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Chemistry LibreTexts

19.E: Exercises

Additional Exercises

  1. For this nucleic acid segment,

    1.jpg
    1. classify this segment as RNA or DNA and justify your choice.
    2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
  2. For this nucleic acid segment,

    2.jpg
    1. classify this segment as RNA or DNA and justify your choice.
    2. determine the sequence of this segment, labeling the 5′ and 3′ ends.
  3. One of the key pieces of information that Watson and Crick used in determining the secondary structure of DNA came from experiments done by E. Chargaff, in which he studied the nucleotide composition of DNA from many different species. Chargaff noted that the molar quantity of A was always approximately equal to the molar quantity of T, and the molar quantity of C was always approximately equal to the molar quantity of G. How were Chargaff’s results explained by the structural model of DNA proposed by Watson and Crick?

  4. Suppose Chargaff (see Exercise 3) had used RNA instead of DNA. Would his results have been the same; that is, would the molar quantity of A approximately equal the molar quantity of T? Explain.

  5. In the DNA segment

    5′‑ATGAGGCATGAGACG‑3′ (coding strand) 3′‑TACTCCGTACTCTGC‑5′ (template strand)

    1. What products would be formed from the segment’s replication?
    2. Write the mRNA sequence that would be obtained from the segment’s transcription.
    3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 5b?
  6. In the DNA segment

    5′‑ATGACGGTTTACTAAGCC‑3′ (coding strand) 3′‑TACTGCCAAATGATTCGG‑5′ (template strand)

    1. What products would be formed from the segment’s replication?
    2. Write the mRNA sequence that would be obtained from the segment’s transcription.
    3. What is the amino acid sequence of the peptide produced from the mRNA in Exercise 6b?
  7. A hypothetical protein has a molar mass of 23,300 Da. Assume that the average molar mass of an amino acid is 120.

    1. How many amino acids are present in this hypothetical protein?
    2. What is the minimum number of codons present in the mRNA that codes for this protein?
    3. What is the minimum number of nucleotides needed to code for this protein?
  8. Bradykinin is a potent peptide hormone composed of nine amino acids that lowers blood pressure.

    1. The amino acid sequence for bradykinin is arg-pro-pro-gly-phe-ser-pro-phe-arg. Postulate a base sequence in the mRNA that would direct the synthesis of this hormone. Include an initiation codon and a termination codon.
    2. What is the nucleotide sequence of the DNA that codes for this mRNA?
  9. A particular DNA coding segment is ACGTTAGCCCCAGCT.

    1. Write the sequence of nucleotides in the corresponding mRNA.
    2. Determine the amino acid sequence formed from the mRNA in Exercise 9a during translation.
    3. What amino acid sequence results from each of the following mutations?

      1. replacement of the underlined guanine by adenine
      2. insertion of thymine immediately after the underlined guanine
      3. deletion of the underlined guanine
  10. A particular DNA coding segment is TACGACGTAACAAGC.

    1. Write the sequence of nucleotides in the corresponding mRNA.
    2. Determine the amino acid sequence formed from the mRNA in Exercise 10a during translation.
    3. What amino acid sequence results from each of the following mutations?

      1. replacement of the underlined guanine by adenine
      2. replacement of the underlined adenine by thymine
  11. Two possible point mutations are the substitution of lysine for leucine or the substitution of serine for threonine. Which is likely to be more serious and why?

  12. Two possible point mutations are the substitution of valine for leucine or the substitution of glutamic acid for histidine. Which is likely to be more serious and why?

Answers

  1.  

    1. RNA; the sugar is ribose, rather than deoxyribose
    2. 5′‑GUA‑3′
 
  1. In the DNA structure, because guanine (G) is always paired with cytosine (C) and adenine (A) is always paired with thymine (T), you would expect to have equal amounts of each.

 
  1.  

    1. Each strand would be replicated, resulting in two double-stranded segments.
    2. 5′‑AUGAGGCAUGAGACG‑3′
    3. met-arg-his-glu-thr
 
  1.  

    1. 194
    2. 194
    3. 582
 
  1.  

    1. 5′‑ACGUUAGCCCCAGCU‑3′
    2. thr-leu-ala-pro-ala
      1. thr-leu-thr-pro-ala
      2. thr-leu-val-pro-ser
      3. thr-leu-pro-gin
 
  1. substitution of lysine for leucine because you are changing from an amino acid with a nonpolar side chain to one that has a positively charged side chain; both serine and threonine, on the other hand, have polar side chains containing the OH group.