book 3 (transcription & translation)

Question 1

describe the process of transcription in prokaryotes

Answer 1

[Step 1 : Initiation of transcription]
1. A detachable sigma subunit in the RNA polymerase of bacteria recognises and binds to the
promoter.
2. RNA polymerase unwinds and unzips the DNA double helix. Hydrogen bonds between
complementary bases of the double helix are broken.
3. One of the exposed strands of DNA is used as a template for transcription. Binding of RNA
polymerase determines which strand of the double helix is used as the template.
4. Transcription of the template strand begins.

[Step 2 : Elongation of transcription]
1. As RNA polymerase moves along the template strand of DNA, it continues to unwind the double
helix, exposing 10 to 20 DNA bases for base-pairing with RNA nucleotides.
2. RNA polymerase adds the corresponding free ribonucleoside triphosphates complementary to
the DNA bases on the template strand.
3. Free ribonucleoside triphosphates (ATP, UTP, CTP, GTP) are added to the 3’ OH group of the
growing RNA chain.
4. The new RNA strand is thus synthesised in the 5’ to 3’ direction.
5. Adjacent ribonucleotides are joined together by phosphodiester bonds, forming a continuous
mRNA chain.

[Step 3 : Termination of transcription]
1. RNA polymerase reaches a termination sequence on the DNA.
2. The transcribed region on the RNA which is complementary to the termination sequence on the
DNA template forms a hairpin loop. This causes the RNA polymerase to detach from the DNA
and release the RNA transcript.
3. These mRNA molecules are ready for translation as soon as they are released from the DNA
templates.
* Simultaneous transcription and translation occurs in prokaryotes due to a lack of the nuclear
envelope.

Question 2

describe the process of transcription in eukaryotes

Answer 2

[Step 1 : Initiation of transcription]
1. A transcription factor recognises and binds to TATA box found within the promoter.
2. Binding initiates a series of interactions between multiple transcription factors at the same site.
3. RNA polymerase II binds to the promoter of the DNA template with the aid of transcription factors.
4. This complex of transcription factors at the promoter causes the DNA double helix to unwind and
unzip and prevent the two DNA strands from re-annealing. Hydrogen bonds between
complementary bases of the double helix are broken.
5. One of the exposed strands of DNA is used as a template for transcription. Binding of RNA
polymerase II determines which strand of the DNA double helix is used as the template.
6. The completed assembly of transcription factors and RNA polymerase II bound to the promoter is
known as the transcription initiation complex
7. Transcription of the template strand begins.

[Step 2 : Elongation of transcription]
1. As RNA polymerase II moves along the template strand of DNA, it continues to unwind the double
helix, exposing 10 to 20 DNA bases for base-pairing with RNA nucleotides.
2. RNA polymerase II adds the corresponding free ribonucleoside triphosphates complementary
to the DNA bases on the template strand.
3. Free ribonucleoside triphosphates (ATP, UTP, CTP, GTP) are added to the 3’ OH group of the
growing RNA chain.
4. The new RNA strand is thus synthesised in the 5’ to 3’ direction.
5. Adjacent ribonucleotides are joined together by phosphodiester bonds, forming a continuous
mRNA chain.

[Step 3 : Termination of transcription]
1. Transcription proceeds until RNA polymerase II reaches a termination sequence on the DNA.
2. Once the RNA polymerase II transcribes a DNA sequence (TTATTT) which codes for a
polyadenylation signal (AAUAAA) on the pre-mRNA.
3. 10 to 35 nucleotides downstream from this polyadenylation signal, the pre-mRNA is cleaved (by
proteins)
4. Newly formed pre-mRNA released from the growing RNA chain and DNA rewinds to form the
double helix.

Question 3

state the processes post-transcriptional modifications

Answer 3

1. the addition of a methylated GTP to the first base of the transcript to form a 5’ cap
2. introns and pre-mRNA being excised while exons are simultaneously being ligated together
3. the addition of many adenine nucleotides to the 3’ end to form a poly(A) tail

Question 4

explain how the correct amino acid is joined to the tRNA

Answer 4

• via amino acid activation whereby it is the attachment of an amino acid to it’s corresponding tRNA with the specific anti-codon, catalysed by aminoacyl tRNA synthetases
• the active site of aminoacyl tRNA synthetases has a specific 3D conformation that is complementary to the conformation of only one type of amino acid and the corresponding tRNA molecule
• synthetase then catalyse the covalent attachment of the specific amino acid to the tRNA at the 5’-CCA-3’ attachment site

Question 5

explain the initiation process of translation for eukaryotes

Answer 5

1. Initiation factors and the initiator tRNA which carries the amino acid methionine associates with
   the small ribosomal subunit forming a pre-initiation complex.
2. The pre-initiation complex then moves downstream along the mRNA until it reaches the start
   codon, AUG which signals the start of translation.
3. The initiator tRNA, in the pre-initiation complex has the anti-codon UAC, binds to the start codon,
   AUG on the 5’ end of the mRNA via complementary base-pairing
4. Hydrolysis of GTP provides energy for the process.
5. The large ribosomal subunit then attaches to the small ribosomal subunit in the pre-initiation
   complex, releasing the initiation factors to form the translation initiation complex.
6. The initiator tRNA is positioned at the Peptidyl-tRNA binding site (P-site) of the large ribosomal
   subunit. The Aminoacyl-tRNA binding site (A-site) is exposed for incoming aminoacyl-tRNA
   in the elongation of polypeptide chain.