Chapter 7- translation Flashcards

1
Q

Know the definition of translation

A

the sequence of bases in mRNA specifies the order in which amino acids are added to a polypeptide

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2
Q

Know the factors required for translation

A

mRNA
initiation factors
elongation factors
aminoacyl tRNA synthetases
tRNA
ribosome (rRNA + ribosomal proteins)

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3
Q

Know the structure of the ribosome and describe the three binding sites for tRNAs

A

Structure: small & large subunit
- large subunit has 3 binding sites:
-A (aminoacyl) site
-P (peptidyl) site
E (exit) site

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4
Q

Describe the properties of codons and reading frames

A

codons: group of 3 adjacent nucleotides in an mRNA that code for an amino acid
- ribosome determines the correct reading frame for the codons
reading frame: region where the ribosome begins reading the sequence of nucleotides

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5
Q

Compare eukaryotic and prokaryotic ribosomes

A

Eukaryotic ribosomes are larger than prokaryotic ribosomes

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6
Q

*Know the structure of tRNA and describe key features of the tRNA molecule

A

structure: cloverleaf; small; 70-90 nucleotides; each have their own self-pairing structure
-anticodon loop
key features: CCA at their 3’ end with the 3’ -OH fo the A as an attachment for an amino acid

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7
Q

Know the function of tRNA Synthetase and describe its role in charging tRNAs

A

function: connect specific amino acids to specific tRNA molecules
- a tRNA without an amino acid attached is uncharged
- a tRNA with an amino acid attached is charged

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8
Q

Describe the base pairing properties between codons and anticodons

A
  • first base in the codon mRNA (5’) pairs with the last base in the anticodon of the tRNA (3’) : antiparallel
  • codons, specify amino acids according to the genetic code
  • AUG = the codon that initiates translation; corresponds to the amio acid methionine
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9
Q

Describe the degeneracy of the genetic code and know how to read the codon chart

A

degeneracy=many amino acids are specified by more than one codon; redundant

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10
Q

Compare translation initiation in prokaryotes vs eukaryotes

A

-Prokaryotes
-mRNAs lack a 5’ cap; the initiation complex forms at one or more internal sequences in the mRNA (Shine-Dalgarno sequence)
-Shine-Dalgarno sequence is where initiation takes place

-Eukaryotes
-initiation complex forms at the 5’ cap and scans along the mRNA until it reaches the start codon AUG
- 5’cap is where initiation takes place

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11
Q

Describe the process of translation initiation and the role of initiation factors

A
  • initiation factors
    - bind to the 5’ cap of the mRNA
    - recruit the small ribosomal subunit
    - bring up a transfer RNA charged with Methionine
    -initiation complex moves along the mRNA until it reaches the AUG start codon
    -when the first AUG codon is reached:
    -large ribosomal subunit joins the complex
    -initiation factors are released
    • the next tRNA is ready to join the ribosome in the A site
      -Once the tRNA is in the A site the bond connecting the Met to its tRNA is transferred to the amino group of the amino acid in the A site
      -PEPTIDE BOND is formed
    • new peptide is attached to the tRNA in the A site
    • formation of a peptide bond requires proteins in the large subunit, but the rRNA in the large subunit is the catalyst
      -the ribosome shifts one codon to the right
      -moves uncharged tRNA (Met) to the E site
      -moves the peptide-bearing tRNA to the P site
      -frees the A site for the next charged tRNA in the line to come in
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12
Q

Describe the process of translation elongation

A

-the ribosome moves one codon farther along the mRNA
-tRNA in the E site is ejected
-tRNA in the P site is moved to the E site
-tRNA in the A site is moved to the P site
- A site is open and available for the next tRNA

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13
Q

Describe the process of translation termination and the role of release factors

A

-elongation continues until a stop codon
-stop codons: UAA, UAG, UGA
-release factor protein binds the A site of the ribosome
- causes the bond connecting the polypeptide to the tRNA in the P site to break
-the breaking of the bond creates the carboxyl terminus of the polypeptide

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