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Chapter 10: Problem 16

What might Watson and Crick have concluded had Chargaffs data from a single source indicated the following? $$\begin{array}{cccc}\mathbf{A} & \mathbf{T} & \mathbf{G} & \mathbf{C} \\\\\% & 29 & 19 & 21 & 31\end{array}$$ Why would this conclusion be contradictory to Wilkins's and Franklin's data?

Short Answer

Expert verified
Answer: The contradiction Watson and Crick might have faced is that the given Chargaff's data did not support the base pairing rule where the amounts of A equal the amounts of T and G equal C, which is critical for the double helix structure of DNA. This was in opposition to the data from Wilkins and Franklin, which strongly supported the double helix structure with base pairs in a 1:1 proportion.

Step by step solution

01

Analyze the given Chargaff's data

Observe the given percentages of the four nucleotides: Adenine (A): 29% Thymine (T): 19% Guanine (G): 21% Cytosine (C): 31%
02

Apply Chargaff's base pairing rule

According to Chargaff's base pairing rule, the amounts of Adenine (A) should equal the amount of Thymine (T); likewise, the amounts of Guanine (G) should equal the amount of Cytosine (C). However, in the given data, A is not equal to T, and G is not equal to C: A \% = 29\% \neq 19\% = T \% G \% = 21\% \neq 31\% = C \%
03

Understand the contradiction with Wilkins's and Franklin's data

Wilkins's and Franklin's data showed X-ray diffraction patterns, which strongly supported the idea of a double helix structure of DNA with base pairs (A-T and G-C) in a 1:1 proportion. However, with the given Chargaff's data where A is not equal to T and G is not equal to C, the double helix structure would not be supported.
04

Conclusion

Watson and Crick might have concluded that Chargaff's base pairing rules were not valid for the given data, as the percentages of A and T, and G and C are unequal. This would have created a contradiction with Wilkins's and Franklin's data, which strongly supported the double helix structure of DNA with base pairs in a 1:1 proportion.

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Most popular questions from this chapter

Electrophoresis is an extremely useful procedure when applied to analysis of nucleic acids as it can resolve molecules of different sizes with relative ease and accuracy. Large molecules migrate more slowly than small molecules in agarose gels. However, the fact that nucleic acids of the same length may exist in a variety of conformations can often complicate the interpretation of electrophoretic separations. For instance, when a single species of a bacterial plasmid is isolated from cells, the individual plasmids may exist in three forms (depending on the genotype of their host and conditions of isolation): superhelical/supercoiled (form I), nicked/open circle (form \(\mathrm{II}\) ), and linear (form III). Form I is compact and very tightly coiled, with both DNA strands continuous. Form II exists as a loose circle because one of the two DNA strands has been broken, thus releasing the supercoil. All three have the same mass, but each will migrate at a different rate through a gel. Based on your understanding of gel composition and DNA migration, predict the relative rates of migration of the various DNA structures mentioned above.

In this chapter, we first focused on the information that showed DNA to be the genetic material and then discussed the structure of DNA as proposed by Watson and Crick. We concluded the chapter by describing various techniques developed to study DNA. Along the way, we found many opportunities to consider the methods and reasoning by which much of this information was acquired. From the explanations given in the chapter, what answers would you propose to the following fundamental questions: (a) How were scientists able to determine that DNA, and not some other molecule, serves as the genetic material in bacteria and bacteriophages? (b) How do we know that DNA also serves as the genetic material in eukaryotes such as humans? (c) How was it determined that the structure of DNA is a double helix with the two strands held together by hydrogen bonds formed between complementary nitrogenous bases? (d) How do we know that G pairs with C and that A pairs with \(\mathrm{T}\) as complementary base pairs are formed? (e) How do we know that repetitive DNA sequences exist in eukaryotes?

Why were \(^{32} \mathrm{P}\) and \(^{35} \mathrm{S}\) chosen for use in the Hershey-Chase experiment? Discuss the rationale and conclusions of this experiment.

If the GC content of a DNA molecule is \(60 \%\), what are the molar percentages of the four bases (G, C, T, A)?

Adenine may also be named 6 -amino purine. How would you name the other four nitrogenous bases, using this alternative system? (O is indicated by "oxy-," and CH by "methyl.")
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