Presentation for Bella Mahl 2024-03-28-24-MW-Overview-Bella.pptx
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.
