Erwin Chargaff’s 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’s rule:

G = C and A = T

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

(b) Does Chargaff’s 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’s 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?

Short Answer

Expert verified

(a) No, there is no relation between the amounts of guanine and adenine.

(b) Yes, Chargaff’s rule states that sum of purine residues equals the sum of the pyrimidine residues.

(c) Chargaff’s rule apply only to double-stranded DNA.

Step by step solution

01

Step-1.: Find Chargaff’s rule implies that equal amounts of guanine and adenine

Chargaff’s rule implies that total number of purine bases is equal to total number of pyrimidine bases, that is, number of guanine units is equal to cytosine units present and number of adenine units equal to thymine units. DNA composition is different in different species. Chargaff’s rule do not apply that number of adenine units are equal to number of guanine units.

02

Step-2.: Find Chargaff’s rule implies that the sum of the purine residues equals the sum of the pyrimidine residues

In double stranded structure of DNA,

A = T = 1

G = C = 1

Percentage of C + G does not necessarily equal to percentage of A + T.

Sum of purine residues is equal to sum of pyrimidine residues. In a double stranded DNA, number of purines and pyrimidines exist in 1:1 ratio as purines bond with pyrimidines in DNA.

03

Step-3: Find Chargaff’s rule applies only to double-stranded DNA

Chargaff’s rule apply only to double-stranded DNA. For each G in a strand, there exists a complementary C in the opposing strand but there is no relation between G and C in the same strand. Complementary base pairing is not applicable in case of RNA as it is single stranded, thus Chargaff’s rule is not applied to RNA.

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Most popular questions from this chapter

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.

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.

An important protecting group developed specifically for polyhydroxy compounds like nucleosides is the tetraisopropyl-disiloxanyl group, abbreviated TIPDS, that can protect two alcohol groups in a molecule.

(a) The TIPDS group is somewhat hindered around the Siatoms by the isopropyl groups. Which OHis more likely to react first with TIPDS chloride? Show the product with the TIPDS group on one oxygen.

(b) Once the TIPDS group is attached at the first oxygen, it reaches around to the next closest oxygen. Show the final product with two oxygens protected.

(c) The unprotected hydroxy group can now undergo reactions without affecting the protected oxygens. Show the product after the protected nucleoside from (b) is treated with tosyl chloride and pyridine, followed by NaBr, ending with deprotection with Bu4NF.

Question:

a) Draw D-allose, the C3 epimer of glucose.

b) Draw D-talose, the C2 epimer of D-galactose.

c) Draw D-idose, the C3 epimer of D-talose. Now compare your answers with Figure 23-3.

d) Draw the C4 “epimer” of D-xylose. Notice that this “epimer” is actually an L-series sugar, and we have seen its enantiomer. Give the correct name for this L-series sugar.

Ruff degradation of D-arabinose gives D-erythrose. The Kiliani-Fischer synthesis converts D-erythrose to a mixture of D-arabinose and D-ribose. Draw out these reactions and give the structure of D-ribose.

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