Knowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why. Figure 17.11RNA processing: addition of the 5 cap and poly-A tail.A modified guanine nucleotideadded to the 5' end50-250 adenine nucleotidesadded to the 3' endDNATRANSCRIPTIONRegion that includesprotein-coding segmentsPolyadenylation signal3'RNA PROCESSING Pre-MRNA5'G-P-P-P-AAAAAAMRNAAAUAAAStart codon Stop codon-TRANSLATIONRibosome5 Сap5'UTR3' UTRPoly-A tailPPolypeptide

Genetic codon It is a group of three nucleotide that code for amino acids. They are total 64…

Knowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why.

Knowing that the genetic code is almost universal, a scientist uses molecular biological methods to insert the human β-globin gene (Shown in Figure 17.11) into bacterial cells, hoping the cells will express it and synthesize functional β-globin protein. Instead, the protein produced is nonfunctional and is found to contain many fewer amino acids than does β-globin made by a eukaryotic cell. Explain why.

Biology Today and Tomorrow without Physiology (MindTap Course List)

5th Edition

ISBN:9781305117396

Author:Cecie Starr, Christine Evers, Lisa Starr

Publisher:Cecie Starr, Christine Evers, Lisa Starr

Chapter7: Gene Expression And Control

Section: Chapter Questions

Problem 1VQ

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Similar questions

An unprocessed pre-mRNA has the following structure. Which of the following is not a possible size (in bp) of the mature mRNA? 205bp 180bp 150bp 100bp
Coding DNA 5’- GTG ACT CGT TGT GCC ATT GCA GCT AAA CAC TTC GAG CCC TGT- 3’ mRNA 5’- GUG ACU CGU UGU GCC AUU GCA GCU AAA CAC UUC GAG CCC UGU- 3’ What is the polypeptide sequence?
Below is the DNA sequence of a patient with overlapping genes (a single mRNA has multiple initiation points for translation) for two different proteins (DADαs and AMA): 5’- GTCCCAACCATGCCCACCGATCTTCCGCCTGCTTCTGAAGATGCGGGCCCAGGGAAATCTCTAACG-3’ 1. Indicate the DNA sequence coding for RNA. 2. Indicate the amino acid sequence of each of them.
On the mRNA codon table below, the first nucleotide in mRNA is to the left, the second is above, and third is to the right. On the sequence, the 5’cap is indicated by (5’). The poly (A) tail is not shown. Use the codon table to translate this short mRNA. Mark the codons and write the amino acid sequence beneath them. (5’)CGUUACAAUGUAUCGCGCGGUACUCGGCAAAGUGCCCUGAAUAGAGUUGGUA(3’) In a previous round of replication, DNA polymerase made a mistake and added a C on what is now the DNA template strand. In the space on the mRNA sequence below, write the added base. Mark the codons again and write the amino acid sequence beneath them. What do you observe? (5’)CGUUACAAUGUAU CGCGCGGUACUCGGCAAAGUGCCCUGAAUAGAGUUGGUA 3’
49Eukaryotic RNA polymerases can be distinguished by their sensitivities to ( ). 50Each bacterial mRNA open reading frame has its own ( ) codon. 51Two nuclear splicing intermediates resemble ______________. A.helicesb.lariatsc.hairpinsD.projections
Select the correct statement(s) about transfer RNAs. Codon recognition occurs through specific base-pairing by the anticodon arm. Amino acids are linked to the T-loop. They are transcribed by RNA Polymerase III. There is strict matching requirements at the third position, else the tRNA cannot interact with the codon.
The sequence below shows the non-coding strand from the whole of the transcribed region of a very short gene. 5’-GGCTTCTTTAGTACTGGCCAGTGGGATCCAAGTAGGCTGCCATTTCGT-3’ Write out the sequence of the mRNA from this gene in the orientation 5′ → 3′ and, using the genetic code (see Fig. 1. overleaf) deduce the amino acid sequence of the peptide it encodes (NB you should read about the operation of the genetic code prior to attempting this question).
What will be the overall anti-codon sequence in tRNA for this mRNA? 5’-GUAGCCUUAUCUAGCGAUCACCGUCCGUAUUACUAGUGGCCAGACUCUUUUCACCAUGUAUAGUUG-3’
mRNA sequence of A gene Find 5’ UTR and 3’UTR of mRNA 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCCUACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACGACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGCUGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGAGGUAGAAGCCGCUGGGGCUUGGGGCU-3’
Translate the following mRNA into protein, starting from the first initiation codon: 5'-CCGAUGCCAUGGCAGCUCGGUGUUAC AAGGCUUGCAUCAGUACCAGUUUGAAUCC-3'
mRNA sequence of A gene Write the amino acid sequence of the gene A. 5’ AAACUGUGACUGAACCUCAAACCCCAAACCAGCCCGAGGAGAACCACAUUCUCCCAGGGA CCCAGGGCGGGCCGUGACCCCUGCGGCGGAGAAGCCUUGGAUAUUUCCACUUCAGAAGCCUACUGGGGAAGGCUGAGGGGUCCCAGCUCCCCACGCUGGCUGCUGUGCAGAUGCUGGACGACAGAGCCAGGAGGGAGGCCGCCAAGAAGGAGAAGGUAGAGCAGAUCCUGGCAGAGUUCCAGCUGCAGGAGGAGGACCUGAAGAAGGUGAUGAGACGGAUGCAGAAGGAGAUGGACCGCGGCCUGAGGUAGAAGCCGCUGGGGCUUGGGGCU-3’
In panel a (see picture), a tRNA is already bound to the initiator codon at the start of the mRNA strand. The anticodon sequence on this tRNA that complements the mRNA codon is A. CCCB. UACC. CAC