According to the endosymbiotic hypothesis- an aerobic prokaryote precu.docx
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According to the endosymbiotic hypothesis, an aerobic prokaryote precursor was engulfed by another leading to some organelle formation. With this, gene transfer had to occur from the precursor cell's nucleus to the host cell. A gene, Cox2, in some species, is found in the mitochondrial DNA while in other species, is found in the nuclear DNA. The initial transcript of the mitochondrial Cox2 gene is modified by RNA editing, where some specific cytosines are changed to uracils. Why do you think this is significant and how might it allow you to discover whether the informational intermediate in the transfer was DNA or RNA? 5 points .
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// array
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int rear;
int size;
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Deque(int size)
{
front = -1;
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void insertfront(int key);
void insertrear(int key);
void deletefront();
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};
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bool Deque::isFull()
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push(int n, Deque &dq) and pop(Deque &dq) functions simulating stack operations.
keep it simple and to the instructions and code provided please! and thank you!
// C++ implementation of De-queue using circular
// array
#include<iostream>
using namespace std;
// Maximum size of array or Dequeue
#define MAX 100
// A structure to represent a Deque
class Deque
{
int arr[MAX];
int front;
int rear;
int size;
public :
Deque(int size)
{
front = -1;
rear = 0;
this->size = size;
}
// Operations on Deque:
void insertfront(int key);
void insertrear(int key);
void deletefront();
void deleterear();
bool isFull();
bool isEmpty();
int getFront();
int getRear();
};
// Checks whether Deque is full or not.
bool Deque::isFull()
{
return ((front == 0 && rear == size-1)||
front == rear+1);
}
// Checks whether Deque is empty or not.
bool Deque::isEmpty ()
{
return (front == -1);
}
// Inserts an element at front
void Deque::insertfront(int key)
{
// check whether Deque if full or not
if (isFull())
{
cout << "Overflow\n" << endl;
return;
}
// If queue is initially empty
if (front == -1)
{
front = 0;
rear = 0;
}
// front is at first position of queue
else if (front == 0)
front = size - 1 ;
else // decrement front end by '1'
front = front-1;
// insert current element into Deque
arr[front] = key ;
}
// function to inset element at rear end
// of Deque.
void Deque ::insertrear(int key)
{
if (isFull())
{
cout << " Overflow\n " << endl;
return;
}
// If queue is initially empty
if (front == -1)
{
front = 0;
rear = 0;
}
// rear is at last position of queue
else if (rear == size-1)
rear = 0;
// increment rear end by '1'
else
rear = rear+1;
// insert current element into Deque
arr[rear] = key ;
}
// Deletes element at front end of Deque
void Deque ::deletefront()
{
// check whether Deque is empty or not
if (isEmpty())
{
cout << "Queue Underflow\n" << endl;
return ;
}
// Deque has only one element
if (front == rear)
{
front = -1;
rear = -1;
}
else
// back to initial position
if (front == size -1)
front = 0;
else // increment front by '1' to remove current
// front value from Deque
front = front+1;
}
// Delete element at rear end of Deque
void Deque::deleterear()
{
if (isEmpty())
{
cout << " Underflow\n" << endl ;
return ;
}
// Deque has only one element
if (front == rear)
{
front = -1;
rear = -1;
}
else if (rear == 0)
rear = size-1;
else
rear = rear-1;
}
// Returns front element of Deque
int Deque::getFront()
{
// check whether Deque is empty or not
if (isEmp.
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Activity- Simulating meiosis 1. Cut out the chromosomes on the second to last page of this lab. One color of chromosomes is inherited from one parent, and the other is inherited from the other parent. Use a paper clip to keep identical sister chromatids together. 2. Use the circles representing cells provided on the last two pages. Meiosis I Prophase I 1. Put all four of the chromosomes that you cut out and place them in the large oval cell labeled A pole and B pole. 2. Bring the members of the homologous pairs together forming a tetrad- Why do you think these are called a tetrad? 3. Crossing-over is the exchange of genetic material between two homologous chromosomes. It creates genetic recombination during meiosis. Simulate crossingover by cutting off one yellow and blue circle from two nonsister chromatids of a single homologous pair and reattaching the circles on with tape. Why did we exchange from nonsister and not from sister chromatids? MetaphaseI Move the homologous chromosome pairs to the equator (middle of the cell) in such a way that the members of each pair can be moved towards opposite sides of the cell AnaphaseI Separate the each member of the homologous pair, moving them toward opposite poles of the cell TelophaseI Do the 2 chromosomes inherited from the father have to stay together after meiosis I? So, what are two ways that meiosis can create genetic recombination? Interkinesis Interkinesis is the periods of time between meiosis I and meiosis II. Meiosis II Prophase II Choose the chromosomes from each pole to represent those in the new parental cell undergoing meiosis II. Place them inside the top two cells on the next page. Metaphase II Move the chromosomes (made out of the sister chromatids) to the equator of this cell. How many chromosomes are at this equator? Anaphase II Pull apart the two sister chromatids Telophase II Once the sister chromatids are apart they are now each called a chromosome. Move each chromosome to its separate pole. Now these chromosomes are in their own cell in a new nucleus. Conclusions Do the new nuclei contain a haploid or diploid chromosome number compared to the original cell? Suppose you had worked with both daughter cells after meiosis I. How many cells would be present when meiosis II was complete? How many chromosomes are in the parental cell undergoing meiosis II (at the beginning of meiosis II? How many chromosomes are in the daughter cell (at the end of meiosis II)? How is this possible? Explain below. To summarize meiosis: the original parent cell has the diploid ( 2 n ) number of chromosomes, and the daughter cells at the end of meiosis II have the number of chromosomes. Meiosis is cell division in which the chromosome number does what?
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