old UEC syllabus

1. Chapter 19
Heredity and Variation
2018
Ong Yee Sing

2. Learning objectives
• 1. to recognize basic genetic materials.
• 2. To understand the structure and replication of DNA molecule.
• 3. To understand the gene expression and the synthesis of proteins.
• 4. Explanation of common genetic terms.
• 5. To comprehensive the Mendel’s Law of segregation.
• 6. To recognize the relative to dominance i.e. complete dominance, incomplete dominance and
codominance.
• 7 To understand the Mendel’ s Law of independent assortment.
• 8. To understand the phenomena of linkage and crossing-over.
• 9. To understand sex determination of human sex and sex-link inheritance.
• 10. To understand the genetic variation that caused by gene mutation, gene recombination and
chromosome mutation.
• 11 To understand the development of biotechnology and the application.
• 12. To recognize the genetic engineering and the process of DNA recombination.

3. 19.1 Basic genetic materials

4. 19.1.1 DNA is the main heredity
material

5. 19.1.1.1 Transformation
experiment

6. How to prove that DNA carries the inherited
materials?
• Griffith's experiment of bacterial transformation
• Avery–MacLeod–McCarty experiment of digestion
• Hershey–Chase experiment of transduction

7. Griffith's experiment
• Griffith's experiment, reported in 1928 by Frederick Griffith.
• Griffith used two strains of pneumococcus (Streptococcus
pneumoniae) bacteria which infect mice.
• The harmful form S-strain
• The safe form R-strain

8. The harmful form S-strain
• Virulent
• Encapsulated
• Produces shiny and smooth
colonies
Convex, smooth colonies of Streptococcus
pneumoniae after 48 hours, anaerobic
atmosphere, 37°C. Smooth form of colonies (S
form).
Columbia sheep blood agar.

9. The safe form R-strain
• Non-virulent
• Non-encapsulated
• Produces dull rough colonies
autolysed colonies with depressed centers and undulated
margins typical of R-phase colonies of pneumococci.

10. • In this experiment,
bacteria from the III-S
strain were killed by heat,
and their remains were
added to II-R strain
bacteria.
• While neither alone
harmed the mice, the
combination was able to
kill its host.

11. • Griffith was also able to
isolate both live II-R and
live III-S strains of
pneumococcus from the
blood of these dead mice.
• Griffith concluded that the
type II-R had been
"transformed" into the
lethal III-S strain by a
"transforming principle"
that was somehow part of
the dead III-S strain
bacteria.

13. What is the
transforming particles?
• Avery and colleagues
identified the transforming
molecule as DNA using a
process of elimination.
• In lab experiments,
extracts from a virulent
bacterial strain were
treated with enzymes that
destroyed proteins, RNA,
or DNA.

14. Avery–MacLeod–
McCarty experiment
• The treated extracts were
then mixed with non-
virulent bacterial cultures
to determine whether
transformation occurred.

15. What is the transforming particles?
• When proteins or RNA were
destroyed in the virulent
bacterial extracts,
transformation still occurred,
and virulent cells appeared
in non-virulent bacterial
cultures.
• However, when DNA was
destroyed in the virulent
bacterial extracts,
transformation did not
occur, and non-virulent
cultures remained non-
virulent.

18. Hershey–Chase experiment
• The Hershey–Chase experiments were a
series of experiments conducted in 1952
by Alfred Hershey and Martha Chase that
helped to confirm that DNA is genetic
material.
• In their experiments, Hershey and Chase
showed that when bacteriophages, which
are composed of DNA and protein, infect
bacteria, their DNA enters the host
bacterial cell, but most of their protein
does not.
A bacteriophage, also known informally as a
phage, is a virus that infects and replicates
within Bacteria and Archaea.

19. Hershey–Chase experiment
• Viruses (T2 bacteriophage) were grown
in one of two isotopic mediums in
order to radioactively label a specific
viral component
• Viruses grown in radioactive sulfur (35S)
had radiolabelled proteins (sulfur is
present in proteins but not DNA)
• Viruses grown in
radioactive phosphorus (32P) had
radiolabeled DNA (phosphorus is
present in DNA but not proteins)

20. • The viruses were
then allowed to
infect a bacterium
(E. coli) and then
the virus and
bacteria were
separated via
centrifugation

21. • The larger bacteria
formed a solid
pellet while the
smaller viruses
remained in the
supernatant
• The bacterial pellet
was found to be
radioactive when
infected by the 32P–
viruses (DNA) but
not the 35S–viruses
(protein)

22. • This demonstrated
that DNA, not
protein, was the
genetic material
because DNA was
transferred to the
bacteria

24. 19.1.2 The structure and
replication of DNA molecule

25. DNA
• DNA (deoxyribonucleic acid)
is the genetic material of
most living organisms.
• DNA determines the heredity
characteristics by controlling
protein synthesis in cells.
• DNA can be found in the
nucleus, chloroplast and
mitochondria of eukaryotes.

26. Structure of DNA
• Nucleosides核苷 =
nucleobase (nitrogenous
base) + five-carbon sugar
(either ribose or
deoxyribose)
• Nucleotide核苷酸 =
nucleobase + a five-
carbon sugar + phosphate
group

27. Purines嘌呤and pyrimidines嘧啶
• Purines = two rings
• adenine (A)
• guanine (G)
• Pyrimidines = one ring
• Cytosine (C)
• Thymine (T) in DNA
• Uracil (U) in RNA

28. • Nucleotides are joined by
phosphodiester bonds to form long
chain of polynucleotide strand.

29. • DNA is a polymer consisting of two
polynucleotides in which the two
chains form the uprights run in
opposite directions.
• The strands of DNA are said to be
“anti-parallel".
• The two polynucleotide strains are
coiled in a right handed spiral to
form a double helix.

30. principle of
complementary base
pairing
• The base pairs are held together by
hydrogen bonds between specific
pairs of bases.
• cytosine(C) ≡ guanine (G)
• adenine (A) = thymine (T)

32. Structure of chromosomes

33. Nucleic acid structure
• Nucleic acid structure is often divided
into four different levels: primary,
secondary, tertiary and quaternary.

34. Primary structure
• Primary structure consists of a linear
sequence of nucleotides that are linked
together by phosphodiester bonds.
• It is this linear sequence of nucleotides that
make up the Primary structure of DNA or
RNA.
• Nucleotides consist of 3 components:
• Nitrogenous base
• 5-carbon sugar which is called deoxyribose
(found in DNA) and ribose (found in RNA).
• One or more phosphate groups.

35. Secondary structure
• Secondary structure is the set of
interactions between bases, i.e., which
parts of strands are bound to each other.
• In DNA double helix, the two strands of
DNA are held together by hydrogen
bonds.

36. Tertiary structure
• Tertiary structure refers to the locations
of the atoms in three-dimensional space.

37. Quaternary structure
• DNA quaternary structure is
used to refer to the binding
of DNA to histones to form
nucleosomes, and then their
organisation into higher-
order chromatin fibres.

38. Nucleosomes
• DNA + histones组蛋白 = nucleosomes核体
• Histones are positively charged proteins with arginine and lysine-
rich, strongly alkalic regions。
• Histone binds well to the negatively charged DNA molecule.
• A nucleosome has eight histone proteins. Hence it may also be refer
as a histone octamer.
• Adjacent nucleosomes are
connected via linker DNA.
• This ultimately produces the
11 nm fiber, which is
traditionally described as
“beads on a string”.

39. Further condensation
• The 11 nm nucleosome fiber
undergoes additional folding to
form a 30 nm fiber om a tube
like manner of solenoid螺线管.
• The solenoid further
condensed into a 400nm
supercoil structure.
• Lastly the structure condensed
into chromatin.

41. • DNA also has primary, secondary, tertiary, and quaternary structures.
• (1) The primary structure of DNA is the primitive linear deoxyribonucleotide chain composed of deoxyribonucleotides in order. The
deoxyribonucleotides contain different nitrogenous base (A/T/C/G), deoxyribose and one or more phosphate groups, and are
linked by phosphodiester bonds.
• (2) DNA's secondary structure is a double helix formed on the basis of two complementary polynucleotide strands by hydrogen
bonds in base pairs. Actually, the stronger forces holding the two strands together are stacking interactions between the bases.
These stacking interactions are stabilized by Van der Waals forces and hydrophobic interactions, and show a large amount of local
structural variability. The two polynucleotide chains form the right hand double helix with the same rotation around the same
common axis. The diameter of the spiral is 2.0nm. The sugar phosphate backbones of two polynucleotide chains are located at the
outer side of the double helix, while bases are located at the inner side. There are two grooves in the double helix, which are
called major groove and minor groove based on their relative size. The axial distance between adjacent base pairs is 0.34nm, and
the wheelbase of each helix is 3.4nm.
• (3) On the basis of the double helix, DNA molecules twist the coil further in order to form a super helix to compress the volume.
The resultant is called tertiary structure of DNA. The tertiary arrangement has large-scale folding taking into consideration into
geometrical and steric constraints. Usually, it includes A-DNA, B-DNA and Z-DNA in space.
• (4) DNA’s quaternary structure is similar to that of the protein to some extent. The quaternary structure of nucleic acids is the
interaction between nucleic acids and other molecules. For example, the chromatin is formed by interaction between DNA and
small histones.

43. DNA replication
Understand the basic rules governing DNA
replication
Introduce proteins that are typically involved in
generalised replication
Reference: Any of the recommended texts
Optional
Nature (2003) vol 421,pp431-435
http://www.bath.ac.uk/bio-sci/cbt/
http://www.dnai.org/lesson/go/2166/1973
http://www.bath.ac.uk/bio-
sci/hejmadi/BB10006/DNA%20replication%2005-
06%20web.ppt

45. Four requirements for DNA to be
genetic material
Must carry information
• Cracking the genetic code
Must replicate
• DNA replication
Must allow for information to change
• Mutation
Must govern the expression of the phenotype
• Gene function

46. DNA Replication
Process of duplication of the entire genome prior to
cell division
Biological significance
• extreme accuracy of DNA replication is necessary in
order to preserve the integrity of the genome in
successive generations
• In eukaryotes , replication only occurs during the S
phase of the cell cycle.
• Replication rate in eukaryotes is slower resulting in
a higher fidelity/accuracy of replication in
eukaryotes

47. The mechanism of DNA replication
Arthur Kornberg, a Nobel prize winner and other
biochemists deduced steps of replication
• Initiation
• Proteins bind to DNA and open up double helix
• Prepare DNA for complementary base pairing
• Elongation
• Proteins connect the correct sequences of nucleotides into a
continuous new strand of DNA
• Termination
• Proteins release the replication complex

48. Initiation
• Ligase binds to DNA
and unwind the
double helix

49. Elongation
• Primase reads DNA template
from 3’ to 5’ direction to
produce RNA primers from 5’
to 3’ direction using
complimentary base pairing
• DNA polymerase reads DNA
template from 3’ to 5’
direction to produce new DNA
strand primers from 5’ to 3’
direction using free
nucleotides.

50. Leading stand
• The leading strand is
produced on the template
parent strand of DNA
which runs in the 3' to 5'
direction toward the fork.
• It's able to be replicated
continuously by DNA
polymerase.

51. Leading stand
• The lagging strand is
produced on the template
parent strand of DNA
which runs in the 5‘ to 3’
direction toward the fork.
• The lagging strand is
synthesized in short,
separated segments called
Okazaki fragment.

53. Removal of RNA primers
• To form a continuous lagging
strand of DNA, the RNA
primers must eventually be
removed from the Okazaki
fragments and replaced with
DNA.
• This is done by another DNA
polymerase.

54. Ligation
• DNA ligase joins pieces of DNA by
catalyzing the formation of
phosphodiester bonds.

55. Termination
• Termination requires
that the progress of the
DNA replication fork
must stop or be
blocked.

56. DNA replication
Steps Enzyme Result
Unwinding of DNA Helicase Hydrogen bonds of the double-helix breaks and DNA
unwinds into two anti-parallel strands
Formation of RNA primer RNA polymerase /
primase
Single-stranded DNA is used as a template to synthesize
small RNA primers.
Synthesis of DNA DNA polymerase
III / δ
DNA synthesis is carried out in the 5 '→3' direction, and
DNA was synthesized at the end of the RNA primer using
single-stranded DNA as a template.
Hydrolysed of RNA
primer
DNA polymerase
I / ε
Remove the RNA primers through hydrolysation and fill the
gaps with DNA.
Linkage of DNA
fragments
Ligase The DNA fragments (including the Okazaki fragments) are
linked together.

57. Basic rules of replication
A. Semi-conservative
B. Starts at the ‘origin’
C. Synthesis always in the 5-3’ direction
D. Can be uni or bidirectional
E. Semi-discontinuous
F. RNA primers required

58. DNA replication
3 possible
models

59. Semi-conservative
replication:
One strand of duplex
passed on unchanged
to each of the daughter
cells. This 'conserved'
strand acts as a
template for the
synthesis of a new,
complementary strand
by the enzyme DNA
polymerase

60. How do we know that DNA replication is semiconservative?
Meselson-Stahl experiments

61. • The second generation will only contains DNA with 14N or
14N/15N. As the number of generation increases, the band
of 14N-DNA increases, while the 14N/15N DNA band fades.

63. B) Starts at origin
Initiator proteins identify specific base sequences on
DNA called sites of origin
Prokaryotes – single origin site E.g E.coli - oriC
Eukaryotes – multiple sites of origin (replicator)
E.g. yeast - ARS (autonomously replicating sequences)
Prokaryotes Eukaryotes

64. In what direction does DNA replication occur?
Where does energy for addition
of nucleotide come from?
What happens if a base
mismatch occurs?
C) Synthesis is ALWAYS in the 5’-3’ direction

65. Why does DNA replication only occur in the 5’ to 3’ direction?
Should be PPP here

66. D) Uni or bidirectional
Replication forks move in one or opposite directions

67. E) Semi-discontinuous replication
Anti parallel strands replicated simultaneously
Leading strand synthesis continuously in 5’– 3’
Lagging strand synthesis in fragments in 5’-3’

68. Semi-discontinuous replication
New strand synthesis always in the 5’-3’ direction

69. F) RNA primers
required

70. What kind of enzyme synthesizes the
new DNA strand?
1) RNA polymerase
2) DNA Polymerase
3) Primase
4) Helicase
5) Topoisomerase

71. In what direction is the newly synthesized
DNA produced?
1. 5'-3'
2. 3'-5'
3. In the direction of the major groove
4. Both 5'-3' and 3'-5' depending on which strand is
being replicated

72. What is the sequence (1 to 4) in which these
proteins function during DNA replication
• ____ RNA primase
• ____ DNA ligase
• ____ DNA polymerase
• ____ DNA helicase1
2
3
4

73. Why is an RNA primer necessary for DNA
replication?
A. The RNA primer is necessary for the activity of DNA ligase.
B. The RNA primer creates the 5’ and 3’ ends of the strand.
C. DNA polymerase can only add nucleotides to RNA
molecules.
D. DNA polymerase can only add nucleotides to an existing
strand

75. The central dogma of molecular biology
• The central dogma of
molecular biology explains
the flow of genetic
information within a cell.
• DNA codes for RNA via the
process of transcription
(occurs within the nucleus)
• RNA codes for protein via
the process of translation
(occurs at the ribosomes)

77. Transcription
• Transcription is the process by
which an RNA sequence is
produced from a DNA
template.
• RNA polymerase separates the
DNA strands and synthesises a
complementary RNA copy
from one of the DNA strands
called the template strand.
(sense)
(antisense)

78. mRNA
• The RNA polymerase
syntheses messenger
RNA (mRNA).
• mRNAs convey genetic
information from DNA
to the ribosome,
where they specify the
amino acid sequence
of the protein products
of gene expression.

79. Pre-mRNA
• RNA polymerase
transcribes primary
transcript mRNA
(known as pre-mRNA)
has to be processed,
mature mRNA.

80. mRNA processing
• Addition of a 5' cap to
the beginning of the
RNA
• Addition of a poly-A
tail to the end of the
RNA
• Chopping out
of introns, and pasting
together of the
remaining sequences
(exons)

81. 5’ cap
• The 5’ cap is added to the first
nucleotide of the pre-mRNA.
• The cap is a modified guanine (G)
nucleotide
• The 5’ cap protects the transcript
from being broken down.
• It also helps the ribosome attach to
the mRNA.

82. RNA splicing
• In RNA splicing, specific parts of
the pre-mRNA, called introns are
recognized and removed by a
protein-and-RNA complex called
the spliceosome.
• The pieces of the RNA that are
not chopped out are
called exons. The exons are
pasted together by the
spliceosome to make the final,
mature mRNA that is shipped out
of the nucleus.
• A key point here is that
it's only the exons of a gene that
encode a protein.

83. Alternative splicing
• In alternative splicing,
one pre-mRNA may be
spliced in either of two
or more than two
different ways.
• Through alternative
splicing, eukaryotes
can encode more
different proteins than
the number of genes
in our DNA.

84. Poly-A tail
• When a sequence called a
polyadenylation signal shows up in
an RNA molecule during
transcription, an enzyme chops the
RNA in two at that site.
• Another enzyme adds about 100 -
200 adenine (A) nucleotides to the
cut end, forming a poly-A tail.
• The tail makes the transcript more
stable and helps it get exported
from the nucleus to the cytosol.

87. Genetic code
• The genetic code is the set of rules used by
living cells to translate information
encoded within genetic material (DNA or
mRNA sequences) into proteins.
• The code defines how sequences of
nucleotide triplets, called codons, specify
which amino acid will be added next during
protein synthesis.

88. Start Codon起始密码子
• The start codon is
the first codon of
a messenger RNA
(mRNA) transcript
translated by a
ribosome.
• The start
codon always
codes for
methionine in
eukaryotes.

89. Stop codon终止密码子
• A stop codon (or
termination codon)
signals a
termination of
translation into
proteins.
• Stop codons signal
the termination of
this process by
binding release
factors, which cause
the ribosomal
subunits to
disassociate,
releasing the amino
acid chain.

90. Reading Frame阅读框架
• Reading frame is a way of dividing the sequence of nucleotides in a
nucleic acid (DNA or RNA) molecule into a set of consecutive, non-
overlapping triplets.
• The ribosome reads the mRNA from 5’ to 3’ end.
• There are always three possible reading frames, but generally only
one that is biologically significant, i.e. can be transcribed into protein.
• An open reading frame (ORF) is a reading frame that has the
potential to be transcribed into RNA and translated into protein.

92. Reading direction
• RNA polymerase reads DNA from
3’ to 5’ direction, producing
mRNA from 5’ to 3’ direction.
• Ribosome reads mRNA from 5’ to
3’ direction, producing proteins
from the N-terminus to C-
terminus.

93. Translation
• Translation is the process in
which ribosomes in the
cytoplasm or ER synthesize
proteins after the process of
transcription of DNA to RNA in
the cell's nucleus.
• In translation, messenger RNA
(mRNA) is decoded in a
ribosome to produce a
specific amino acid chain, or
polypeptide.

94. transfer RNA
• A transfer RNA (tRNA) carries an
amino acid to the ribosome as
directed by a 3-nucleotide sequence
(codon) in a messenger RNA (mRNA).
• Each tRNA contains a distinct
anticodon triplet sequence that can
form 3 complementary base pairs to
one or more codons for an amino acid.

95. Three phases of translation
• Initiation
• Elongation
• Termination

96. Initiation
• The small ribosomal subunit binds to the 5’-end of the mRNA and
moves along it until it reaches the start codon (AUG).
• Next, the appropriate tRNA molecule bind to the codon via its
anticodon (according to complementary base pairing).
• Finally, the large ribosomal subunit aligns itself to the tRNA molecule
at the P site and forms a complex with the small subunit.

97. Elongation
• A second tRNA molecule
pairs with the next codon
in the ribosomal A site.
• The amino acid in the P
site is covalently attached
via a peptide bond
(condensation reaction) to
the amino acid in the A
site.
• The tRNA in the P site is
now deacylated (no amino
acid), while the tRNA in
the A site carries the
peptide chain.

98. Elongation
• The ribosome moves
along the mRNA strand by
one codon position (in a
5’ → 3’ direction).
• The deacylated tRNA
moves into the E site and
is released, while the
tRNA carrying the peptide
chain moves to the P site.
• Another tRNA molecules
attaches to the next codon
in the now unoccupied A
site and the process is
repeated.

99. Elongation
A ribosome translating a protein that is secreted into the
endoplasmic reticulum. tRNAs are colored dark blue.

100. Termination
• Elongation and translocation continue in a repeating cycle until the
ribosome reaches a stop codon.
• These codons do not recruit a tRNA molecule, but instead recruit a
release factor that signals for translation to stop.
• The polypeptide is released and the ribosome disassembles back into
its two independent subunits.

101. Comparison of transcription and translation

102. Comparison of DNA replication, transcription
and translation

103. Summary