15.E: Exercises

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Additional Exercises

For this nucleic acid segment,

a. classify this segment as RNA or DNA and justify your choice. b. determine the sequence of this segment, labeling the 5′ and 3′ ends. (Use Figure 15.1.1 to identify each base.)
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?
In the DNA segment

5′‑ATGAGGCATGAGACG‑3′ (coding strand) 3′‑TACTCCGTACTCTGC‑5′ (template strand)
a. What products would be formed from the segment’s replication?; b. Write the mRNA sequence that would be obtained from the segment’s transcription. c. What is the amino acid sequence of the peptide produced from the mRNA in part b?
A particular DNA coding segment is 5'-ACGTTAGCCCCAGCT-3'.
A) Write the sequence of nucleotides in the corresponding mRNA. B) Determine the amino acid sequence formed from the mRNA in part a during translation.; B) What amino acid sequence results from each of the following mutations? a. deletion of the underlined guanine; b. insertion of thymine immediately after the underlined guanine; c. replacement of the underlined guanine by adenine

Answers

a. RNA; the sugar is ribose, rather than deoxyribose; b. 5′‑GUA‑3′

2. 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.

3. a. Each strand would be replicated, resulting in two double-stranded segments.

(More specifically, the original coding strand, 5′‑ATGAGGCATGAGACG‑3′, would be paired with a new copy of the template strand, 3′‑TACTCCGTACTCTGC‑5′.

The original template strand, 3′‑TACTCCGTACTCTGC‑5′, would be paired with a new copy of the coding strand, 5′‑ATGAGGCATGAGACG‑3′.)

b. mRNA = 5'-AUGAGGCAUGAGACG-3' c. peptide = met-arg-his-glu-thr

A) mRNA = 5′‑ACGUUAGCCCCAGCU‑3′; B) Thr-Leu-Ala-Pro-Ala; C) a. Thr-Leu-Pro-Gln-(incomplete triplet codon), b. Thr-Leu-Val-Pro-Ser-(incomplete codon), c. Thr-Leu-Ser-Pro-Ala