Chapter 17 - From Gene to Protein

Question 1

Gene Expression

Answer 1

the process by which DNA directs the synthesis of proteins by transcribing a mRNA sequence and translating it to produce a polypeptide chain

Question 2

Differences in Transcription and Translation between Eukaryotes and Prokaryotes

Answer 2

• Bacteria do not have nuclei therefore there is no membrane separating DNA and mRNA from ribosomes
  
  • Allows mRNA synthesis to begin while it is being transcribed
  • Eukaryotic cells have an envelope separating the two processes over space and time
  • Eukaryotic cells also create a primary transcript which has to undergo modifications to produce final functional RNA transcript

Question 3

Genetic Code

Answer 3

the genetic instructions for a polypeptide chain are written in the DNA as a series of nonoverlapping three nucleotide words
- These are translated into 3 nonoverlapping words in mRNA before translated into an amino acid

Question 4

Important Features of the genetic Code

Answer 4

• Two codons can code for 1 amino acid = redundancy
  ○ Codons often differ only in the third nucleotide base
  • 1 codon cannot code for 2 amino acids = no ambiguity
    ○ Each codon codes for a specific amino acid

Question 5

Components of Transcription

Answer 5

RNA polymerase pries the two strands of DNA apart and joins together RNA nucleotides in a 5’-3’ direction
○ Don’t need a primer and can start from scratch

Promotor: specific sequences of nucleotides which mark where transcription of a gene begins and ends and where the RNA polymerase attaches to

Transcription Unit: the stretch of DNA downstream from the promotor that is transcribed into the RNA molecule

Terminator: sequence which signals the end of transcription

Question 6

Transcription Initiation Complex

Answer 6

the complete assembly of transcription factors and RNA polymerase bound to a promotor
- TATA box is the sequence of DNA which helps form

Question 7

Stages of Transcription

Answer 7

1. Initiation:
   
   • RNA polymerase binds in a specific orientation on the promotor after transcription factors are attached allowing it to unwind the DNA strands
     ○ Transcription factors in eukaryotes guide the binding of polymerase and initiation of transcription.
   2. Elongation:
      - As polymerase moves downstream along the DNA, it untwists the helix exposing 10-20 nucleotides at a time and adds complementary nucleotides to the 3’ end of the growing RNA molecule
      - The growing RNA strand peels away and the DNA strand reforms a double helix
   3. Termination:
      - In bacteria, transcription goes through a terminator sequence, causing the polymerase to detach and release the RNA transcript which then undergoes further modifications
   • In eukaryotes, RNA polymerase 2 transcribes a polyadenylation signal sequence which is a stretch of 6 RNA nucleotides
   • This results in enzymatic proteins cutting the RNA transcript free from the polymerase about 10-35 nucleotides further downstream, releasing pre-mRNA.
   • The enzymes then catch up to the polymerase and it dissociates from DNA

Question 8

RNA processing

Answer 8

the modification of RNA primary transcripts including splicing of introns, joining of exons and alterations of 5’ and 3’ ends to produce a mature mRNA

- 5' receives a 5' cap: a modified guanine nucleotide, after the initiation of transcription

- 3' receives a poly A tail: sequence of 50-250 adenine nucleotides, at the termination of transcription

Question 9

Functions of end RNA processing modifications

Answer 9

• Cap:
  ○ Facilitate the export of mature mRNA out of the nucleus
  ○ Help ribosomes attach to the 5’ end of the mRNA molecule
  
  • Poly A Tail:
    Help protect the mRNA from degradation by hydrolytic enzymes due to mRNA’s temporary nature

Question 10

RNA splicing

Answer 10

the removal of introns in DNA and the joining together of exons
- Introns: noncoding segments
- Exons: coding segments

     - Carried out by a spliceosome

Question 11

Importance of exons

Answer 11

1. A single gene can encode for more than one kind of polypeptide based on which segments are treated as exons during RNA processing
   Alternative RNA splicing: different mRNA molecules are produced from the same primary transcript due to different segments being treated as introns and extrons

Domains: a discrete structural and functional region of a protein
2. Different exons can code for different domains of a protein, creating different active sites and catalyzing different reactions

3. Introns also facilitate the evolution of new proteins because of exon shuffling where introns increase the probability of crossing over between the exons of alleles  Creates no combinations of exons and therefore proteins with different structures and functions

Question 12

Transfer RNA

Answer 12

transfers an amino acid from the cytoplasmic pool to a growing polypeptide chain
- bear a specific amino acid at one end of its 3d shape, while at the other end is a nucleotide triplet: anticodon,

Question 13

Structure of tRNA

Answer 13

Consists of a single RNA strand which has both 5’ and 3’ ends folded and located near one end of the structure
○ 3’ end is the attachment site for each amino acid
○ The loop on the opposite side of the structure contains the anticodon

Question 14

Aminoacyl-tRNA synthetase:

Answer 14

an enzyme joining the correct amino acid to tRNA
○ There are 20 different Aminoacyl-tRNA synthetases one for each amino acid, which binds that amino acid to an appropriate tRNA
○ Catalyzes the covalent attachment of the amino acid in a process driven by ATP hydrolysis resulting in a charged tRNA

Question 15

Wobble

Answer 15

flexibility in base pairing rules in which the nucleotide at the 5’ end of a tRNA anticodon can form hydrogen bonds with more than one kind of base pair at the 3’ end of a codon

• there are not enough types of tRNA molecules for each one to be specific to each possible codon so flexible base pairing occurs / wobble

Question 16

Ribosomes

Answer 16

• Ribosomes facilitate the specific coupling of tRNA anticodons with mRNA codons during protein synthesis
  
  • They are composed of a larger and smaller subunits, each made up of proteins and a ribosomal RNA
  • The ribosome holds the mRNA and tRNA in close proximity so the next amino acid can be added to the carboxyl end of the growing polypeptide
  • It then catalyzes the formation of the peptide bond
  • Before passing the polypeptide through the exit tunnel in the larger subunit

Question 17

tRNA binding sites at a ribosome

Answer 17

• A site: holds the tRNA molecule carrying the next amino acid
  
  • • P site: holds the tRNA carrying the growing polypeptide chain
  • E Site: where discharged tRNA’s leave the ribosome

Question 18

Stages of Translation: Initiation

Answer 18

• A small ribosomal subunit binds to the mRNA followed by the specific initiator tRNA molecule, carrying the first amino acid methionine
  ○ The mRNA sequence is bound at the specific RNA sequence, just upstream of the start codon
  ○ The initiator tRNA molecule and small subunit is bound to the 5’ cap of mRNA and scans downstream until it reaches the start codon
  
  • The larger subunit then binds to the tRNA molecule at the P site to form a Translation initiation Complex
  • this is brough together by initiation factors
    This leaves the tRNA in the P site, and the A site is vacant for the next tRNA molecule

Question 19

Stages of Translation: Elongation

Answer 19

• Amino acids are added one by one to the previous amino acid at the C-Terminus end through the assistance of elongation factor proteins
  1. Codon recognition requires hydrolysis of GTP to increase accuracy and efficiency
  2. rRNA molecule in large subunit catalyzes peptide bond formation between growing polypeptide and amino group of new amino acid.
  Another GTP is hydrolyzed to provide energy for translocation of tRNA molecules between A, P and E sites as well as provides energy to break hydrogen bonds between tRNA molecule in E site and release it.

Question 20

Stages of Translation: Termination

Answer 20

• Elongation continues until a stop codon is read and brought to the A site, bringing release factors to the A site
  
  • This results in the addition of a water molecule instead of an amino acid, which hydrolyzes the bond between completed polypeptide and tRNA in P site, releasing the polypeptide
  • Hydrolysis of 2 more GTP molecules occur in order for the release of remaining tRNA molecules breakdown of the ribosome subunits

Question 21

Post Translational Modifications

Answer 21

• modification of certain amino acids by the attachment of sugars, lipids, phosphate groups or other additions in order for the protein to do its job
  
  • The removal of one or more amino acids by enzymes from the amino end of the chain
  • The chain can be enzymatically cleaved into two or more pieces in order for them to be activated
  • Two or more polypeptide chains synthesized separately can come together for quaternary structure.

Question 22

Relationship between location of ribosome and its purpose

Answer 22

• The location of the ribosome determines where the synthesized polypeptide chain goes
  ○ Free ribosomes suspended in the cytosol synthesize proteins which function in the cytosol
  ○ Bound ribosomes attached to the ER make proteins for the endomembrane system

Question 23

Signal Peptide

Answer 23

a sequence of amino acids which targets the ribosome to bind to the ER
- Polypeptide synthesis always begins in the cytosol on a free ribosome unless the growing polypeptide cues the ribosome to attach to the ER

Question 24

Signal Recognition Particle

Answer 24

a protein RNA complex which recognizes and binds to the ribosome carrying the signal peptide as it emerges, dragging the ribosome to the ER by binding now to a receptor protein on the ER

Question 25

Polyribosomes

Answer 25

a group of several ribosomes attached to, and translating, the same messenger RNA molecule
- allows a single mRNA is used to make many copies of a polypeptide simultaneously

Question 26

Substitution

Answer 26

the replacement of one nucleotide and its partner with another pair of nucleotides
- Some have no effect due to the redundancy of the nuclear code(more than one codon translates for the same amino acid)

Question 27

Types of substitution Mutations

Answer 27

• Silent Mutation: a substitution mutation with no observable effect due to the changed codon coding for the same amino acid
  • Missense Mutations: a substitution mutation resulting in a codon which codes for a different amino acid
    ○ Can have little effect on the protein as it may have similar properties or be in a region of the protein which is not essential
    ○ Can be detrimental if affecting an area crucial for the protein function e.g. the active site
  • Nonsense mutation: a substitution mutation resulting in a codon coding for a stop amino acid, resulting in shorter and nonfunctional proteins

Question 28

Insertions and Deletions

Answer 28

the addition or loss of nucleotide pairs in a sequence
Can alter the reading frame of the genetic message; frameshift mutation

Question 29

Spontaneous Mutations

Answer 29

when an incorrect nucleotide is added to a growing chain and is mismatched with the nucleotide base on the complementary strand, resulting in an incorrect base and a faulty template strand if not fixed by DNA proofreading

Question 30

Mutagen

Answer 30

a chemical or physical agent that interacts with DNA and can cause a mutation