In prokaryotes, a search for genes in a DNA sequenceinvolves scanning the DNA sequence for long openreading frames (that is, reading frames uninterruptedby stop codons). What problem can you see with thisapproach in eukaryotes?

In prokaryotes, a search for genes in a DNA sequenceinvolves scanning the DNA sequence for long openreading frames (that is, reading frames uninterruptedby stop codons). What problem can you see with thisapproach in eukaryotes?

Biology: The Dynamic Science (MindTap Course List)

4th Edition

ISBN:9781305389892

Author:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Publisher:Peter J. Russell, Paul E. Hertz, Beverly McMillan

Chapter19: Genomes And Proteomes

Section: Chapter Questions

Problem 1ITD: Below is a sequence of 540 bases from a genome. What information would you use to find the...

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Consider the following DNA template: 5’-AAGAGGTTCCAATGCAGGCACTCACCAACTCTTAAATAAA-3’ 3’-TTCTCCAAGGTTACGTCCGTGAGTGGTTGAGAATTTATTT-5’ If the bottom DNA strand is used as template to transcribe mRNA, predict the amino acid sequence that would result from the process of translation. Met-Ala-Leu-Thr-Gln-Glu-Gly Met-Gly-Ser-Leu-Asn-Ser-Gln Met-Thr-Asn-Ser-Leu-Ala-Gln Met-Gln-Ala-Leu-Thr-Asn-Ser Met-Glu-Ala-His-Trp-Ser-Tyr
Where and for what reason (s) are you likely to observe the letter "N" in your Sanger sequence reads? Select all that apply. Near the end of a read (which is typically about 500-700 bp in length); DNA polymerase will nearly always incorporate a ddNTP into a read when it reaches this length. Anywhere a mutation leads to incorporation of a translation stop codon in the DNA; DNA polymerase is unable to incorporate ddNTPs into this sequence Anywhere the candidate differs from the reference sequence; any deviation from the reference sequence will inhibit the ability of the sequence analyzer to correctly interpret the base In the first 25-100 base pairs of the read; ddNTPs are unlikely to be incorporated in this region because they are greatly outnumbered by dNTPs.
Below is a sequence of 540 bases from a genome. What information would you use to find the beginnings and ends of open reading frames? How many open reading frames can you find in this sequence? Which open reading frame is likely to represent a protein- coding sequence, and why? Which are probably not functioning protein-coding sequences, and why? Note: for simplicitys sake, analyze only this one strand of the DNA double helix, reading from left to right, so you will only be analyzing three of the six reading frames shown in Figure 19.4.
Consequences of the Wobble Hypothesis Point out why Cricks wobble hypothesis would allow fewer than 61 anticodons to be used to translate the 61 sense codons. How might wobble tend to accelerate the rate of translation?
Suppose your supervisor is working on a molecule “X” with UniProtKB accession number P18564. Being a team member in the project, you have been asked to provide the following information with high accuracy? I. Chromosome number: II. Protein size: III. Number of exons: IV. Stop codon: V. Size of the longest exon in nucleotides
The following DNA sequences found on the sense strand belong to the same eukaryotic gene: Sequence 1: 5'-GATTCAATAAAGCTCAGATCGCTCACGTCGCGACTC-3' Sequence 2: 5'-TCCGAGGTCACTAGATACTCGTCGATCGTATAAATG-3' a) Which sequence is likely to be found upstream from the coding sequence? Justify your answer. b) Which sequence is likely to be found downstream from the coding sequence? Justify your answer. c) Which sequence will not be transcribed into an mRNA transcript? Justify your answer.
Shown below is an R loop prepared for electron microscopy by annealing a purified eukaryotic messenger RNA with DNA from a genomic clone containing the full-length gene corresponding to the mRNA. (a) How many exons does the gene contain? How many introns? (b) Where in this structure would you expect to find a 5′,5′-internucleotide bond? Where would you expect to find a polyadenylic acid sequence?
Could a single nucleotide deletion restore the function of a protein-coding gene interrupted by the insertion of a 4-nt sequence? Explain.
The sequence of the coding strand of a bacterial gene is given below. The positions of the first nine bases are numbered for your convenience. A missense mutation was introduced at position seven where the C was changed to a T resulting a mutant gene. 123456789 5'- ATGGCCCGACCGCAACTTTTCCGAGCTCTGGTGTCTGCGCAGTGACC-3 a. Write the template DNA (complementary strand) sequence for the wild type gene above b. Write the DNA sequence of the mutant gene (Both DNA strands) c. Write the sequence of mRNA produced from the mutant gene d. Write the sequence of the mutant protein using the codon usage table provided in the end of this document.
Recall that constructs used for floxing a gene contain,within one of the gene’s introns, two loxP sites flanking a gene for neomycin resistance (Fig. 18.11a). AloxP site is only 34 base pairs long, as shown in thefollowing figure.ATAACTTCGTATA ATGTATGC TATACGAAGTTATInverted repeat Spacer Inverted repeatExplain how you could use PCR to generate a neomycin resistance gene flanked by loxP sites, starting witha plasmid containing a neorgene. If you had the intronof the target gene cloned in a plasmid vector, howcould you insert your PCR product into the intron?
Posttranscriptional modifications include adding a 5’ cap and a 3’ tail removing a 5’ cap and a 3’ tail removing of exons addition of exons
Assume a bacterial gene underwent a mutation, where a thymine base from an early portion of the coding sequence of the DNA is replaced with a cytosine (as illustrated below). Original sequence (coding strand): AGTTCCTACAAAATGGAGCTGTCTTGGCATGTAGTCTTT ...[Sequence continues with another 80 bases] New sequence: AGTTCCCACAAAATGGAGCTGTCTTGGCATGTAGTCTTT...[Sequence continues with another 80 bases] UAC encodes tyrosine, CAC encodes histine, per the coding table. (This question can be answered without use of the code table, but it is provided here as a resource.) What would the expected result of such a mutation be on the final protein product of the mutated gene (compared to the original, non-mutant product)? The protein will be very different from the original version, and likely non-functional. The protein will be cut short, ending after the first amino acid. There will be no protein produced at all. No change – the protein will be the same.…