Explain how the sea urchin and salmon data demonstrate both of Chargaff’s rules.

Short Answer

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Both of Chargaff's rules are supported by data on sea urchins and salmon. It is because adenine (A) and thymine (T), and guanine (G) and cytosine (C) are present in equal amounts.

It shows equality in quantity and 1:1 ratio between the bases A and T and between the bases G and C in sea urchin and salmon.

Step by step solution

01

Chargaff’s rule

Chargaff’s rule states that there is equality in the number of complementary bases. It implies that an equal quantity of bases adenine(A) and thymine(T) exists. Similarly, an equal quantity of bases guanine(G) and cytosine(C) exists.

As the quantity of purine is equal to pyrimidine, the stoichiometric ratio between them should be 1:1. It implies that A:T or G:C should be 1:1.

02

Sea urchin demonstrates Chargaff’s rules

In sea urchin, the base percentage of adenine and thymine are given as 32.8 and 32.1, respectively. The base percentage of guanine and cytosine are given as 17.7 and 17.3, respectively. It shows that adenine and thymine amounts and guanine and cytosine amounts are close and almost equal.

The ratio of adenine and thymine, which is 32.8:32.1, is approximately 1:1. The ratio of guanine and cytosine, which is 17.7:17.3, is approximately 1:1.

03

Salmon demonstrates Chargaff’s rules 

In salmon, the base percentage of adenine and thymine are given as 29.7 and 29.1, respectively. The base percentage of guanine and cytosine are given as 20.8 and 20.4, respectively. It shows that adenine equals thymine amounts and guanine equals cytosine amounts.

The ratio of adenine and thymine, which is 29.7:29.1, is approximately 1:1. The ratio of guanine and cytosine, which is 20.8:20.4, is approximately 1:1.

Therefore, the base percentages in sea urchin and salmon satisfy Chargaff’s rules: the amount of purine and pyrimidine are equal, and the complementary bases have a 1:1 stoichiometric ratio.

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Most popular questions from this chapter

A biochemist isolates, purifies, and combines in a test tube a variety of molecules needed for DNA replication. When she adds some DNA to the mixture, replication occurs, but each DNA molecule consists of a normal strand paired with numerous segments of DNA a few hundred nucleotides long. What has she probably left out of the mixture?

  1. DNA polymerase
  2. DNA ligase
  3. Okazaki fragments
  4. primase

In a nucleosome, the DNA is wrapped around

  1. histones
  2. ribosomes
  3. polymerase molecules
  4. a thymine dimer

E.coli grown on a 15N medium is transferred to a 14N medium and allowed to grow for two more generations (two rounds of DNA replication). DNA extracted from these cells is centrifuged. What density distribution of DNA would you expect in this experiment?

  1. one high-density and one low-density band
  2. one intermediate density band
  3. one high density and one intermediate-density band
  4. one low-density and one intermediate-density band

What two properties, one structural and one functional, distinguish heterochromatin from euchromatin?

Model building can be an important part of the scientific process. The illustration shown above is a computer-generated model of a DNA replication complex. The parental and newly synthesized DNA strands are colour coded differently, as are each of the following three proteins DNA pol III, the sliding clamp, and single-stranded binding protein.

  1. Using what you've learned in this chapter to clarify this model, label each DNA strand and protein.
  2. Draw an arrow to indicate the overall direction of DNA replication.
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