Lecture 7 - From DNA to Protein: Gene Expression

Question 1

Central Dogma

Answer 1

DNA (transcription) -> RNA (translation) -> Polypeptide

Question 2

Some viruses have ___ as their main basis of information storage.

Answer 2

RNA

RNA (transcription) -> RNA (translation) -> Polypeptide

Question 3

One gene-one enzyme hypothesis was validated by the study of _________.

Answer 3

Alkaptonuria

• mutation that causes accumulation of homogentistic acid.
• when the allele is mutated, the enzyme to break down homogentistic acid is inactive.

Question 4

Testing the one gene-one enzyme hypothesis required model organisms which have these four characteristics (4)

Answer 4

• easy to grow in lab
• short generation times
• easy to manipulate genetically
• produce large numbers of progeny
  ex. pea plant, mouse, drosophila, bread mold

Question 5

Testing the one gene-one enzyme hypothesis by comparing ______ strains with the wild type (3)

Answer 5

mutant

• mutations were induced with X-rays
• the mutant strains needed additional nutrients
• they were able to determine the metabolic pathway of arginine.

Question 6

Two main players of transcription (2)

Answer 6

1. RNA: like DNA, is formed from nucleotides (has ribose sugar, uracil instead of thymine)
2. RNA polymerase: catalyzes transcription

Question 7

There are ___ types of RNA polymerases in Eukaryotes:

Answer 7

three

1. RNA polymerase I transcribes genes encoding ribosomal RNAs.
2. RNA polymerase II transcribes genes encoding proteins.
3. RNA polymerase III transcribes genes encoding transfer and ribosomal RNAS (tRNA & rRNA).

Question 8

RNA polymerase catalysis reaction (4 main points)

Answer 8

• 5’-3’ direction
• copies only one strand at a time
• processive - one enzyme-template binding results in polymerization of hundreds of RNA bases
• does not need a primer.

Question 9

Transcription Initiation (2)

Answer 9

• RNA polymerase will bind to the promoter (promoter is a sequence of DNA in front of the gene of interest)
• This binding will determine directionality and which of the two DNA strands is going to be transcribed.

Question 10

Transcription (2)

Answer 10

• template is the sequence of DNA being copied.
• non-template is the sequence of DNA that codes for the mRNA being produced. also, non-template has promoter and has identical base sequence to RNA (except for T for U).

Question 11

Transcription Elongation (2)

Answer 11

• RNA polymerase unwinds DNA about 10 base pairs at a time; reads template in 3’ to 5’ direction.
• RNA polymerases do not proofread and correct mistakes.

Question 12

Transcription termination (2)

Answer 12

• specified by a specific DNA sequence.
• In prokaryotes, helper protein binds to the transcript and causes it to detach from the DNA. The transcript forms a loop (hairpin) and falls away from the DNA.

Question 13

Transcription of eukaryotes require RNA processing (3 steps)

Answer 13

• splicing
• capping
• polyadenylation

Question 14

Eukaryotic mRNA contains ____.

Answer 14

introns (INTervening regiONS)

Question 15

Splicing (2)

Answer 15

• splicing is the process by which introns are removed to produce a mature mRNA.
• process is performed by the spliceosome complex which is a mixture of RNA and protein.

Question 16

Spliceosome assembly (4)

Answer 16

• specific sequences on the pre-mRNA identify the presence of an intron.
• snRNPS (small nuclear ribonucleoproteins) bind to these sequences and recruit components of the spliceosome.
• the hydroxyl (-OH) group of the new 3’ end and the phosphate (PO4) group of the new 5’ end react to connect exons, relieving the intron and spliceosome complex to create mature mRNA.
• the excised intron and spliceosome are degraded.

Question 17

mRNA capping (3)

Answer 17

• a 5’-cap is added at the 5’ end
• the cap is a modified GTP which facilitates mRNA binding to a ribosome and helps translation initiation.
• also protects mRNA from being digested by ribonucleases.

Question 18

Polyadenylation (2)

Answer 18

• a poly A tail is added at the 3’ end. This is an addition of ATP at the end of the mRNA.
• The tail assists in export from the nucleus and is important for stability of the mRNA.

Question 19

Translation has __ players.

Answer 19

1. mRNA: codes for the protein that is being produced. also note that mRNA is the product of transcription.
2. tRNA: brings the amino acids and reads the code presented by mRNA.
3. rRNA: ribosomal RNA that makes up the ribosome which catalyzes the translation reaction.

Question 20

tRNAs (4)

Answer 20

• each tRNA will bind to a specific enzyme (aminoacyl tRNA synthetase) that will catalyze the attachment of a specific amino acid.
• the tRNA recognizes the codon of the mRNA
• binds to the ribosome
• has attached specific amino acid (the amino acid attachment site is always CCA)

Question 21

tRNA charging (2)

Answer 21

• charging requires ATP; a high-energy bond forms between the amino acid and the tRNA.
• this energy is later used to form the peptide bond.

Question 22

Decoding the code (2)

Answer 22

• Nirenberg and Matthaei used simple artificial mRNAs of known composition to identify the polypeptide that resulted.
• this led to the identification of the first three codons.

Question 23

The code (3)

Answer 23

• AUG is the start codon: initiation signal for translation
• Three different stop codons: termination signals, including UAA, UAG, and UGA
• For most amino acids, there is more than one codon: the genetic code is redundant.

Question 24

Wobble - a redundant code (3)

Answer 24

• Wobble: Specificity for the base at the 3’ end of the codon is not always observed.
• Example: Codons for alanine - GCA, GCC, and GCU - are recognized by the same tRNA.
• As a result of wobble there are fewer tRNA species, but the genetic code is not ambiguous.

Question 25

The ribosome (3)

Answer 25

- catalyzes translation - complex composed of rRNA and proteins (catalysis is performed by the rRNA) - ribosome reads the mRNA 5'-3' direction and catalyzes the reaction from the N- to the C-terminal.

Question 26

There are ___ subunits of the ribosome.

Answer 26

two: large and small

Question 27

There are ___ binding sites on the large subunit of the ribosome.

Answer 27

three 1. A (aminoacyl tRNA) - binds with anticodon of charged tRNA 2. P (peptidyl tRNA) - where tRNA adds its amino acid to the growing chain 3. E (exit site) - where tRNA sits before being released from the ribosome.

Question 28

Translation initiation (2)

Answer 28

Prokaryotes: - rRNA binds to the Shine-Dalgarno sequence of the mRNA - An initiation complex forms - a charged tRNA and small ribosomal subunit, both bound to mRNA Eukaryotes: - ribosome binds to the 5' cap to initiate translation

Question 29

Translation elongation (3 steps)

Answer 29

1. Incoming charged tRNA pairs with the codon at the A-site 2. The free end of the amino acid on the A site reacts with the polypeptide chain, releasing the tRNA on the P site; ribosome shifts 3. Shifting of ribosome opens a new A site.

Question 30

Translation termination (2)

Answer 30

- translation ends when a stop codon enters the A site | - Stop codons bind a protein release factor which hydrolyzes bond between the polypeptide and the tRNA in the P site.

Question 31

Proofreading mechanism (3)

Answer 31

- ribosome checks for correct codon-anti codon pairing. - charged tRNA binds to EF-Tu (elongation factor) - If correct codon-anti codon match occurs, EF-Tu releases the charged tRNA and reaction proceeds.

Question 32

Protein processing: Post-translation modifications (3)

Answer 32

Most polypeptides are modified after translation: not all proteins have methionine (AUG) as the first AA. This is due to post-translation modification. - Phosphorylation: addition of phosphate groups catalyzed by protein kinases (the charged phosphate groups change the conformation and may expose active sites or binding sites) - Glycosylation: addition of sugars to form glycoproteins. - Proteolysis: polypeptide is cut by proteases (e.g., signal sequence is removed) - Pro-Opiomelancortin (POMC) is a pro-hormone.

Question 33

Differences between prokaryotes and eukaryotes (2)

Answer 33

- eukaryotes have a nucleus so transcription occurs inside the nucleus; translation occurs in the cytoplasm. prokaryotes' transcription and translation occur in cytoplasm. - eukaryotes mRNA gets capped, spliced, and polyadenylated.

Question 34

Mature mRNA leaves the nucleus through ______ ___.

Answer 34

nuclear pores.

Question 35

Targeting (3)

Answer 35

- protein sequences (targeting sequences) can be recognized by targeting proteins to bring the newly synthesized protein to the right place (e.g. nuclear localization sequence, export sequence) - ER sequence will target the protein to be synthesized in the ER - In the ER, the protein can be targeted to a specific region (plasma membrane, vesicles, lysosome)