Gene X, Y and Z DNA Transcription and Translation Questions

BIO152 Cell Biology Worksheet 7 Gene X, Y and Z DNA transcription and translation Questions 1. 3’-… CAA ACG GAG GAG CTT TAA CCT TAA…-5’ The above is a non-template (coding region) DNA strand sequence of Gene X. What is the mRNA sequence? Provide and label the 5’ to 3’ orientation of the transcript. First, you need to write the complementary DNA strand (which will then become the template sequence): 5’-...GTT TGC CTC CTC GAA ATT GGA ATT...-3’ Second, you need to transcribe the template DNA sequence to mRNA sequence. Remember that the DNA dependent RNA polymerase will use U instead of T in RNA. 3’...CAA ACG GAG GAG CUU UAA CCU UAA...-5’ Next, rewrite the transcript to show 5’ to 3’ (flip around) orientation for convention. Remember not to change any bases. 5’-...AAU UCC AAU UUC GAG GAG GCA AAC...-3’ 2. What is the protein sequence for this transcript, assuming that the first codon is the start of the protein? Label the ends for amino and carboxy termini. Use the Triplet Code table provided on the back page. Now translate the mRNA transcript into the amino acid sequence. (NH2-)....Asn Ser Asn Phe Glu Glu Ala Asn....(-COOH) 1 3. If the following point mutation occurred in Gene X over time and changed the sequence, what effect would this have on the protein? Mutant Gene X sequence: 3’-… CAA ACG GAG GAG CTT TAA TCT TAA…-5’ Work through the sequence conversion as above (see Q1) to transcribe the DNA into mRNA and then translate into the amino acid sequence. First, you need to write the complementary DNA strand (which will then become the template sequence): 5’-...GTT TGC CTC CTC GAA ATT AGA ATT...-3’ Second, you need to transcribe the template DNA sequence to mRNA sequence. Remember that the DNA dependent RNA polymerase will use U instead of T in RNA. 3’...CAA ACG GAG GAG CUU UAA UCU UAA...-5’ Rewrite to show 5’ to 3’ (flip around) orientation for convention. Remember not to change any bases. 5’-...AAU UCU AAU UUC GAG GAG GCA AAC...-3’ You should deduce that the mutant code still produces a serine amino acid and is therefore identical to the original gene. Thus, this specific mutation does not alter the protein sequence or function. (NH2-)....Asn Ser Asn Phe Glu Glu Ala Asn....(-COOH) Notice the degeneracy in certain genetic codons (different triplet codes for the same amino acid). 2 4. Consider the following DNA sequence for a coding region of Gene Y. 5’-…ATG TGC ATG CCC GAA GCT GGA GGG ACT…-3’ What is the mRNA sequence? Provide and label the 5’ to 3’ orientation of the transcript. 3’....UAC ACG UAC GGG CUU CGA CCU CCC UGA...5’ 5’...AGU CCC UCC AGC UUC GGG CAU GCA CAU...3’ 5. What is the protein sequence for this transcript? (NH2-)...Ser Pro Ser Ser Phe Gly His Ala His...(-COOH) 6. If the following point mutations occurred in Gene Y, accumulating over time, what effect would this have on the protein? Mutant Gene Y sequence: 5’-…ATG TGC ATG TCA GAA GCT GGA GGG ACT…-3’ 3’...UAC ACG UAC AGU CUU CGA CCU CCC UGA...5’ Rewrite orientation 5’...AGU CCC UCC AGC UUC UGA CAU GCA CAU...3’ (NH2-)...Ser Pro Ser Ser Phe (-COOH) 3 7. What type of mutation was introduced into Gene Y? Stop codon = nonsense mutation Note: A truncated (short) gene of only five amino acids in length will be produced. Certain genetic diseases are due to nonsense mutations in genes, giving rise to mutant proteins, which lack functionality. 8. 3’-… TAC GCT GAA CCA TTT GGG CCC CAT TAG…-5’ The above is a DNA coding strand sequence of Gene Z. What is the mRNA sequence? Provide and label the 5’ to 3’ orientation of the transcript. Remember, first you need to write the complementary DNA strand (which will then become the template sequence): 5’-...ATG CGA CTT GGT AAA CCC GGG GTA ATC...-3’ Second, you need to transcribe the template DNA sequence to mRNA sequence. Remember that the DNA dependent RNA polymerase will use U instead of T in RNA. 3’...UAC GCU GAA CCA UUU GGG CCC CAU UAG ...-5’ Next, rewrite the transcript to show 5’ to 3’ (flip around) orientation for convention. Remember not to change any bases. 5’-...GAU UAC CCC GGG UUU ACC AAG UCG CAU...-3’ 4 9. What is the protein sequence for this transcript, assuming that the first codon is the start of the protein? Label the ends for amino and carboxy termini. Use the Triplet Code table provided on the back page. Now translate the mRNA transcript into the amino acid sequence. (NH2-)....Asp Tyr Pro Gly Phe Thr Lys Ser His....(-COOH) 5 10. If the following point mutation occurred in Gene Z over time and changed the sequence, what effect would this have on the protein? Mutant Gene Z sequence: 3’-… TAC GCT GAA CCA AAT GGG CCC CAT TAG…-5’ Work through the sequence conversion as above (see Q1) to transcribe the DNA into mRNA and then translate into the amino acid sequence. Remember, first you need to write the complementary DNA strand (which will then become the template sequence): 5’-...ATG CGA CTT GGT TTA CCC GGG GTA ATC...-3’ Second, you need to transcribe the template DNA sequence to mRNA sequence. Remember that the DNA dependent RNA polymerase will use U instead of T in RNA. 3’...UAC GCU GAA CCA AAU GGG CCC CAU UAG ...-5’ Next, rewrite the transcript to show 5’ to 3’ (flip around) orientation for convention. Remember not to change any bases. 5’-...GAU UAC CCC GGG UAA ACC AAG UCG CAU...-3’ Now translate the mRNA transcript into the amino acid sequence. (NH2-)....Asp Tyr Pro Gly(-COOH) Therefore, a stop codon generated a shortened Gene Z, which will likely give rise to a faulty gene product (or non-functional protein). 6 7