Open reading frames... correspond to introns, which are not read by the ribosome during translation correspond to contiguous fragments of DNA sequence that do not contain a stop codon when read in a particular frame correspond to contiguous fragments of DNA sequence that do not contain a stop codon when read in any of six frames are often rich in acetylated histones which allow transcription occur when fragments of DNA sequence are highly similar between two species are recognized by ribosomes to initiate translation

Open reading frames... correspond to introns, which are not read by the ribosome during translation correspond to contiguous fragments of DNA sequence that do not contain a stop codon when read in a particular frame correspond to contiguous fragments of DNA sequence that do not contain a stop codon when read in any of six frames are often rich in acetylated histones which allow transcription occur when fragments of DNA sequence are highly similar between two species are recognized by ribosomes to initiate translation

Name: Open reading frames...correspond to introns, which are not read by th
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Description: Open reading frames...correspond to introns, which are not read by the ribosome during translationcorrespond to contiguous fragments of DNA sequence that do not contain a stop codon when read in a particular framecorrespond to contiguous fragments o

Biochemistry

6th Edition

ISBN:9781305577206

Author:Reginald H. Garrett, Charles M. Grisham

Publisher:Reginald H. Garrett, Charles M. Grisham

Chapter29: Transcription And The Regulation Of Gene Expression

Section: Chapter Questions

Problem 1P

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Similar questions

. What is the minumum number of tRNA molecules that a cell must contain in order to translate all 61 sense codons?
a. Use the genetic code provided in class to predict the sequence of protein formed from the following RNA strand. CGCUACAUCUUU b. If a gene mutation results in a frame shift, meaning the RNA sequence being read shifts rover by a base, and the translation machinery codes, instead, using the nucleotide sequence below, what is the amino acid sequence in the resulting portion of protein? GCUACAUCUUUA c. If a gene mutation results in a point mutation, meaning a single nucleotide is converted to a different nucleotide base, and the translation machinery codes, instead, using the nucleotide sequence below, what is the amino acid sequence in the resulting portion of protein? CGCUCCAUCUUU d. Which mutation is more likely to result in a slightly altered, but still functional, protein, the frame shift mutation or the point mutation? Briefly explain your answer.
Match the term to its best definition. molecule with high-energy bond that carries amino acid to translation site part of a gene that gets translated into a protein primary transcript in eukaryotes that needs to be processed single stranded nucleic acid that helps both in transcription and in splicing nucleic acid that makes up 50% of each ribosome sequence of DNA upstream from the gene that binds RNA polymerase before transcription post processing molecule that carries instructions for making a protein from the nucleus to a ribosome in the cytoplasm part of a gene that gets spliced out during pre mRNA processing
. In bacterial genes, as soon as any partial mRNA transcriptis produced by the RNA polymerase system, the ribosome assembles on it and starts translating. Draw a diagram of this process, identifying 5′ and 3′ ends of mRNA,the COOH and NH2 ends of the protein, the RNA polymerase, and at least one ribosome. Why couldn’t this system work in eukaryotes?
E32. In the technique of DNase I footprinting, the binding of a protein to a region of DNA protects that region from digestion by DNase I by blocking the ability of DNase I to gain access to the DNA. In the DNase I footprinting experiment shown here, a researcher began with a sample of cloned DNA 400 bp in length. This DNA contained a eukaryotic promoter for RNA polymerase II. The assembly of general transcription factors and RNA polymerase II at the core promoter is described in Chapter 12 (see Figure 12.14). For the sample loaded in lane 1, no proteins were added. For the sample loaded in lane 2, the 400-bp fragment was mixed with RNA polymerase II plus TFIID and TFIIB. 2 400 350 250 175 50 Which region of this 400-bp fragment of DNA is bound by RNA polymerase II and TFIID and TFIIB? || III ||| | ||||
DNA: 5’-CTCTACTATAAACTCAATAGGTCC-3’ Draw a box around the sequence where RNA polymerase will bind to the DNA. What is this sequence called? Will transcription start at this sequence, to the left of this sequence (“upstream”) or, to the right of this sequence (“downstream”)? Draw a small arrow above the DNA strand where transcription will begin. Which DNA strand will RNA polymerase transcribe? Highlight this strand with your highlighter. (Hint: RNA pol is similar to DNA pol because it can only make new RNA in the 5’ to 3’ direction. Draw in an arrow to show the direction that RNA polymerase will move along the DNA strand.
. The genetic code is thought to have evolved to maximize genetic stability by minimizing the effect on protein function of most substitution muta- tions (single-base changes). We will use the six arginine codons to test this idea. Consider all of the substitutions that could affect all of the six arginine codons. (a) How many total mutations are possible? (b) How many of these mutations are "silent," in the sense that the mutant codon is changed to another Arg codon? (c) How many of these mutations are conservative, in the sense that an Arg codon is changed to a functionally similar Lys codon?
I've attached the table of transcription ans translation for a DNA and Bees work, Genes A and B are exons while C is an intron. Gene A has a silent mutation and Gene B has a nonsense mutation. Please answer the below for me The 3 genes code for different proteins: • Gene A = protein essential for stinger • Gene B = DNA replication enzyme • Gene C = fuzzy hair protein Do you think it matters which protein is mutated? Is one protein more important than another? How would you try to help the bees stay healthy using the information from the mutations?
The diagram below shows an imaginary eukaryotic structural gene containing two exons. The exon nucleotides are numbered beginning at the transcription start site and a portion of the intron is not shown to save space: Help Center? transcription start site promoter U STACAGTATAAATGAATTAATTGACGTATGTCAATCGGTAAGT...TCAGGTACT U UUU} Phe UUG} Leu exon 1 3 ATGTCATATTTACTTAATTAACTGCATACAGTTAGCCATTCA...AGTCCATGAATGACTTATGTGCGGTTATTTACTGAT... Second letter C Predict the amino acid sequence of the polypeptide encoded by this structural gene. The genetic code is provided below.
. Some tRNAs can recognize more than one codon because there is a relaxation of the complementation rule of base pairing between the anticodon and codon in the third position (of the codon sequence). This relaxation is called
. The human gene for ß2 lens crystallin has the components listed below. The numbers represent nucleotidepairs that make up the particular component. Assumefor simplicity that no alternative splicing is involved.5′ UTR 1741st exon 1191st intron 5322nd exon 3372nd intron 14313rd exon 2083rd intron 3804th exon 4444th intron 995th exon 5463′ UTR 715Answer the following questions about the ß2 lenscrystallin gene, primary transcript, and gene product.Questions asking where should be answered with oneof the 11 components from the list or with None.Assume poly-A tails contain 150 As.a. How large is the ß2 lens crystallin gene in bp (basepairs)?b. How large is the primary transcript for ß2 lenscrystallin in bases?c. How large is the mature mRNA for ß2 lens crystallin in bases?d. Where would you find the base pairs encoding theinitiation codon?e. Where would you find the base pairs encoding thestop codon?f. Where would you find the base pairs encoding the5′ cap?g. Where would you find the base…
. The human gene for ß2 lens crystallin has the components listed below. The numbers represent nucleotidepairs that make up the particular component. Assumefor simplicity that no alternative splicing is involved.5′ UTR 1741st exon 1191st intron 5322nd exon 3372nd intron 14313rd exon 2083rd intron 3804th exon 4444th intron 995th exon 5463′ UTR 715Answer the following questions about the ß2 lenscrystallin gene, primary transcript, and gene product.Questions asking where should be answered with oneof the 11 components from the list or with None.Assume poly-A tails contain 150 As.a. How large is the ß2 lens crystallin gene in bp (basepairs)?b. How large is the primary transcript for ß2 lenscrystallin in bases?c. How large is the mature mRNA for ß2 lens crystallin in bases?d. Where would you find the base pairs encoding theinitiation codon?e. Where would you find the base pairs encoding thestop codon?f. Where would you find the base pairs encoding the5′ cap?g. Where would you find the base…