Biology 2e
2nd Edition
ISBN: 9781947172517
Author: Matthew Douglas, Jung Choi, Mary Ann Clark
Publisher: OpenStax
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Chapter 15, Problem 19CTQ
Imagine if there were 200 commonly occurring amino acids instead of 20. Given what you know about the genetic code, what would be the shortest possible codon length? Explain.
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Chapter 15 Solutions
Biology 2e
Ch. 15 - Figure 15.11 A scientist splices a eukaryotic...Ch. 15 - Figure 15.13 Errors in splicing are implicated in...Ch. 15 - Figure 15.16 Many antibiotics inhibit bacterial...Ch. 15 - The AUC and AUA codons in mRNA both specify...Ch. 15 - How many nucleotides are in 12 mRNA codons? 12 24...Ch. 15 - Which event contradicts the central dogma of...Ch. 15 - Which subunit of the E. coli polymerase confers...Ch. 15 - The -10 and -35 regions of prokaryotic promoters...Ch. 15 - Three different bacteria species have the...Ch. 15 - Which feature of promoters can be found in both...
Ch. 15 - What transcripts will be most affected by low...Ch. 15 - How do enhancers and promoters differ? Enhancers...Ch. 15 - Which pre-mRNA processing step is important for...Ch. 15 - What processing step enhances the stability of...Ch. 15 - A scientist identifies a pre-mRNA with the...Ch. 15 - The RNA components of ribosomes are synthesized in...Ch. 15 - In any given species, there are at least how many...Ch. 15 - A scientist introduces a mutation that makes the...Ch. 15 - Imagine if there were 200 commonly occurring amino...Ch. 15 - Discuss how degeneracy of the genetic code makes...Ch. 15 - A scientist sequencing itiRNA identifies the...Ch. 15 - If mRNA is complementary to the DNA template...Ch. 15 - In your own words, describe the difference between...Ch. 15 - A fragment of bacterial DNA reads: 3’...Ch. 15 - A scientist observes that a cell has an RNA...Ch. 15 - Chronic lymphocytic leukemia patients often harbor...Ch. 15 - Transcribe and translate the following DNA...Ch. 15 - Explain how single nucleotide changes can have...Ch. 15 - A normal mRNA that reads 5’ -...
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- Below is the 5’–3’ strand of a double-stranded DNA molecule with the following nucleotide sequences:5’ C C T A T G C A G T G G C C A T A T T C C A A A G C A T A G C 3’ 1. If the RNA synthesized above (item #3) is a functional mRNA and all the nucleotides belong to an exon,a. how many codons are present in this mRNA?b. how many codons actually code for a protein in this mRNA?c. what stop codon is present in this mRNA?arrow_forwardBelow is the 5’–3’ strand of a double-stranded DNA molecule with the following nucleotide sequences:5’ C C T A T G C A G T G G C C A T A T T C C A A A G C A T A G C 3’ 1. If the RNA synthesized above (item #1) is a functional mRNA and all the nucleotides belong to an exon,a. how many codons are present in this mRNA?b. how many codons actually code for proteins in this mRNA?c. what stop codon is present in this mRNA?arrow_forwardDNA A G T A C C G G G C A A A C T G C A T T G T G mRNA U C A U G G C C C G U U U G A C G U A A C A C Use the "Genetic Code Chart" to determine the sequence of amino acids in your polypeptide chain. Remember to START translation at the start codon by adding a Methionine and STOP translating when you reach a stop codon.arrow_forward
- The genetic information contained in DNA consists of a linear sequence of coding units known as codons. Each codon consists of three adjacent DNA nucleotides that corresponto a single amino acid in a protien. The E.coli DNA molecule contains 4.70 x 10^6 base pairs. Determine the number of codons that can be present. Assuming that the average protein in E.coli consists of a chain of 400 amino acids, calculate the maximum number of protiens that can be coded by an E.coli DNA molecule.arrow_forwardConsider the following wild-type double-stranded DNA sequence: 5' TATGAA AGT3 non-transcribed strand (sense strand) 5' 3' ATACTTTCA transcribed strand In the space below, write ONE of the possible DNA sequences of the transcribed strand shown above that results from BOTH a single substitution mutation of the first codon following the start codon that would also cause a nonsense mutation. Use the mRNA codon chart in the Appendix of your manual to help you. Answer: Checkarrow_forwardThe genetic code is thought to have evolved to maximize genetic stability by minimizing the effect on protein function of most substitution mutations (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 mutantcodon is changed to another Arg codon?(c) How many of these mutations are conservative, in the sense that an Argcodon is changed to a functionally similar Lys codon?arrow_forward
- Please explain why it is useful that our RNA is read in codons. Imagine a hypothetical scenario where there are 95 amino acids and only 6 nucleotides available. Calculate how many nucleotides per codon would be required to code for all 95 amino acids. Show and explain your work.arrow_forwardThe sequence of a polypeptide is determined by the order of codons that specify the amino acids in the polypeptide. How many different sequences of codons can specify the polypeptide sequence methionine-histidine-lysine? (Use the table to find the number of possibilities.) SECOND BASE UAU UACFTyrosine (Tyr) UAA -Stop codon UAG -Stop codon UUUL UGU Cysteine (Cys) UCU uc UCA FSerine (Ser) uca Uuc Phenylalanine (Phe) UUAL Leucine (Leu) CAU CAC CAA Glutamine (Gin) CAGF UGA -Stop codon uaa -Tryptophan (Trp) CGU сос CGA FArginine (Arg) CU CU Histidine (His) CuA FLeucine (Leu) Cua) Proline (Pro) CCA cca AAU Asparagine (Asn) AGU Serine (Ser) AGC AUU ACU ACC Threonine (Thr) AACF AAA AAGLysine (Lys) AUC Fisoleucine (lle) AUA Methionine (Met) AUG - Start codon ACA ACG AGA AGGFArginine (Arg) GU GACAspartic acid (Asp) GGA GAA Glutamic acid (Glu) Gaa) GcU -Valine (Val) G GUA GCA FAlanine (Ala) Glycine (Gly) 8. 1 4 THIRD BASE 2. FIRST BASEarrow_forwardUse the genetic code table to determine the amino acid sequence of the given message strand of DNA below, from N to C terminal. All introns were removed in this sequence for simplicity. Write the amino acids in their 3-letter abbreviation and separate with a dash.arrow_forward
- Decode the unknown word using the genetic code and fill out the missing nucleotides. Translate the generated mRNA sequence using the universal genetic code table and the information derived from the previous steps. Determine the unknown word by arranging the single letter amino acid code of the hypothetical protein produced bound by the start and stop codon.arrow_forwardTranslate the following RNA sequence by using the genetic below. Start at the beginning of the sequence and don't worry about start and stop codons. Write out the sequence using the single letter code. (This table displays the amino acids in a single-letter code instead of a three-letter code. Each codon is found by matching the first position on the left of the chart, second position at the top, and last position at the right. For example, the codon CAG gives the amino acid "Q") 5' UCAACUGCGAAUCUGGAAUAU 3'arrow_forwardTranslate the following RNA sequence by using the genetic below. Start at the beginning of the sequence and don't worry about start and stop codons. Write out the sequence using the single letter code. (This table displays the amino acids in a single-letter code instead of a three-letter code. Each codon is found by matching the first position on the left of the chart, second position at the top, and last position at the right. For example, the codon CAG gives the amino acid "Q") 5' UCAACUGCGAAUCUGGAAUAU 3'arrow_forward
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