3.3 & 7.1 DNA Structure

01/11/2012 § 1 Comment

I know we’re not exactly done with the previous unit (Molecules!) but here we go with our brand new unit: DNA. I actually love anything that has to do with DNA because it all just makes perfect, logical sense. But then I looked at the essay questions.

Spoilers: there are seventeen. SEVENTEEN.

First of all, DNA is a polymer of nucleotides, which makes it a molecule constructed out of relatively simple subunits. The structure of DNA consists of a sugar, a phosphate group, and a (nitrogenous) base. DNA (or deoxyribonucleic acid) is an example of a nucleic acid, which is what a polymer of nucleotides is. RNA, or ribonucleic acid is another example of a nucleic acid. The difference between the two is that the sugar for DNA is deoxyribose while the sugar for RNA is simply ribose. Diagrams of DNA structure are usually shown as pentagons (for the sugar), circles (for the phosphate group) and rectangles (for the bases).

The structure of DNA involves alternating the sugar molecule and the phosphate group that become the backbone for the nitrogenous bases. These backbones are single strands and DNA consists of two strands that are held together with hydrogen bonds (so slightly weak but strong enough) and arranged in an anti-parallel fashion.

There are four nitrogenous bases that make up DNA. These are adenine, cytosine, guanine, and thymine, more informally known as A, C, G and T. There are actually two kinds of bases, purines and pyrimidines. The purines are adenine and guanine. The pyrimidines are cytosine and thymine. A purine is always partnered (through hydrogen bonds) with a pyrimidine, but more specifically:

  • adenine (a purine) is always with thymine (a pyrimidine)
  • guanine (a purine) is always with cytosine (a pyrimidine)

The two strands are complementary because of this rule that A is always with T and G is always with C. When the DNA needs to be replicated, the pattern will always be the same for the newly replicated strand of DNA. The sequence on one DNA strand will determine the other strand, and though they’re different, the coding is exactly the same – more on that later.

Momentary Historical Flashback: Scientists Rosalind Franklin (of King’s College, London), and James Watson and Francis Crick (of Cambridge University) discovered evidence that DNA takes up the structure of a double helix.

The discovery of DNA’s structure also led to other deductions:

  • the two nucleotide polymers coil around each other into the double helix
  • the strands are directed at a 3′ to 5′ and 5′ to 3′ direction – this has to do with the type of carbon molecule that ends the strand.

Remember in one of our earlier units between eukaryotes and prokaryotes when we learned that the DNA of prokaryotes are naked while the DNA of eukaryotes aren’t? Or, in more specific terms, the DNA of eukaryotes are associated with proteins while those of prokaryotes are not? Yeeeaaaahh.

These proteins are called histones. DNA wraps around nucleosomes, which are groups of eight histones that form an octomer (monomer, dimer, polymer, octomer, you catch my drift?). These nucleosomes are what gives DNA the image of a string of beads under an electron micrograph. The levels of organization within chromosomes can be seen in the image below.

Nucleosomes have two main functions:

  1. they supercoil DNA and are what make chromosomes shorter and fatter to prepare for mitosis and meiosis
  2. they regulate transcription, which is the expression of genes within a cell

Essay Questions

  1. Most of the DNA of a human cell is contained in the nucleus. Distinguish between unique and highly repetitive sequences in nuclear DNA. (5)
  2. Draw a labelled diagram to show four DNA nucleotides, each with a different base, linked together in two strands. (5)
  3. Explain the structure of the DNA double helix, including its subunits and the way in which they are bonded together. (8)
  4. Outline the structure of the nucleosomes in eukaryotic chromosomes. (4)
  5. State a role for each of four different named enzymes in DNA replication. (6)
  6. Explain the process of DNA replication. (8)
  7. Explain how the process of DNA replication depends on the structure of DNA. (9)
  8. Describe the genetic code. (6)
  9. Discuss the relationship between genes and polypeptides. (5)
  10. Explain briefly the advantages and disadvantages of the universality of the genetic code to humans. (4)
  11. Compare the processes of DNA replication and transcription. (9)
  12. Distinguish between RNA and DNA. (3)
  13. Describe the roles of mRNA, tRNA and ribosomes in translation. (6)
  14. Outline the structure of tRNA. (5)
  15. Outline the structure of a ribosome. (4)
  16. Explain the process of translation. (9)
  17. Compare DNA transcription with translation. (4)

DATA BASED QUESTIONS

Page 58, Chargaff’s data

1. Compare the base composition of Mycobacterium tuberculosis (a prokaryote) with the base composition of the eukaryotes shown in the table.

The contents of adenine and thymine (15.1 and 14.6 respectively) in Mycobacterium tuberculosis are less so than the contents of the eukaryotes in the graph (which are all at least above 27 – for both adenine and thymine). The contents of guanine and cytosine (34.9 and 35.4 respectively) in Mycobacterium tuberculosis are more so than the contents of the eukaryotes in the graph (which are all at least below 23 – for both guanine and cytosine). The contents are opposites of each other.

2. Calculate the base ratio A+G/T+C, for humans and for Mycobacterium tuberculosis. Show your working.

A + G = 15.1 + 34.9 = 50

T + C = 35.4 + 14.6 = 50

A+G/T+C = 50/50 = 1/1 = 1

3. Evaluate the claim that in the DNA of eukaryotes and prokaryotes the amount of adenine and thymine are equal and the amounts of guanine and cytosine are equal.

This statement is false because Mycobacterium tuberculosis (as stated in question #1) clearly has less adenine and thymine than the eukaryotes in this set of data. Also, Mycobacterium tuberculosis has more guanine and cytosine than any of the eukaryotes in the data. The amount of adenine and thymine, guanine and cytosine are not equal in eukaryotes and prokaryotes.

4. Explain the ratios between the amounts of bases in eukaryotes and prokaryotes in terms of the structure of DNA.

Eukaryotes – since the amount of adenine and thymine are larger so than the amount of guanine and cytosine, the structure of DNA of eukaryotes have more nitrogenous bases of adenine and thymine than bases of guanine and cytosine. 

Prokaryotes – since the amount of guanine and cytosine are larger so than the amount of adenine and thymine (seen in Mycobacterium tuberculosis), then the structure of DNA in prokaryotes holds more guanine and cytosine nitrogenous bases than it holds adenine and thymine nitrogenous bases.

5. Suggest reasons for the difference in the base composition of bacteriophage T2 and the polio virus.

Bacteriophage T2 and polio viruses are two different viruses that do different things. T2 causes whatever disease or infection it causes while the polio virus obviously causes polio. Because they have different purposes, they require a different base composition in their DNA. Other than their purpose, they require a different DNA composition to build their structure.

§ One Response to 3.3 & 7.1 DNA Structure

  • cafergy says:

    Grade 5 A consistent and thorough understanding of the required knowledge and skills, and the ability to apply them in a variety of situations.

    DBQ p. 58, Chargaff’s Data
    1. The quantities of the 4 bases are similar across all of the eukaryotes:
    • The relative quantities of bases in Mycobacterium are distinct from the eukaryotes
    • Mycobacterium has less adenine and thymine, but more guanine and cytosine
    • The amount of adenine approximately equals the amount of thymine in both
    • The amount of guanine is approximately equal to the amount of cytosine in both

    2. 1.00 in both cases

    3. Within experimental error, the data supports the hypothesis.

    4. Complementary base pairing between A and T would mean that they would need to be present in equal quantities; same argument for C and G

    5. • Polio virus is a single-stranded RNA virus
    • Bacteriophage T2 is a double-stranded DNA virus

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