Assignment-Synthesis of a Protein - Lab Report Example

Sequence A: TCTTCCCTCCTAAACGTTCAACCGGTTCTTAATCCGCCGCCAGGGCCCCGCCCCTCAGAAGTTGGT DNA: TCT/TCC/CTC/CTA/AAC/GTT/CAA/CCG/GTT/CTT/AAT/CCG/CCG/CCA/GGG/CCC/CGC/CCC/TCA/GAA/GTT/GGTmRNA:UCU/UCC/CUC/CUA/AAC/GUU/CAA/CCG/GUU/CUU/AAU/CCG/CCG/CCA/GGG/CCC/CGC/CCC/UCA/GAA/GUU/GGU66 nucleotides1 UCUUCCCUCC UAAACGUUCA ACCGGUUCUU AAUCCGCCGC CAGGGCCCCG 51 CCCCUCAGAA GUUGGUNote: The numbers 1 and 51 represent the nucleotide positions. Example, at position 51, the nucleotide base is C or CytosineWhen codons 24 to 66, including codons 66 are removed, the sequence reads,UCUUCCCUCCUAAACGUUCAACC UCU/ UCC/ CUC/ CUA/ AAC/ GUU/ CAA/ CC Amino acid sequence: Serine (S)/ Serine (S)/ Leucine (L)/ Leucine (L)/ Aspargine (N)/ Valine (V) / Glutamine (Q)Therefore, resulting protein sequence is,SSLLNVQSequence B: TCAGACGTTTTTGCCCCGTAACAACTTGTTACAACATGGTCATAAACGTCAGAGATGGTCATGAATCTCTTAACTDNA: TCA/GAC/GTT/TTT/GCC/CCG/TAA/CAA/CTT/GTT/ACA/ACA/TGG/TCA/TAA/ACG/TCA/GAG/ATG/GTC/ATG/AAT/CTC/TTA/ACTmRNA:UCA/GAC/GUU/UUU/GCC/CCG/UAA/CAA/CUU/GUU/ACA/ACA/UGG/UCA/UAA/ACG/UCA/GAG/AUG/GUC/AUG/AAU/CUC/UUA/ACU75 nucleotides1 UCAGACGUUU UUGCCCCGUA ACAACUUGUU ACAACAUGGU CAUAAACGUC 51 AGAGAUGGUC AUGAAUCUCU UAACUWhen codons 24 to 66, including codons 66 are removed, the sequence reads,UCAGACGUUU UUGCCCCGUA ACACUCUUAA CUUCA/ GAC/ GUU/ UUU/ GCC/ CCG/ UAA/ CAC/ UCU/ UAA/ CUAmino acid sequence: Serine (S)/ Aspartic acid (D)/ Valine (V) /Phenylalanine (F) / Alanine (A) / Proline (P)/ STOP CODON/Histidine (H)/ Serine (S)/ STOP CODONTherefore, resulting protein sequence is,SDVFAP.HS.(.

= stop codon)Sequence C: TACAAAAAGACCTCACATGTAACACCCCAACTCCGCGTGAAGAACCAAACAGCGATACCGCTCCCGAAAAAGATATGGGDNA: TAC/AAA/AAG/ACC/TCA/CAT/GTA/ACA/CCC/CAA/CTC/CGC/GTG/AAG/AAC/CAA/ACA/GCG/ATA/CCG/CTC/CCG/AAA/AAG/ATA/TGG/GmRNA:UAC/AAA/AAG/ACC/UCA/CAU/GUA/ACA/CCC/CAA/CUC/CGC/GUG/AAG/AAC/CAA/ACA/GCG/AUA/CCG/CUC/CCG/AAA/AAG/AUA/UGG/G79 nucleotides 1 UACAAAAAGA CCUCACAUGU AACACCCCAA CUCCGCGUGA AGAACCAAAC 51 AGCGAUACCG CUCCCGAAA AAGAUAUGGGWhen codons 24 to 66, including codons 66 are removed, the sequence reads,UACAAAAAGA CCUCACAUGU AACAAA AAGAUAUGGGUAC/ AAA/ AAG/ ACC/ UCA/ CAU/ GUA/ ACA/ AAA/ AGA/ UAU/ GGGAmino acid sequence: Tyrosine (Y)/ Lysine (K)/ Lysine (K) /Threonine (T) / Serine (S) / Histidine (H)/ Valine (V)/ Threonine (T) / Lysine (K) / Arginine (R)/ Tyrosine (Y)/ Glycine (G)Therefore, resulting protein sequence is,YKKTSHVTKRYGAnalysis Answers:1.

The beginning sequence can be identified using the start codon. The start codon is typically AUG in eukaryotes (or ATG in DNA; this also encodes methionine). In addition to AUG, alternative start codons, mainly GUG and UUG are used in prokaryotes. The yellow highlighted ones are the start codons.2. The end sequence can be identified using either one of the three the stop codons namely, UAA (Ochre), UAG (Amber), UGA (Opal). The blue highlighted ones are the stop codons.3. The translated protein of Sequence A (SSLLNVQ) contains 7 amino acids.

The translated proteins of Sequence B (SDVFAP.HS.) contain 6 and 2 amino acids respectively. The translated protein of Sequence C (YKKTSHVTKRYG) contains 12 amino acids.4. If a particular protein was absent, it would lead to errors in protein synthesis. Errors in protein synthesis disrupt cellular fitness, cause disease phenotypes, leads to loss of function of the protein (non-functional proteins) and protein misfolding. Protein misfolding destabilizes membranes and induces chronic stress.

Errors in protein synthesis produce polypeptides displaying a gain of toxic function which may confer an alternate or pathological function on a normal, folded protein. In case of enzymes, the hydrophobic core provides structural stability for the molecule and amino acid changes may result in unstable protein product that is temperature sensitive. As the catalytic site of the enzyme is extremely sensitive, a single point mutation (either a deletion or insertion or substitution) may completely abolish function.5. The above sequences represent those of eukaryotes as the intron or the non-coding region (codons from 24 to 66, including 66) had to be removed before translating it into a polypeptide or a protein.

Whether sequence is eukaryotic or prokaryotic, it can be identified by the presence of introns and exons. In the case of prokaryotes, the initial RNA molecule or initial transcript is equivalent to the final mature RNA. In most eukaryotic genes, the initial transcript is processed so that the mature RNA is different. Most eukaryotes have genes that contain introns that do not code for polypeptides. However, prokaryotic genes do not contain introns. Eukaryotic RNAs, after transcription, have the intron sequences that are removed/ spliced in order to produce the final mature RNA.

The sequences in eukaryotes that are represented in the final RNA and code for amino acids in a polypeptide are called exons.6. The tRNA anticodon sequence that would build this protein is complementary to the mRNA sequence. The tRNA anticodon bonds to a given mRNA codon, by hydrogen bonding between A-U pairs and C-G pairs.The tRNA anticodon sequence to the mRNA sequence, UCUUCCCUCCUAAACGUUCAACC is: AGAAGGGAGGATTTGCAAGTTGGThe tRNA anticodon sequence to the mRNA sequence, UCAGACGUUUUUGCCCCGUAACACUCUUAACU is: AGTCTGCAAA AACGGGGCAT TGTGAGAATTGAThe tRNA anticodon sequence to the mRNA sequence, UACAAAAAGACCUCACAUGUAACAAAAAGAUAUGGG is: ATGTTTTTCTGGAGTGTACATTGTTTTTCTATACCC7.

The significance and function of the codons 24 to 66 (GGUUCUUAAUCCGCCGCCAGGGCCCCGCCCCUCAGAAGUUGGU)The above sequence represents the intron or the non-protein-coding region of the initial mRNA transcript. The intron sequences are spliced in order to produce the final mature RNA in eukaryotes. Introns play important roles in gene regulation. One of the functions of introns is that they regulate the frequency of transcription (messenger RNA synthesis) of the neighboring exons.8. In prokaryotes, starting with the amino acid sequence of protein, it is possible to obtain the same DNA nucleotide sequence as the encoding DNA does not contain introns.

However, in eukaryotes, it is not possible to obtain the same DNA nucleotide sequence as the encoding DNA is typically discontinuous: stretches of encoding DNA (called exons) are interspersed with long stretches of non-encoding DNA (called introns). After the DNA is transcribed into a string of RNA and before the RNA is translated into protein; the introns are edited/ spliced out.