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