AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf
•
0 likes•2 views
Answer: Answer: Part B). The F2 progeny contain recombinants only when F1 females are test crossed; Some of the F2 progeny classes require crossing over; others do not. Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio. Explanation: To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing over. A balanced, or chromosomal translocation is a condition in which part of a chromosome has broken off and reattached in another location. In other words, it means that sections of two chromosomes have switched places. Solution Answer: Answer: Part B). The F2 progeny contain recombinants only when F1 females are test crossed; Some of the F2 progeny classes require crossing over; others do not. Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio. Explanation: To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing over. A balanced, or chromosomal translocation is a condition in which part of a chromosome has broken off and reattached in another location. In other words, it means that sections of two chromosomes have switched places..
Report
Share
Report
Share
Download to read offline
Recommended
Help if you can Genetics question. thanks beforehand! You continue y.pdf
Help if you can Genetics question. thanks beforehand! You continue your experiments with the
black mutation on chromosome II. You cross the reciprocal translocation strain to the black pure
line to generate F_1 flies that are both translocation heterozygotes and Bb heterozygotes. As
before, you testcross F_1 males and females in two separate experiments. What can you
conclude from these results? Select all statements that are true. The F_2 progeny in both
experiments contain recombinants. The F_2 progeny contain recombinants only when F?_1
females are testcrossed. None of the F_2 progeny classes require crossing over. Some of the
F_2 progeny classes require crossing over, others do not. Chromosome II is one of the
chromosomes involved in the reciprocal balanced translocation.
Solution
Since the recombinants can be formed only when there is cross over in atleast one the parent; the
F2 progeny from cross between F1 males and pure black females do not have recombinants.
Only the F2 progeny from the cross between F1 females and pure black males have the
recombinants as cross over occurs in F1 female. The F2 progeny has both parental types and
recombinants; formation of parental types does not require the crossover. Therefore, some of the
F2 progeny class require cross over while others do not. Since the gene is present on the
chromosome II, it is involved in the reciprocal translocation.
Correct statements are
1: The F2 progeny has recombinants only when F1 females are testcrossed
2: Some of the F2 progeny classes require crossing over; others donot.
3: Chromosome II is one of the chromosome for reciprocal translocation..
1. Escherichia coli have F plasmid which by replication gets transfe.pdf
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+.
Solution
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Bacterial Conjugation (Genetic recombination in Bacteria)
Introduction, discovery, f-plasmid and its properties, bacterial conjugation steps, Hfr strain.
For 12th, M.Sc, B.Sc students.
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
HFR cells are formed from homologous recombination due to integratio.pdf
HFR cells are formed from homologous recombination due to integration of the conjugate
plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another
cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell.
Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its
chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it
can donate to another cell through conjugation. Both Hfr & F+ cells are competent for
conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell.
Interrupted mating:
F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously
originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F-
factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This
structure is going to promote high frequency recombination through interrupted mating into the
recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a
starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin
antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting
conjugation. This is completely based on the streptomycin-sensitivity allele (str-s) in the Hfr
strains.
On the otherhand, streptomycin-resistance allele (str-r) presence in the recipient is employed
considerably to specifically destroy the Hfr donor cells after conjugation.
Solution
HFR cells are formed from homologous recombination due to integration of the conjugate
plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another
cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell.
Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its
chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it
can donate to another cell through conjugation. Both Hfr & F+ cells are competent for
conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell.
Interrupted mating:
F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously
originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F-
factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This
structure is going to promote high frequency recombination through interrupted mating into the
recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a
starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin
antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting
conjugation. This is completely based on the st.
Recommended
Help if you can Genetics question. thanks beforehand! You continue y.pdf
Help if you can Genetics question. thanks beforehand! You continue your experiments with the
black mutation on chromosome II. You cross the reciprocal translocation strain to the black pure
line to generate F_1 flies that are both translocation heterozygotes and Bb heterozygotes. As
before, you testcross F_1 males and females in two separate experiments. What can you
conclude from these results? Select all statements that are true. The F_2 progeny in both
experiments contain recombinants. The F_2 progeny contain recombinants only when F?_1
females are testcrossed. None of the F_2 progeny classes require crossing over. Some of the
F_2 progeny classes require crossing over, others do not. Chromosome II is one of the
chromosomes involved in the reciprocal balanced translocation.
Solution
Since the recombinants can be formed only when there is cross over in atleast one the parent; the
F2 progeny from cross between F1 males and pure black females do not have recombinants.
Only the F2 progeny from the cross between F1 females and pure black males have the
recombinants as cross over occurs in F1 female. The F2 progeny has both parental types and
recombinants; formation of parental types does not require the crossover. Therefore, some of the
F2 progeny class require cross over while others do not. Since the gene is present on the
chromosome II, it is involved in the reciprocal translocation.
Correct statements are
1: The F2 progeny has recombinants only when F1 females are testcrossed
2: Some of the F2 progeny classes require crossing over; others donot.
3: Chromosome II is one of the chromosome for reciprocal translocation..
1. Escherichia coli have F plasmid which by replication gets transfe.pdf
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+.
Solution
1. Escherichia coli have F plasmid which by replication gets transferred to another bacteria. Also
the transfer is assisted by formation of proteins which are itslef formed by F plasmid. These
proteins form tube like structure which helps in transfer of single stranded DNA to another
bacteria.
F plasmid has several homologous regions to the chromosomes which helps in recombination
process which is known as single cross over and thus F plasmid is integrated into the
chromosomes. The bacterial cell which consists of F plasmid also is known as Hfr strain.
This strains sometimes transfer parts of of its chromosome to the cell which do not have F
plasmid and thus is known as F- cell. Here at this place phenomenon of double cross over occurs
which results in conversion of the F- cell to F+..
Conjuagtion types and application
this presentation basically focusses on the types of conjugation namely Hfr and F' conjugation.
Bacterial Conjugation (Genetic recombination in Bacteria)
Introduction, discovery, f-plasmid and its properties, bacterial conjugation steps, Hfr strain.
For 12th, M.Sc, B.Sc students.
Conjugation: Discovery, F+, F- and Hfr conjugation, F- genetic crosses
These slides contain information about bacterial conjugation.
HFR cells are formed from homologous recombination due to integratio.pdf
HFR cells are formed from homologous recombination due to integration of the conjugate
plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another
cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell.
Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its
chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it
can donate to another cell through conjugation. Both Hfr & F+ cells are competent for
conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell.
Interrupted mating:
F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously
originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F-
factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This
structure is going to promote high frequency recombination through interrupted mating into the
recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a
starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin
antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting
conjugation. This is completely based on the streptomycin-sensitivity allele (str-s) in the Hfr
strains.
On the otherhand, streptomycin-resistance allele (str-r) presence in the recipient is employed
considerably to specifically destroy the Hfr donor cells after conjugation.
Solution
HFR cells are formed from homologous recombination due to integration of the conjugate
plasmid into bacterial chromosome. Hfr cells may donate genes from its chromosomes to another
cell. Hfr cells do not possess a complete \"circular chromosome\" i.e. plasmid as F+ cell.
Therefore, the major events of Hfr (high frequency recombinant) cells contain F factor in its
chromosomal DNA often referred as \"episome\". F+ cell possess F factor known as plasmid & it
can donate to another cell through conjugation. Both Hfr & F+ cells are competent for
conjugation through sex pili or mating bridge & finally transfer F factor into the recipient cell.
Interrupted mating:
F factor of HFR cells is mainly pertaining to a class of conjugate plasmids which meticulously
originated to control sexual functions in prokaryotes via fertility inhibition. The origin of F-
factor is mainly with most common functional segments such as gene traJ, Ori, OriC. This
structure is going to promote high frequency recombination through interrupted mating into the
recipient cell. The structure of this F factor contains OriT for origin of transfer and acts as a
starting point for gene transmission to the bacterium of F- (in which no F factor). A streptomycin
antibiotic containing medium is used to kill the remaining Hfr donor cells after interrupting
conjugation. This is completely based on the st.
5 If an organism were heterozygous for 10 loci how many di.pdf
5. If an organism were heterozygous for 10 loci, how many different combinations could the
organism produce in its gametes? Show all your work to receive full credit. 6. In tomato plants, the
production of red fruit color is under the control of an allele R. Yellow. tomatoes are rr. The
dominant phenotype for fruit shape is under the control of an allele T. which produces two lobes.
Multilobed fruit, the recessive phenotype, have the phenotype tt. Two different crosses are made
between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios
to determine the genotypes and phenotypes of each parent. (a) Cross 1 progeny: 3/8 two-lobed,
red 3/8 two-lobed, yellow 1/8 multilobed, red 1/8 multilobed, yellow What are the genotypes and
phenotypes of the parents in this cross?3. A tropicat fish fancier produced an albino zebra fish. it
lacked blue stripes characteristic of the normal zebra fish. When she crossed a normal zebra fish
to the albino, all the offspring were narmat. When she crossed two of the normal F1 offspring, y
were abbino and % were normal. When an albino was crossed to the normal offspring, half were
albino and half were normal. What is the genotype and phenotype of each fish mentioned in this
problem? 4. In pea plants, the appearance of flowers along the main stem is a dominant
phenotype called "axial" and is controlled by an allele T. The recessive phenotype, produced by an
allele t, has flowers only at the end of the stem called "terminal". Pod form displays a dominant
phenotype "inflated", controlled by allele F, and a recessive "constricted" form produced by allele f
. A pure-breeding axial, constricted plant is crossed to one that is purebreeding terminal, inflated.
A. The F1 progeny of this cross are allowed to self-fertilize. What is the expected phenotypic
distribution of the F2 progeny? List the genotypes and phenotypes of the f1 generation and the
phenotypes of the F2 progeny. B. If an F1 plant from the initial cross described above, is crossed
to a plant that is terminal, constricted, what is the expected phenotypic distribution among the
resulting progeny?.
1. VacuolesThe vacuole is an organelle in plant cells which stores.pdf
1. Vacuoles
The vacuole is an organelle in plant cells which stores water and nutrients for the cell. It takes in
and release the water as per cell\'s requirement.
2. . In plasmolysed Elodea leaf the gap between the cell wall and the cell membrane is filled with
0.6M sucrose solution.
Solution
1. Vacuoles
The vacuole is an organelle in plant cells which stores water and nutrients for the cell. It takes in
and release the water as per cell\'s requirement.
2. . In plasmolysed Elodea leaf the gap between the cell wall and the cell membrane is filled with
0.6M sucrose solution..
1. B) - Two independent properties serve to specify the state.2. B.pdf
1. B) - Two independent properties serve to specify the state.
2. B) - The locus of all points where the second law is satisfied.
3. B) - Is the p-T coordinates at which solid, liquid and vapour co-exist in equlibrium.
Solution
1. B) - Two independent properties serve to specify the state.
2. B) - The locus of all points where the second law is satisfied.
3. B) - Is the p-T coordinates at which solid, liquid and vapour co-exist in equlibrium..
(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4.
Solution
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4..
I have inserted spaces. Consider f(x) belonging to f(X) thi.pdf
I have inserted spaces.
Consider f(x) belonging to f(X)
this means : x belongs to X
But X is a subset of Y
So, x belongs to Y
So, f(x) belongs to f(Y).
Hence all f(x) in f(X) belong to f(Y)
So, f(X) is a subset of f(Y)
Example: f(x) = sin(x) : R -> R
X = [0,2pi]
Y = [-2pi,2pi]
f(X) = [-1,1]
f(Y) = [-1,1]
So, f(X) = f(Y)
Solution
I have inserted spaces.
Consider f(x) belonging to f(X)
this means : x belongs to X
But X is a subset of Y
So, x belongs to Y
So, f(x) belongs to f(Y).
Hence all f(x) in f(X) belong to f(Y)
So, f(X) is a subset of f(Y)
Example: f(x) = sin(x) : R -> R
X = [0,2pi]
Y = [-2pi,2pi]
f(X) = [-1,1]
f(Y) = [-1,1]
So, f(X) = f(Y).
C program that prompts user to enter two floating point t.pdf
/**
C program that prompts user to enter two floating point
type values and print the product of two values upt to 3
decimal places
*/
//program1.c
//header files
#include
#include
//main function
int main()
{
//declare two double type variables
double num1;
double num2;
//set a double variable,product=0
double product=0;
printf(\"Enter num1 and num2: \");
//prompt for num1 and num2 values from keyboard
scanf(\"%lf %lf\",&num1,&num2);
//calculate product
product=num1*num2;
//print product of two numbers and limit product to 3 decimal places
printf(\"The product of %lf and %lf is %5.3lf \ \",num1,num2,product);
getch();
return 0;
}
Sample output:
Enter num1 and num2: 2.25 3.25
The product of 2.250000 and 3.250000 is 7.313
------------------------------------------------------------------------------------
//2.
/**
C program that prompts user to enter integer values
until user enters 999 to stop reading. Then prit
number of even and odd to console.
*/
//program2.c
//header files
#include
#include
//main function
int main()
{
//set SENTINAL=999;
int SENTINAL=999;
//declare integer variable
int value;
//set even and odd=0
int even=0;
int odd=0;
printf(\"Enter a vlaue or 999 to stop reading input \");
scanf(\"%d\",&value);
//read until value is not 999
while(value!=SENTINAL)
{
//check if value is even
if(value%2==0)
//increment even by one
even++;
else
//otherwise increment odd by one
odd++;
printf(\"Enter a vlaue or 999 to stop reading input \");
//read a value
scanf(\"%d\",&value);
}
printf(\"Even numbers : %d\ \",even);
printf(\"Odd numbers : %d\ \",odd);
//pause program output on console
getch();
return 0;
}
Sample output:
Enter a vlaue or 999 to stop reading input 1
Enter a vlaue or 999 to stop reading input 2
Enter a vlaue or 999 to stop reading input 3
Enter a vlaue or 999 to stop reading input 4
Enter a vlaue or 999 to stop reading input 5
Enter a vlaue or 999 to stop reading input 6
Enter a vlaue or 999 to stop reading input 7
Enter a vlaue or 999 to stop reading input 8
Enter a vlaue or 999 to stop reading input 9
Enter a vlaue or 999 to stop reading input 10
Enter a vlaue or 999 to stop reading input 999
Even numbers : 5
Odd numbers : 5
------------------------------------------------------------------------------------------------------------
//3.
/**
C program that prompts length in inches and prints the length
in miles, yards, feet to console.
*/
//program3.c
//header files
#include
#include
//function prototype
void measure(int inch,int &m,int &y,int &f, int &i);
//main function
int main()
{
int miles,yards,feet,inches=0;
int len;
printf(\"Enter the length (inches): \");
//prompt for len
scanf(\"%d\",&len);
//calling measure method
measure(len,miles,yards,feet,inches);
//print len
printf(\"Total length : %d\ \", len);
//print miles,yards, feet and inches
printf(\"Miles : %d\ \", miles);
printf(\"Yards : %d\ \", yards);
printf(\"Feet : %d\ \", feet);
printf(\"Inches : %d\ \", inches);
//pause program output on console
getch();
return.
The carbonyl functional group in glucose is an al.pdf
The carbonyl functional group in glucose is an aldehyde while the fructose contains
a ketone
Solution
The carbonyl functional group in glucose is an aldehyde while the fructose contains
a ketone.
Well to find the pH at the equivalance point. Acc.pdf
Well to find the pH at the equivalance point. According to mychemistry book(
Chemistry: A Molecular Approach BY Nivaldo J. Tro,et. all.) . pH= pKa at the equivalnce point.
Solution
Well to find the pH at the equivalance point. According to mychemistry book(
Chemistry: A Molecular Approach BY Nivaldo J. Tro,et. all.) . pH= pKa at the equivalnce point..
The weakest base in the reaction is A the enolat.pdf
The weakest base in the reaction is: A the enolate in the reaction is: C
Solution
The weakest base in the reaction is: A the enolate in the reaction is: C.
so take an aliquot of the upper layer and add a f.pdf
so take an aliquot of the upper layer and add a few drops of water. if it\'s miscible,
it\'s water. if not, it\'s the ether.
Solution
so take an aliquot of the upper layer and add a few drops of water. if it\'s miscible,
it\'s water. if not, it\'s the ether..
moles of NH4Cl formed = 20.050 = 0.1 moles so we.pdf
moles of NH4Cl formed = 2*0.050 = 0.1 moles so weight = 0.1 * 53.49 = 5.35 so
D is answer
Solution
moles of NH4Cl formed = 2*0.050 = 0.1 moles so weight = 0.1 * 53.49 = 5.35 so
D is answer.
Latency is a measure of time delay experienced in.pdf
Latency is a measure of time delay experienced in a system, the precise definition
of which depends on the system and the time being measured.
Solution
Latency is a measure of time delay experienced in a system, the precise definition
of which depends on the system and the time being measured..
Entropy is the measure of disorder. Therefore, th.pdf
Entropy is the measure of disorder. Therefore, the choice should be (D) because the
entropy of a system increases when two systems come in contact. Here, the two systems are two
liquids.
Solution
Entropy is the measure of disorder. Therefore, the choice should be (D) because the
entropy of a system increases when two systems come in contact. Here, the two systems are two
liquids..
dydx = 1x dy = dxx y = ln x + c .pdf
dy/dx = 1/x dy = dx/x y = ln x + c
Solution
dy/dx = 1/x dy = dx/x y = ln x + c.
Ø Napoleonic era brought some relief to the faltering ottoman empire.pdf
Ø Napoleonic era brought some relief to the faltering ottoman empire .it distracted Russia from
further advances. Napoleon I himself invaded Egypt, but his army was trapped when the British
sank the French fleet. A peace interlude in 1803 allowed the army to return to France.
Ø To secure his own domination and to render the rest Europe virtually powerless, napoleon
established an alliance with Russia by concluding the treaty of titlist in 1807. Russia undertook
to aid napoleon in his war against Britain ,in turn the emperor of Russia would receive the
ottoman territories of Moldavia and Wallachia if the sultan refused to surrender these territories,
France and Russia were to attack the empire, and the ottoman in Europe were to be partitioned
between the two allies.
Ø The napoleon scheme threatened threated not only the sultan, but also Britain, Austria and
Prussia, which was almost powerless in the face of such a potent alliance. The alliance naturally
proved accommodating to the au strains , who hoped that a joint Franco –Russian attack which
would probably have utterly divested the ottoman empire, could be prevented by diplomacy, but
if diplomatic measures failed , the Austrian minister klemens wenzel, prince von Metternich
decided that he would support the partition of ottoman empire a solution disadvantages to
Austria. But not as dangerous as a complete Russian takeover of southeastern Europe.
Ø The Napoleonic era (1799–1815) brought some relief to the faltering Ottoman Empire. It
distracted Russia from further advances. Napoleon I himself invaded Egypt, but his army was
trapped there when the British sank the French fleet. A peace interlude in 1803 allowed the army
to return to France.
Solution
Ø Napoleonic era brought some relief to the faltering ottoman empire .it distracted Russia from
further advances. Napoleon I himself invaded Egypt, but his army was trapped when the British
sank the French fleet. A peace interlude in 1803 allowed the army to return to France.
Ø To secure his own domination and to render the rest Europe virtually powerless, napoleon
established an alliance with Russia by concluding the treaty of titlist in 1807. Russia undertook
to aid napoleon in his war against Britain ,in turn the emperor of Russia would receive the
ottoman territories of Moldavia and Wallachia if the sultan refused to surrender these territories,
France and Russia were to attack the empire, and the ottoman in Europe were to be partitioned
between the two allies.
Ø The napoleon scheme threatened threated not only the sultan, but also Britain, Austria and
Prussia, which was almost powerless in the face of such a potent alliance. The alliance naturally
proved accommodating to the au strains , who hoped that a joint Franco –Russian attack which
would probably have utterly divested the ottoman empire, could be prevented by diplomacy, but
if diplomatic measures failed , the Austrian minister klemens wenzel, prince von Metternich
decided that he would.
YES=Cathode rays have mass. .yes=Matter contains positive and nega.pdf
YES=Cathode rays have mass.
.yes=Matter contains positive and negative charge.
YES=.Particles of the cathode rays are fundamental to all matter.
.yes= An atom is divisible.
Solution
YES=Cathode rays have mass.
.yes=Matter contains positive and negative charge.
YES=.Particles of the cathode rays are fundamental to all matter.
.yes= An atom is divisible..
More Related Content
Similar to AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf
5 If an organism were heterozygous for 10 loci how many di.pdf
5. If an organism were heterozygous for 10 loci, how many different combinations could the
organism produce in its gametes? Show all your work to receive full credit. 6. In tomato plants, the
production of red fruit color is under the control of an allele R. Yellow. tomatoes are rr. The
dominant phenotype for fruit shape is under the control of an allele T. which produces two lobes.
Multilobed fruit, the recessive phenotype, have the phenotype tt. Two different crosses are made
between parental plants of unknown genotype and phenotype. Use the progeny phenotype ratios
to determine the genotypes and phenotypes of each parent. (a) Cross 1 progeny: 3/8 two-lobed,
red 3/8 two-lobed, yellow 1/8 multilobed, red 1/8 multilobed, yellow What are the genotypes and
phenotypes of the parents in this cross?3. A tropicat fish fancier produced an albino zebra fish. it
lacked blue stripes characteristic of the normal zebra fish. When she crossed a normal zebra fish
to the albino, all the offspring were narmat. When she crossed two of the normal F1 offspring, y
were abbino and % were normal. When an albino was crossed to the normal offspring, half were
albino and half were normal. What is the genotype and phenotype of each fish mentioned in this
problem? 4. In pea plants, the appearance of flowers along the main stem is a dominant
phenotype called "axial" and is controlled by an allele T. The recessive phenotype, produced by an
allele t, has flowers only at the end of the stem called "terminal". Pod form displays a dominant
phenotype "inflated", controlled by allele F, and a recessive "constricted" form produced by allele f
. A pure-breeding axial, constricted plant is crossed to one that is purebreeding terminal, inflated.
A. The F1 progeny of this cross are allowed to self-fertilize. What is the expected phenotypic
distribution of the F2 progeny? List the genotypes and phenotypes of the f1 generation and the
phenotypes of the F2 progeny. B. If an F1 plant from the initial cross described above, is crossed
to a plant that is terminal, constricted, what is the expected phenotypic distribution among the
resulting progeny?.
Similar to AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf (7)
5 If an organism were heterozygous for 10 loci how many di.pdf
5 If an organism were heterozygous for 10 loci how many di.pdf
More from aoneonlinestore1
1. VacuolesThe vacuole is an organelle in plant cells which stores.pdf
1. Vacuoles
The vacuole is an organelle in plant cells which stores water and nutrients for the cell. It takes in
and release the water as per cell\'s requirement.
2. . In plasmolysed Elodea leaf the gap between the cell wall and the cell membrane is filled with
0.6M sucrose solution.
Solution
1. Vacuoles
The vacuole is an organelle in plant cells which stores water and nutrients for the cell. It takes in
and release the water as per cell\'s requirement.
2. . In plasmolysed Elodea leaf the gap between the cell wall and the cell membrane is filled with
0.6M sucrose solution..
1. B) - Two independent properties serve to specify the state.2. B.pdf
1. B) - Two independent properties serve to specify the state.
2. B) - The locus of all points where the second law is satisfied.
3. B) - Is the p-T coordinates at which solid, liquid and vapour co-exist in equlibrium.
Solution
1. B) - Two independent properties serve to specify the state.
2. B) - The locus of all points where the second law is satisfied.
3. B) - Is the p-T coordinates at which solid, liquid and vapour co-exist in equlibrium..
(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4.
Solution
(NH4)2SO4 is a soluble salt and is fully ionized in solution:
(NH4)2SO4 => 2 NH4+ + SO42-
NH4+ is a weak acid and ionizes further: NH4+<=> NH3 + H+
The H+ formed can react with C2O42- to form the weak acid oxalic acid (H2C2O4):
C2O42- + 2 H+ <=> H2C2O4
Removal of C2O42- causes the equilibrium: ZnC2O4 <=> Zn2+ + C2O42-
to shift to the right according to Le Chatelier\'s principle and more ZnC2O4 dissolves.
Thus the solubility of ZnC2O4 increases in the presence of (NH4)2SO4..
I have inserted spaces. Consider f(x) belonging to f(X) thi.pdf
I have inserted spaces.
Consider f(x) belonging to f(X)
this means : x belongs to X
But X is a subset of Y
So, x belongs to Y
So, f(x) belongs to f(Y).
Hence all f(x) in f(X) belong to f(Y)
So, f(X) is a subset of f(Y)
Example: f(x) = sin(x) : R -> R
X = [0,2pi]
Y = [-2pi,2pi]
f(X) = [-1,1]
f(Y) = [-1,1]
So, f(X) = f(Y)
Solution
I have inserted spaces.
Consider f(x) belonging to f(X)
this means : x belongs to X
But X is a subset of Y
So, x belongs to Y
So, f(x) belongs to f(Y).
Hence all f(x) in f(X) belong to f(Y)
So, f(X) is a subset of f(Y)
Example: f(x) = sin(x) : R -> R
X = [0,2pi]
Y = [-2pi,2pi]
f(X) = [-1,1]
f(Y) = [-1,1]
So, f(X) = f(Y).
C program that prompts user to enter two floating point t.pdf
/**
C program that prompts user to enter two floating point
type values and print the product of two values upt to 3
decimal places
*/
//program1.c
//header files
#include
#include
//main function
int main()
{
//declare two double type variables
double num1;
double num2;
//set a double variable,product=0
double product=0;
printf(\"Enter num1 and num2: \");
//prompt for num1 and num2 values from keyboard
scanf(\"%lf %lf\",&num1,&num2);
//calculate product
product=num1*num2;
//print product of two numbers and limit product to 3 decimal places
printf(\"The product of %lf and %lf is %5.3lf \ \",num1,num2,product);
getch();
return 0;
}
Sample output:
Enter num1 and num2: 2.25 3.25
The product of 2.250000 and 3.250000 is 7.313
------------------------------------------------------------------------------------
//2.
/**
C program that prompts user to enter integer values
until user enters 999 to stop reading. Then prit
number of even and odd to console.
*/
//program2.c
//header files
#include
#include
//main function
int main()
{
//set SENTINAL=999;
int SENTINAL=999;
//declare integer variable
int value;
//set even and odd=0
int even=0;
int odd=0;
printf(\"Enter a vlaue or 999 to stop reading input \");
scanf(\"%d\",&value);
//read until value is not 999
while(value!=SENTINAL)
{
//check if value is even
if(value%2==0)
//increment even by one
even++;
else
//otherwise increment odd by one
odd++;
printf(\"Enter a vlaue or 999 to stop reading input \");
//read a value
scanf(\"%d\",&value);
}
printf(\"Even numbers : %d\ \",even);
printf(\"Odd numbers : %d\ \",odd);
//pause program output on console
getch();
return 0;
}
Sample output:
Enter a vlaue or 999 to stop reading input 1
Enter a vlaue or 999 to stop reading input 2
Enter a vlaue or 999 to stop reading input 3
Enter a vlaue or 999 to stop reading input 4
Enter a vlaue or 999 to stop reading input 5
Enter a vlaue or 999 to stop reading input 6
Enter a vlaue or 999 to stop reading input 7
Enter a vlaue or 999 to stop reading input 8
Enter a vlaue or 999 to stop reading input 9
Enter a vlaue or 999 to stop reading input 10
Enter a vlaue or 999 to stop reading input 999
Even numbers : 5
Odd numbers : 5
------------------------------------------------------------------------------------------------------------
//3.
/**
C program that prompts length in inches and prints the length
in miles, yards, feet to console.
*/
//program3.c
//header files
#include
#include
//function prototype
void measure(int inch,int &m,int &y,int &f, int &i);
//main function
int main()
{
int miles,yards,feet,inches=0;
int len;
printf(\"Enter the length (inches): \");
//prompt for len
scanf(\"%d\",&len);
//calling measure method
measure(len,miles,yards,feet,inches);
//print len
printf(\"Total length : %d\ \", len);
//print miles,yards, feet and inches
printf(\"Miles : %d\ \", miles);
printf(\"Yards : %d\ \", yards);
printf(\"Feet : %d\ \", feet);
printf(\"Inches : %d\ \", inches);
//pause program output on console
getch();
return.
The carbonyl functional group in glucose is an al.pdf
The carbonyl functional group in glucose is an aldehyde while the fructose contains
a ketone
Solution
The carbonyl functional group in glucose is an aldehyde while the fructose contains
a ketone.
Well to find the pH at the equivalance point. Acc.pdf
Well to find the pH at the equivalance point. According to mychemistry book(
Chemistry: A Molecular Approach BY Nivaldo J. Tro,et. all.) . pH= pKa at the equivalnce point.
Solution
Well to find the pH at the equivalance point. According to mychemistry book(
Chemistry: A Molecular Approach BY Nivaldo J. Tro,et. all.) . pH= pKa at the equivalnce point..
The weakest base in the reaction is A the enolat.pdf
The weakest base in the reaction is: A the enolate in the reaction is: C
Solution
The weakest base in the reaction is: A the enolate in the reaction is: C.
so take an aliquot of the upper layer and add a f.pdf
so take an aliquot of the upper layer and add a few drops of water. if it\'s miscible,
it\'s water. if not, it\'s the ether.
Solution
so take an aliquot of the upper layer and add a few drops of water. if it\'s miscible,
it\'s water. if not, it\'s the ether..
moles of NH4Cl formed = 20.050 = 0.1 moles so we.pdf
moles of NH4Cl formed = 2*0.050 = 0.1 moles so weight = 0.1 * 53.49 = 5.35 so
D is answer
Solution
moles of NH4Cl formed = 2*0.050 = 0.1 moles so weight = 0.1 * 53.49 = 5.35 so
D is answer.
Latency is a measure of time delay experienced in.pdf
Latency is a measure of time delay experienced in a system, the precise definition
of which depends on the system and the time being measured.
Solution
Latency is a measure of time delay experienced in a system, the precise definition
of which depends on the system and the time being measured..
Entropy is the measure of disorder. Therefore, th.pdf
Entropy is the measure of disorder. Therefore, the choice should be (D) because the
entropy of a system increases when two systems come in contact. Here, the two systems are two
liquids.
Solution
Entropy is the measure of disorder. Therefore, the choice should be (D) because the
entropy of a system increases when two systems come in contact. Here, the two systems are two
liquids..
dydx = 1x dy = dxx y = ln x + c .pdf
dy/dx = 1/x dy = dx/x y = ln x + c
Solution
dy/dx = 1/x dy = dx/x y = ln x + c.
Ø Napoleonic era brought some relief to the faltering ottoman empire.pdf
Ø Napoleonic era brought some relief to the faltering ottoman empire .it distracted Russia from
further advances. Napoleon I himself invaded Egypt, but his army was trapped when the British
sank the French fleet. A peace interlude in 1803 allowed the army to return to France.
Ø To secure his own domination and to render the rest Europe virtually powerless, napoleon
established an alliance with Russia by concluding the treaty of titlist in 1807. Russia undertook
to aid napoleon in his war against Britain ,in turn the emperor of Russia would receive the
ottoman territories of Moldavia and Wallachia if the sultan refused to surrender these territories,
France and Russia were to attack the empire, and the ottoman in Europe were to be partitioned
between the two allies.
Ø The napoleon scheme threatened threated not only the sultan, but also Britain, Austria and
Prussia, which was almost powerless in the face of such a potent alliance. The alliance naturally
proved accommodating to the au strains , who hoped that a joint Franco –Russian attack which
would probably have utterly divested the ottoman empire, could be prevented by diplomacy, but
if diplomatic measures failed , the Austrian minister klemens wenzel, prince von Metternich
decided that he would support the partition of ottoman empire a solution disadvantages to
Austria. But not as dangerous as a complete Russian takeover of southeastern Europe.
Ø The Napoleonic era (1799–1815) brought some relief to the faltering Ottoman Empire. It
distracted Russia from further advances. Napoleon I himself invaded Egypt, but his army was
trapped there when the British sank the French fleet. A peace interlude in 1803 allowed the army
to return to France.
Solution
Ø Napoleonic era brought some relief to the faltering ottoman empire .it distracted Russia from
further advances. Napoleon I himself invaded Egypt, but his army was trapped when the British
sank the French fleet. A peace interlude in 1803 allowed the army to return to France.
Ø To secure his own domination and to render the rest Europe virtually powerless, napoleon
established an alliance with Russia by concluding the treaty of titlist in 1807. Russia undertook
to aid napoleon in his war against Britain ,in turn the emperor of Russia would receive the
ottoman territories of Moldavia and Wallachia if the sultan refused to surrender these territories,
France and Russia were to attack the empire, and the ottoman in Europe were to be partitioned
between the two allies.
Ø The napoleon scheme threatened threated not only the sultan, but also Britain, Austria and
Prussia, which was almost powerless in the face of such a potent alliance. The alliance naturally
proved accommodating to the au strains , who hoped that a joint Franco –Russian attack which
would probably have utterly divested the ottoman empire, could be prevented by diplomacy, but
if diplomatic measures failed , the Austrian minister klemens wenzel, prince von Metternich
decided that he would.
YES=Cathode rays have mass. .yes=Matter contains positive and nega.pdf
YES=Cathode rays have mass.
.yes=Matter contains positive and negative charge.
YES=.Particles of the cathode rays are fundamental to all matter.
.yes= An atom is divisible.
Solution
YES=Cathode rays have mass.
.yes=Matter contains positive and negative charge.
YES=.Particles of the cathode rays are fundamental to all matter.
.yes= An atom is divisible..
TWEEN ENGINE1.Tween engine is universal.2.Reusability of tween .pdf
TWEEN ENGINE:
1.Tween engine is universal.
2.Reusability of tween engine is possible to animate our projects in java .
3.It is generic(collective) and tells us to acess the features.
4.It supports all the interpolation functions.
5.Designed especially for the games.
6.By default they are pooled hence they can be used for android game development.
7.Single line is enough to create and start a normal interpolation.
8.yoyo style option is present to repeat the tweens and timelines.
Solution
TWEEN ENGINE:
1.Tween engine is universal.
2.Reusability of tween engine is possible to animate our projects in java .
3.It is generic(collective) and tells us to acess the features.
4.It supports all the interpolation functions.
5.Designed especially for the games.
6.By default they are pooled hence they can be used for android game development.
7.Single line is enough to create and start a normal interpolation.
8.yoyo style option is present to repeat the tweens and timelines..
To identify the identity of Variable Interest Entity , first of all .pdf
To identify the identity of Variable Interest Entity , first of all we will check the snature, extent
and scope of the entity.
We will check the following
Solution
To identify the identity of Variable Interest Entity , first of all we will check the snature, extent
and scope of the entity.
We will check the following.
technology is the application of scientific knowledge. technology is.pdf
technology is the application of scientific knowledge. technology is neutral by itself. Its effects
on environment depends on the people who handle it. technology had its ability to mitigate many
problems of world. it can reproduce the lost species and balance the ecosystem. it can fasten the
growth of trees and decrease the carnbon dioxide levels in the world. it can stop the spread of
epidemics that can eliminate a particular species from world. it can detect meteors and comets
and save world from destruction.
as i said its effects depend on people who handle it. Most of the people so called technocrats use
it for momentary monetarygains. this had led the world to the brisk of chaos. they focus on
technology producing gadgets that make people happy for short term and failing to understand
the long term effects
Solution
technology is the application of scientific knowledge. technology is neutral by itself. Its effects
on environment depends on the people who handle it. technology had its ability to mitigate many
problems of world. it can reproduce the lost species and balance the ecosystem. it can fasten the
growth of trees and decrease the carnbon dioxide levels in the world. it can stop the spread of
epidemics that can eliminate a particular species from world. it can detect meteors and comets
and save world from destruction.
as i said its effects depend on people who handle it. Most of the people so called technocrats use
it for momentary monetarygains. this had led the world to the brisk of chaos. they focus on
technology producing gadgets that make people happy for short term and failing to understand
the long term effects.
b)The cations on the left side act as oxidants. C.pdf
b)The cations on the left side act as oxidants. Cations on the left are easily
oxidisable and hence are reducing agents
Solution
b)The cations on the left side act as oxidants. Cations on the left are easily
oxidisable and hence are reducing agents.
More from aoneonlinestore1 (20)
1. VacuolesThe vacuole is an organelle in plant cells which stores.pdf
1. VacuolesThe vacuole is an organelle in plant cells which stores.pdf
1. B) - Two independent properties serve to specify the state.2. B.pdf
1. B) - Two independent properties serve to specify the state.2. B.pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
(NH4)2SO4 is a soluble salt and is fully ionized in solution(NH4).pdf
I have inserted spaces. Consider f(x) belonging to f(X) thi.pdf
I have inserted spaces. Consider f(x) belonging to f(X) thi.pdf
C program that prompts user to enter two floating point t.pdf
C program that prompts user to enter two floating point t.pdf
The carbonyl functional group in glucose is an al.pdf
The carbonyl functional group in glucose is an al.pdf
Well to find the pH at the equivalance point. Acc.pdf
Well to find the pH at the equivalance point. Acc.pdf
The weakest base in the reaction is A the enolat.pdf
The weakest base in the reaction is A the enolat.pdf
so take an aliquot of the upper layer and add a f.pdf
so take an aliquot of the upper layer and add a f.pdf
moles of NH4Cl formed = 20.050 = 0.1 moles so we.pdf
moles of NH4Cl formed = 20.050 = 0.1 moles so we.pdf
Latency is a measure of time delay experienced in.pdf
Latency is a measure of time delay experienced in.pdf
Entropy is the measure of disorder. Therefore, th.pdf
Entropy is the measure of disorder. Therefore, th.pdf
Ø Napoleonic era brought some relief to the faltering ottoman empire.pdf
Ø Napoleonic era brought some relief to the faltering ottoman empire.pdf
YES=Cathode rays have mass. .yes=Matter contains positive and nega.pdf
YES=Cathode rays have mass. .yes=Matter contains positive and nega.pdf
TWEEN ENGINE1.Tween engine is universal.2.Reusability of tween .pdf
TWEEN ENGINE1.Tween engine is universal.2.Reusability of tween .pdf
To identify the identity of Variable Interest Entity , first of all .pdf
To identify the identity of Variable Interest Entity , first of all .pdf
technology is the application of scientific knowledge. technology is.pdf
technology is the application of scientific knowledge. technology is.pdf
b)The cations on the left side act as oxidants. C.pdf
b)The cations on the left side act as oxidants. C.pdf
Recently uploaded
Overview on Edible Vaccine: Pros & Cons with Mechanism
This ppt include the description of the edible vaccine i.e. a new concept over the traditional vaccine administered by injection.
Unit 2- Research Aptitude (UGC NET Paper I).pdf
This slide describes the research aptitude of unit 2 in the UGC NET paper I.
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
In this webinar you will learn how your organization can access TechSoup's wide variety of product discount and donation programs. From hardware to software, we'll give you a tour of the tools available to help your nonprofit with productivity, collaboration, financial management, donor tracking, security, and more.
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
Biological Screening of Herbal Drugs in detailed.
Biological screening of herbal drugs: Introduction and Need for
Phyto-Pharmacological Screening, New Strategies for evaluating
Natural Products, In vitro evaluation techniques for Antioxidants, Antimicrobial and Anticancer drugs. In vivo evaluation techniques
for Anti-inflammatory, Antiulcer, Anticancer, Wound healing, Antidiabetic, Hepatoprotective, Cardio protective, Diuretics and
Antifertility, Toxicity studies as per OECD guidelines
Unit 8 - Information and Communication Technology (Paper I).pdf
This slides describes the basic concepts of ICT, basics of Email, Emerging Technology and Digital Initiatives in Education. This presentations aligns with the UGC Paper I syllabus.
Best Digital Marketing Institute In NOIDA
Safalta Digital marketing institute in Noida, provide complete applications that encompass a huge range of virtual advertising and marketing additives, which includes search engine optimization, virtual communication advertising, pay-per-click on marketing, content material advertising, internet analytics, and greater. These university courses are designed for students who possess a comprehensive understanding of virtual marketing strategies and attributes.Safalta Digital Marketing Institute in Noida is a first choice for young individuals or students who are looking to start their careers in the field of digital advertising. The institute gives specialized courses designed and certification.
for beginners, providing thorough training in areas such as SEO, digital communication marketing, and PPC training in Noida. After finishing the program, students receive the certifications recognised by top different universitie, setting a strong foundation for a successful career in digital marketing.
Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Pavilions
NGV Architecture Commission Competition
MPavilion Commission Competition
Other Pavilion Designs
The Challenger.pdf DNHS Official Publication
Read| The latest issue of The Challenger is here! We are thrilled to announce that our school paper has qualified for the NATIONAL SCHOOLS PRESS CONFERENCE (NSPC) 2024. Thank you for your unwavering support and trust. Dive into the stories that made us stand out!
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Event Link:-
https://meetups.mulesoft.com/events/details/mulesoft-mysore-presents-exploring-gemini-ai-and-integration-with-mulesoft/
Agenda
● Java 17 Upgrade Overview
● Why and by when do customers need to upgrade to Java 17?
● Is there any immediate impact to upgrading to Mule Runtime 4.6 and beyond?
● Which MuleSoft products are in scope?
For Upcoming Meetups Join Mysore Meetup Group - https://meetups.mulesoft.com/mysore/
YouTube:- youtube.com/@mulesoftmysore
Mysore WhatsApp group:- https://chat.whatsapp.com/EhqtHtCC75vCAX7gaO842N
Speaker:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Organizers:-
Shubham Chaurasia - https://www.linkedin.com/in/shubhamchaurasia1/
Giridhar Meka - https://www.linkedin.com/in/giridharmeka
Priya Shaw - https://www.linkedin.com/in/priya-shaw
Shyam Raj Prasad-
https://www.linkedin.com/in/shyam-raj-prasad/
Honest Reviews of Tim Han LMA Course Program.pptx
Personal development courses are widely available today, with each one promising life-changing outcomes. Tim Han’s Life Mastery Achievers (LMA) Course has drawn a lot of interest. In addition to offering my frank assessment of Success Insider’s LMA Course, this piece examines the course’s effects via a variety of Tim Han LMA course reviews and Success Insider comments.
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk from the OECD’s Centre for Educational Research and Innovation presents at the Ask an Expert Webinar: How can education support child empowerment?
Thesis Statement for students diagnonsed withADHD.ppt
Presentation required for the master in Education.
Acetabularia Information For Class 9 .docx
Acetabularia acetabulum is a single-celled green alga that in its vegetative state is morphologically differentiated into a basal rhizoid and an axially elongated stalk, which bears whorls of branching hairs. The single diploid nucleus resides in the rhizoid.
"Protectable subject matters, Protection in biotechnology, Protection of othe...
Protectable subject matters, Protection in biotechnology, Protection of other biological materials, Ownership and period of protection
Recently uploaded (20)
Overview on Edible Vaccine: Pros & Cons with Mechanism
Overview on Edible Vaccine: Pros & Cons with Mechanism
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Welcome to TechSoup New Member Orientation and Q&A (May 2024).pdf
Azure Interview Questions and Answers PDF By ScholarHat
Azure Interview Questions and Answers PDF By ScholarHat
Unit 8 - Information and Communication Technology (Paper I).pdf
Unit 8 - Information and Communication Technology (Paper I).pdf
Digital Artifact 2 - Investigating Pavilion Designs
Digital Artifact 2 - Investigating Pavilion Designs
Multithreading_in_C++ - std::thread, race condition
Multithreading_in_C++ - std::thread, race condition
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Mule 4.6 & Java 17 Upgrade | MuleSoft Mysore Meetup #46
Francesca Gottschalk - How can education support child empowerment.pptx
Francesca Gottschalk - How can education support child empowerment.pptx
Thesis Statement for students diagnonsed withADHD.ppt
Thesis Statement for students diagnonsed withADHD.ppt
"Protectable subject matters, Protection in biotechnology, Protection of othe...
"Protectable subject matters, Protection in biotechnology, Protection of othe...
AnswerAnswerPart B).The F2 progeny contain recombinants only.pdf
- 1. Answer: Answer: Part B). The F2 progeny contain recombinants only when F1 females are test crossed; Some of the F2 progeny classes require crossing over; others do not. Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio. Explanation: To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing over. A balanced, or chromosomal translocation is a condition in which part of a chromosome has broken off and reattached in another location. In other words, it means that sections of two chromosomes have switched places. Solution Answer: Answer: Part B). The F2 progeny contain recombinants only when F1 females are test crossed; Some of the F2 progeny classes require crossing over; others do not. Chromosome ll is one of the chromosomes involved in the reciprocal balanced translocatio. Explanation: To produce phenotypes, sepi, fully fertile and wild-type, semi-sterile are not require crossing over while to produce phenotypes, sepia, semi-sterile and wile-type, fully fertile require crossing over. A balanced, or chromosomal translocation is a condition in which part of a chromosome has broken off and reattached in another location. In other words, it means that sections of two chromosomes have switched places.