This document discusses nucleic acid structure and gene expression. It begins by introducing nucleotides, which are the monomers that make up nucleic acids like DNA and RNA. They consist of a nitrogenous base, a 5-carbon sugar, and phosphate groups. The document then summarizes the central dogma of molecular biology, which states that DNA is transcribed into RNA which is translated into protein. It describes the structure of DNA as a double helix with complementary base pairing, and RNA as generally single-stranded. The document concludes by discussing DNA replication and the cell cycle.
Nucleic acids are macromolecules that store genetic information and enable protein production. Nucleic acids include DNA and RNA. These molecules are composed of long strands of nucleotides. Nucleotides are composed of a nitrogenous base, a five-carbon sugar, and a phosphate group.
Nucleic acids are macromolecules that store genetic information and enable protein production. Nucleic acids include DNA and RNA. These molecules are composed of long strands of nucleotides. Nucleotides are composed of a nitrogenous base, a five-carbon sugar, and a phosphate group.
Introduction to molecular genetic. ADN an ARN, structure and function.
Central dogma of molecular biology. Replication, transcription and translation.
Mutation, causes and classification
A brief introduction to human genetics. Relevant to medical students i.e biochem, anatomy and physiology students.
It might be very short but it is also helpful.
Introduction to molecular genetic. ADN an ARN, structure and function.
Central dogma of molecular biology. Replication, transcription and translation.
Mutation, causes and classification
A brief introduction to human genetics. Relevant to medical students i.e biochem, anatomy and physiology students.
It might be very short but it is also helpful.
Similar to Nucleic Acid Chemistry Of Purine and Pyrimidine Ribonucleotide Part I MD1 By Dr Mohamed Abdelbaky.ppt (20)
micro teaching on communication m.sc nursing.pdfAnurag Sharma
Microteaching is a unique model of practice teaching. It is a viable instrument for the. desired change in the teaching behavior or the behavior potential which, in specified types of real. classroom situations, tends to facilitate the achievement of specified types of objectives.
Title: Sense of Smell
Presenter: Dr. Faiza, Assistant Professor of Physiology
Qualifications:
MBBS (Best Graduate, AIMC Lahore)
FCPS Physiology
ICMT, CHPE, DHPE (STMU)
MPH (GC University, Faisalabad)
MBA (Virtual University of Pakistan)
Learning Objectives:
Describe the primary categories of smells and the concept of odor blindness.
Explain the structure and location of the olfactory membrane and mucosa, including the types and roles of cells involved in olfaction.
Describe the pathway and mechanisms of olfactory signal transmission from the olfactory receptors to the brain.
Illustrate the biochemical cascade triggered by odorant binding to olfactory receptors, including the role of G-proteins and second messengers in generating an action potential.
Identify different types of olfactory disorders such as anosmia, hyposmia, hyperosmia, and dysosmia, including their potential causes.
Key Topics:
Olfactory Genes:
3% of the human genome accounts for olfactory genes.
400 genes for odorant receptors.
Olfactory Membrane:
Located in the superior part of the nasal cavity.
Medially: Folds downward along the superior septum.
Laterally: Folds over the superior turbinate and upper surface of the middle turbinate.
Total surface area: 5-10 square centimeters.
Olfactory Mucosa:
Olfactory Cells: Bipolar nerve cells derived from the CNS (100 million), with 4-25 olfactory cilia per cell.
Sustentacular Cells: Produce mucus and maintain ionic and molecular environment.
Basal Cells: Replace worn-out olfactory cells with an average lifespan of 1-2 months.
Bowman’s Gland: Secretes mucus.
Stimulation of Olfactory Cells:
Odorant dissolves in mucus and attaches to receptors on olfactory cilia.
Involves a cascade effect through G-proteins and second messengers, leading to depolarization and action potential generation in the olfactory nerve.
Quality of a Good Odorant:
Small (3-20 Carbon atoms), volatile, water-soluble, and lipid-soluble.
Facilitated by odorant-binding proteins in mucus.
Membrane Potential and Action Potential:
Resting membrane potential: -55mV.
Action potential frequency in the olfactory nerve increases with odorant strength.
Adaptation Towards the Sense of Smell:
Rapid adaptation within the first second, with further slow adaptation.
Psychological adaptation greater than receptor adaptation, involving feedback inhibition from the central nervous system.
Primary Sensations of Smell:
Camphoraceous, Musky, Floral, Pepperminty, Ethereal, Pungent, Putrid.
Odor Detection Threshold:
Examples: Hydrogen sulfide (0.0005 ppm), Methyl-mercaptan (0.002 ppm).
Some toxic substances are odorless at lethal concentrations.
Characteristics of Smell:
Odor blindness for single substances due to lack of appropriate receptor protein.
Behavioral and emotional influences of smell.
Transmission of Olfactory Signals:
From olfactory cells to glomeruli in the olfactory bulb, involving lateral inhibition.
Primitive, less old, and new olfactory systems with different path
Explore natural remedies for syphilis treatment in Singapore. Discover alternative therapies, herbal remedies, and lifestyle changes that may complement conventional treatments. Learn about holistic approaches to managing syphilis symptoms and supporting overall health.
Report Back from SGO 2024: What’s the Latest in Cervical Cancer?bkling
Are you curious about what’s new in cervical cancer research or unsure what the findings mean? Join Dr. Emily Ko, a gynecologic oncologist at Penn Medicine, to learn about the latest updates from the Society of Gynecologic Oncology (SGO) 2024 Annual Meeting on Women’s Cancer. Dr. Ko will discuss what the research presented at the conference means for you and answer your questions about the new developments.
These lecture slides, by Dr Sidra Arshad, offer a quick overview of physiological basis of a normal electrocardiogram.
Learning objectives:
1. Define an electrocardiogram (ECG) and electrocardiography
2. Describe how dipoles generated by the heart produce the waveforms of the ECG
3. Describe the components of a normal electrocardiogram of a typical bipolar leads (limb II)
4. Differentiate between intervals and segments
5. Enlist some common indications for obtaining an ECG
Study Resources:
1. Chapter 11, Guyton and Hall Textbook of Medical Physiology, 14th edition
2. Chapter 9, Human Physiology - From Cells to Systems, Lauralee Sherwood, 9th edition
3. Chapter 29, Ganong’s Review of Medical Physiology, 26th edition
4. Electrocardiogram, StatPearls - https://www.ncbi.nlm.nih.gov/books/NBK549803/
5. ECG in Medical Practice by ABM Abdullah, 4th edition
6. ECG Basics, http://www.nataliescasebook.com/tag/e-c-g-basics
New Directions in Targeted Therapeutic Approaches for Older Adults With Mantl...i3 Health
i3 Health is pleased to make the speaker slides from this activity available for use as a non-accredited self-study or teaching resource.
This slide deck presented by Dr. Kami Maddocks, Professor-Clinical in the Division of Hematology and
Associate Division Director for Ambulatory Operations
The Ohio State University Comprehensive Cancer Center, will provide insight into new directions in targeted therapeutic approaches for older adults with mantle cell lymphoma.
STATEMENT OF NEED
Mantle cell lymphoma (MCL) is a rare, aggressive B-cell non-Hodgkin lymphoma (NHL) accounting for 5% to 7% of all lymphomas. Its prognosis ranges from indolent disease that does not require treatment for years to very aggressive disease, which is associated with poor survival (Silkenstedt et al, 2021). Typically, MCL is diagnosed at advanced stage and in older patients who cannot tolerate intensive therapy (NCCN, 2022). Although recent advances have slightly increased remission rates, recurrence and relapse remain very common, leading to a median overall survival between 3 and 6 years (LLS, 2021). Though there are several effective options, progress is still needed towards establishing an accepted frontline approach for MCL (Castellino et al, 2022). Treatment selection and management of MCL are complicated by the heterogeneity of prognosis, advanced age and comorbidities of patients, and lack of an established standard approach for treatment, making it vital that clinicians be familiar with the latest research and advances in this area. In this activity chaired by Michael Wang, MD, Professor in the Department of Lymphoma & Myeloma at MD Anderson Cancer Center, expert faculty will discuss prognostic factors informing treatment, the promising results of recent trials in new therapeutic approaches, and the implications of treatment resistance in therapeutic selection for MCL.
Target Audience
Hematology/oncology fellows, attending faculty, and other health care professionals involved in the treatment of patients with mantle cell lymphoma (MCL).
Learning Objectives
1.) Identify clinical and biological prognostic factors that can guide treatment decision making for older adults with MCL
2.) Evaluate emerging data on targeted therapeutic approaches for treatment-naive and relapsed/refractory MCL and their applicability to older adults
3.) Assess mechanisms of resistance to targeted therapies for MCL and their implications for treatment selection
Acute scrotum is a general term referring to an emergency condition affecting the contents or the wall of the scrotum.
There are a number of conditions that present acutely, predominantly with pain and/or swelling
A careful and detailed history and examination, and in some cases, investigations allow differentiation between these diagnoses. A prompt diagnosis is essential as the patient may require urgent surgical intervention
Testicular torsion refers to twisting of the spermatic cord, causing ischaemia of the testicle.
Testicular torsion results from inadequate fixation of the testis to the tunica vaginalis producing ischemia from reduced arterial inflow and venous outflow obstruction.
The prevalence of testicular torsion in adult patients hospitalized with acute scrotal pain is approximately 25 to 50 percent
These simplified slides by Dr. Sidra Arshad present an overview of the non-respiratory functions of the respiratory tract.
Learning objectives:
1. Enlist the non-respiratory functions of the respiratory tract
2. Briefly explain how these functions are carried out
3. Discuss the significance of dead space
4. Differentiate between minute ventilation and alveolar ventilation
5. Describe the cough and sneeze reflexes
Study Resources:
1. Chapter 39, Guyton and Hall Textbook of Medical Physiology, 14th edition
2. Chapter 34, Ganong’s Review of Medical Physiology, 26th edition
3. Chapter 17, Human Physiology by Lauralee Sherwood, 9th edition
4. Non-respiratory functions of the lungs https://academic.oup.com/bjaed/article/13/3/98/278874
Tom Selleck Health: A Comprehensive Look at the Iconic Actor’s Wellness Journeygreendigital
Tom Selleck, an enduring figure in Hollywood. has captivated audiences for decades with his rugged charm, iconic moustache. and memorable roles in television and film. From his breakout role as Thomas Magnum in Magnum P.I. to his current portrayal of Frank Reagan in Blue Bloods. Selleck's career has spanned over 50 years. But beyond his professional achievements. fans have often been curious about Tom Selleck Health. especially as he has aged in the public eye.
Follow us on: Pinterest
Introduction
Many have been interested in Tom Selleck health. not only because of his enduring presence on screen but also because of the challenges. and lifestyle choices he has faced and made over the years. This article delves into the various aspects of Tom Selleck health. exploring his fitness regimen, diet, mental health. and the challenges he has encountered as he ages. We'll look at how he maintains his well-being. the health issues he has faced, and his approach to ageing .
Early Life and Career
Childhood and Athletic Beginnings
Tom Selleck was born on January 29, 1945, in Detroit, Michigan, and grew up in Sherman Oaks, California. From an early age, he was involved in sports, particularly basketball. which played a significant role in his physical development. His athletic pursuits continued into college. where he attended the University of Southern California (USC) on a basketball scholarship. This early involvement in sports laid a strong foundation for his physical health and disciplined lifestyle.
Transition to Acting
Selleck's transition from an athlete to an actor came with its physical demands. His first significant role in "Magnum P.I." required him to perform various stunts and maintain a fit appearance. This role, which he played from 1980 to 1988. necessitated a rigorous fitness routine to meet the show's demands. setting the stage for his long-term commitment to health and wellness.
Fitness Regimen
Workout Routine
Tom Selleck health and fitness regimen has evolved. adapting to his changing roles and age. During his "Magnum, P.I." days. Selleck's workouts were intense and focused on building and maintaining muscle mass. His routine included weightlifting, cardiovascular exercises. and specific training for the stunts he performed on the show.
Selleck adjusted his fitness routine as he aged to suit his body's needs. Today, his workouts focus on maintaining flexibility, strength, and cardiovascular health. He incorporates low-impact exercises such as swimming, walking, and light weightlifting. This balanced approach helps him stay fit without putting undue strain on his joints and muscles.
Importance of Flexibility and Mobility
In recent years, Selleck has emphasized the importance of flexibility and mobility in his fitness regimen. Understanding the natural decline in muscle mass and joint flexibility with age. he includes stretching and yoga in his routine. These practices help prevent injuries, improve posture, and maintain mobilit
TEST BANK for Operations Management, 14th Edition by William J. Stevenson, Ve...kevinkariuki227
TEST BANK for Operations Management, 14th Edition by William J. Stevenson, Verified Chapters 1 - 19, Complete Newest Version.pdf
TEST BANK for Operations Management, 14th Edition by William J. Stevenson, Verified Chapters 1 - 19, Complete Newest Version.pdf
Knee anatomy and clinical tests 2024.pdfvimalpl1234
This includes all relevant anatomy and clinical tests compiled from standard textbooks, Campbell,netter etc..It is comprehensive and best suited for orthopaedicians and orthopaedic residents.
Pulmonary Thromboembolism - etilogy, types, medical- Surgical and nursing man...VarunMahajani
Disruption of blood supply to lung alveoli due to blockage of one or more pulmonary blood vessels is called as Pulmonary thromboembolism. In this presentation we will discuss its causes, types and its management in depth.
Lung Cancer: Artificial Intelligence, Synergetics, Complex System Analysis, S...Oleg Kshivets
RESULTS: Overall life span (LS) was 2252.1±1742.5 days and cumulative 5-year survival (5YS) reached 73.2%, 10 years – 64.8%, 20 years – 42.5%. 513 LCP lived more than 5 years (LS=3124.6±1525.6 days), 148 LCP – more than 10 years (LS=5054.4±1504.1 days).199 LCP died because of LC (LS=562.7±374.5 days). 5YS of LCP after bi/lobectomies was significantly superior in comparison with LCP after pneumonectomies (78.1% vs.63.7%, P=0.00001 by log-rank test). AT significantly improved 5YS (66.3% vs. 34.8%) (P=0.00000 by log-rank test) only for LCP with N1-2. Cox modeling displayed that 5YS of LCP significantly depended on: phase transition (PT) early-invasive LC in terms of synergetics, PT N0—N12, cell ratio factors (ratio between cancer cells- CC and blood cells subpopulations), G1-3, histology, glucose, AT, blood cell circuit, prothrombin index, heparin tolerance, recalcification time (P=0.000-0.038). Neural networks, genetic algorithm selection and bootstrap simulation revealed relationships between 5YS and PT early-invasive LC (rank=1), PT N0—N12 (rank=2), thrombocytes/CC (3), erythrocytes/CC (4), eosinophils/CC (5), healthy cells/CC (6), lymphocytes/CC (7), segmented neutrophils/CC (8), stick neutrophils/CC (9), monocytes/CC (10); leucocytes/CC (11). Correct prediction of 5YS was 100% by neural networks computing (area under ROC curve=1.0; error=0.0).
CONCLUSIONS: 5YS of LCP after radical procedures significantly depended on: 1) PT early-invasive cancer; 2) PT N0--N12; 3) cell ratio factors; 4) blood cell circuit; 5) biochemical factors; 6) hemostasis system; 7) AT; 8) LC characteristics; 9) LC cell dynamics; 10) surgery type: lobectomy/pneumonectomy; 11) anthropometric data. Optimal diagnosis and treatment strategies for LC are: 1) screening and early detection of LC; 2) availability of experienced thoracic surgeons because of complexity of radical procedures; 3) aggressive en block surgery and adequate lymph node dissection for completeness; 4) precise prediction; 5) adjuvant chemoimmunoradiotherapy for LCP with unfavorable prognosis.
Nucleic Acid Chemistry Of Purine and Pyrimidine Ribonucleotide Part I MD1 By Dr Mohamed Abdelbaky.ppt
1. Nucleic Acid,
Chemistry Of Purine and
Pyrimidine Ribonucleotide
Dr MOHAMED ABDELBAKY
ASSOCIATE PROFESSOR-BIOCHEMISTRY
TAU : An Innovative Centre of Excellence in Higher Medical Education & Research
1
2. Learning Objective
At end of the lecture, students can:
Explain nucleotide structure and nomenclature.
Describe organization of DNA versus RNA.
Understand general features of a chromosome.
2
3. 3
Central Dogma Of Molecular Biology
• An organism must be able to store and preserve its
genetic information, pass that information to future
generations.
• The future generations express that information as
it carries out all the processes of life.
• The major steps involved in handling genetic
information are illustrated by the central dogma of
molecular biology.
4. 4
• Genetic information is stored in the base sequence
of DNA molecules.
• Ultimately, during the process of gene expression,
this information is used to synthesize all the proteins
made by an organism.
• A gene is a unit of the DNA that encodes a
particular or RNA molecule protein.
6. 6
Gene Expression and DNA Replication
• When cells divide, each daughter cell must
receive an accurate copy of the genetic
information.
• DNA replication is the process in which each
chromosome is duplicated before cell division.
7. 7
• Transcription, the first stage in gene
expression, involves transfer of information
found in a double-stranded DNA molecule to
the base sequence of a single-stranded RNA
molecule (messenger RNA; mRNA).
• From a messenger RNA; mRNA, the process
known as translation converts the
information in the RNA base sequence to the
amino acid sequence of a protein.
8. 8
Gene Expression DNA Replication
Produces all the proteins required by
an organism.
Duplicates the chromosomes before
cell division.
Transcription of DNA: RNA copy of a
small section of a chromosome
(average size of human gene, 104–
105 nucleotide pairs).
DNA copy of entire chromosome
(average size of human chromosome,
108 nucleotide pairs).
Transcription occurs in the nucleus
throughout interphase.
Occurs during S-phase.
Translation of RNA (protein synthesis)
occurs in the cytoplasm throughout
the cell cycle.
Replication in nucleus.
9. 9
The cycle cycle describes the timing of some events in a
eukaryotic cell.
The M phase (mitosis) is the time in which the cell divides
to form 2 daughter cells.
Interphase describes the time between 2 cell divisions or
mitoses.
Gene expression occurs throughout all stages of
interphase.
The Cell Cycle
10. 10
1. G1 phase (gap 1):
o Is a period of cellular growth before DNA synthesis.
o Cells that have stopped cycling, such as muscle and
nerve cells, are said to be in a special state called G0.
2. S phase (DNA synthesis):
Is the period of time during which DNA replication
occurs.
At the end of S phase, each chromosome has doubled
its DNA content and is composed of 2 identical sister
chromatids linked at the centromere.
3. G2 phase (gap 2):
Is a period of cellular growth after DNA synthesis and it is
preceding mitosis.
Replicated DNA is checked for any errors before cell division.
Interphase is subdivided as follows:
12. 12
Many chemotherapeutic agents function by targeting specific
phases of the cell cycle.
Methotrexate, 5-fluorouracil, Hydroxyurea targeting S-
phase.
Bleomycin targeting G2 phase.
Paclitaxel, Vincristine, vinblastine targeting M phase.
Cyclophosphamide, Cisplatin are non cell-cycle
specific
Medical Correlation
13. 13
Control of the cell cycle is accomplished at
checkpoints between the various phases by
proteins such as cyclins and cyclin-
dependent kinases.
These checkpoints ensure that cells will not
enter the next phase of the cycle until the
molecular events in the previous cell cycle
phase are true and completed.
Control Of The Cell Cycle
14. 14
Reverse transcription, which produces DNA copies
of an RNA, is more commonly associated with life
cycles of retroviruses.
Synthesis of DNA from an RNA template is
catalyzed by reverse transcriptase.
Reverse transcription has a limited role in human
cells, where it plays a role in amplifying certain
highly repetitive sequences in the DNA.
Reverse transcription
15. 15
1.Retroviruses, replicate and express their genome
through a DNA intermediate (an integrated provirus).
2.Retroviruses contain RNA as their genetic
material.
3.The retroviral RNA serves as a template for the
synthesis of DNA by reverse transcriptase.
4.The DNA that is generated can be inserted into
the genome (chromosomes) of the host cell and
be expressed.
Reverse transcription
16. 16
Nucleotide Structure
Nucleic acids, are classified according to the
pentose they contain:
1. RNA (ribonucleic acid) contains ribose
sugar.
2. DNA (deoxyribonucleic acid) contains
deoxyribose sugar.
17. 17
Bases
There are 2 types of nitrogen-containing bases commonly found
in nucleotides: purines and pyrimidines.
1. Purines:
Contain 2 rings in their structure.
The purines commonly found in nucleic acids are adenine
(A) and guanine (G); both are found in DNA and RNA.
Other purine nucleotides not usually found in nucleic
acids, include xanthine, hypoxanthine, and uric acid.
2. Pyrimidines
They have only 1 ring.
Cytosine (C) is present in both DNA and RNA.
Thymine (T) is usually found only in DNA.
Uracil (U) is found only in RNA.
19. 19
Nucleosides and Nucleotides
Nucleosides are formed by covalently linking a
base to the number 1 carbon of a sugar.
Nucleotides are formed when 1 or more
phosphate groups is attached to the 5′ carbon
of a nucleoside.
Nucleoside di- and triphosphates are high-
energy compounds because of the hydrolytic
energy associated with the acid anhydride
bonds.
23. 23
Nucleic Acids
•Nucleic acids are polymers of nucleotides joined by
3′, 5′-phosphodiester bonds.
•The phosphodiester bonds; a phosphate group links
the 3′ carbon of a sugar to the 5′ carbon of the next
sugar in the chain.
•Each strand has a distinct 5′ end and 3′ end, and thus
has polarity.
•A phosphate group is often found at the 5′ end, and a
hydroxyl group is often found at the 3′ end.
24. 24
• The base sequence of a nucleic acid strand is written
by convention, in the 5′→3′ direction (left to right).
• According to this convention, the sequence of the
strand on the left (in figure) must be written 5′-TCAG-3′
or TCAG.
• If written backward (from 3’ to 5’), the ends must be
labeled: 3′-GACT-5′.
• The positions of phosphates may be shown:
pTpCpApG.
• In DNA, a “d” (deoxy) may be included: dTdCdAdG.
25. 25 N
H3C
O
O
O
5´CH2
5´CH
CH3
3´
2 O
3´
O
3´
3´
O
O
O P O
O
T
N
N
N
A
O H
N H N
OH
O 5´CH2
O
P O
O
5´CH2
O P O
O
O
O
3´
O
5´CH2
P
O
O
O
OH
O P O
O
O
O P O
O
H
N
N
O
3´
O
3´
3´
C
N
N
N
G
N
N
N
N
N H O
5´CH2
O
O
5´CH2
O
P O
O
5´CH2
O
H
G
N
C
N H N
N H O
H
O H N
H
N
O
A
N
T
H
N H O
N H N
3´
3´
5´
5´
N H N
O H N
H
N
5′-TCAG-3′
or TCAG.
26. 26
O
Adenine
NH2
H
H
N N
N
N
O
O P O
CH2 O– O
5' end
Cytosine
O
N
H
NH2
N
O H
O P O
C
H2 O–
O
Thymine
O
O
H3C N H
N
Guanine
O
H
H
NH2
N N
N
N
O
O
Phosphate
backbone
O
O H
O P O
C
H2 O–
Phosphodies
ter bond
O P O CH2
O–
3' end
A segment of a polynucleotide strand. This strand contains thymine and
exclusively deoxyribose, so it is a segment of DNA. The phosphodiester bonds
which link the 3 ́ and 5 ́ carbons of the sugars.
27. 27
•DNA contains the bases adenine (A), guanine (G), cytosine
(C), and thymine (T).
•RNA contains A, G, and C, but has uracil (U) instead of
thymine (T).
•The genetic message resides in the sequence of bases
along the polynucleotide chain.
•In DNA, two polynucleotide chains are joined by pairing
between their bases (adenine with thymine and guanine with
cytosine), and they form a double helix.
•One chain runs in a 5′ to 3′ direction and the other runs 3′ to 5′.
28. 28
Antiparallel strands of DNA. That the strands run in
opposite directions, as determined by the hydroxyl groups
on carbons 3 and phosphate group on carbon 5 of the
deoxyribose (the 3´ and 5´ carbons).
29. 29
•DNA molecules in eukaryotes interact with histones
to form strands of nucleosomes, which wind into more
tightly coiled structures.
•RNA is single-stranded, but the strands loop back on
themselves and the bases pair: guanine with cytosine
and adenine with uracil.
•This allows the RNA to form a three-dimensional
structure that can be recognized by specific proteins
and enzymes.
30. 30
•mRNA has a cap at the 5′ end and a poly(A) tail at
the 3′ end.
•rRNA has extensive base-pairing.
•tRNA forms a cloverleaf structure that contains many
unusual nucleotides and an anticodon.
• In eukaryotes, DNA is generally double-stranded
(dsDNA) and RNA is generally single-stranded
(ssRNA).
• Exceptions occur in certain viruses, some of
which have ssDNA genomes and some of which
have dsRNA genomes.
32. 32
DNA Structure
Some of the features of double-stranded
DNA include:
The 2 strands are antiparallel (opposite in
direction).
The 2 strands are complementary:
• A always pairs with T (2 hydrogen bonds).
• G always pairs with C (3 hydrogen bonds).
• Thus, the base sequence on one strand
defines the base sequence on the other
strand.
33. 33
Because of the specific base pairing, the
amount of A equals the amount of T.
The amount of G equals the amount of C.
The total purines equals total pyrimidines.
The Chargaff’s Rules:
34. 34
Using Chargaff’s Rules
In dsDNA (or dsRNA) (ds = double-stranded) % A = %
T (% U)
And %G = %C
% purines = % pyrimidines
A sample of DNA has 10% G; what is the % T?
10% G + 10% C = 20%
therefore, % A + % T must total 80% 40% A and 40%
T
Ans: 40% T
With minor modification (substitution of U for T) these
rules also apply to dsRNA.
35. 35
Most DNA occurs in nature as a right-handed double-
helical molecule known as Watson-Crick DNA or B-DNA.
The hydrophilic sugar-phosphate backbone of each strand is
on the outside of the double helix.
The hydrogen-bonded base pairs are stacked in the center of
the molecule.
There are about 10 base pairs per complete turn of the helix.
A rare left-handed double-helical form of DNA that rich in G-
C– sequences; is known as Z-DNA.
The biologic function of Z-DNA is unknown, but may be
related to gene regulation.
37. 37
Denaturation And Renaturation of DNA
Double-helical DNA can be denatured by conditions
that disrupt hydrogen bonding and base stacking,
resulting in the “melting” of the double helix into two
single strands that separate from each other.
No covalent bonds are broken in this process.
Causes of DNA Denaturation:
Heat
• Alkaline pH
• Chemicals such as formamide and urea
38. 38
Denatured single-stranded DNA can be
renatured (annealed) if the denaturing
condition is slowly removed.
For example, if a solution containing heat-
denatured DNA is slowly cooled, the two
complementary strands can become base-
paired again.
40. 40
The renaturation or annealing of complementary
DNA strands is an important step in probing a
Southern blot and in performing the polymerase
chain reaction (PCR).
In these techniques, a well-characterized probe
DNA is added to a mixture of target DNA molecules.
The mixed sample is denatured and then renatured
and when probe DNA binds to target DNA
sequences of sufficient complementarity, the
process is called hybridization.
41. 41
Organization Of DNA
Large DNA molecules must be packaged in specific
way that they can fit inside the cell and still be
functional.
1. Supercoiling
o Mitochondrial DNA and the DNA of most
prokaryotes are closed circular structures.
o These molecules may exist as relaxed circles or as
supercoiled structures in which the helix is twisted
around itself in 3-dimensional space.
42. 42
o Supercoiling results from strain on the molecule
caused by under- or over- winding the double helix:
1- Negatively supercoiled
DNA is formed if the DNA is wound more loosely than
in Watson-Crick DNA and this form is required for
most biologic reactions.
2- Positively supercoiled
DNA is formed if the DNA is wound more tightly than in
Watson-Crick DNA.
43. 43
o Topoisomerases are enzymes that can change
the amount of supercoiling in DNA molecules.
o They make transient breaks in DNA strands by
alternately breaking and resealing the sugar-
phosphate backbone.
o For example, in Escherichia coli, DNA gyrase
(DNA topoisomerase II) can introduce negative
supercoiling into DNA.
44. 44
2. Nucleosomes And Chromatin
Nuclear DNA in eukaryotes is found in chromatin
associated with histones proteins.
The basic packaging unit of chromatin is the
nucleosome.
Histones are rich in lysine and arginine, which
give a positive charge on the proteins.
Two copies each of histones H2A, H2B, H3, and
H4 aggregate to form the histone octamer (8
protein units).
45. 45
DNA is wound around the outside of this
octamer to form a nucleosome; is referred to
as a 10nm chromatin fiber (a series of
nucleosomes is sometimes called “beads on
a string”).
Histone H1 is associated with the linker DNA
found between nucleosomes to help
package them into a solenoid-like structure
(coiled shape), which is a thick 30-nm fiber.
46. 46
Nucleosomes and Chromatin
Expanded view of a
nucleosome
+HI Without HI
Expanded view
30 nm
10 nm
Sensitive to
nuclease
H1
H4 H3
H2B
H2A
H4
H2B
H3
H2A
High-Yield
Nucleosome and Nucleofilament Structure
in Eukaryotic DNA
47. 47
Each eukaryotic chromosome in G0 or G1 contains
one linear molecule of double-stranded DNA.
Cells in interphase contain 2 types of chromatin:
a) Euchromatin (more opened and available
for gene expression).
Euchromatin generally corresponds to the
nucleosomes (10-nm fibers) loosely associated
with each other.
48. 48
DNA
Linker DNA
Core histones (H2A, H2B,
H3, and H4)
Histone H1
Nucleosome core
The solenoid
A poly-nucleosome, indicating the histone cores and
linker DNA. The DNA is depicted in blue, whereas the
histones are depicted as light brown spheres
49. 49
b) Heterochromatin (much more highly condensed
and associated with areas of the chromosomes that
are not expressed).
Heterochromatin is more highly condensed,
producing interphase heterochromatin as well as
chromatin characteristic of mitotic chromosomes.
50. 50
More active
DNA double helix 10 nm chromatin 30 nm chromatin
(nucleosomes) (nucleofilament)
30 nm fiber forms loops attached Higher order to scaffolding
proteins packaging
Euchromatin Heterochromatin
Less active
51. Text Books Required
Thomas M Devlin
Lippincott's
D M Vasudevan
Harpers
Lieberman & Peet
Thompson & Thompson
51
52. Questions
52
1. A double-stranded RNA genome isolated from
a virus in the stool of a child with
gastroenteritis was found to contain 20%
uracil. What is the percentage of guanine in
this genome?
A. 15
B. 25
C. 30
D. 75
E. 85
53. 53
1. A double-stranded RNA genome isolated from
a virus in the stool of a child with
gastroenteritis was found to contain 20%
uracil. What is the percentage of guanine in
this genome?
A. 15
B. 25
C. 30
D. 75
E. 85
54. 54
1. Answer: C.
U = A = 20%.
Since C+ G = 60%, G = 30%.
Alternatively, U = A = 20%, then U + A = 40% C
+ G = 60%, and G = 30%.
55. 55
2. Endonuclease activation and chromatin
fragmentation are characteristic features of
eukaryotic cell death by apoptosis. Which of
the following chromosome structures would
most likely be degraded first in an apoptotic
cell?
A. Barr body
B. 10-nm fiber
C. 30-nm fiber
D. Centromere
E. Heterochromatin
56. 56
2. Endonuclease activation and chromatin
fragmentation are characteristic features of
eukaryotic cell death by apoptosis. Which of
the following chro- mosome structures would
most likely be degraded first in an apoptotic
cell?
A. Barr body
B. 10-nm fiber
C. 30-nm fiber
D. Centromere
E. Heterochromatin
57. 57
2. Answer: B. The more “opened” the DNA,
the more sensitive it is to enzyme attack.
The 10-nm fiber is the most open structure
listed.
The endonuclease would attack the region of
unprotected DNA between the nucleosomes.
58. 58
3. A medical student working in a molecular
biology laboratory is asked by her mentor to
determine the base composition of an
unlabeled nucleic acid sample left behind by a
former research technologist. The results of her
analysis show 20% adenine, 50% cytosine, 15%
thymine and 15% guanine. What is the most
likely source of the nucleic acid in this sample?
A. Bacterial chromosome
B. Viral genome
C. Bacterial plasmid
D. Mitochondrial chromosome
E. Nuclear chromosome
59. 59
3. A medical student working in a molecular
biology laboratory is asked by her mentor to
determine the base composition of an
unlabeled nucleic acid sample left behind by a
former research technologist. The results of her
analysis show 20% adenine, 50% cytosine, 15%
thymine and 15% guanine. What is the most
likely source of the nucleic acid in this sample?
A. Bacterial chromosome
B. Viral genome
C. Bacterial plasmid
D. Mitochondrial chromosome
E. Nuclear chromosome
60. 60
3. Answer: B. A base compositional analysis that
deviates from Chargaff’s rules (%A = %T, %C = %G)
is indicative of single-stranded, not double- stranded,
nucleic acid molecule. All options listed except B are
examples of circular (choices A, C and D) or linear
(choice E) DNA double helices. Only a few viruses
(e.g. parvovirus) have single-stranded DNA.
61. 61
4. The sequence of part of a DNA strand is
the following:–ATTCGATTGCCCACGT–.
When this strand is used as a template for
DNA synthesis, the product will be which
one of the following?
A. TAAGCTAACGGGTGCA
B. UAAGCUAACGGGUGCA
C.ACGUGGGCAAUCGAAU
D.ACGTGGGCAATCGAAT
E. TGCACCCGTTAGCTTA
62. 62
4. The sequence of part of a DNA strand is
the following:–ATTCGATTGCCCACGT–.
When this strand is used as a template for
DNA synthesis, the product will be which
one of the following?
A. TAAGCTAACGGGTGCA
B. UAAGCUAACGGGUGCA
C.ACGUGGGCAAUCGAAU
D.ACGTGGGCAATCGAAT
E. TGCACCCGTTAGCTTA
63. 63
4. The ansWer is D ACGTGGGCAATCGAAT;
DNA replication will be complementary to the
template, and antiparallel. Reading from the
5′ end of the template, the product will be 3′-
TGCACCCGTTAGCTTA-5’. Recall that uracil
(U) is not placed into DNA by DNA
polymerase.
64. 64
5. The procedure of Southern blotting involves treatment
of the solid support (nitrocellulose) containing the DNA
with NaOH to denature the double helix. Treatment of a
Northern blot with NaOH, however, will lead to the
hydrolysis of the nucleic acid on the filter paper. This is
due to which major chemical feature of the nucleic acids
involved in a Northern blot?
A. The presence of thymine
B. The presence of uracil
C. The presence of a 2′-hydroxyl group
D. The presence of a 3′-hydroxyl group
E. The presence of a 3′–5′ phosphodiester linkage
65. 65
5. The procedure of Southern blotting involves treatment
of the solid support (nitrocellulose) containing the DNA
with NaOH to denature the double helix. Treatment of a
Northern blot with NaOH, however, will lead to the
hydrolysis of the nucleic acid on the filter paper. This is
due to which major chemical feature of the nucleic acids
involved in a Northern blot?
A. The presence of thymine
B. The presence of uracil
C. The presence of a 2′-hydroxyl group
D. The presence of a 3′-hydroxyl group
E. The presence of a 3′–5′ phosphodiester linkage
66. 66
6. An African native who is going to college in the
United States experiences digestive problems
(bloating, diarrhea, and flatulence) whenever she
eats foods containing milk products. She is most
likely deficient in splitting which type of chemical
bond?
A. A sugar bond
B. An ester linkage
C. A phosphodiester bond
D. An amide bond
E. A glycosidic bond
67. 67
6. An African native who is going to college in the
United States experiences digestive problems
(bloating, diarrhea, and flatulence) whenever she
eats foods containing milk products. She is most
likely deficient in splitting which type of chemical
bond?
A. A sugar bond
B. An ester linkage
C. A phosphodiester bond
D. An amide bond
E. A glycosidic bond
68. 68
7. A single-stranded DNA molecule contains
20%A, 25%T, 30%G, and 25%C. When the
complement of this strand is synthesized, the T
content of the resulting duplex will be which
one of the following?
A. 20%
B. 22.5%
C. 25%
D. 27.5%
E. 30%
69. 69
7. A single-stranded DNA molecule contains
20%A, 25%T, 30%G, and 25%C. When the
complement of this strand is synthesized, the T
content of the resulting duplex will be which
one of the following?
A. 20%
B. 22.5%
C. 25%
D. 27.5%
E. 30%
70. 70
7. The answer is B: 22.5%. The given strand of DNA
contains 25%T; the complementary strand will contain
20%T (this must be equivalent to the content of A in
the given strand, since A and T base pair, and [A] = [T]
in duplex DNA). For the entire duplex then, the T con-
tent is the average of 25% and 20%, or 22.5% for the
duplex. The [A] in the duplex will also be 22.5%
(again, since [A] = [T]), and the concentrations of [G]
and [C] will each be 27.5% for the duplex.