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Chapter 23: Q55P (page 1255)

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?

Short Answer

Expert verified

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

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

(c) Chargaff鈥檚 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鈥檚 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鈥檚 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鈥檚 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鈥檚 rule is not applied to RNA.

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Most popular questions from this chapter

Emil Fischer synthesized L-gulose, an unusual aldohexose that reduces to give D-glucitol. Suggest a structure for this L sugar, and show how L-gulose gives the same alditol as D-glucose. (Hint: D-Glucitol has -CH2OHgroups at both ends. Either of these primary alcohol groups might have come from reduction of an aldehyde.)

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.

After a series of Kiliani鈥揊ischer syntheses on (+)-glyceraldehyde, an unknown sugar is isolated from the reaction mixture. The following experimental information is obtained:

(1) Molecular formula C6H12O6

(2) Undergoes mutarotation.

(3) Reacts with bromine water to give an aldonic acid.

(4) Reacts with HNO3 to give an optically active aldaric acid.

(5) Ruff degradation followed by HNO3 oxidation gives an optically inactive aldaric acid. (6) Two Ruff degradations followed by HNO3 oxidation give meso-tartaric acid.

(7) When the original sugar is treated with CH3I and Ag2O, a pentamethyl derivative is formed. Hydrolysis gives a tetramethyl derivative with a free hydroxy group on C5.

(a) Draw a Fischer projection for the open-chain form of this unknown sugar. Use Figure 23-3 to name the sugar.

(b) Draw the most stable conform

Predict the products obtained when D-galactose reacts with each reagent.

(a) Br2 and H2O

(b) NaOH,H2O

(c)CH3OH, H+

(d) Ag(NH3)+2OH+

(e) H2, Ni

(f) excess Ac2O and pyridine

(g) excess CH3I ,Ag2O

(h) NaBH4

(i) Br2 , H2O then H2O2 and Fe2(SO4)3

(j) (1) KCN/HCN; (2) H2 Pd/BaSO4; (3) H3O+

(k)excess HIO4

(a) Which of the D-aldopentoses will give optically active aldaric acids on oxidation with HNO3 ?

(b) Which of the D-aldotetroses will give optically active aldaric acids on oxidation withHNO3 ?

(c) Sugar X is known to be a D-aldohexose. On oxidation with HNO3 , X gives an optically inactive aldaric acid. When X is degraded to an aldopentose, oxidation of the aldopentose gives an optically active aldaric acid. Determine the structure of X.

(d) Even though sugar X gives an optically inactive aldaric acid, the pentose formed by degradation gives an optically active aldaric acid. Does this finding contradict the principle that optically inactive reagents cannot form optically active products?

(e) Show what products results if the aldopentose formed from degradation of X is further degraded to an aldotetrose. DoesHNO3 oxidize this aldotetrose to an optically active aldaric acid?
See all solutions

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