Transcription and Translation

Transcription

Structures of DNA and RNA

DNA contains deoxyribose while RNA contains ribose;

DNA is double-stranded while RNA is single-stranded;

the base thymine found in DNA is replaced by uracil in RNA;

one form of DNA (double helix) but several forms of RNA (tRNA, mRNA and rRNA);

Transcription:

Transcription: Process of copying a sequence of DNA bases to mRNA

Introns and exons

Translation

Translation: decoding of mRNA at a ribosome to produce a polypeptide

TOK Linnkk.

Translation (Cont..)

Initiation

Transcription initiation complex -the area where transcription factors and RNA polymerase are bound to the promoter

TATA box -promoter DNA sequence-the actual sequence is 5'-TATAAA-3' -RNA polymerase binding site

After polymerase is bound to the promoter DNA, the two DNA strands unwind and the enzyme starts transcribing the template strand

Elongation

A. RNA polymerase moves along DNA templateB. It unwinds 10-20 DNA bases at a timeC. RNA polymerase adds nucleotides in the 5’→3’

direction D. As RNA polymerase moves along, the DNA double

helix reformsE. The new section of RNA ‘peels away’ as the

double helix reforms

Translocation

Termination

A. Transcription stops when RNA polymerase reaches a section of DNA called the terminator

B. Terminator sequence = AAUAAAC. Next, the RNA strand is released and RNA

polymerase dissociates from the DNAD. The RNA strand will go through more processing

Transcription (termination)

A. Transcription stops when RNA polymerase reaches a section of DNA called the terminator

B. Terminator sequence = AAUAAAC. Next, the RNA strand is released and RNA

polymerase dissociates from the DNAD. The RNA strand will go through more processing

Sense vs. Antisense DNA strandsA. The DNA double helix has two strands

B. Only one of them is transcribedC. The transcribed strand is the antisense strandD. The non transcribed strand is the sense strandE. mRNA is complementary to the anitsense strand

Sense vs. Antisense DNA strandsF. The 5’ end of the RNA nucleotides are added to the

3’ end of the growing chainG. RNA nucleotides are linked together in the same

fashion as DNA molecules

RNA splicing (in eukaryotes)

A. In eukaryotes RNA transcripts have long non-coding stretches of nucleotides-these regions will not be translated

B. The non-coding sections are dispersed between coding sectionsC. Introns-non-coding sections of nucleic acid found between coding regionsD. Exons-coding regions of nucleic acids

(eventually these are expressed as amino acids)

RNA splicing (in eukaryotes)

E. RNA polymerase transcribes introns and exons,-this is pre-mRNA

F. Pre-mRNA never leaves the cell’s nucleusG. The introns are excised and exons are joined together to form mRNAH. pre-mRNAI. Mature mRNA

Translation

A. Translation-forming of a polypeptide-uses mRNA as a template for a.a. sequence-4 steps (initiation, elongation, translocation and termination)-begins after mRNA enters cytoplasm-uses tRNA (the interpreter of mRNA)

TranslationB. Ribosomes

-made of proteins and rRNA-each has a large and small subunit-each has three binding sites for tRNA on its surface-each has one binding site for mRNA-facilitates codon and anticodon bonding-components of ribosomes are made in the nucleus and exported to the cytoplasm where they join to form one functional unit

8. Translation

B. Ribosomes (continued)-the three tRNA binding sites are:

1. A site=holds tRNA that is carrying the next amino acid to be added

2. P site= holds tRNA that is carrying the growing polypeptide chain

3. E site= where discharged tRNAs leave the ribosome

Ribosomal structure

EP A

Large subunit

Peptidyl-tRNA binding site

Aminoacyl-tRNA binding site

mRNA

5’

Exit site

Small subunit

3’

TranslationC. The genetic code

Four RNA nucleotides are arranged 20 different ways to make 20 different amino acids

Nucleotide bases exist in triplets Triplets of bases are the smallest units that can code for an a.a. 3 bases = 1 codon = 1 a.a. There are 64 possible codes (64=43)

Translation

C. The genetic code Most of the 20 a.a. have between 2 and 4 possible

codes The mRNA base triplets are codons In translation the codons are decoded into amino acids

that make a polypeptide chain It takes 300 nucleotides to code for a polypeptide

made of 100 amino acids (Why?)

Translation

C. The genetic code (continued) 61 of 64 codons code for a.a. Codon AUG has two functions

-codes for amino acid methionine (Met)-functions as a start codon

mRNA codon AUG starts translation The three ‘unaccounted for’ codons act as stop codons

(end translation)

Translation

D. How it worksDNA (antisense)

A C C A A A C C GmRNA (transcription)

U G G U U U G G Cpolypeptide (translation)

Trp - Phe - Gly-

TranslationE. More on tRNA

tRNA is transcribed in the nucleus and must enter the cytoplasm

tRNA molecules are used repeatedly Each tRNA molecule links to a particular mRNA

codon with a particular amino acid When tRNA arrives at the ribosome it has a

specific amino acid on one end and an anticodon on the other

Anticodons (tRNA) bond to codons (mRNA)p. 304 (red book)

Where the a.a. attaches

Hydrogen bonds

Anticodon

=

tRNA diagrams

Although we draw tRNA in a clover shape it’s true 3-D conformation is L-shaped.

Translation (Initiation)

A. Initiation1. Brings together mRNA, tRNA (w/ 1st a.a.) and ribosomal subunits2. Small ribosomal subunit binds to mRNA and an initiator tRNA

-start codon= AUG-start anticodon-UAC-small ribosomal subunit attaches to 5’ end of

mRNA

#9. Translation (Initiation)B. Initiation

2. (continued)-downstream from the 5’ end is the start codon AUG

(mRNA)-the anticodon UAC carries the a.a. Methionine

3.After the union of mRNA, tRNA and small subunit, the large ribosomal subunit attaches4. Initiation is complete

Translation (Initiation)

B. Initiation5. The intitiator tRNA and a.a. will sit in the P site of the large ribosomal subunit6. The A site will remain vacant and ready for the aminoacyl-tRNA

Translation (Initiation)

Translation (Elongation)A. Amino acids are added one by one to the

first amino acid (remember, the goal is to make a polypeptide)

B. Step 1- Codon recognitiona. mRNA codon in the A site forms hydrogen bonds with the tRNA

anitcodon

C. Step 2- Peptide bond formationa. The ribosome catalyzes the formation of the peptide bonds between

the amino acids (the one already in place and the one being added)b. The polypeptide extending from the P site moves to the A site to

attach to the new a.a.

Translocation

A. The tRNA w/ the polypeptide chain in the A site is translocated to the P site

B. tRNA at the P site moves to the E site and leaves the ribosome

C. The ribosome moves down the mRNA in the 5’→3’ direction

Termination

A. Happens at the stop codonB. Stop codons are UAA, UAG and

UGA -they do not code for a.a.C. C. The polypeptide is freed from

the ribosome and the rest of the translation assembly comes apart

Gene expression

A. Jacob and Monad (1961)-studied control of protein synthesis in E. coli and lactose digesting enzymes-found that E. coli do not produce lactose digesting enzymes when grown in a medium without lactose-when bacteria were placed in a lactose environment, enzymes were found within minutes

Gene expression

B. Genes can be switched on or off as necessary-a gene that is ‘on’ will be transcribed-in E.coli, the enzyme lactase will be produced if the gene is ‘on’-if the gene is ‘off’ mRNA will not be created and translation can not occur

Gene expression

C. The operon model-proposed by Jacob and Monad-explains how genes switch on and off-operon=promoter, operator and structural genes-lac operon is found in E.coli

Gene expressionD. The lac operon

Gene expression

D. The lac operon (no lactose)-lactose is absent, repressor is active, operon is off, no mRNA is produced, RNA polyermase cannot bind because it is blocked by the repressor that has bound to the operator

Gene expression

D. The lac operon (lactose is present)-lactose is present, repressor is inactive, operon is on, mRNA is transcribed, RNA polymerase binds to operator-an isomer of lactose binds to the repressor and changes its shape-this prevents it from binding to the operator-lactase is produced