What is a gene? A sequence of DNA nucleotides that specifies the primary structure of a polypeptide...

What is a gene?

• A sequence of DNA nucleotides that specifies the primary structure of a polypeptide chain (tells the cell how to make it)

• Genes-made of nucleotides• Proteins-made of amino acids• How does a nucleotide code (in the

nucleus) specify an amino acid sequence (in the cytoplasm)?

The Central Dogma

• DNA is transcribed into RNA-characteristics of RNA

• RNA is translated into protein

• Advantages

• Exceptions

LE 17-4

DNAmolecule

Gene 1

Gene 2

Gene 3

DNA strand(template)

3

TRANSCRIPTION

Codon

mRNA

TRANSLATION

Protein

Amino acid

35

5

The Genetic code-characteristics

• Triplet (3 nucleotides=codon=info for a specific amino acid);64 different codons (3 are stop codons)

• Universal• Redundant (61 codons-20 amino acids)-

variability in third nucleotide of codon. Advantages of a redundant code?

• Non-overlapping • Exceptions (ciliates; mito/chloroplasts)

LE 17-5Second mRNA base

Fir

st

mR

NA

ba

se

(5 e

nd

)

Th

ird

mR

NA

ba

se

(3 e

nd

)

Figure 17-06

Gene Expression

• If a gene is transcribed and the m-rna is translated (the gene is expressed); a protein is made. This often changes the phenotype of the cell that produces the protein.

• Differential gene expression is involved in embryonic development and cell specialization.

• Totipotency-each cell has the genetic information for an entire organism.

• Differential gene expression results in cell specialization (differentiation)

• Hormones often play a role in gene expression

Transcription

• The first step in gene expression• Takes place in the nucleus• Requirements• A. RNA nucleotides• B. DNA template (gene)• C. Enzymes (RNA polymerase)• Only one of the two DNA strands is copied

(template strand)

LE 17-7a-2

Promoter

53

35

35

53

Transcription unit

DNA

Initiation

Start pointRNA polymerase

UnwoundDNA

RNAtran-script

Template strandof DNA

LE 17-7a-3

Promoter

53

Transcription unit

35DNA

Start pointRNA polymerase

Initiation

35

53

UnwoundDNA

RNAtran-script

Template strandof DNA

Elongation

RewoundDNA

35

53 3

5

RNAtranscript

LE 17-7a-4Promoter

35

Transcription unit

DNA

InitiationRNA polymerase

Start point

Template strandof DNA

RNAtran-script

UnwoundDNA

Elongation

3

3

53

5

5

3 5

RewoundDNA

5 3

35 35

RNAtranscript Termination

35

5 3Completed RNA transcript

LE 17-7b

ElongationNon-templatestrand of DNA

RNApolymerase

RNA nucleotides

3 end3

5

5

Newly madeRNA

Templatestrand of DNA

Direction of transcription(“downstream”)

LE 17-8

Promoter

53

35

TATA box Start point

Transcriptionfactors

53

35

Several transcriptionfactors

Additional transcriptionfactors

RNA polymerase IITranscription factors

RNA transcript

53

355

Transcription initiation complex

Eukaryotic promoters

TemplateDNA strand

Transcription-some important details

• Rate-30-60 nucleotides/second• RNA polymerase (Many forms in eucaryotes, 3

basic types in bacteria: type I transcribes r-rna, type II-mrna, types III-trna)

• Promotors-(approximately 100 nucleotides)-strong and weak promotors

• Eukaryotes-transcription factors needed to help RNA polymerase to bind to TATA box (region of promotor 25 nucleotides upstream from initiation site).

RNA products of transcription

• m-rna

• t-rna

• r-rna

• sn-RNA (small nuclear)

• mi-Rna (micro)

• Si-rna (small interfering)

Not all genes code for proteins (m-rna)-Rrna and t-rna are obvious examples

• Actually, recent discoveries indicate that a large part of the eukaryotic genome is non-coding RNA-Introns

• Small rna (micro rna and small interfering rna)-play a crucial role in the regulation of gene expression involving both transcription and translation. Rna interference (Rnai)

• We’ll talk about regulation of gene expression in Chapter 15.

Ribosomal RNA and ribosomes

• R-rna; one of two important components of ribosomes (other is protein-some of the proteins are enzymes). 60% r-rna; 40% protein.

• Ribosomes consist of 2 subunits • Ribosomes needed to translate proteins• “workbench of protein synthesis”• Position t-rna (which is attached to a specific

amino acid) on the codon of a m-rna• Result is the synthesis of a protein (whose

amino acid sequence is specified by the m-rna; which is transcribed from a gene)

LE 17-16b

P site (Peptidyl-tRNAbinding site)

E site (Exit site)

mRNAbinding site

A site (Aminoacyl-tRNA binding site)

Largesubunit

Smallsubunit

Schematic model showing binding sites

E P A

LE 17-16a

tRNAmolecules

Exit tunnelGrowingpolypeptide

Largesubunit

mRNA 3

Computer model of functioning ribosome

Smallsubunit

5

E P A

Ribosomal –rna processing

T-rna

• Single polynucleotide chain folded into a complex 3-D shape (inter-chain H bonding). 75-80 nucleotides in length

• Binds a specific amino acid (involvement of amino-acyl-trna-synthetase

• Attaches to a specific m-rna codon via its anticodon

• How many different t-rna’s are there? 61? Actually only 45 (wobble)

LE 17-14a

Amino acidattachment site

Hydrogenbonds

3

5

Two-dimensional structureAnticodon

Amino acidattachment site

35

Hydrogenbonds

Anticodon Anticodon

Symbol used in this bookThree-dimensional structure

3 5

“Charging” t-rna with its specific amino acid

• “charging” enzyme-amino acyl t-rna synthetase (20 different enzymes)

• Requires ATP

LE 17-15Amino acid Aminoacyl-tRNA

synthetase (enzyme)

Pyrophosphate

Phosphates

tRNA

AMP

Aminoacyl tRNA(an “activatedamino acid”)

Messenger Rna (m-rna)

• Contains the information for the primary sequence of a polypeptide chain

• Consists of codons

• Binds to ribosomes

• T-rna binds to m-rna (codon/anticodon)

LE 17-13

Polypeptide

tRNA withamino acidattached

Ribosome

tRNA

Anticodon

35

mRNA

Aminoacids

Codons

Translation

• Codons (m-rna) read by ribosomes/t-rna• Polypeptide chain produced• 3 steps in translation-• A. initiation• B. elongation• C. termination• Translation is a process that consumes a

tremendous amount of energy (ATP and GTP)

LE 17-16c

Amino end

mRNA

5

3

Growing polypeptide

Next amino acidto be added topolypeptide chain

tRNA

Codons

Schematic model with mRNA and tRNA

E

Translation-Initiation

• Initiation codon is AUG

• T-rna that bonds to AUG has an anticodon UAC-this carries the amino acid methionine

• Requires a GTP molecule

• Requires proteins called initiation factors.

LE 17-17

Met

GTPInitiator tRNA

mRNA

53

mRNA binding site

Smallribosomalsubunit

Start codon

P site

5 3

Translation initiation complex

E A

Largeribosomalsubunit

GDP

Met

Translation-Elongation

• The elongation cycle takes about 60 milliseconds

• During elongation, one m-rna codon is read and then the ribosomes moves down the message to the next codon.

• Binding of incoming t-rna to the A site of the ribosome requires a GTP

• Translocation-requires a GTP

LE 17-18

Ribosome ready fornext aminoacyl tRNA

mRNA

5

Amino endof polypeptide

E

Psite

Asite

3

2

2 GDP

E

P A

GTP

GTP

GDP

E

P A

E

P A

Translation-Termination

• When the ribosome reaches a termination codon, it causes the m-rna/ribosome complex to separate

• No t-rna binds to the termination codon.

• Release factors

• Newly made polypeptide chain is released (folds into its characteristic 3-D shape)

LE 17-19

Releasefactor

Stop codon(UAG, UAA, or UGA)

5

3

5

3

5

Freepolypeptide

3

When a ribosome reaches a stopcodon on mRNA, the A site of theribosome accepts a protein calleda release factor instead of tRNA.

The release factor hydrolyzes thebond between the tRNA in theP site and the last amino acid of thepolypeptide chain. The polypeptideis thus freed from the ribosome.

The two ribosomal subunitsand the other componentsof the assembly dissociate.

Summary of energy demands for protein synthesis

• A rough estimate is that for every amino acid incorporated into a polypeptide chain, 3 ATP/GTP are consumed

A. Charging the amino acid (1 ATP)B. Binding of incoming t-rna into the A site (1

GTP)C. Translocation (1 GTP)D. So a small protein (120 amino acids in length)

would cost the cell 360 ATP/GTP to make (the equivalent of 12 glucose molecules going through aerobic cell respiration)

Polyribosomes

• A single ribosome can translate an average-sized polypeptide in about 1 minute

• Several ribosomes can translate the same message one after the other.

• Increases the efficiency of protein production

LE 17-20a

Incomingribosomalsubunits

Growingpolypeptides

Completedpolypeptide

Start ofmRNA(5 end)

End ofmRNA(3 end)

Polyribosome

An mRNA molecule is generally translated simultaneouslyby several ribosomes in clusters called polyribosomes.

LE 17-20b

Ribosomes

mRNA

This micrograph shows a large polyribosome in a prokaryotic cell (TEM).m0.1

M-rna modifications

• Eukaryotic M-rna is modified extensively after transcription (while its still in the nucleus)

• These modifications include

A.Polyadenylation-added to 3’ end of m-rna

B. 5’ cap

C. Intron removal

M-RNA modifications

• Poly A tail

• A. added to the 3’ end of the m-rna

• B.30-200 Adenine nucleotides

• C. roles-regulation of transport of m-rna out of the nucleus; regulation of degradation of m-rna in the cytoplasm; helps m-rna attach to small ribosomal subunit

M-RNA modifications (continued)

• 5’ cap

• A. Modified guanine nucleotide stuck onto 5’ end of m-rna

• B. Roles- positioning of m-rna on small ribosomal subunit in initiation; protects m-rna from degradation

LE 17-9

5Protein-coding segment

5 Start codon Stop codon Poly-A tail

Polyadenylation signal

5 3Cap UTR UTR

Introns

• Many eukaryotic genes have nucleotide sequences that don’t code for amino acids (Introns)

• Introns separate coding sequences (exons). Split genes

• Introns must be removed from the m-rna before it is translated (introns have nucleotide sequences that indicate splicing sites)

• Splicesomes are molecular machines that remove introns from m-rna

LE 17-11-1

Exon 15

Intron

Other proteins

Protein

snRNA

snRNPs

RNA transcript (pre-mRNA)

Spliceosome

Exon 2

LE 17-11-2

5

Spliceosome

Cut-outintron

mRNA

Exon 1 Exon 25

Spliceosomecomponents

Significance of introns

• Why would chromosomes carry around extra DNA that isn’t used in the final m-rna?

A. Expensive to maintain (energy). B. Splicing out introns is a risky business

(what if it’s done incorrectly)C. With these disadvantages, there must be

an advantage or natural selection would not favor this arrangement

Benefits of Introns

• Evolution of protein diversity

• One gene can be alternatively spliced in a number of different ways to form several different types of m-rna (alternative splicing)

• Human antibody genes-about 500 genes can code for billions of different antibody molecules because of alternative splicing.

Figure 15.12

DNA

PrimaryRNAtranscript

mRNA or

Exons

Troponin T gene

RNA splicing

1 2 3 4 5

1 2 3 5 1 2 4 5

1 2 3 4 5

Summary of Transcription and Translation

Mutation

• An alteration in the nucleotide sequence of a DNA molecule (chromosome)

• Chromosomal mutations (duplications; deletions; inversions)

• Point mutations-alterations of one or a few nucleotides in a gene

Point mutations

• Spontaneous mutations

• Induced mutations

• Consequences of mutations-

• A. no effect-”silent mutations”

• B. harmful mutations-(may be lethal)

• C. beneficial mutations (rare)

Spontaneous mutations

• Base pairing errors; why aren’t they corrected by DNA repair enzymes?

• Effects:• A. no effect-silent mutation (redundancy of

genetic code; alteration of a non-critical amino acid)

• B. Positive effect-rare• C. negative effect-missense mutations;

nonsense mutations

LE 17-24a

Wild-type

mRNA

Protein

Amino end Carboxyl end

Stop

35

LE 17-24bBase-pair substitution

No effect on amino acid sequenceU instead of C

Stop

MissenseA instead of G

Stop

NonsenseU instead of A

Stop

Sickle cell anemia

• Results of a spontaneous missense mutation

• Result-altered hemoglobin molecule

• Effect-Depends on the environmental conditions and number of copies of the defective gene you inherited.

LE 17-23

Wild-type hemoglobin DNA

mRNA

3 5 53

5 3 35

Mutant hemoglobin DNA

mRNA

Normal hemoglobin Sickle-cell hemoglobin

Induced mutations

• Caused by environmental damage• Radiation (UV)- T-T dimers; excision

repair enzymes; xerdoerma pigmentosa • Chemicals-Common result-base pair

addition or deletion• Result of addition or deletion (frame shift

mutation)-missense or nonsense • Worst scenario-addition/deletion of 1 or 2

nucleotides at the beginning of a gene

LE 17-25

Base-pair insertion or deletion

Frameshift causing immediate nonsense

Extra U

MissingFrameshift causingextensive missense

Insertion or deletion of 3 nucleotides:no frameshift but extra or missing amino acid

Missing

Stop

Stop

Amino end Carboxyl end

Stop

Wild type

mRNA

Protein5 3

Mutations and Cancer

• Many mutations make cells cancerous

• 90% of known carcinogens are mutagens

• Ames test-screens potential chemicals for being carcinogens by seeing if they are mutagens

• Bacteria are the test subjects in the Ames test.

Employee Resource Manual

• Plasmids

• Restriction Endonucleases

• Agarose Gel Electrophoresis

Plasmids

• Small extrachromosomal pieces of DNA found in some bacterial species

• May carry additional genes (such as antibiotic resistance)

• Can be genetically modified and used as vectors for genetic engineering

PUC 18-Plasmid

Restriction Endonucleases

• Produced by some bacteria as a defense against virus infection

• Cleave DNA at specific bases sequences (different recognition site for each different enzyme)

• Can be used to join DNA from 2 different sources (plasmid DNA and genomic DNA)

ECOR1

Agarose Gel Electrophoresis

• Separates DNA based upon size differences

• DNA is pulled through a gel by an electric current

• (-) charged DNA is pulled to the positive pole of the apparatus.

• Smaller pieces of DNA migrate through the gel faster than larger pieces of DNA

Agarose Gel Electrophoresis

Procell

Procell in Action

What is your first job assignment?

• Clone the H gene (use a bacteria to make copies of the gene for us)

What kind of bacteria do we use to clone the H gene?

• E.coli (lacZ(-), amp sensitve)

Where is the H gene located?

• Lambda virus

How do you get the cloned gene into the bacteria so the bacteria can copy it?

• Transform E.coli (lacZ(-), amp sensitve) with PUC 18-lambda plasmid (heat shock and osmotic shock)

Lambda virus genes have

been inserted into the plasmids here

How do you get lambda genes into PUC 18 plasmid?

• Incubate PUC-18 and lambda with EcoRI, ligate products

How many different plasmids do you get when you mix PUC 18 and lambda, both of which have been

ECOR1 and then ligated?

7

Would all 7 of the plasmids be recombinant (have lambda DNA)?

No!

How do you tell if bacteria have been transformed successfully with PUC-18

plasmid?

• They will grow on amp agar.

How can you distinguish whether plasmids that transformed bacteria were recombinant (lambda and PUC-18) or nonrecombinant (pUC-18 only)?

Plate the transformed cells on Xgal-amp agar

E.Coli transformed With nonrecombinant Plasmids (PUC-18)

E.Coli transformed With recombinant Plasmids (PUC-18/lambda)