DNA & RNA

The Central Dogma

*Enduring understanding 3.A: Heritable information provides for continuity of life.

-Essential knowledge 3.A.1: DNA, and in some cases RNA, is the primary source of heritable information.

a. Genetic information is transmitted from one generation to the next through DNA or RNA. Evidence of student learning is a demonstrated understanding of each of the following:

• 1. Genetic information is stored in and passed to subsequent generations through DNA molecules and, in some cases, RNA molecules.
• 2. Noneukaryotic organisms have circular chromosomes, while eukaryotic organisms have multiple linear chromosomes, although in biology there are exceptions to this rule.
• 3. Prokaryotes, viruses and eukaryotes can contain plasmids, which are small extra-chromosomal, double-stranded circular DNA molecules.
• 4. The proof that DNA is the carrier of genetic information involved a number of important historical experiments. These include:

i. Contributions of Watson, Crick, Wilkins, and Franklin on the structure of DNA

ii. Avery-MacLeod-McCarty experiments

iii. Hershey-Chase experiment

• 5. DNA replication ensures continuity of hereditary information.

i. Replication is a semiconservative process; that is, one strand serves as the template for a new, complementary strand.

ii. Replication requires DNA polymerase plus many other essential cellular enzymes, occurs bidirectionally, and differs in the production of the leading and lagging strands.

• 6. Genetic information in retroviruses is a special case and has an alternate flow of information: from RNA to DNA, made possible by reverse transcriptase, an enzyme that copies the viral RNA genome into DNA. This DNA integrates into the host genome and becomes transcribed and translated for the assembly of new viral progeny. [See also 3.C.3]
• b. DNA and RNA molecules have structural similarities and differences that define function. [See also 4.A.1] Evidence of student learning is a demonstrated understanding of each of the following:
  • 1. Both have three components — sugar, phosphate and a nitrogenous base — which form nucleotide units that are connected by covalent bonds to form a linear molecule with 3' and 5' ends, with the nitrogenous bases perpendicular to the sugar-phosphate backbone.
  • 2. The basic structural differences include:
    • i. DNA contains deoxyribose (RNA contains ribose).
    • ii. RNA contains uracil in lieu of thymine in DNA.
    • iii. DNA is usually double stranded, RNA is usually single stranded.
    • iv. The two DNA strands in double-stranded DNA are antiparallel in directionality.
  • 3. Both DNA and RNA exhibit specific nucleotide base pairing that is conserved through evolution: adenine pairs with thymine or uracil (A-T or A-U) and cytosine pairs with guanine (C-G). i. Purines (G and A) have a double ring structure. ii. Pyrimidines (C, T and U) have a single ring structure.
  • 4. The sequence of the RNA bases, together with the structure of the RNA molecule, determines RNA function.
    • i. mRNA carries information from the DNA to the ribosome.
    • ii. tRNA molecules bind specific amino acids and allow information in the mRNA to be translated to a linear peptide sequence.
    • iii. rRNA molecules are functional building blocks of ribosomes.
    • iv. The role of RNAi includes regulation of gene expression at the level of mRNA transcription.
• c. Genetic information flows from a sequence of nucleotides in a gene to a sequence of amino acids in a protein. Evidence of student learning is a demonstrated understanding of each of the following:
  • 1. The enzyme RNA-polymerase reads the DNA molecule in the 3' to 5' direction and synthesizes complementary mRNA molecules that determine the order of amino acids in the polypeptide.
  • 2. In eukaryotic cells the mRNA transcript undergoes a series of enzyme-regulated modifications.
  • 3. Translation of the mRNA occurs in the cytoplasm on the ribosome.
  • 4. In prokaryotic organisms, transcription is coupled to translation of the message. Translation involves energy and many steps, including initiation, elongation and termination.
    • i. The mRNA interacts with the rRNA of the ribosome to initiate translation at the (start) codon.
    • ii. The sequence of nucleotides on the mRNA is read in triplets called codons.
    • iii. Each codon encodes a specific amino acid, which can be deduced by using a genetic code chart. Many amino acids have more than one codon.
    • iv. tRNA brings the correct amino acid to the correct place on the mRNA.
    • v. The amino acid is transferred to the growing peptide chain.
    • vi. The process continues along the mRNA until a “stop” codon is reached.
    • vii. The process terminates by release of the newly synthesized peptide/protein.
• d. Phenotypes are determined through protein activities.
• e. Genetic engineering techniques can manipulate the heritable information of DNA and, in special cases, RNA.
• f. Illustrative examples of products of genetic engineering.
  • Learning Objectives:
    • LO 3.1 The student is able to construct scientific explanations that use the structures and mechanisms of DNA and RNA to support the claim that DNA and, in some cases, that RNA are the primary sources of heritable information. [See SP 6.5]
    • LO 3.2 The student is able to justify the selection of data from historical investigations that support the claim that DNA is the source of heritable information. [See SP 4.1]
    • LO 3.3 The student is able to describe representations and models that illustrate how genetic information is copied for transmission between generations. [See SP 1.2]
    • LO 3.4 The student is able to describe representations and models illustrating how genetic information is translated into polypeptides. [See SP 1.2]
    • LO 3.5 The student can justify the claim that humans can manipulate heritable information by identifying at least two commonly used technologies. [See SP 6.4]
    • LO 3.6 The student can predict how a change in a specific DNA or RNA sequence can result in changes in gene expression. [See SP 6.4]