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The recombinant plasmid method is used for identifying sequences that can serve as origins of replication. The DNA of interest is fragmented into pieces that are ligated into plasmids, and the plasmids are transformed into host cells. To form colonies, the plasmids must contain a sequence that can serve as an origin of replication. The recombinant plasmid method has worked for identifying E . coli and yeast origins, but not origins for higher eukaryotes like humans. What is the most likely explanation for this observation? The ends of human origins of replication are unable to be ligated into plasmids. Human chromosomes do not contain origins of replication. Human origins of replication are probably determined by a chromatin structure unable to form in this assay. Legally speaking, human cells cannot be used in a laboratory. .
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Replicative mechanisms of Plasmid and Circular Genomes
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extra-chromosomal-inheritance[1].pptx.pdfpdf
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
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Slide 6: Mechanisms of Extrachromosomal Inheritance
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Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
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Slide 8: Importance of Extrachromosomal Inheritance
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Slide 9: Recent Research and Advances
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Slide 10: Conclusion
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Basic genetics /certified fixed orthodontic courses by Indian dental academy
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EXCEPT:a. Mitochondria and plastids contain circular DNA molecules similar to the circular
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can be used to transcribe and translate DNA into proteinse. Mitochondria and plastids have a
double membrane (inner and outer)
Solution
Out of the above options , the correct will be
B. Mitochondria and plastids a have their genetic material in the form of DNA not RNA.
They donot have any such machinary for their expression , so this option is incorrect regarding
endosymbiosis.
Otherwise rest all the sentenses are true.
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layer..
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The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and
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Similar to The recombinant plasmid method is used for identifying sequences that (1).docx
extra-chromosomal-inheritance[1].pptx.pdfpdf
Slide 1: Title Slide
Extrachromosomal Inheritance
Slide 2: Introduction to Extrachromosomal Inheritance
Definition: Extrachromosomal inheritance refers to the transmission of genetic material that is not found within the nucleus.
Key Components: Involves genes located in mitochondria, chloroplasts, and plasmids.
Slide 3: Mitochondrial Inheritance
Mitochondria: Organelles responsible for energy production.
Mitochondrial DNA (mtDNA): Circular DNA molecule found in mitochondria.
Inheritance Pattern: Maternally inherited, meaning it is passed from mothers to all their offspring.
Diseases: Examples include Leber’s hereditary optic neuropathy (LHON) and mitochondrial myopathy.
Slide 4: Chloroplast Inheritance
Chloroplasts: Organelles responsible for photosynthesis in plants.
Chloroplast DNA (cpDNA): Circular DNA molecule found in chloroplasts.
Inheritance Pattern: Often maternally inherited in most plants, but can vary in some species.
Examples: Variegation in plants, where leaf color patterns are determined by chloroplast DNA.
Slide 5: Plasmid Inheritance
Plasmids: Small, circular DNA molecules found in bacteria and some eukaryotes.
Features: Can carry antibiotic resistance genes and can be transferred between cells through processes like conjugation.
Significance: Important in biotechnology for gene cloning and genetic engineering.
Slide 6: Mechanisms of Extrachromosomal Inheritance
Non-Mendelian Patterns: Do not follow Mendel’s laws of inheritance.
Cytoplasmic Segregation: During cell division, organelles like mitochondria and chloroplasts are randomly distributed to daughter cells.
Heteroplasmy: Presence of more than one type of organellar genome within a cell, leading to variation in expression.
Slide 7: Examples of Extrachromosomal Inheritance
Four O’clock Plant (Mirabilis jalapa): Shows variegated leaves due to different cpDNA in leaf cells.
Petite Mutants in Yeast: Result from mutations in mitochondrial DNA affecting respiration.
Slide 8: Importance of Extrachromosomal Inheritance
Evolution: Provides insight into the evolution of eukaryotic cells.
Medicine: Understanding mitochondrial inheritance helps in diagnosing and treating mitochondrial diseases.
Agriculture: Chloroplast inheritance can be used in plant breeding and genetic modification.
Slide 9: Recent Research and Advances
Gene Editing: Techniques like CRISPR-Cas9 are being used to edit mitochondrial and chloroplast DNA.
Therapies: Development of mitochondrial replacement therapy (MRT) for preventing mitochondrial diseases.
Slide 10: Conclusion
Summary: Extrachromosomal inheritance involves the transmission of genetic material outside the nucleus and plays a crucial role in genetics, medicine, and biotechnology.
Future Directions: Continued research and technological advancements hold promise for new treatments and applications.
Slide 11: Questions and Discussion
Invite Audience: Open the floor for any questions or further discussion on the topic.
Basic genetics /certified fixed orthodontic courses by Indian dental academy
Welcome to Indian Dental Academy
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and offering a wide range of dental certified courses in different formats.
Indian dental academy has a unique training program & curriculum that provides students with exceptional clinical skills and enabling them to return to their office with high level confidence and start treating patients
State of the art comprehensive training-Faculty of world wide repute &Very affordable.
All of these are examples of evidence used to support the hypothesis.pdf
All of these are examples of evidence used to support the hypothesis of endosymbiosis,
EXCEPT:a. Mitochondria and plastids contain circular DNA molecules similar to the circular
chromosomes of bacteriab. Mitochondria and plastids, like prokaryotes, only contain genetic
material in the form of RNAc. Mitochondria and plastids can replicate themselves in a process
similar to prokaryotesd. Mitochondria and plastids have their own ribosomes and enzymes that
can be used to transcribe and translate DNA into proteinse. Mitochondria and plastids have a
double membrane (inner and outer)
Solution
Out of the above options , the correct will be
B. Mitochondria and plastids a have their genetic material in the form of DNA not RNA.
They donot have any such machinary for their expression , so this option is incorrect regarding
endosymbiosis.
Otherwise rest all the sentenses are true.
They have circular genome like Prokaryotes, and can replicate and also have double membrane
layer..
Bacterial plasmid
A bacterial plasmid is a short, usually circular, and double-stranded segment of DNA that is found in the cytoplasm separate from the main bacterial chromosome. This presentation contains plasmid features, replication, classification and its uses.
5. Architecture of chromosome.pdf
Detail study about Cromosome Structure, function of chomoses and types of chromosome
Nuclear Genomes(Short Answers and questions)
1. What did researchers find when they sequenced the centromeres of Arabidopsis? Why was this finding surprising?
Ans: Before the Arabidopsis sequences were obtained it was thought that these repeat sequences were by far the principal component of centromeric DNA. However, Arabidopsis centromeres also contain multiple copies of genome-wide repeats, along with a few genes, the latter at a density of 7–9 per 100 kb compared with 25 genes per 100 kb for the noncentromeric regions of Arabidopsis chromosomes. The discovery that centromeric DNA contains genes was a big surprise because it was thought that these regions were genetically inactive.
2. What differences in gene distribution and repetitive DNA content are seen when yeast and human chromosomes are compared?
Ans. A typical region of a human chromosome will have few genes (most of which will contain introns), several repeated sequences, and a large amount of nonrepetitive, nongenic DNA. Yeast chromosomes have higher gene densities, with very few genes containing introns, and have few repeated sequences and much less nongenic DNA.
3. The human genome contains about 50,000 fewer genes than was predicted by many researchers. Why were these initial predictions so high?
Ans. These early estimates were high because they were based on the supposition that, in most cases, a single gene specifies a single mRNA and a single protein. According to this model, the number of genes in the human genome should be similar to the number of proteins in human cells, leading to the estimates of 80,000–100,000. The discovery that the number of genes is much lower than this indicates that alternative splicing, the process by which exons from a pre-mRNA are assembled in different combinations so that more than one protein can be coded by a single gene is more prevalent than was originally appreciated.
4. What are the different methods used to catalog genes? What are the advantages or disadvantages of these methods?
Ans. Gene catalogs can be based on the known functions of genes, but such catalogs are incomplete because in most genomes many genes have unknown functions. Gene catalogs that are based on the identities of protein domains coded by genes are more comprehensive as these include many genes whose specific functions are unknown.
5. What is the function of the different genes in the human globin gene families?
Ans. The globins are the blood proteins that combine to make hemoglobin, each molecule of hemoglobin being made up of two a-type and two b-type globins.The a-globin cluster is located on chromosome 16 and the b-cluster on chromosome 11. Both clusters contain genes that are expressed at different developmental stages and each includes at least one pseudogene. Note that expression of the a-type gene x2 begins in the embryo and continues during the fetal stage; there is no fetal-specific a-type globin. The q pseudogene is expressed but its protein product is inactive. None of the other p
Chromosomes
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and
offering a wide range of dental certified courses in different formats.for more details please visit
www.indiandentalacademy.com
Dna n gene
The Indian Dental Academy is the Leader in continuing dental education , training dentists in all aspects of dentistry and
offering a wide range of dental certified courses in different formats.for more details please visit
www.indiandentalacademy.com
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Basic genetics /certified fixed orthodontic courses by Indian dental academy
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All of these are examples of evidence used to support the hypothesis.pdf
All of these are examples of evidence used to support the hypothesis.pdf
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int Card::getValue() {
if (rank == 'A') {
return 1;
}
else if (rank == '2') {
return 2;
}
else if (rank == '3') {
return 3;
}
else if (rank == '4') {
return 4;
}
else if (rank == '5') {
return 5;
}
else if (rank == '6') {
return 6;
}
else if (rank == '7') {
return 7;
}
else if (rank == '8') {
return 8;
}
else if (rank == '9') {
return 9;
}
else if (rank == 'K') {
return 10;
}
else if (rank == 'Q') {
return 10;
}
else {
return 10;
}
}
//Show a carrd representation
void Card::showCard() {
cout << rank << suit<<".";
}
Deck.h
//Create a deck of cards
#ifndef DECK_H
#define DECK_H
#include "Card.h"
class Deck {
//Member variables
private:
Card deck[52];
int cardsCnt;
//Member variables
public:
//Constructor
Deck();
//Create a fresh deck
void refreshDeck();
//Get a card from top
Card deal();
//Shuffle
void shuffle();
//Number of cards left in deck
int cardsLeft();
//Display deck
void displayDeck();
};
#endif // !DECK_H
Deck.cpp
//Implementation of deck class
#include "Deck.h"
//Constructor
Deck::Deck() {
char ranks[] = { 'A','1','2','3','4','5','6','7','8','9','J','Q','K' };
char suits[] = { 'S','H','D','C' };
int k = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 13; j++)
{
deck[k++] = Card(ranks[j], suits[i]);
}
}
cardsCnt = 52;
}
//Create a fresh deck
void Deck::refreshDeck() {
char ranks[] = { 'A','1','2','3','4','5','6','7','8','9','J','Q','K' };
char suits[] = { 'S','H','D','C' };
int k = 0;
for (int i = 0; i < 4; i++) {
for (int j = 0; j < 13; j++)
{
deck[k++] = Card(ranks[j], suits[i]);
}
}
cardsCnt = 52;
}
//Get a card from top
Card Deck::deal() {
Card c=deck[cardsCnt - 1];
cardsCnt--;
return c;
}
//Shuffle
void Deck::shuffle() {
srand(0);
for (int i = 0; i <cardsCnt; i++)
{
int r = i + (rand() % (52 - i));
Card temp = deck[i];
deck[i] = deck[r];
deck[r] = temp;
}
}
//Number of cards left in deck
int Deck::cardsLeft() {
return cardsCnt;
}
//Display deck
void Deck :: displayDeck() {
for (int i = 0; i < 52; i++) {
if (i % 13 == 0 && i != 0) {
cout << endl;
deck[i].showCard();
cout << " ";
}
else {
deck[i].showCard();
cout << " ";
}
}
}
Main.cpp
#include "Deck.h"
#include <stack>
//Create object of Deck class
Deck deck;
//Funct.
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