Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens.
Solution
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens..
A dam can also be used to collect or store water which can be evenly distributed between locations. Dams generally serve the primary purpose of retaining water, while other structures such as floodgates or levees (also known as dikes) are used to manage or prevent water flow into specific land regions. The earliest known dam is the Jawa Dam in Jordan, dating to 3,000 BC
#includestdio.h#includestdlib.h structure of a stack node .pdfankitcomputer11
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdfankitcomputer11
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
More Related Content
Similar to Cones in gymnosprems are homologous to gametophyte of seedless plant.pdf
A dam can also be used to collect or store water which can be evenly distributed between locations. Dams generally serve the primary purpose of retaining water, while other structures such as floodgates or levees (also known as dikes) are used to manage or prevent water flow into specific land regions. The earliest known dam is the Jawa Dam in Jordan, dating to 3,000 BC
#includestdio.h#includestdlib.h structure of a stack node .pdfankitcomputer11
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct sNode* new_node =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_node == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
/* put in the data */
new_node->data = new_data;
/* link the old list off the new node */
new_node->next = (*top_ref);
/* move the head to point to the new node */
(*top_ref) = new_node;
}
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref)
{
int res;
struct sNode *top;
/*If stack is empty then error */
if(*top_ref == NULL)
{
printf(\"Stack overflow \ \");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->data;
*top_ref = top->next;
free(top);
return res;
}
}
/* Driver function to test anove functions */
int main()
{
/* Create a queue with items 1 2 3*/
struct queue *q = (struct queue*)malloc(sizeof(struct queue));
q->stack1 = NULL;
q->stack2 = NULL;
enQueue(q, 1);
enQueue(q, 2);
enQueue(q, 3);
/* Dequeue items */
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
printf(\"%d \", deQueue(q));
getchar();
}
Solution
#include
#include
/* structure of a stack node */
struct sNode
{
int data;
struct sNode *next;
};
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data);
/* Function to pop an item from stack*/
int pop(struct sNode** top_ref);
/* structure of queue having two stacks */
struct queue
{
struct sNode *stack1;
struct sNode *stack2;
};
/* Function to enqueue an item to queue */
void enQueue(struct queue *q, int x)
{
push(&q->stack1, x);
}
/* Function to dequeue an item from queue */
int deQueue(struct queue *q)
{
int x;
/* If both stacks are empty then error */
if(q->stack1 == NULL && q->stack2 == NULL)
{
printf(\"Q is empty\");
getchar();
exit(0);
}
/* Move elements from satck1 to stack 2 only if
stack2 is empty */
if(q->stack2 == NULL)
{
while(q->stack1 != NULL)
{
x = pop(&q->stack1);
push(&q->stack2, x);
}
}
x = pop(&q->stack2);
return x;
}
/* Function to push an item to stack*/
void push(struct sNode** top_ref, int new_data)
{
/* allocate node */
struct s.
Solution Since we know that the subshell with .pdfankitcomputer11
Solution
: Since we know that the subshell with : n=2 and l=1 is the 2p subshell; Similarly : if n=3 and
l=0, it is the 3s subshell and so on that\'s why it is n=3, l=2 and m_l=0 Please rate.
let V be nonempty open subset of X then V inters.pdfankitcomputer11
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X
Solution
let V be nonempty open subset of X then V intersection U is nonempty because U
is dense in X therefore V is not contained in X/U so X/U is nowhere dense in X.
Look at the periodic table. When Be loses its f.pdfankitcomputer11
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be
Solution
Look at the periodic table. When Be loses it\'s first electron it now only has 3
electroms (Just like Li) when it loses it\'s second electron in only has 2 electrons (just like He)
when it loses it\'s third electron it goes from 1s2 to 1s1 therefore the 1s orbital loses an electron
in the third ionization step for Be.
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14])
Solution
taking L.T.,,,,,,,,,,,,Y(s)[s^2+4s+14]=L(T(t))...........for periodic functions the trans
form is given as ........inte(0 to T)(T(t))exp(-st)*dt/[1-exp(-sT)].........solving we get the the
integral as [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]...........putting it in the equation
we get [(1+exp(-s)-2*exp(-s/2)/s^2.+ (exp(-s)-exp(-s/2))/s]/([s^2+4s+14]).
the salt is neutral in charge. OS of Ca is +2, so.pdfankitcomputer11
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2
Solution
the salt is neutral in charge. OS of Ca is +2, so (NO3)2 bears -2 charges, so NO3
bears -1 charge, we know O holds -2 charge Thus, (the charge of N) + -2x3 =-1 so, N = +5 Ca
+2, N +5, O -2.
nonmetals, metals, metalloids, and gases. They ar.pdfankitcomputer11
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life
Solution
nonmetals, metals, metalloids, and gases. They are important because they are
involved in everyday life.
In thermodynamics, a state function, function of .pdfankitcomputer11
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni)
Solution
In thermodynamics, a state function, function of state, state quantity, or state
variable is a property of a system that depends only on the current state of the system, not on the
way in which the system acquired that state (independent of path). A state function describes the
equilibrium state of a system. For example, internal energy, enthalpy, and entropy are state
quantities because they describe quantitatively an equilibrium state of a thermodynamic system,
irrespective of how the system arrived in that state. In contrast, mechanical work and heat are
process quantities because their values depend on the specific transition (or path) between two
equilibrium states. The following are considered to be state functions in thermodynamics: Mass
(m) Energy (E) Enthalpy (H) Internal energy (U) Gibbs free energy (G) Helmholtz free energy
(A) Exergy Entropy (S) Pressure (p) Temperature (T) Volume (V) Chemical composition
Specific volume (v) or its reciprocal Density (?) Fugacity Altitude Particle number (ni).
The process of protein synthesis and secretion starts from the endop.pdfankitcomputer11
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis).
Solution
The process of protein synthesis and secretion starts from the endoplasmic reticulum. The
polypeptide chain is formed in the ribosomes, and then transferred to the Golgi apparatus, where
the final modifications are done. The Golgi apparatus sorts the modified poly peptide molecules
and packages into new transport vesicles which reach their particular destinations. The
tranlocons transport the polypeptides with a signal sequence to the interior of the endoplasmic
reticulum. Multiple control mechanisms of the endoplasmic reticulum ensures that only properly
folded proteins to export towards the Golgi apparatus.
The plasma membrane proteins synthesized in the endoplasmic reticulum (ER). The proteins
generally contain endoplasmic reticulum signal sequence at their N-terminus, which directs them
to the rough endoplasmic reticulum. After completion of translation process in the RER, the
polypeptide chains get inserted into the ER membrane or into the lumen of ER.
Some proteins move to the transport vesicles, which fuse together to form cis- Golgi vesicles, the
cis-form again moves to trans-form. During this process, many proteins undergo modifications
particularly at oligosaccharide chains. From the trans-Golgi vesicles, proteins move to cell
membrane through small transportation vesicles (exocytosis)..
The IUPAC names of the given compounds is 2-chloro-1-butanol.pdfankitcomputer11
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol
Solution
The IUPAC names of the given compounds is:
2-chloro-1-butanol
2-methoxycyclohexanol.
Solution - (c) - 20001 SharesCalculation of Number of shares to be.pdfankitcomputer11
Solution
- (c) - 20001 Shares
Calculation of Number of shares to be controlled to guarantee the election to be board.
= [1/(Number of Seats of Board of Directors + 1)*Shares Outstanding] + 1
= [1/(5+1)*120000 + 1]
= [(1/6)*120000] + 1
= 20001 Shares..
Move decimal to behind the first number4.71Round4.714.71 X 1.pdfankitcomputer11
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal)
Solution
Move decimal to behind the first number
4.71
Round
4.71
4.71 X 10^2 (2=the number of spaces you had to move the decimal).
Lewis acid is the electron pair acceptorLewis base is the electron.pdfankitcomputer11
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base
Solution
Lewis acid is the electron pair acceptor
Lewis base is the electron pair donar
a) OH- + Al(OH)3 -----> Al(OH)4-
Al(OH)3 is the lewis acid
OH- is the lewis base
b) SO3 + H2O --------> H2SO4
H2O is the lewis acid
SO3 is the lewis base
H2O is the lewis acid
SO3 is the lewis base
c) Co3+ + 6NH3 -----> Co(NH3)63+
Co 3+ is the lewis acid
NH3 is the lewis base
Co 3+ is the lewis acid
NH3 is the lewis base.
first treat with alcoholic KOH to get cyclo hexene and then treat it.pdfankitcomputer11
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol
Solution
first treat with alcoholic KOH to get cyclo hexene and then treat it with alkaline KMnO4 to get
trans cyclohxane -1,2 -diol.
An azeotrope is a mixture of two or more liquids .pdfankitcomputer11
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4
Solution
An azeotrope is a mixture of two or more liquids in such a ratio that its composition
cannot be changed by simple distillation. This occurs because, when an azeotrope is boiled, the
resulting vapor has the same ratio of constituents as the original mixture. Because their
composition is unchanged by distillation, azeotropes are also called constant boiling mixtures.
Common drying agenets for organic compounds : Alcohols : Anhydrous K2CO3, CaSO4 or
Mg2SO4 and CaO Alkylhalides/aryl halides : Anhydrous CaCl2,CaSO4 or Mg2SO4 and
phosphoric acid Saturated aromatic hydrocarbonds : CaCl2,CaSO4 and phosphoric oxide
Aldehydes : CaSO4,Mg2SO4 Na2SO4 Ketones : CaSO4,MgSO4 ,NaSO4 or K2CO3 Organic
bases : Solid KOH/NaOH ,CaO or Barium oxide Organic acids : Calcium sulphate, Mg2SO4.
Here we have sodium bisulfate NaHSO4, as you have shown the ions pr.pdfankitcomputer11
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :)
Solution
Here we have sodium bisulfate: NaHSO4, as you have shown the ions present are of equal
charge. Thus, there are equal amounts of both ions.
Recall molarity,
molarity = mole / volume
Thus, we can calculate the number of moles:
M = mol / V
0.55 = mol / 1.32L
Solving for moles gives: mole = 0.726 moles.
Since we established earlier, both species (Na+ and HSO4- are present in equal amounts), thus
moles (Na+) = moles (HSO4-) = 0.726 moles
Please don\'t forget to rate, I hope this helped! :).
The Roman Empire A Historical Colossus.pdfkaushalkr1407
The Roman Empire, a vast and enduring power, stands as one of history's most remarkable civilizations, leaving an indelible imprint on the world. It emerged from the Roman Republic, transitioning into an imperial powerhouse under the leadership of Augustus Caesar in 27 BCE. This transformation marked the beginning of an era defined by unprecedented territorial expansion, architectural marvels, and profound cultural influence.
The empire's roots lie in the city of Rome, founded, according to legend, by Romulus in 753 BCE. Over centuries, Rome evolved from a small settlement to a formidable republic, characterized by a complex political system with elected officials and checks on power. However, internal strife, class conflicts, and military ambitions paved the way for the end of the Republic. Julius Caesar’s dictatorship and subsequent assassination in 44 BCE created a power vacuum, leading to a civil war. Octavian, later Augustus, emerged victorious, heralding the Roman Empire’s birth.
Under Augustus, the empire experienced the Pax Romana, a 200-year period of relative peace and stability. Augustus reformed the military, established efficient administrative systems, and initiated grand construction projects. The empire's borders expanded, encompassing territories from Britain to Egypt and from Spain to the Euphrates. Roman legions, renowned for their discipline and engineering prowess, secured and maintained these vast territories, building roads, fortifications, and cities that facilitated control and integration.
The Roman Empire’s society was hierarchical, with a rigid class system. At the top were the patricians, wealthy elites who held significant political power. Below them were the plebeians, free citizens with limited political influence, and the vast numbers of slaves who formed the backbone of the economy. The family unit was central, governed by the paterfamilias, the male head who held absolute authority.
Culturally, the Romans were eclectic, absorbing and adapting elements from the civilizations they encountered, particularly the Greeks. Roman art, literature, and philosophy reflected this synthesis, creating a rich cultural tapestry. Latin, the Roman language, became the lingua franca of the Western world, influencing numerous modern languages.
Roman architecture and engineering achievements were monumental. They perfected the arch, vault, and dome, constructing enduring structures like the Colosseum, Pantheon, and aqueducts. These engineering marvels not only showcased Roman ingenuity but also served practical purposes, from public entertainment to water supply.
Operation “Blue Star” is the only event in the history of Independent India where the state went into war with its own people. Even after about 40 years it is not clear if it was culmination of states anger over people of the region, a political game of power or start of dictatorial chapter in the democratic setup.
The people of Punjab felt alienated from main stream due to denial of their just demands during a long democratic struggle since independence. As it happen all over the word, it led to militant struggle with great loss of lives of military, police and civilian personnel. Killing of Indira Gandhi and massacre of innocent Sikhs in Delhi and other India cities was also associated with this movement.
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The French Revolution, which began in 1789, was a period of radical social and political upheaval in France. It marked the decline of absolute monarchies, the rise of secular and democratic republics, and the eventual rise of Napoleon Bonaparte. This revolutionary period is crucial in understanding the transition from feudalism to modernity in Europe.
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Cones in gymnosprems are homologous to gametophyte of seedless plant.pdf
1. Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens.
Solution
Cones in gymnosprems are homologous to gametophyte of seedless plant
Seedless plants do not produce flowers or grow from seeds. They have two stages in their
lifecycle
1) spores producing
2) sex cells (sperm and eggs) producing stage
Gymnosperms (greek meaning naked seeds) are seed-bearing vascular plants in which the seeds
(ovule) are not inside an ovary rather they are present on scales or leaves often modified to cone.
Eg. Pine tree.
When you see a pine tree the main plant is sporophyte (diploid stage). The gametophyte (haploid
stage) form is confined to the cones which are either female cones having ovule or male cones
having pollens.
