Translation

Question 1

Gene

Answer 1

Specific sequence of nucleotides in a DNA molecule which codes for a specific sequence of amino acids in one polypeptide chain → gene product

Question 2

How do genes determine phenotypic characteristics

Answer 2

DNA (gene) → sequence of codons on mRNA → sequence of amino acids (R-groups) → unique conformation of a protein → function → phenotype

Question 3

Gene expression

Answer 3

• Flow of genetic information from DNA to protein

- Transcription + translation

Question 4

Central Dogma (3)

Answer 4

1. Replication → DNA-directed DNA synthesis
2. Gene expression/protein synthesis
3. Reverse transcription

Question 5

Genetic code

Answer 5

• 20 different amino acids
• 3 nucleotides code for 1 amino acid
• AUG → start codon
• UAA, UAG, UGA → stop codon

Question 6

Features of genetic code

Answer 6

1. Triplet code → triplet of nucleotides in mRNA → codons → amino acid
2. Universal → same codon codes for same a.a. in all organisms
3. Degenerate (redundant) → same amino acid may be coded for by several codons
4. Non-overlapping
5. Continuous
6. Includes start and stop codon → reading frame

Question 7

Translation

Answer 7

Process by which sequence of ribonucleotides in an mRNA molecule is converted into a sequence of amino acids in a polypeptide chain

Question 8

Translation steps (4)

Answer 8

1. Amino acid activation
2. Initiation
3. Elongation and translocation
4. Termination

Question 9

1. Amino acid activation

Answer 9

• Each amino acid covalently attached to 3’ CCA stem of specific tRNA with specific anticodon → amino-acyl tRNA
• Catalysed by specific aminoacyl-tRNA synthetase
• Each enzyme has active site complementary in conformation and charge to:
  1) specific amino acid
  2) unique identity sites at 3’CCA stem and anticodon on tRNA
  → double specificity
• Requires ATP

Question 10

2. Initiation (Eukaryotes)

Answer 10

1. Initiation factors facilitate the binding of the small ribosomal subunit to both mRNA and initiator tRNA.

• Initiation factors and initiator tRNA (carrying methionine) bind to small ribosomal subunit
• Small ribosomal subunit then recognises and binds to the 5’ 7 methylguanosine cap of the mRNA → moves in the 5’ to 3’ direction along the mRNA in search of the start codon
• Initiator tRNA associates with the start codon by cbp

Question 11

2. Initiation (Prokaryotes)

Answer 11

1. Initiation factors facilitate the binding of the small ribosomal subunit to both mRNA and initiator tRNA.
   - Initiation factors bind to the small ribosomal subunit and facilitate its binding to Shine-Dalgarno sequence so that the start codon can be correctly positioned before the initiator tRNA and large ribosomal subunit bind

Question 12

2. Initiation (Eukaryotes and prokaryotes)

Answer 12

2. Large ribosomal subunit binds, completing the ribosome → forms translation initiation complex
3. The initiator tRNA will be positioned at the P site (peptidyl-tRNA binding site)
4. The A site (aminoacyl-tRNA binding site) will be vacant for the addition of the next aminoacyl tRNA molecule
5. GTP is required for the initiation stage

Question 13

3. Elongation and translocation

Answer 13

a) Codon recognition
b) Peptide bond formation
c) Translocation

Question 14

a) Codon recognition

Answer 14

Anticodon of incoming aminoacyl tRNA complementary base pairs with mRNA codon in A site by forming H bonds

Question 15

b) Peptide bond formation (2)

Answer 15

1. Peptidyl transferase in large ribosomal subunit catalyses peptide bond formation between amino acid carried by tRNA in A site and methionine/amino acid at carboxyl end of growing polypeptide chain carried by tRNA in the P site.
2. The methionine/amino acid dissociates from the (initiator) tRNA it was bound to.

Question 16

c) Translocation (5)

Answer 16

1. Ribosome shifts one codon down mRNA in 5’ to 3’ direction → polypeptide synthesised from amino to carboxyl end
2. The tRNA from the P site is shifted to the E site (exit site) and released into cytosol.
3. The peptidyl-tRNA with growing polypeptide is translocated from A site to P site.
4. Empty A site is ready to receive the next incoming aminoacyl tRNA, with anticodon complementary to mRNA codon exposed at A site.
5. The process is repeated until a stop codon is reached

Question 17

4. Termination (4)

Answer 17

1. When the stop codon (UAG, UAA, UGA) reaches the A site, release factors enter the A site.
2. Binding of the release factors causes the hydrolysis of the bond between the polypeptide chain and the tRNA at the P site.
3. The polypeptide is released from the ribosome as it completes its folding into it secondary and tertiary structure.
4. The ribosome disassembles into its subunits.

Question 18

Eukaryotes vs prokaryotes

Answer 18

• In eukaryotes, transcription takes place in the nucleus and the pre-mRNA undergoes post-transcriptional modification within nucleus before being transported to the cytoplasm for translation.
• In prokaryotes, mRNA can be translated while being transcribed.

Question 19

Types of mutations (2)

Answer 19

1. Gene mutations

2. Chromosomal aberration (under mitosis and meiosis)

Question 20

Gene mutations

Answer 20

• Alteration in the sequence of nucleotides which may change the sequence of amino acids in a polypeptide chain
• May change the 3D shape of the protein → affecting protein function → affect phenotype of organism
• Can be caused by mutagen → chemical/physical agent that interacts with DNA

Question 21

Types of gene mutations (4)

Answer 21

1. Substitution → nucleotide replaced by a different nucleotide
2. Inversion → segment of nucleotide sequences separates from the allele and rejoins at the original position but it is inverted
3. Insertion → one or several nucleotides are inserted in a sequence
4. Deletion → one or several nucleotides are deleted from a sequence

Question 22

Outcomes of mutations (4)

Answer 22

1. Frame-shift mutation
2. Silent mutation
3. Missense mutation
4. Nonsense mutation

Question 23

1. Frame-shift mutation

Answer 23

• Due to insertion/deletion of no. of nucleotides not divisible by 3
• Disrupts reading frame
• Produces different and non-functional polypeptide

Question 24

2. Silent mutation

Answer 24

• Point mutation → does not change amino acid sequence in a polypeptide → same polypeptide synthesised
• Can occur in the either coding or non-coding regions
• If mutation occurs in coding sequence → degeneracy of genetic code → >1 codon can code for the same amino acid
• If the mutation occurs in the non-coding region

Question 25

3. Missense mutation

Answer 25

- Point mutation → single nucleotide change → codon that codes for different amino acid - If the new amino acid has similar biochemical properties (e.g. charge, size) to the one that was replaced, the mutation is conservative, else non-conservative

Question 26

4. Nonsense mutation

Answer 26

- Point mutation → premature stop codon | - Polypeptide to be truncated and non-functional

Question 27

Sickle-cell anaemia

Answer 27

- Single base substitution in gene coding for β-globin chain - CTC → CAC (DNA template strand) - 6th codon GAG → GUG (mRNA) - Glutamate → valine (amino acid) - HbA → HbS (protein)

Question 28

Effect of change

Answer 28

- Charged and hydrophilic glutamate → non-polar and hydrophobic valine in HbS. - At [O₂], HbS undergoes conformation change → hydrophobic areas on different HbS stick together - Polymerisation of HbS → formation of abnormal, rigid, rod-like fibres - Shape of red blood cell distorted → sickle shaped.

Question 29

Effects of disease

Answer 29

- Sickle RBC more fragile and break easily - Shortage of RBC → poor O₂ transport - Leads to anaemia, lack of energy and heart failure. - Sickle RBC may also lodge in small blood vessels → interfere with blood circulation - Lead to organ damage.

Question 30

Inheritance of mutation

Answer 30

- Homozygous recessive disorder - Sufferers need 2 copies of mutant form of gene - Heterozygous → sickle cell trait

Question 31

Structure of tRNA

Answer 31

1. Single-stranded RNA 2. Folds back upon itself and held in shape by hydrogen bonding between complementary base pairs at certain regions to form a 3D L-shaped structure 3. 3 bases form an anticodon 4. 3’ end with CCA stem is attachment site for a specific amino acid that corresponds to anticodon

Question 32

Role of tRNA (2)

Answer 32

1. Bring in specific amino acids in a sequence corresponding to the sequence of codons in mRNA to the growing polypeptide 2. Can facilitate translation due to: a) ability to bind to a specific single amino acid b) ability of anticodon to base-pair with mRNA codon

Question 33

Role of rRNA (4)

Answer 33

1. Associates with a set of proteins to form ribosomes (70/80S) 2. Main constituent of the interface between the large and small subunits of the ribosome → small ribosomal subunit can bind to the mRNA → cbp can occur between rRNA in mRNA binding site and mRNA. 3. Main constituent of the P and A site on large ribosomal subunit → enables binding of aminoacyl-tRNAs to P and A site 4. An rRNA molecule (peptidyl transferase) on large ribosomal subunit catalyses formation of peptide bonds between the amino group of the new amino acid in the A site and the carboxyl end of the growing polypeptide in the P site.