STD_LOGIC;
begin
clock<= load and clk;
u3:D_Ff port map(clock,ClrN,D3,Q3);
u2:D_Ff port map(clock,ClrN,D2,Q2);
u1:D_Ff port map(clock,ClrN,D1,Q1);
U0:D_Ff port map(clock,ClrN,D0,Q0);
end Behavioral;
for D-Flipflop:
entity D_FF is
Port ( clock,ClrN : in STD_LOGIC;
D : in STD_LOGIC;
Q : out STD_LOGIC);
end D_FF;
architecture Behavioral of D_FF is
begin
process(clk,ClrN)
begin
if(ClrN = \'0\') then
if(falling_edge(clk)) then
Q <= D;
end if;
else
Q <= \'0\';
end if;
end process;
end Behavioral;.
Why do carrier-mediated processes become saturated (choose the best .pdf
Why do carrier-mediated processes become saturated (choose the best answer)? Question 8
options:
A:The carriers are very specific.
B:The number of carriers in a membrane is limited.
C:The concentration gradient is too strong.
D:The amount of ATP in the cell is limited.
E:The concentration gradient is too weak
Solution
Answer is E.
It is a passive process in which the solutes move down the concentration gradient..
Why is it that linked lists are better when it comes to needing freq.pdf
Why is it that linked lists are better when it comes to needing frequent add(), remove(), and get()
operations?
Solution
1.Linked list permits the constant time insertions or removals with iterators.
2.The items are able to add or remove from the center of the list.
3.interpretating the starting size of an array is not needed.
4.insert and delete manipulations are simply accomplished in a linkedlist.
5.allocation and deallocation is possible at the time of programm running.
6.the main advantage of using linkedlist is reusing of current iterators to insert and delete
element..
Which of the following isare characteristic of the parasympathetic .pdf
Which of the following is/are characteristic of the parasympathetic nervous system, but not of
the sympathetic nervous system?
A ganglia in or near target tissues
B nicotinic receptors on postganglionic neurons
C muscarinic receptors on some target tissues
D 1 receptors on some target tissues
E 1 receptors on some target tissues
Solution
A. ganglia in or near target tissues. The parasympathetic nervous system is a slower system that
moves along longer pathways. Preganglionic fibres from the medulla or spinal cord project
ganglia close to the target organ. They create a synapse, which eventually creates the desired
response..
What statistic is often interpreted as providing an estimate of the .pdf
What statistic is often interpreted as providing an estimate of the average difference between a
statistic and a parameter?
Solution
The standard deviation, which finds the root of the mean squared-difference. It is interpreted as a
measure of the average difference between the sample mean (statistic) and population mean
(parameter)..
The waiting time X until delivery of a new component for an indus.pdf
The waiting time X until delivery of a new component for an industrial operation is uniformly
distributed over the interval from 1 to 5 days. The cost of this delay is given by probability
density function for Y.
Solution
uniform distribution over the interval from 1 to 5
x = 1 / (5-1 ) = 1/4
y = 2* (1/4)^2 + 3
y = 3.125.
The Latin term a priori describes the origin of knowledge developed .pdf
The Latin term a priori describes the origin of knowledge developed rationally, and the term a
posteriori describes knowledge developed through observation and experience. What is produced
through a priori rational thought that applies to ethics? nalysis systems Principles of ethics
Concern for outcomes Emotions Subjective sense of identity
Solution
please explain the question.
Suppose X and Y are two independent random variables with the follow.pdf
Suppose X and Y are two independent random variables with the following moment generating
functions: Mx(t)= e^(-t+(t^2)), My(t)= e^(2t+1/2(t^2))
Let Z = X
Solution
Given Mx(t) = e-t+t^2)
= 1-t+t^2)+(-t+t^2)^2/2!+....
= 1-t+t^2+t^2/2+t^3+t^4/2+...
Coefficient of t = -1
Mean of x = -1
E(X^2) = coeff of t^2 =3/2
Variance x = 3/2-1 = 1/2
My(t) = e^(2t+t^2/2)
= 1+2t+t^2/2 +4t^2+...
Mean of Y = 2
E(Y^2) = 9/2
Var(Y) = 0.5
Hence Var (z)
= Var(x-5y) = Var(x)+25Var(Y)
= 1/2 + 25(0.5)
=0.5+12.5 = 13..
The (circle one meanmedian) would be a better measure of central ten.pdf
The (circle one mean/median) would be a better measure of central tendency if you are
describing the annual incomes of all people n Minnesota.
Solution
vaiables like Annual incomes have a lower limit (zero) but no upper limit, and thus have huge
outliers on right. Median is a better measure of central tendency for such skewed distributions..
Summary of Ethics and Information Security (one page)Solution.pdf
Summary of Ethics and Information Security (one page)
Solution
Ethics and Information Security
here Ethics is nothing the principles and standards that guide our behavior towards other people
secondly Intellectual property is creative work in physical form
Copyright is nothing but the legal protection afforded an expression of an idea, such as song,
video game, and sometypes of proprietary documents
Pirated software is the unauthorized use, duplication, distribution, or sale of copyrighted
software
Counterfeit software is a software that is manufactured to look like the real thing and sold as
such
here Privacy include
1. Right to be left alone
2. Control your own personal possessions
3. Not to be observedwithout your consent
Epolicies are policies and procedures that address the ethical use of computers and internet usage
in thebusiness environment
and CIO Magazine’s Size Principles for Ethical Information Management
finally Information is a valuable corporate assets and should be managed as such, like cash,
facilities, orany other corporate asset2.
The CIO is steward of corporate information and is responsible for managing it over its life
cycle-from its generation to its appropriate destruction3.
The CIO is responsible for controlling access to and use of information, as determined
bygovernmental regulation and corporate policy4.
The CIO is responsible for preventing the inappropriate destruction of information5.
The CIO is responsible for bringing technological knowledge to the development of
informationmanagement practices and policies6.
The CIO should partner with executive peers to
develop and execute the organization’s
information management policies
Ethical computer use policy
contains general principles to guide computer user behavior. Example : theethical computer use
policy might explicitly state that users should refrain from playing computer gamesduring
working hours
Information privacy policy
contains general principles regarding information privacy
Acceptable use policy (AUP)
is a policy that a user must agree to follow in order to be provided accessto a network or to the
internet.
Nonrepudiation is nothing but a contractual stipulation to ensure that ebusiness participants do
not deny their onlineactions.
suppose f is continuous on the open interval from negative infinity .pdf
suppose f is continuous on the open interval from negative infinity to positive infinity and
f(1)=1. I was given the hint that this problem involves using the concepts of limits, derivative,
integral, L\'Hopital\'s rule, chain rule, and the fundamental theorem of calculus.
Solution
apply l\' hospital -sinx^2 * 2x*f(cosx^2)/4x^3 -sinx^2**f(cosx^2)/2x^2 put the
limit -1/2.
So I already have most of the code and now I have to1. create an .pdf
So I already have most of the code and now I have to:
1. create an index file to keep track of all the inventory IDs and their locations
2. modify my class to be able to display any record with a given ID without searching through
the entire file
// This program displays the contents of the inventory file.
#include
#include
#include
using namespace std;
class DB{
// Declaration of InventoryItem structure
class InventoryItem
{
public:
char Id[5];
char desc[31];
int qty;
float price;
};
string DBname;
fstream inventory;// (DBname, ios::in | ios::binary);
public: DB(string N){ DBname = N; }// constructor
private:void Input(InventoryItem &Inv)
{
cout << \"Please type Id\" << endl;
cin >> Inv.Id;
cout << \"Please type Desc\" << endl;
cin >> Inv.desc;
cout << \"Please type qty\" << endl;
cin >> Inv.qty;
cout << \"Please type price\"<> Inv.price;
}
public: void Create(int Nrec)
{
inventory.open(DBname, ios::out | ios::binary);
// fstream inventory(\"Inventory.dat\", ios::out | ios::binary);
InventoryItem record = { \"\", \" \", 0, 0.0 };
// Write the blank records
for (int count = 0; count < Nrec; count++)
{
Input(record);
cout << \"Now writing record \" << count << endl;
inventory.write(reinterpret_cast(&record), sizeof(record));
}
// Close the file.
inventory.close();
return ;
}
public: void Display()
{
inventory.open(DBname, ios::in | ios::binary);
InventoryItem record = { \"\",\"\", 0, 0.0 };
// Now read and display the records
inventory.read(reinterpret_cast(&record),
sizeof(record));
while (!inventory.eof())
{
cout << \"Inventory Id: \";
cout << record.Id << endl;
cout << \"Description: \";
cout << record.desc << endl;
cout << \"Quantity: \";
cout << record.qty << endl;
cout << \"Price: \";
cout << record.price << endl << endl;
inventory.read(reinterpret_cast(&record), sizeof(record));
}
// Close the file.
inventory.close();
return;
}
};
int main()
{
DB Mydb(\"Inventory.dat\"); // declare a database
Mydb.Create(3); // load data
cout << \"***** display ***\" << endl;
Mydb.Display(); // print entire database
// implement the following function
// Mydb.Show(\"AB001\"); // display record with given ID
char C; cin >> C; return 0;
}
Solution
Programming Code in C++
#include
#include
#include
#include
#include
#include
using namespace std;
//Structure.
struct InvStruct
{
string iName;
int iQuantity;
double iPrice;
};
const int MAXSIZE = 9;
void addToInventory(InvStruct aList[], int& aSize);
void displayInventory(const InvStruct aList[], int aSize);
void saveToFile(const InvStruct aList[], int aSize);
void openAFile(InvStruct aList[], int& aSize);
char menuResponse();
int main(int argc, char *argv[])
{
InvStruct iRecords[MAXSIZE];
int noOfRec = 0;
bool reRun = true;
do
{
cout << \"Inventory has \" << noOfRec << \" items\" << endl;
switch (menuResponse())
{
case \'A\':
addToInventory(iRecords, noOfRec);
break;
case \'D\':
displayInventory(iRecords, noOfRec);
break;
case \'O\':
openAFile(iRecords, noOfRec);
break;
case \'S\':
saveToFile(iRecords, noOfRe.
Question 48 of 75. Cynthia (9) is the stepdaughter of the taxpayer .pdf
Question 48 of 75. Cynthia (9) is the stepdaughter of the taxpayer She is a Brazilan cfizen and
resident The Elane (19) is less than a half-tme student who earns S4000Ste mined te t1payers
daughter and provided less than 50% of her own support she is not eated taxpayer paid over half
her support and lived with her for seven months Cymthia dld not contribute to her own support O
Kenny (21) is the taxpayer\'s nephew and a tus tme student He lived with the taxpayer for thiree
months before moving into another Mana (5) is the daughter of the taxpayer with whom she hed
for seven morts nte arrent yea Maa s anurnamed u steen who ad not provide over hat hei om
upport aunts home The other aunt is willing to give the dependency exomption to the taxpayer
Solution.
1)Why does a fetal pig have anything in its intestine2) Are there.pdf
1)Why does a fetal pig have anything in its intestine?
2) Are there any differences between the thyroid gland, thymus gland, and spleen in a male and
female pig? Briefly explain.
3) What is the significance of the different length:volume ratios of the digestive organs in a male
and female pig? Explain you answer.
4) Is there a reason why a fetal pig\'s intestinal contents are a certain color?
Please help ):
Solution
Ans:
1) Why does a fetal pig have anything in its intestine?
The fetal pig\'s digestive organs are well developed before birth, although it does not ingest food.
These organs include the esophagus, stomach, small and large intestines. Mesenteries serve to
connect the organs of the fetal pig together. In order for digestion to occur, the fetal pig would
have to ingest food. Instead, it gains much needed nutrition from the mother pig via the umbilical
cord. In the adult pig, food will follow the general flow through the esophagus, which can be
located behind the tracheae. From the oral cavity, the esophagus leads to the stomach, small
intestine, and large intestine. Other organs developing during fetal pig development such as the
gallbladder, pancreas and spleen are all critical in contributing to the overall flow of the digestive
system because they contain digestive enzymes that will perform chemical digestion of food.
After food is digested and nutrients are absorbed, the food follows through the large intestine and
solid wastes are excreted through the anus. In the fetal pig however, the metabolic wastes are
sent back to the mother through the umbilical cord where the mother excretes the wastes. Other
remaining wastes remain in the fetal pig until birth.
2) Are there any differences between the thyroid gland, thymus gland, and spleen in a male and
female pig? Briefly explain.
Thyroid gland: Lying atop the trachea, locate the pinkish-brown, V shaped structure called the
thyroid gland. This gland secretes hormones that control growth and metabolism. Difference
between male and femalepigs was detected in the follicular hyperplasia and hypertrophy of the
thyroid gland
Thymus gland: Thymus was a large bilobed structure situated in the superior mediastinum. The
difference between the histological appearance of male and female is in the female the lobules
are highly vascular and Hassall’s corpuscles are either absent or only one is present. In the male
2-3 Hassall’s corpuscles are seen in the lobules. There by showing some sex difference.
Spleen: a flattened organ that lies across the stomach and toward the extreme left side of the pig.
The spleen stores blood and is not part of the digestive system. On the underside of the spleen,
locate the splenic artery. There is no difference in spleen between males and females, but the
stomach and small intestine were heavier in females than in males.
3) What is the significance of the different length:volume ratios of the digestive organs in a male
and female pig? Explain you answer.
A range of .
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