Microtubule organizing center is required to concentrate all mitocho.pdf
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Microtubule organizing center is required to concentrate all mitochondria around the basal body of the flagellum. Solution Microtubule organizing center is required to concentrate all mitochondria around the basal body of the flagellum..
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SH bond is more polar Solution .pdf
number ogf nitrate ion = 4.5*3*6.023*10^23 = 8.13 x10^24 ions
Solution
number ogf nitrate ion = 4.5*3*6.023*10^23 = 8.13 x10^24 ions.
It is ionic solid (compound). It will be more sol.pdf
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar
Solution
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar.
IUPAC name 1-ethoxy-2-methylpropane other name.pdf
IUPAC name: 1-ethoxy-2-methylpropane other name: isobutylethyl ether
Solution
IUPAC name: 1-ethoxy-2-methylpropane other name: isobutylethyl ether.
Evaporation at 20°C is negligible. HgO decomposes.pdf
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus.
Solution
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus..
E) Mg is the reducing agent and Cu^2+ is the oxid.pdf
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct
Solution
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct.
Recommended
SH bond is more polar Solution .pdf
number ogf nitrate ion = 4.5*3*6.023*10^23 = 8.13 x10^24 ions
Solution
number ogf nitrate ion = 4.5*3*6.023*10^23 = 8.13 x10^24 ions.
It is ionic solid (compound). It will be more sol.pdf
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar
Solution
It is ionic solid (compound). It will be more soluble in water as CCl4 is non-polar.
IUPAC name 1-ethoxy-2-methylpropane other name.pdf
IUPAC name: 1-ethoxy-2-methylpropane other name: isobutylethyl ether
Solution
IUPAC name: 1-ethoxy-2-methylpropane other name: isobutylethyl ether.
Evaporation at 20°C is negligible. HgO decomposes.pdf
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus.
Solution
Evaporation at 20°C is negligible. HgO decomposes on exposure to light, on
heating above 500°C producing highly toxic fumes including mercury and oxygen, which
increases fire hazard. Reacts violently with reducing agents, chlorine, hydrogen peroxide,
magnesium (when heated), disulfur dichloride and hydrogen trisulfide. Shock-sensitive
compounds are formed with metals and elements such as sulfur and phosphorus..
E) Mg is the reducing agent and Cu^2+ is the oxid.pdf
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct
Solution
E) Mg is the reducing agent and Cu^2+ is the oxidizing agent. is correct.
boiling point KCl CH2O CO2 note KCl is ioni.pdf
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction.
Solution
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction..
A is right option beacuse H2O will attack on carb.pdf
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water
Solution
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water.
Fifi eventhough is a male; it is possible to clone it to have a dupl.pdf
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father.
Solution
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father..
Please post the picture , it is not visible to me.pdf
Please post the picture , it is not visible to me.
Solution
Please post the picture , it is not visible to me..
one. If there was more than one proton, they woul.pdf
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons)
Solution
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons).
Moles of Cu present = 0.57663.5 = 0.00907 1L ha.pdf
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L
Solution
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L.
In R2, the only non-trivial subspaces are lines p.pdf
In R2, the only non-trivial subspaces are lines passing through the origin.
Solution
In R2, the only non-trivial subspaces are lines passing through the origin..
first one is far more acidic due to high inductiv.pdf
first one is far more acidic due to high inductive effect of three F s
Solution
first one is far more acidic due to high inductive effect of three F s.
Wait I am doing it ...please rate 5 ..Ill post the answer in comme.pdf
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments....
Solution
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments.....
When ever you impelemented an interface we must define the abstract .pdf
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TODO Auto-generated method stub
return null;
}
public boolean remove(Object arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean removeAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean retainAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public int size() {
// TODO Auto-generated method stub
return 0;
}
public Object[] toArray() {
// TODO Auto-generated method stub
return null;
}
public T[] toArray(T[] arg0) {
// TODO Auto-generated method stub
return null;
}
}
now use this program and test run y
Solution
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TO.
The primary role of adaptation of flowers, fruits or seeds remain to.pdf
The primary role of adaptation of flowers, fruits or seeds remain to promote pollination in plants
for which the plants acquire multiple type of adaptations.
Please find the answers below:
A. Role of flowers in adaptation: Flowers play a crucial role in adaptation by acquiring
differntial colors in order to attract insects which helps in cross pollination. Also, the flowers
might acquire changes in shape of their petals by exaggerating the length and location of ovary
so that the stigma might be more protruded in nature in order to aid pollination. Alternatively,
petals might be altogether absent in order to absolutely promote pollination by insects or wind.
B. Role of fruits in adaptation: A fruiting body can help in adaptation of a plant by many means.
The fruiting body might be reduced in size in order to promote easy dispersal by air. Small fruit
size aids in dispersal of seeds in many plants since a ripened ovary or fruit shrinks in size thus
revealing the seed for dispersal. Also, under harsh conditions, the fruit might convert into a
protective structure which can aid sustanance of the seed until favourable conditions arrive.
Acquisition of bright colors also helps to attract insects and animals which in turn promotes
dispersal of seeds.
C. Role of seeds in adaptation: Some plants are evolved with attachment of feathery structures on
the seeds which further promote dispersal by wind. Some aquatic plants may acquire changes in
shapes of seeds to promote buoyancy so that the seed might float on the water surface and do not
sink. Structural modification of seeds is realised in draught conditions and flood conditions when
hard external coating is formed around the seed to prevent undue loss or influx of water thus
protecting the germplasm.
Thus, structural and functional modification of plant organs help to disperse the seeds in plants
and help in adaptation under differential environmental conditions.
Solution
The primary role of adaptation of flowers, fruits or seeds remain to promote pollination in plants
for which the plants acquire multiple type of adaptations.
Please find the answers below:
A. Role of flowers in adaptation: Flowers play a crucial role in adaptation by acquiring
differntial colors in order to attract insects which helps in cross pollination. Also, the flowers
might acquire changes in shape of their petals by exaggerating the length and location of ovary
so that the stigma might be more protruded in nature in order to aid pollination. Alternatively,
petals might be altogether absent in order to absolutely promote pollination by insects or wind.
B. Role of fruits in adaptation: A fruiting body can help in adaptation of a plant by many means.
The fruiting body might be reduced in size in order to promote easy dispersal by air. Small fruit
size aids in dispersal of seeds in many plants since a ripened ovary or fruit shrinks in size thus
revealing the seed for dispersal. Also, under harsh conditions, the fruit might conver.
The origins of algebra can be traced to the ancient Babylonians, who.pdf
The origins of algebra can be traced to the ancient Babylonians, who developed a positional
number system that greatly aided them in solving their rhetorical algebraic equations. The
Babylonians were not interested in exact solutions but approximations, and so they would
commonly use linear interpolation to approximate intermediate values.
quadratic equations in ancient and Medieval times
Solution
The origins of algebra can be traced to the ancient Babylonians, who developed a positional
number system that greatly aided them in solving their rhetorical algebraic equations. The
Babylonians were not interested in exact solutions but approximations, and so they would
commonly use linear interpolation to approximate intermediate values.
quadratic equations in ancient and Medieval times.
Synapomorphies are defined as ancestral characters shared by two or .pdf
Synapomorphies are defined as ancestral characters shared by two or more species.
The chordata word has emerged from characters which are being shared between organisms
indicating common ancestry. There are two well known ancestral characters common between
majority of the chordates i.e. notochord and pharyngeal Slits.
Notochord: This is common between all the chordates at different stages of growth and
development. It represents like a flexible rod that helps in the movement of the chordates.
Pharyngeal Slits: These are openings present in pharynx or throat region. These slits have
evolved differently in various chordates such as used to filter food particles in older chordates, in
other species of chordates, slits are present only during the embryonic stages.
Solution
Synapomorphies are defined as ancestral characters shared by two or more species.
The chordata word has emerged from characters which are being shared between organisms
indicating common ancestry. There are two well known ancestral characters common between
majority of the chordates i.e. notochord and pharyngeal Slits.
Notochord: This is common between all the chordates at different stages of growth and
development. It represents like a flexible rod that helps in the movement of the chordates.
Pharyngeal Slits: These are openings present in pharynx or throat region. These slits have
evolved differently in various chordates such as used to filter food particles in older chordates, in
other species of chordates, slits are present only during the embryonic stages..
Peptide signals are signals in mitochondria as wll as chloroplast fo.pdf
Peptide signals are signals in mitochondria as wll as chloroplast for incoming molecule.
Nuclear localization signals (NLS) or nuclear export signals (NES) are the signals for the
nucleus.
Solution
Peptide signals are signals in mitochondria as wll as chloroplast for incoming molecule.
Nuclear localization signals (NLS) or nuclear export signals (NES) are the signals for the
nucleus..
Phosphorescence is longer-lived than fluorescence. The transisition .pdf
Phosphorescence is longer-lived than fluorescence. The transisition probablility is much greater
for fluorescence so the life time is 1.0 x 10^5 longer for phosphorescence because the probablitiy
for the electron to go to the lower state is 1.0 x 10^5 lower.
1.0 X 10^-9 s (1.0 x 10^5) = 1 x 10-4 s
Solution
Phosphorescence is longer-lived than fluorescence. The transisition probablility is much greater
for fluorescence so the life time is 1.0 x 10^5 longer for phosphorescence because the probablitiy
for the electron to go to the lower state is 1.0 x 10^5 lower.
1.0 X 10^-9 s (1.0 x 10^5) = 1 x 10-4 s.
Element symbol is Ge All other elements in its g.pdf
Element symbol is: Ge All other elements in its group are C, Si, Sn, Pb
Solution
Element symbol is: Ge All other elements in its group are C, Si, Sn, Pb.
In Caenorhabditis elegans the two sexual phenotypes that occur are m.pdf
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence.
Solution
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence..
What is the purpose of studying mathematics.pptx
Students often ask about what the purpose is for their learning. This PowerPoint highlights some really important reasons to study Mathematics.
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boiling point KCl CH2O CO2 note KCl is ioni.pdf
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction.
Solution
boiling point: KCl > CH2O > CO2 note: KCl is ionic compound, which is of high
boiling point and high melting point characters. CH2O are polar molecule, which can form
intermolecular hydrogen bonds CO2 is nonpolar molecule with a very weak intermolecular
interaction..
A is right option beacuse H2O will attack on carb.pdf
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water
Solution
A is right option beacuse H2O will attack on carbonil group and formed O- will
accpet H+ from water.
Fifi eventhough is a male; it is possible to clone it to have a dupl.pdf
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father.
Solution
Fifi eventhough is a male; it is possible to clone it to have a duplicate of it. Egg cell has to be
selected from a female pet. Nucleus of egg cell to be removed. Donor somatic cell of Fifi has to
be selected. Donor somatic cell of Fifi has the nucleus. Donor cell of Fifi (with nucleus) has to be
bombarded into an egg cell (without nucleus). The fused cell divides normally. The Embryo
developed thus has to be placed in the uterus of the female. The embryo develops in the womb of
the female into a lovable pet Fifi, which is just a copy of its father..
Please post the picture , it is not visible to me.pdf
Please post the picture , it is not visible to me.
Solution
Please post the picture , it is not visible to me..
one. If there was more than one proton, they woul.pdf
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons)
Solution
one. If there was more than one proton, they wouldn\'t be hydrogen. (Isotopes are
varying numbers of neutrons).
Moles of Cu present = 0.57663.5 = 0.00907 1L ha.pdf
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L
Solution
Moles of Cu present = 0.576/63.5 = 0.00907 1L have HNO3 have = 16 moles
Thus Volume required = (1/16)*( 0.00907) = 0.000566875 L.
In R2, the only non-trivial subspaces are lines p.pdf
In R2, the only non-trivial subspaces are lines passing through the origin.
Solution
In R2, the only non-trivial subspaces are lines passing through the origin..
first one is far more acidic due to high inductiv.pdf
first one is far more acidic due to high inductive effect of three F s
Solution
first one is far more acidic due to high inductive effect of three F s.
Wait I am doing it ...please rate 5 ..Ill post the answer in comme.pdf
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments....
Solution
Wait I am doing it ...please rate 5 ..I\"ll post the answer in comments.....
When ever you impelemented an interface we must define the abstract .pdf
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TODO Auto-generated method stub
return null;
}
public boolean remove(Object arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean removeAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean retainAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public int size() {
// TODO Auto-generated method stub
return 0;
}
public Object[] toArray() {
// TODO Auto-generated method stub
return null;
}
public T[] toArray(T[] arg0) {
// TODO Auto-generated method stub
return null;
}
}
now use this program and test run y
Solution
When ever you impelemented an interface we must define the abstract methods in that interface
in inheritance class. so that we must define the all methods in that interface (collection)
solved problem:
import java.util.Collection;
import java.util.Iterator;
public class PartialArrayMultiset implements Collection
{
protected E[] _store;
protected int _size;
protected long _modCount;
public PartialArrayMultiset()
{
_modCount=0;
clear();
}
public void clear()
{
_store=(E[])(new Object[16]);
_size=0;
_modCount+=1;
}
public boolean add(E e)
{
if(_size==_store.length)
return false;
_store[_size]=e;
_size++;
return true;
}
public void removeAtIndex(int i)
{
if(i<0 || i>= _size)
throw new ArrayIndexOutOfBoundsException(i);
E oldValue=_store[i];
for(int hole=i;hole<_size-1;hole++)
_store[hole]=_store[hole+1];
_store[_size-1]=null;
_size--;
}
public boolean contains(Object obj)
{
return false;
}
public boolean addAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean containsAll(Collection arg0) {
// TODO Auto-generated method stub
return false;
}
public boolean isEmpty() {
// TODO Auto-generated method stub
return false;
}
public Iterator iterator() {
// TO.
The primary role of adaptation of flowers, fruits or seeds remain to.pdf
The primary role of adaptation of flowers, fruits or seeds remain to promote pollination in plants
for which the plants acquire multiple type of adaptations.
Please find the answers below:
A. Role of flowers in adaptation: Flowers play a crucial role in adaptation by acquiring
differntial colors in order to attract insects which helps in cross pollination. Also, the flowers
might acquire changes in shape of their petals by exaggerating the length and location of ovary
so that the stigma might be more protruded in nature in order to aid pollination. Alternatively,
petals might be altogether absent in order to absolutely promote pollination by insects or wind.
B. Role of fruits in adaptation: A fruiting body can help in adaptation of a plant by many means.
The fruiting body might be reduced in size in order to promote easy dispersal by air. Small fruit
size aids in dispersal of seeds in many plants since a ripened ovary or fruit shrinks in size thus
revealing the seed for dispersal. Also, under harsh conditions, the fruit might convert into a
protective structure which can aid sustanance of the seed until favourable conditions arrive.
Acquisition of bright colors also helps to attract insects and animals which in turn promotes
dispersal of seeds.
C. Role of seeds in adaptation: Some plants are evolved with attachment of feathery structures on
the seeds which further promote dispersal by wind. Some aquatic plants may acquire changes in
shapes of seeds to promote buoyancy so that the seed might float on the water surface and do not
sink. Structural modification of seeds is realised in draught conditions and flood conditions when
hard external coating is formed around the seed to prevent undue loss or influx of water thus
protecting the germplasm.
Thus, structural and functional modification of plant organs help to disperse the seeds in plants
and help in adaptation under differential environmental conditions.
Solution
The primary role of adaptation of flowers, fruits or seeds remain to promote pollination in plants
for which the plants acquire multiple type of adaptations.
Please find the answers below:
A. Role of flowers in adaptation: Flowers play a crucial role in adaptation by acquiring
differntial colors in order to attract insects which helps in cross pollination. Also, the flowers
might acquire changes in shape of their petals by exaggerating the length and location of ovary
so that the stigma might be more protruded in nature in order to aid pollination. Alternatively,
petals might be altogether absent in order to absolutely promote pollination by insects or wind.
B. Role of fruits in adaptation: A fruiting body can help in adaptation of a plant by many means.
The fruiting body might be reduced in size in order to promote easy dispersal by air. Small fruit
size aids in dispersal of seeds in many plants since a ripened ovary or fruit shrinks in size thus
revealing the seed for dispersal. Also, under harsh conditions, the fruit might conver.
The origins of algebra can be traced to the ancient Babylonians, who.pdf
The origins of algebra can be traced to the ancient Babylonians, who developed a positional
number system that greatly aided them in solving their rhetorical algebraic equations. The
Babylonians were not interested in exact solutions but approximations, and so they would
commonly use linear interpolation to approximate intermediate values.
quadratic equations in ancient and Medieval times
Solution
The origins of algebra can be traced to the ancient Babylonians, who developed a positional
number system that greatly aided them in solving their rhetorical algebraic equations. The
Babylonians were not interested in exact solutions but approximations, and so they would
commonly use linear interpolation to approximate intermediate values.
quadratic equations in ancient and Medieval times.
Synapomorphies are defined as ancestral characters shared by two or .pdf
Synapomorphies are defined as ancestral characters shared by two or more species.
The chordata word has emerged from characters which are being shared between organisms
indicating common ancestry. There are two well known ancestral characters common between
majority of the chordates i.e. notochord and pharyngeal Slits.
Notochord: This is common between all the chordates at different stages of growth and
development. It represents like a flexible rod that helps in the movement of the chordates.
Pharyngeal Slits: These are openings present in pharynx or throat region. These slits have
evolved differently in various chordates such as used to filter food particles in older chordates, in
other species of chordates, slits are present only during the embryonic stages.
Solution
Synapomorphies are defined as ancestral characters shared by two or more species.
The chordata word has emerged from characters which are being shared between organisms
indicating common ancestry. There are two well known ancestral characters common between
majority of the chordates i.e. notochord and pharyngeal Slits.
Notochord: This is common between all the chordates at different stages of growth and
development. It represents like a flexible rod that helps in the movement of the chordates.
Pharyngeal Slits: These are openings present in pharynx or throat region. These slits have
evolved differently in various chordates such as used to filter food particles in older chordates, in
other species of chordates, slits are present only during the embryonic stages..
Peptide signals are signals in mitochondria as wll as chloroplast fo.pdf
Peptide signals are signals in mitochondria as wll as chloroplast for incoming molecule.
Nuclear localization signals (NLS) or nuclear export signals (NES) are the signals for the
nucleus.
Solution
Peptide signals are signals in mitochondria as wll as chloroplast for incoming molecule.
Nuclear localization signals (NLS) or nuclear export signals (NES) are the signals for the
nucleus..
Phosphorescence is longer-lived than fluorescence. The transisition .pdf
Phosphorescence is longer-lived than fluorescence. The transisition probablility is much greater
for fluorescence so the life time is 1.0 x 10^5 longer for phosphorescence because the probablitiy
for the electron to go to the lower state is 1.0 x 10^5 lower.
1.0 X 10^-9 s (1.0 x 10^5) = 1 x 10-4 s
Solution
Phosphorescence is longer-lived than fluorescence. The transisition probablility is much greater
for fluorescence so the life time is 1.0 x 10^5 longer for phosphorescence because the probablitiy
for the electron to go to the lower state is 1.0 x 10^5 lower.
1.0 X 10^-9 s (1.0 x 10^5) = 1 x 10-4 s.
Element symbol is Ge All other elements in its g.pdf
Element symbol is: Ge All other elements in its group are C, Si, Sn, Pb
Solution
Element symbol is: Ge All other elements in its group are C, Si, Sn, Pb.
In Caenorhabditis elegans the two sexual phenotypes that occur are m.pdf
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence.
Solution
In Caenorhabditis elegans the two sexual phenotypes that occur are males and hermaphrodites.
Males have testes and the hermaphrodites have both testes and ovaries.
They have only one type of sex chromosomes - X and have autosomes.
Sexes are determined by the ratio of sex chromosomes to autosomes. X chromosomes:autosomes
XX/AA = 1 hermaphrodite ; XO/AA =0.5 males
There are two sexes in mammals - males and female. In mammals, Y chromosome is used for
sex determination. Y chromosomes confer maleness as it has a gene responsible for testes
development of gonads. In its absence the gonads develop into ovaries conferring femaleness.
XX female; XY male
In birds, there are two sexes males and females. Here W chromosome confers femaleness.
ZZ - male; ZW - female.
2.
Reciprocal cross - the cross in which the phenotype of parents are reversed from the previous
original cross.
AA (male) x aa (female) - Original cross
aa(male) x AA(female) - Reciprocal cross
Significance of reciprocal cross - to find out the role of parental sex on pattern of inheritence..
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