/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} 16.1-3ITD : If Chargaff's rule—that the ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

: If Chargaff's rule—that the amount of A equals the amount of T and the amount of C equals the amount of G—is valid, then hypothetically, we could extrapolate this to the combined DNA of all species on Earth (like one huge Earth genome). To see whether the data in the table support this hypothesis, calculate the average percentage for each base in your completed table by averaging the values in each column. Does Chargaff's equivalence rule still hold true?

Source of DNA

Base Percentage

Adenine

Guanine

Cytosine

Thymine

Sea urchin

32.8

17.7

17.3

32.1

Salmon

29.7

20.8

20.4

29.1

Wheat

28.1

21.8

22.7

27.4

E. coli

24.7

26.0

25.3

24.0

Human

30.4

19.8

19.8

30.1

Ox

29.0

21.0

21.0

29.0

Short Answer

Expert verified

Source of DNA

Base Percentage

Adenine

Guanine

Cytosine

Thymine

Sea urchin

32.8

17.7

17.3

32.1

Salmon

29.7

20.8

20.4

29.1

Wheat

28.1

21.8

22.7

27.4

E. coli

24.7

26.0

25.3

24.0

Human

30.4

19.8

19.8

30.1

Ox

29.0

21.0

21.0

29.0

Average %

29.11

21.18

21.08

28.61

The average percentage for the bases adenine is 29.11%, guanine is 21.18%, cytosine is 21.08%, and thymine is 28.61%.

Yes, Chargaff's equivalence rule still holds for all the purines and pyrimidines average percentages. It is because the average from the table follows Chargaff's A = T and G = C equivalence relationships.

Step by step solution

01

Chargaff's equivalence rule

Chargaff's equivalence rule states that the amount of complementary purine and pyrimidine must be equal. Consequently, DNA must have a 1:1 stoichiometric ratio for complementary purines and pyrimidines.

It implies that [A]=[T], [G]=[C], A+G=C+T,A:T=1:1andG:C=1:1

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The elongation of the leading strand during DNA synthesis

  1. progress away from the replication fork.
  2. occurs in the 3’→5’direction.
  3. produces Okazaki fragments.
  4. depends on the action of DNA polymerase.

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

In his work with pneumonia-causing bacteria and mice, Griffith found that

  1. the protein coat from pathogenic cells was able to transform non-pathogenic cells.
  2. heat-killed pathogenic cells caused pneumonia.
  3. some substances from pathogenic cells were transferred to non-pathogenic cells, making them pathogenic.
  4. the polysaccharide coat of bacteria caused pneumonia.

What is the basis for the difference in how the leading and lagging strands of DNA molecules are synthesized?

  1. Origins of replication occur only at the 5’ end.
  2. Helicases and single-strand bindingproteins work at the 5’end.
  3. DNA polymerase can join new nucleotides only to the 3’end of a pre-existing strand, and the strands are antiparallel.
  4. DNA ligase works only in the 3’5’ direction.

Some bacteria may be able to respond to environmental stress by increasing the rate at which mutations occur during cell division. How might this be accomplished? Might there be an evolutionary advantage to this ability? Explain.

See all solutions

Recommended explanations on Biology Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.