nucleic acids

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Nucleic Acids

Nucleic Acids

Structures of Nucleic Acids

DNA Replication

RNA and Transcription

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Nucleotides

Nucleic acids consist of nucleotides that have a sugar, nitrogen base, and phosphate

nucleoside

Sugar

Base

PO4

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Nitrogen-Containing Bases

N

N

N

N

H

NH2

N

N

O

CH3

O

H

H

N

N

N

N

O

H

NH2

H

N

N

NH2

CH3

O

H

N

N

O

CH3

O

H

H

adenine (A) thymine (T)

guanine (G) cytosine (C) uracil (U)

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Sugars

O OHCH2

OHOH

HO HO O OHCH2

OH

ribose deoxyribose

(no O)

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Nucleosides in DNA

Base SugarNucleoside

Adenine (A) Deoxyribose Adenosine

Guanine (G) Deoxyribose Guanosine

Cytosine (C) Deoxyribose Cytidine

Thymine (T) Deoxyribose Thymidine

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Nucleosides in RNA

Base SugarNucleoside

Adenine (A) ribose Adenosine

Guanine (G) ribose Guanosine

Cytosine (C) ribose Cytidine

Uracil (U) ribose Uridine

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Example of a Nucleoside

O

OH

N

N

NH2

O

CH2OP

O

O-

O-

deoxyctyidine monophosphate (dCMP)

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Nucleotides in DNA and RNA

DNAdAMP Deoxyadenosine monophosphatedGMP Deoxyguanosine monophosphatedCMP Deoxycytidine monophosphatedTMP Deoxythymidine monophosphate

RNAAMP adenosine monophosphateGMP guanosine monophosphateCMP cytidine monophosphateUMP uridine monophosphate

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Structure of Nucleic Acids

• Polymers of four nucleotides• Linked by alternating sugar-phosphate bonds• RNA: ribose and A, G, C, U• DNA: deoxyribose and A,G,C,T

nucleotide nucleotide nucleotide nucleotide

P sugar

base

P sugar

base

P sugar

base

P sugar

base

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Nucleic Acid Structure

3,5-phosphodiester bond

O

N

N

NH2

O

CH2OP

O

O-

O-

OH

O

N

N

NH2

CH2OP

O

O-

OH

O

N

N

AMP

CMP

3

5

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Double Helix of DNA

• DNA contains two strands of nucleotides• H bonds hold the two strands in a double-

helix structure• A helix structure is like a spiral stair case• Bases are always paired as A–T and G-C• Thus the bases along one strand

complement the bases along the other

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Complementary Base Pairs

•Two H bonds for A-T•Three H bonds for G-C

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Double Helix of DNA

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Learning Check NA1

Write the complementary base sequence for the matching strand in the following DNA section:

-A-G-T-C-C-A-A-T-G-C-

• • • • • • • • • • • • • • • • • • • •

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Solution NA1

Write the complementary base sequence for the matching strand in the following DNA section:

-A-G-T-C-C-A-A-T-G-C- • • • • • • • • • •

• • • • • • • • • •

-T-C-A-G-G-T-T-A-C-G-

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DNA Replication

• DNA in the chromosomes replicates itself every cell division

• Maintains correct genetic information• Two strands of DNA unwind• Each strand acts like a template• New bases pair with their complementary base• Two double helixes form that are copies of

original DNA

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DNA Unwinds

G- -C

A- -T

C- -G

T- -A

G-CA-TC-GT-A

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DNA Copied with Base Pairs

Two copies of original DNA strand

G-C G-CA-T A-TC-G C-GT-A G-A

Nucleic Acid Chemistry

Where the info is…interpreting the blueprint

Central Dogma

DNA ---------------- RNA-------------- protein

Replication

transcription translation

Central Dogma

• Replication– DNA making a copy of itself

• Making a replica

• Transcription– DNA being made into RNA

• Still in nucleotide language

• Translation– RNA being made into protein

• Change to amino acid language

Replication

• Remember that DNA is self complementary

• Replication is semiconservative– One strand goes to next generation– Other is new

• Each strand is a template for the other– If one strand is 5’ AGCT 3’– Other is: 3’ TCGA 5’

Replica

• Write the strand complementary to:

3’ ACTAGCCTAAGTCG 5’

Answer

Replication is Semiconservative

Replication

• Roles of enzymes– Topoisomerases– Helicase– DNA polymerases– ligase

• DNA binding proteins– DNA synthesis

• Leading strand• Lagging strand

Replication

Replication

• Helix opens– Helicase

• Causes supercoiling upstream– Topoisomerases (gyrase)

• DNA Binding Proteins– Prevent reannealing

Replication

Replication

• Leading strand– 3’ end of template– As opens up, DNA polymerase binds– Makes new DNA 5’ - 3’

• Same direction as opening of helix• Made continuously

Replication

Replication

• Lagging strand– 5’ end of template

• Can’t be made continuously as direction is wrong

– RNA primer– New DNA made 5’ 3’

• Opposite direction of replication• Discontinuous

– Okazaki fragments

• Ligase closes gaps

Transcription• DNA template made into RNA copy

– Uracil instead of Thymine

• One DNA strand is template– Sense strand

• Other is just for replication – Antisense (not to be confused with

nonsense!)

• In nucleus– nucleoli

Transcription

• From following DNA strand, determine RNA sequence

3’ GCCTAAGCTCA 5’

Answer

Transcription

Transcription

• DNA opens up– Enzymes?

• RNA polymerase binds – Which strand?– Using DNA template, makes RNA

• 5’-3’• Raw transcript called hnRNA

TranscriptionHow does RNA polymerase know where to

start?

upstream promotor sequences

Pribnow Box

TATA box

RNA polymerase starts transcription X nucleotides downstream of TATA box

Introns and Exons

• Introns– Intervening sequences– Not all DNA codes for protein– Regulatory info, “junk DNA”

• Exons– Code for protein

Processing of hnRNA into mRNA

• 3 steps– Introns removed

• Self splicing

– 5’ methyl guanosine cap added– Poly A tail added

• Moved to cytosol for translation

Processing of hnRNA into mRNA

Translation

• RNA -- Protein– Change from nucleotide language to amino

acid language

• On ribosomes

• Vectorial nature preserved– 5’ end of mRNA becomes amino terminus

of protein– Translation depends on genetic code

Genetic Code

• Nucleotides read in triplet “codons”– 5’ - 3’

• Each codon translates to an amino acid• 64 possible codons

– 3 positions and 4 possiblities (AGCU) makes 43 or 64 possibilities

– Degeneracy or redundancy of code• Only 20 amino acids• Implications for mutations

Genetic Code

Genetic Code

• Not everything translated• AUG is start codon

– Find the start codon

• Also are stop codons• To determine aa sequence

– Find start codon– Read in threes– Continue to stop codon

Translation

• Steps:– Find start codon (AUG) – After start codon, read codons, in threes– Use genetic code to translate

Translate the following:

GCAGUCAUGGGUAGGGAGGCAACCUGAACCGAC

Answer

Translation Process

• Requires Ribosomes, rRNA, tRNA and, of course, mRNA– Ribosome

• Made of protein and rRNA• 2 subunits• Has internal sites for 2 transfer RNA molecules

Ribosome

Left is cartoon diagram Right is actual picture

Transfer RNA

• Mostly double stranded– Folds back on itself

• Several loops– Anticodon loop

• Has complementary nucleotides to codons

• 3’ end where aa attach

Transfer RNA

Translation

• Initiation– Ribosomal subunits assemble on mRNA– rRNA aids in binding of mRNA

• Elongation– tRNAs with appropriate anticodon loops bind to complex– have aa attached (done by other enzymes)– Amino acids transfer form tRNA 2 to tRNA 1– Process repeats

• Termination– tRNA with stop codon binds into ribosome– No aa attached to tRNA– Complex falls apart

Translation

Translation

• Happening of process (circa 1971)

• http://www.youtube.com/watch?v=u9dhO0iCLww

Mutations

• Changes in nucleotide sequence

• Can cause changes in aa sequence– Degeneracy in genetic code can prevent

• Two types– Point mutations

• Single nucleotide changes

– Frame shift• Insertions or deletions

Point Mutations

• Single nucleotide changes

• Old sequenceAUG GGU AGG GAG GCA ACC UGA ACC GAC

aa: G R E A T

New sequence

AUG GGU AGU GAG GCA ACC UGA ACC GAC

aa: G S E A T

Point mutations

• Depending on change, may not change aa sequence

• Old sequenceAUG GGU AGG GAG GCA ACC UGA ACC GAC

aa: G R E A T

New sequence

AUG GGU AGA GAG GCA ACC UGA ACC GAC

aa: G R E A T

Point Mutations

• Change could make little difference– If valine changed to leucine, both nonpolar

• Change could be huge,– Could erase start codon

• Old sequenceAUG GGU AGG GAG GCA ACC UGA ACC GACaa: G R E A T

New sequenceAUU GGU AGA GAG GCA ACC UGA ACC GACaa: no start codon…protein not made

Point Mutations

• Other possibilities,– Stop codon inserted

• Truncated protein

– Stop codon changed• Extra long protein

• Bottom line,– Depends on what change is

Frame Shift mutations

• Insertions or deletions– Change the reading frame

• Insertion exampleOld sequence

AUG GGU AGG GAG GCA ACC UGA ACC GACaa: G R E A T

New sequenceAUG GGU AGG AGA GGC AAC CUG AAC CGA Caa: G R R G N L N R

Frame Shift Mutations

• Deletion example

• Old sequenceAUG GGU AGG GAG GCA ACC UGA ACC GAC

aa: G R E A T

New sequence Delete second A (Underlined above)

AUG GGU GGG AGG CAA CCU GAA CCG AC

aa: G G R Q P G P

Complementary DNA Strand

Template:

3’ ACTAGCCTAAGTCG 5’

5’ TGATCGGATTCAGC 3’

Back

RNA Transcript

DNA 3’ GCCTAAGCTCA 5’

RNA 5’ CGGAUUCGAGU 3’

Back

Translation AnswerFind start codon

GCAGUCAUGGGUAGGGAGGCAACCUGAACCGAC

Read in threes after that:

AUG GGU AGG GAG GCA ACC UGA ACC GAC

Using Genetic code

AUG GGU AGG GAG GCA ACC UGA ACC GAC

G R E A T stop After stop codon…rest is garbage

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