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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

Expert verified

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

Erwin Chargaff鈥檚 discovery that DNA contains equimolar amounts of guanine and cytosine and also equimolar amounts of adenine and thymine has come to be known as Chargaff鈥檚 rule:

G = C and A = T

(a) Does Chargaff鈥檚 rule imply that equal amounts of guanine and adenine are present in DNA? That is, does G = A?

(b) Does Chargaff鈥檚 rule imply that the sum of the purine residues equals the sum of the pyrimidine residues? That is, does A + G = C + T?

(c) Does Chargaff鈥檚 rule apply only to double-stranded DNA, or would it also apply to each individual strand if the double helical strand were separated into its two complementary strands?

The relative configurations of the stereoisomers of tartaric acid were established by the following synthesis:

(1) D-(+)-glyceraldehydediastereomers A and B (separated)

(2) Hydrolysis of A and B using aqueous Ba(OH)2 gave C and D, respectively.

(3) HNO3 oxidation of C and D gave (-)-tartaric acid and meso-tartaric acid, respectively.

(a) You know the absolute configuration of D-(+)-glyceraldehyde, Use Fischer projections to show the absolute configurations of products A, B, C, and D.

(b) Show the absolute configurations of the three stereoisomers of tartaric acid: (+)-tartaric acid, (-)-tartaric acid, and meso-tartaric acid.

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.

Predict the products formed when the following sugars react with excess acetic anhydride and pyridine.

  1. 伪-顿-驳濒耻肠辞辫测谤补苍辞蝉别
  2. 尾-顿-谤颈产辞蹿耻谤补苍辞蝉别

Like glucose, galactose mutarotates when it dissolves in water. The specific rotation of 伪-D-galactopyranoseis +150.70 , and that of the 尾 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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