Molecular Biology: DNA, gene, chromosome and genome (Learning Objectives) Nucleic acid structure and composition • Compare and contrast the structure of DNA and RNA: features they share and how do they differ? (number of stands, sugar, and nitrogen bases). Which nitrogen bases are found in DNA or RNA. • Learn the base pairing rule its underlying reason and its effect on the DNA structure • Learn the base-pairing rule, its underlying reason and its effect on the DNA structure. • Distinguish between DNA, gene, chromosome and genome. • Explain the DNA structure: number of strands, polarity (5’-3’), complementary strands and their anti-parallel nature. • Distinguish between the strong covalent bonds that hold the sugar-phosphate backbone of nucleic acids and the weaker chemical bonds form between the nitrogen bases and hold the two strands of DNA. L h DNA i k d h ti i id th l i i ti ith hi t • Learn how DNA is packaged as chromatin inside the nucleus in association with histone proteins. DNA replication • Explain the steps for DNA replication: strand separation name of protein enzyme pairing Explain the steps for DNA replication: strand separation, name of protein enzyme, pairing of complementary nucleotides, polymerization, termination. • Distinguish between the DNA template strand and new strand. In what direction is the new strand made? C d t t th l di d l i l th i d DNA t d hi h i • Compare and contrast the leading and lagging newly synthesized DNA strands: which is continuously made and which is discontinuously made? Role of DNA ligase.
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Molecular Biology: DNA, gene, chromosome and genome (Learning Objectives)
Nucleic acid structure and composition• Compare and contrast the structure of DNA and RNA: features they share and how do they
differ? (number of stands, sugar, and nitrogen bases). Which nitrogen bases are found in DNA or RNA.
• Learn the base pairing rule its underlying reason and its effect on the DNA structure• Learn the base-pairing rule, its underlying reason and its effect on the DNA structure.• Distinguish between DNA, gene, chromosome and genome.• Explain the DNA structure: number of strands, polarity (5’-3’), complementary strands and
their anti-parallel nature. p• Distinguish between the strong covalent bonds that hold the sugar-phosphate backbone of
nucleic acids and the weaker chemical bonds form between the nitrogen bases and hold the two strands of DNA. L h DNA i k d h ti i id th l i i ti ith hi t• Learn how DNA is packaged as chromatin inside the nucleus in association with histoneproteins.
DNA replication• Explain the steps for DNA replication: strand separation name of protein enzyme pairingExplain the steps for DNA replication: strand separation, name of protein enzyme, pairing
of complementary nucleotides, polymerization, termination. • Distinguish between the DNA template strand and new strand. In what direction is the new
strand made?C d t t th l di d l i l th i d DNA t d hi h i• Compare and contrast the leading and lagging newly synthesized DNA strands: which is continuously made and which is discontinuously made? Role of DNA ligase.
Molecular Biology: DNA, gene, chromosome and genome (cont’d)Transcription• Explain the purpose for this process and its sub-cellular compartment.
L th th t f t i ti d th bi l l i l d• Learn the three steps of transcription and the bio-molecules involved. • Name the stretch of nucleotides in the DNA that binds RNA polymerase to initiate
transcription.• Name the DNA stretch that causes RNA polymerase to come off DNA terminating
transcription.• Explain the split nature of eukaryotic genes. Distinguish between exon, intron. Which
contains information that will specify the amino acid sequence of the protein product?• Explain RNA processing of eukaryotic cellsExplain RNA processing of eukaryotic cells.Genetic code• Explain the language of nucleic acids: letters and words (nucleotides and codons), number
of nucleotides that make up a codon, and total number of codons. Whi h d k th i iti ti f t l ti h t i id d it if ? H• Which codon marks the initiation of translation, what amino acid does it specify? How many stop codons specify the termination of translation.
Translation or protein synthesis• Where does it take place, what components are necessary, and what are the three steps.• Which biomolecule can interpret the language of nucleic acids into the language of
proteins Mutation• Define the term mutation how do they arise and list the types of mutations What impactDefine the term mutation, how do they arise and list the types of mutations. What impact
might a mutation have on the protein product?• Use the genetic code table to translate a sequence of codons into a sequence of amino
acids and the reverse as well as the sequence of normal and mutated proteins
The DNA of the gene is transcribed into RNA whichTHE FLOW OF GENETIC INFORMATION
The DNA of the gene is transcribed into RNA which is translated into the polypeptide (protein)
DNA
Transcription
RNA
Protein
Translation
Figure 10.6A
Nucleic Acid Chemical Structure
Sugar‐phosphate backbone
DNA and RNA are polymers of nucleotides
A
Sugar phosphate backbone
Phosphate groupNitrogenous base
SugarA
C C Phosphategroup
OH3C C
C
C
CN
HO
Nitrogenous base(A, G, C, or T)DNA
nucleotide
T
G
T
G
O–
OO P CH
2
C CN
C
H O
C
O
HC H HC
H
Thymine (T)
T T
O HSugar(deoxyribose)
DNA nucleotide
DNA polynucleotideFigure 10.2A
DNA has four kinds of nitrogenous bases: A T CDNA has four kinds of nitrogenous bases: A, T, C, and G
O H HN H HN O
CC
C
CC
C
N
C
H
ONH
H3C H
H
N
N OC H C
N
N N
N
C
CC
C H
N
N
C
CN
C HN
CN
H CH
H H H H H
Thymine (T) Cytosine (C) Adenine (A) Guanine (G)
PurinesPyrimidines
Figure 10.2B
RNA is also a nucleic acid withRNA is also a nucleic acid with• a slightly different sugar
• U instead of T
KNitrogenous base (A G C or U) Key
Hydrogen atomCarbon atomNitrogen atomOxygen atom
(A, G, C, or U)Phosphategroup
OH
C
C
C
C
NH
O
Phosphorus atom
O–
OO CH2
C CNH O
O
C H H C
Uracil (U)
P
CC
O
H
C H H
OH
C
H
Figure 10.2C, D
Sugar(ribose)
• Gene: a linear stretch of nucleotides with information• Gene: a linear stretch of nucleotides with information for one product (polypeptide or protein)
• Chromosome: a very long stretch of DNA carrying DNA i l i d i h imany genes. DNA is always associated with protein as
chromatin
DNA Coiling into chromatin and condensed chromosomes
http://www.biostudio.com/demo freeman dna coiling.htmhttp://www.biostudio.com/demo_freeman_dna_coiling.htm
• Genome: totality of DNA in a cell• Genome: totality of DNA in a cell
Erwin Chargaff, 1947Erwin Chargaff, 1947Biochemical analysis of DNA nucleotides from different speciesdifferent speciesA = T & C = G
Human DNA A 30 9%A = 30.9%T = 29.4%C = 19.9%G = 19.8%
DNA is a double stranded helix
DNA Shape
DNA is a double‐stranded helixJames Watson and Francis Crick worked out the three‐dimensional structure of DNA based on work bydimensional structure of DNA, based on work by Rosalind Franklin
Figure 10.3A, B
The structure of DNAThe structure of DNA• two polynucleotide strands wrapped around each other in a double helix
Figure 10.3C Twist
– Covalent bonds hold the sugar phosphate backbone– Hydrogen bonds between bases hold the two strandsy g– Each base pairs with a complementary partner
A with T, and G with C
G CT A
A T
O
OOH
–OP
H2COH
O T A
Hydrogen bond
GG
CCA T
GC
O O–OPO
O O
2C
H2C
O
OP O
–
OO
O
O
T A
G C
CH2
CH
Basepair
GC
T A
T A
O OP–O
–O OPO O
OH2C
O
O
O
O–
O–
OPO
O
OC G
CH2
CH2
A T
A T
G C
O O
OH
H2C O
O
O
O
O–
HO
OP
P
O
OA T
CH2
Figure 10.3D
A TO
Ribbon model Partial chemical structure Computer model
DNA REPLICATIONDNA replication depends on specific base pairing
• Starts with the separation of DNA strands• A protein enzyme uses each strand as a template to• A protein enzyme uses each strand as a template to
assemble new nucleotides into complementary strands
A T
C G
G C
A T
C G
G C
A T
C G
G C
A T
C G
G C
A T
C G
C
Figure 10.4A
A T
T A
A T
T A
A T
T A
C
A
T
AT
A
Parental moleculeof DNA
Both parental strands serve as templates
Two identical daughtermolecules of DNA
Nucleotides
“Build a DNA Molecule”Build a DNA Molecule
http://learn.genetics.utah.edu/units/basics/builddna/
DNA replication is a complex process• the helical DNA molecule must untwist or unwind• several proteins are involved including DNA
Polymerase• it start at specific DNA sequences, origin of replication
G C
A TG CC G
http://www.dnai.org/a/index.htmlA T
C G
AGA
CG
C G
TCT
CG
CG
TAG
C
TAT
AAT
TA
CG
CG
CG
TAG
C
T
A
TA
AT
TA
Figure 10.4B
TA
• Replication of long stretches of DNA• Begins at multiple specific sites on the double helix
Origin of replicationParental strand
Daughter strand
Bubble
Figure 10.5A
Two daughter DNA molecules
Each DNA strand of the double helix is oriented in the opposite direction
5 end 3 end
P HO
A T134
5
1 432
5 end 3 end
P P
C G
21 15 4
P
P
P
P
G C
P P
POHT A
Figure 10.5B3 end 5 end
– The enzyme DNA polymerase uses a single strand and makes a new complementary strand in a 5’ to 3’ directionp y• one daughter strand is made as a continuous piece• the other strand is synthesized as a series of short piece
which are then connected by the enzyme DNA ligase
3DNA polymerase 3
53
5Daughter strandsynthesizedcontinuously
Daughter
Parental DNA
p ymolecule
53
gstrandsynthesizedin pieces
53
Figure 10.5CDNA ligase
Overall direction of replication
TranscriptionProducing copies of the genetic
RNA polymerase
g p gmessages in the form of RNA
DNA of gene
PromoterDNA
TerminatorDNA
1 Initiation
Transcription of a gene
1. DNA: double –stranded
1 Initiation
2. Enzyme: RNA Polymerase
3. Monomers: RNA nucleotides2 Elongation
3. Monomers: RNA nucleotides
4. Steps:
Initiation‐ Start at promoter Growing3 TerminationInitiation Start at promoter
Elongation
Termination Stop at
GrowingRNA
3 Termination
Termination‐ Stop at termination signal Completed RNA RNA
polymeraseFigure 10.9B
– RNA polymerase • unwinds the two DNA strands
• Uses one strand as a template strand
• Polymerizes RNA nucleotides against the template strand following the base pairing rulesfollowing the base pairing rules
– Single‐stranded messenger RNA (mRNA) peels away from the template strandfrom the template strand
– The DNA strands rejoinRNApolymerase
RNA nucleotides
polymerase
T C C A A T
Direction of
ATTGGAT
A U C C Ahttp://www.dnai.org/a/index.html
Direction of transcription
Template Strand of DNA
Newly made RNAFigure 10.9A
Eukaryotic RNA is processed before leaving the nucleus
Split Genes of EukaryotesEukaryotic RNA is processed before leaving the nucleus
– Noncoding segments called introns are spliced out– a cap and a tail are added to the ends
Exon Intron Exon Intron ExonDNA
T i tiCap TranscriptionAddition of cap and tail
RNAtranscript with cap
Introns removedTail
with capand tail
Exons spliced togethermRNA
diCoding sequence Nucleus
CytoplasmFigure 10.10
THE FLOW OF GENETIC INFORMATION FROM DNA TO RNA TO PROTEIN
Th i f ti i d b f DNA• The information carried by sequence of DNA bases constitutes an organism’s genotype
• The DNA genotype is expressed as proteins, which provide the molecular basis for the phenotype
Genetic information written in a code that isGenetic information written in a code that is translated into amino acid sequences
The “words” of the DNA “language” are triplets of b (3 b l ) ll d dbases (3 bases long) called codons
Each codons in a gene specify one amino acid sequence of the polypeptide
Protein synthesis
Translation is the RNA–directed synthesis of a ypolypeptide
Translation of the language of nucleic acids into the language of proteins (amino acids)the language of proteins (amino acids)
One codon ‐‐‐‐‐‐ one amino acid
Gene 1
DNA molecule
Gene 2
Gene 3
DNA strand‐ template
Transcription
A A A C C G G C A A A A
Translation
RNA
Codon
U U U G G C C G U U U U
PolypeptideAmino acidFigure 10.7
In-class activity/Genetic code
Use the genetic code table to answer the followingUse the genetic code table to answer the following questions1. How many codons are there for leu (leucine)?
2 How many codons are there for Met (Methionine)?2. How many codons are there for Met (Methionine)?
3. How many codons are there for Phe (phenylalanine)?
Draw a conclusion about the number of codons for aminoDraw a conclusion about the number of codons for amino acids. 4. How many “stop” codons are there?
Answer the following questions using this genetic code:Answer the following questions using this genetic code:5’-AUGACCCCUUUGUUAUAC-3’
5. How long is this message in nucleotides?
6 Is this the information present in DNA or in mRNA?6. Is this the information present in DNA or in mRNA? Explain your answer
.7. Write down the sequence of amino acids coded for by the above stretch of nucleotidesthe above stretch of nucleotides.
how long is this polypeptide? .
The genetic code U C A G
Second base
The genetic code is the Rosetta stone of life
UUC
UGUUGC
UGA Stop
Phe
Leu
Ser
CysTyr
U UUA
UUU
UCCUCU
UCA
UAC
UAU
UAA Stop
U
C
Astone of life Leu
Leu Pro
His
ArgCCUC
CUU
UCG
CCC
CCU
UAG Stop
CACCAU
UGG Trp
CGC
CGU
G
U
C
Nearly all organisms use
Leu Pro
Asn
Gln
Ser
ArgC
irst base CUG
CUA
AUU
CCGCCA
ACU
CAGCAA
AAU
CGGCGA
AGU
A
G
U
exactly the same genetic code Met or
start
IleThr
Asn
Lys
Ser
Arg
AF AUC
AUG
AUA
ACC
ACCACA
AAC
AAG
AAA
AGC
AGG
AGA
C
A
G
Val Ala
Asp
Glu
GlyGGUC
GUU
GUA
GCCGCU
GCA
GAC
GAU
GAA
GGCGGU
GGA
U
C
A
Figure 10.8A
GluGUG GCG GAG GGG G
An exercise in translating the genetic codeAn exercise in translating the genetic code
Strand to be transcribed (template)
T A C T T C A A A A T C
A T G A A G T T T T A GDNA
Transcription
A U G A A G U U U U A GRNA
S
Translation
Startcondon
Stopcondon
Figure 10.8B Met Lys PhePolypeptide
A codon start codon within the mRNA message marks the translation initiation and a stop codonmarks the translation initiation and a stop codonmarks the end of translation
Start of genetic message
End
Figure 10.13A
TranslationTranslation• Takes place in cytoplasm• Involves:• Involves:
Ribosomes: Two subunits (each made of many proteins & r‐RNA)y p )
mRNAt‐RNA (interpreter)20 amino acids
Transfer RNA molecules serve as interpretersTransfer RNA molecules serve as interpreters during translation
Amino acid attachment site
Hydrogen bond
RNA polynucleotide chain
AnticodonFigure 10.11A
– Each tRNA molecule is a folded molecule bearing a b t i l t ll d ti d dbase triplet called an anticodon on one end
– A specific amino acid is attached to the other end
Amino acidAmino acidattachment site
Anticodon
Figure 10.11B
A ribosome attaches to the mRNA translates itsA ribosome attaches to the mRNA translates its message into a specific polypeptide aided by transfer RNAs (tRNAs)( )
Protein Synthesis animation
http://highered mcgraw‐http://highered.mcgrawhill.com/sites/0072437316/student_view0/chapter15/animations.html#
Translation Intro (Flash Animation)( )
Steps of protein Synthesis1. Initiation
a. Binding of mRNA to small ribosomal subunita. Binding of mRNA to small ribosomal subunitb. Binding of Met‐tRNA to the initiation codon AUG on
mRNAc. Binding of large ribosomal subunit
2. Elongationg
3 Termination3. Termination
mRNA a specific tRNA and the ribosome subunitsmRNA, a specific tRNA, and the ribosome subunits assemble during initiation
LargeMet Met
Initiator tRNA
Large ribosomalsubunit
Initiator tRNA
A siteU A CAU C
A U G A U G
P site
1 2mRNA Small ribosomal
subunit
Startcodon
Figure 10.13B
Steps of protein Synthesis1. Initiation
a. binding of mRNA to small ribosomal subunita. binding of mRNA to small ribosomal subunitb. binding of Met‐tRNA to AUG on mRNAc. Binding of large ribosomal subunit
2. Elongationa. binding of a second tRNA to next codongb. formation of peptide bond c. sliding of ribosome by one codonAmino acids are added to the growing polypeptide chain until a stop codon is reached.
3. Termination
Elongation steps
Polypeptide
P siteAnticodon
Aminoacid
A site
mRNA Codons 1 Codon recognition
mRNAmovement
StStopcodon
New
2Peptide bondformation
NewPeptidebond
Figure 10.14 3 Translocation
Steps of protein Synthesis1. Initiation
a. binding of mRNA to small ribosomal subunitgb. binding of Met‐tRNA to AUG on mRNAc. Binding of large ribosomal subunit
2. Elongationa. binding of a second tRNA to next codonb. formation of peptide bond c. sliding of ribosome by one codonAmino acids are added to the growing polypeptide chain until a stop codon is reached.
3. TerminationDisassembly of the protein synthesis machinery
Review: The flow of genetic information in the cellReview: The flow of genetic information in the cell is DNARNAprotein
The sequence of codons in DNA, via the sequence of d i RNA ll t th i t t fcodons in mRNA spells out the primary structure of
a polypeptide
Mutations are changes in the DNA base sequence• Caused by errors in DNA replication or recombination,
or by mutagens
Mutant hemoglobin DNANormal hemoglobin DNA
C T T C A T
gg
G A A G U A
mRNA mRNA
Normal hemoglobin Sickle‐cell hemoglobin
Glu ValFigure 10.16A
Substituting, inserting, or deleting nucleotides alters a gene with varying effects on the organism. g y g g
Normal gene
mRNA
Met Lys Phe Gly Ala
A U G A A G U U U G G C G C A
Protein
Base substitution
A U G A A G U U U A G C G C A
Met Lys Phe Ser Ala
A U G A A G U U U A G C G C A
Base deletion Missing
A U G A A G U U G G C G C A U
U
Met Lys Leu Ala HisFigure 10.16B
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