Chapter 23: PROBLEM 23.66 (page 1257)

Question: Exposure to nitrous acid (see Section 19-16), sometimes found in cells, can convert cytosine to uracil.

  1. Propose a mechanism for this conversion.
  2. Explain how this conversion would be mutagenic upon replication.
  3. DNA generally includes thymine, rather than uracil(found in RNA). Based on this fact, explain why the nitrous acid-induced mutation of cytosine to uracil is more easily repaired in DNA than it is in RNA.

Short Answer

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Answer

a. The mechanism for the conversion is:

b . Conversion of cytosine to uracil will be mutagenic when a wrong base is inserted in a nucleic acid chain.

c. If cytosine gets diazotized to uracil, the DNA repair enzymes may detect it as a mutation and correct it. In DNA, thymine is used instead of uracil.

Step by step solution

01

Nitrous acid

Nitrous acid

02

Mechanism for the conversion

As diazonium is unstable, thus nitrogen gas leaves quickly, driving the reaction forward. Then, an attack of water as nucleophile occurs at the carbon center. Further on proton abstraction by water, the enol tautomer of uracil forms, which gets converted into its more stable keto tautomer.

The mechanism for the conversion

03

Conversion be mutagenic upon replication.

Generally, in base pairing, cytosine pairs up with guanine. If cytosine is converted to uracil, each replication will not carry over the complement of cytosine, i.e., guanine. Instead, it will carry the complement of uracil, i.e., adenine. Thus, cytosine conversion to uracil will be mutagenic as the wrong base is inserted in a nucleic acid chain

04

Nitrous acid-induced cytosine mutation to uracil is more easily repaired in DNA than in RNA.

In RNA, cytosine conversion to uracil is not detected as a problem because uracil, which is a base, is normally found in RNA and goes unrepaired. In DNA, thymine is found instead of uracil with an extra methyl group. DNA repair enzymes can detect cytosine conversion to uracil as mutation and correct it.

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Most popular questions from this chapter

Fructose is found in many fruits. From memory, draw fructose in

  1. the Fischer projection of the open chain.
  2. The most stable chair conformation of the most stable pyranose anomer.
  3. The Haworth projection of the most stable pyranose anomer

Aldohexoses A and B both undergo Ruff degradation to give aldopentose C. On treatment with warm nitric acid, aldopentose C gives an optically active aldaric acid. B also reacts with warm nitric acid to give an optically active aldaric acid, but A reacts to give an optically inactive aldaric acid. Aldopentose C is degraded to aldotetrose D, which gives optically active tartaric acid when it is treated with nitric acid. Aldotetrose D is degraded to (+)-glyceraldehyde. Deduce the structures of sugars A,B,C and D, and use Figure 23-3 to determine the correct names of these sugars.

. (a) An aliphatic aminoglycoside is relatively stable to base, but it is quickly hydrolyzed by dilute acid. Propose a mechanism for the acid-catalyzed hydrolysis.

(b) Ribonucleosides are not so easily hydrolyzed, requiring relatively strong acid. Using your mechanism for part (a), show why cytidine and adenosine (for example) are not so readily hydrolyzed. Explain why this stability is important for living organisms.

All of the rings of the four heterocyclic bases are aromatic. This is more apparent when the polar resonance forms of the amide groups are drawn, as is done for thymine at left. Redraw the hydrogen-bonded guanine-cytosine and adenine-thymine pairs shown in figure 23-24, using the polar resonance forms of the amides. Show how these forms help to explain why the hydrogen bonds involved in these pairings are particularly strong. Remember that a hydrogen bond arises between an electron-deficient hydrogen atom and electron-rich pair of nonbonding electrons.

Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of α-D-galactopyranoseis +150.70 , and that of the β anomer is +52.80 . When either of the pure anomers dissolves in water, the specific rotation gradually changes to +80.20. Determine the percentages of the two anomers present at equilibrium.

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