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Cell & Mole Bio - HES > Lecture 7 - From DNA to Protein: Gene Expression > Flashcards

Lecture 7 - From DNA to Protein: Gene Expression Flashcards

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Biology 101 flashcards
1
Q

Central Dogma

A

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

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2
Q

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

A

RNA

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

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3
Q

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

A

Alkaptonuria

  • mutation that causes accumulation of homogentistic acid.
  • when the allele is mutated, the enzyme to break down homogentistic acid is inactive.
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4
Q

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

A
  • easy to grow in lab
  • short generation times
  • easy to manipulate genetically
  • produce large numbers of progeny
    ex. pea plant, mouse, drosophila, bread mold
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5
Q

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

A

mutant

  • mutations were induced with X-rays
  • the mutant strains needed additional nutrients
  • they were able to determine the metabolic pathway of arginine.
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6
Q

Two main players of transcription (2)

A
  1. RNA: like DNA, is formed from nucleotides (has ribose sugar, uracil instead of thymine)
  2. RNA polymerase: catalyzes transcription
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7
Q

There are ___ types of RNA polymerases in Eukaryotes:

A

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).
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8
Q

RNA polymerase catalysis reaction (4 main points)

A
  • 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.
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9
Q

Transcription Initiation (2)

A
  • 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.
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10
Q

Transcription (2)

A
  • 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).
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11
Q

Transcription Elongation (2)

A
  • 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.
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12
Q

Transcription termination (2)

A
  • 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.
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13
Q

Transcription of eukaryotes require RNA processing (3 steps)

A
  • splicing
  • capping
  • polyadenylation
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14
Q

Eukaryotic mRNA contains ____.

A

introns (INTervening regiONS)

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15
Q

Splicing (2)

A
  • 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.
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16
Q

Spliceosome assembly (4)

A
  • 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.
17
Q

mRNA capping (3)

A
  • 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.
18
Q

Polyadenylation (2)

A
  • 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.
19
Q

Translation has __ players.

A
  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.
20
Q

tRNAs (4)

A
  • 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)
21
Q

tRNA charging (2)

A
  • 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.
22
Q

Decoding the code (2)

A
  • 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.
23
Q

The code (3)

A
  • 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.
24
Q

Wobble - a redundant code (3)

A
  • 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.
25
Q

The ribosome (3)

A
  • 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.
26
Q

There are ___ subunits of the ribosome.

A

two: large and small

27
Q

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

A

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.
28
Q

Translation initiation (2)

A

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

29
Q

Translation elongation (3 steps)

A
  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.
30
Q

Translation termination (2)

A
  • 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.

31
Q

Proofreading mechanism (3)

A
  • 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.
32
Q

Protein processing: Post-translation modifications (3)

A

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.
33
Q

Differences between prokaryotes and eukaryotes (2)

A
  • 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.
34
Q

Mature mRNA leaves the nucleus through ______ ___.

A

nuclear pores.

35
Q

Targeting (3)

A
  • 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)

Cell & Mole Bio - HES (11 decks)

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