Unit 6 Objectives
CHAPTER
10
Molecular Biology of the Gene
Chapter Objectives
The Structure of the Genetic Material
10.1 Describe the experiments of Griffith, Hershey, and Chase, which supported the idea that DNA
was life’s genetic material.
 Griffith:
o Studying two strains (varieties) of a bacterium:
 A harmless strain
 A pathogenic (disease-causing) strain that causes pneumonia.
o Surprised to find that when he killed the pathogenic bacteria and then mixed the bacterial remains
with living harmless bacteria, some living bacterial cells were converted to the disease-causing
form.
o All of the descendants of the transformed bacteria inherited the newly acquired ability to cause
disease.
o Clearly, some chemical component of the dead bacteria could act as a “transforming factor” that
brought about a heritable change in live bacteria.
 Hershey and Chase:
o Preformed a very convincing set of experiments that showed DNA to be the genetic material of T2,
a virus that infects the bacterium E. coli.
o Viruses that exclusively infect bacteria are called bacteriophages (“bacteria-eaters”), or phages for
short.
o Knew that T2 could reprogram its host cell to produce new phages, but they did not know which
component – DNA or protein – was responsible for this ability
o Found answer by devising an experiment to determine what kinds of molecules the phage
transferred to E. coli during infection.
o Experiment used only a few relatively simple tools: chemicals containing radioactive isotopes; a
radioactivity detector; a kitchen blender; and a centrifuge, a device that spins test tubes to separate
particles of different weights.
o Used different radioactive isotopes to label the DNA and protein inn T2. First they grew T2 with E.
coli in a solution containing radioactive sulfur.
o Protein contains sulfur but DNA does not, so as new phages were made, the radioactive sulfur atoms
were incorporated only into the proteins of the bacteriophage.
o Grew a separate batch of phages in a solution containing radioactive phosphorus.
o Since nearly all the phage’s phosphorus is in DNA, this labeled only the phage DNA.
o This prepared them for their experiment:
 Mix radioactively labeled phages with bacteria. The phages infected the bacterial cells
 Agitate the cultures in a blender to separate the phages outside of the bacteria from the cells
and their contents.
 Centrifuged the mixture so that the bacteria formed a pellet at the bottom of the test tube.
 Finally they measured the radioactivity in the pellet and in the liquid.
o
Found that when the bacteria had been infected with T2 phages containing labeled protein, the
radioactivity ended up mainly in the liquid within the centrifuge tube, which contained phages but
not bacteria.
o Result suggested that the phage protein didn’t enter the cells
o When the bacteria had been infected with phages whose DNA was tagged, then most of the
radioactivity was in the pellet of bacterial cells at the bottom of the centrifuge tube.
o When bacteria were returned to liquid growth medium, they soon lysed or broke open, releasing new
phages with radioactive phosphorus in their DNA but no radioactive sulfur in their proteins.
o The replication cycle of phage was originally outlined by Hershey and Chase:
 Attaches to the host bacterial cell
 Injects its DNA into the host
 The phage DNA directs the host cell to make more phage DNA and proteins; new phages
assemble.
 The cell lyses and releases the new phages.
10.2–10.3 Compare the structures of DNA and RNA.
 Structure of DNA:
o Nucleic acids, consists of long chains (polymers) of chemical units (monomers) called nucleotides.
o Double helix
o Polynucleotide, a nucleotide polymer (chain).
o Each type of DNA nucleotide has a different nitrogen-containing base:
 Adenine (A)
 Cytosine (C)
 Thymine (T)
 Guanine (G)
o Polynucleotides vary in length from long to very long, causing the number of possible
polynucleotides to be enormous
o The nucleotides are joined to one another by covalent bonds between the sugar of one nucleotide
and the phosphate of the next
o The nucleotides are joined to one another by covalent bonds between the sugar of one nucleotide
and the phosphate of the next. This results in a sugar-phosphate backbone, a repeating pattern of
sugar-phosphate-sugar-phosphate.
o The nitrogenous bases are arranged like ribs that project from the backbone.
o The chemical structure of a single nucleotide’s three components”
 The phosphate group has a phosphorus atom (P) at its center and is the source of the word
acid in nucleic acid.
 The sugar has five carbon atoms (four in its ring and one extending above the ring)
 The ring also includes an oxygen atom.
 The sugar is called deoxyribose because, compared with the sugar ribose, it is missing an
oxygen atom.
 Structure of RNA:
o Its sugar is ribose rather than deoxyribose.
o Has an –OH group attached to the C atom at its lower-right corner in its sugar ring
o Has Uracil (U) instead of thymine (T).
10.3 Explain how Chargaff’s rules relate to the structure of DNA.
 His experiments showed that the amount of A=T and C=G in terms of types of DNA.
 The amount of A always equaled that of T
 The amount of C always equaled that of G.
DNA Replication
10.4 Explain how the structure of DNA facilitates its replication.
 Semiconservative model:
o
o
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Half of the parental molecule is maintained (conserved) in each daughter molecule.
The idea that there is specific pairing of bases in DNA was the flash of inspiration that led Watson
and Crick to the correct structure of the double helix
o The specific pairing of complementary bases connects the Daughter strands
10.5 Describe the process of DNA replication. Describe the mechanisms that correct errors caused by
environmental damage or errors from replication.
Begins at special sites called origins of replication (short stretches of DNA having a specific sequence of
nucleotides where proteins attach to the DNA and separate the stands).
Then proceeds in both directions, creating replication “bubble.”
The parental DNA strands (blur) open up as daughter strands (gray) elongate on both sides of each buble.
The DNA molecule of a eukaryotic chromosome has many origins where replication can start
simultaneously.
Thus, hundreds or thousands of bubbles can be present at once, shortening the total time needed for
replication.
Eventually, all the bubbles fuse, yielding two completed daughter DNA molecules.
The DNA ligase is an enzyme that links, or ligates, the Okazaki fragments (new strands of synthesized
daughter strands which can be synthesized in one continuous piece by a DNA polymerase working toward
the forking point of the parental DNA).
DNA polymerases and DNA ligase are also involved in repairing DNA damaged by harmful radiation, such
as ultraviolet light and X-rays, or toxic chemicals in the environment, such as those found in tobacco smoke.
The Flow of Genetic Information from DNA to RNA to Protein
10.6 Describe the locations, reactants, and products of transcription and translation.
10.7–10.8 Explain how the “languages” of DNA and RNA are used to produce
polypeptides.
10.9 Explain how mRNA is produced using DNA.
 mRNA is produced using DNA by an RNA molecule being transcribed form a DNA template by a process
that resembles the synthesis of a DNA strand during DNA replication.
 The two NDA strands must first separate at the place where the process will start.
 Next, only one of the DNA strands serves as a template for the newly forming RNA molecule; the other
strand is unused.
 RNA nucleotides follow the same base-pairing rules that govern DNA replication, except that U, rather than
T, pairs with A,
 The RNA nucleotides are linked by the transcription enzyme RNA polymerase
 The “start transcribing” signal is a nucleotide sequence called a promoter (a specific binding site for RNA
polymerase and determines which of the two strands of the DNA double helix is used as the template in
transcription).
 Three stages of transcription:
o Called initiation; the RNA polymerase attaches to the promoter and the start of RNA synthesis.
o Elongation; the RNA GROWS LONGER. As RNA synthesis continues, the RNA strand peels away
from its DNA template, allowing the two separated DNA strands to come back together in the
region already transcribed.
o Termination; the RNA polymerase reaches a sequence of bases in the DNA template called a
terminator. This sequence signals the end of the gene; at that point, the polymerase molecules
detaches from the RNA molecule and the gene.
 Occurs in the nucleus
10.10 Explain how eukaryotic RNA is processed before leaving the nucleus.
 Called mRNA because it conveys genetic messages from DNA to the translation machinery of the cell.
 Information then translated into polypeptides.
 mRNA molecules must exit the nucleus via the nuclear pores and enter the cytoplasm, where the machinery
for polypeptide synthesis is located.
 Before leaving, eukaryotic transcripts are modified, or processed, in several ways:
o One kind of RNA processing in the addition of extra nucleotides to the ends of the RNA transcript
10.11 Relate the structure of tRNA to its functions in the process of translation.
 It is up to the cell’s molecular interpreters, tRNA molecules, to match amino acids to the appropriate codons
to form the new polypeptide.
 To preform this task, tRNA molecules must carry out two functions:
o Picking up the appropriate amino acids
o Recognizing the appropriate codons in the mRNA.
 The unique structure of rRNA molecules enables them to preform both tasks.
 tRNA molecule is made of a single strand of RNA – one polynucleotide chain – consisting of about 80
nucleotides.
 By twisting and folding upon itself, tRNA forms several double stranded regions in which short stretches of
RNA base pair with other stretches via hydrogen bonds.
 A single-stranded loop at one end of the folded molecule contains a special triplet of bases called an
anticodon.
 The anticodon of tRNA recognizes a particular codon on mRNA by using base-pairing rules.
 At the other end of the tRNA molecule is a site where one specific kind of amino acid can attach.
 X
 One end has amino acid
 Other end has the ant-codon
10.12 Describe the structure and function of ribosomes.
 Ribosomes: structures in the cytoplasm that position mRNA and tRNA close together and catalyze the
synthesis of polypeptides.
o Consists of two subunits, each made up of proteins and a kind of RNA called ribosomal RNA
(rRNA)
 In function:
o Similar in prodaryotes and eukaryotes
o But those of eukaryotes are slightly larger and different in composition.
o The differences are medically significant
o Certain antibiotic drugs can inactivate prokaryotic ribosomes while leacing eukaryotic ribososmes
unaffected.
10.13 Explain how translation begins.
 Translation begins with the stage of initiation.
 This process of polypeptide initiation brings together the mRNA, a tRNA bearing the first amino acid, and the
two subunits of a ribosome.
 Establishes exactly where translation will begin, ensuring that the mRNA codons are translated into the
correct sequence of amino acids.
 Initiation occurs in two steps:
o An mRNA molecule binds to a small ribosomal subunit. A special initiator tRNA binds to the
specific codon, called the start codon, where translation is to begin on the mRNA molecule. The
initiator tRNA carries the amino acid methionine (Met); its anticodon, UAC, binds to the start
codon, AUG.
o A large ribosomal subunit binds to the small one, creating a functional ribosome. The initiator
tRNA fits into one of the two binding sites on the ribosome. This site, called the P site, will hold
the growing polypeptide. The other tRNA binding site, called the A site , is vacant and ready for the
next amino-acid-bearing tRNA.
10.14 Describe the step-by-step process by which amino acids are added to a growing polypeptide
chain.
 Codon recognition:
o The anticodon of an incoming tRNA molecule, carrying its amino acid, pairs with the mRNA codon
in the A site of the ribosome.
 Peptide bond formation:
o The polypeptide separates from the tRNA in the P site and attaches by a new peptide bond to the
amino acid carried by the tRNA in the A site. The ribosome catalyzes formation of the peptide
bond, adding one more amino acid to the growing polypeptide chain
 Translocation:
o The P site tRNA now leaves the ribosome, and the ribosome tranclocates (moves) the remaining
tRNA in the A site, with the growing polypeptide, to the P site. The codon and anticodon remain
hydrogen bonded, and the mRNA and tRNA move as a unit. This movement brings into the A site
the next mRNA codon to be translated, and the process can start again with step 1.
10.15 Diagram the overall process of transcription and translation.
 Look at power point packet given during class.
10.16 Describe the major types of mutations, causes of mutations, and potential
consequences.
 Unregulated alterations in DNA that may or may not affect the phenotype of an individual. Proteins encoded
by these mutated genes are also altered and may fail to function properly.
 Caused by errors in Reproduction/Crossing over or Mutagen
 Examples of Mutagen (Causes)
o U.V. Gamma
o X-Rays
o Smoke, smog, chemical
 Point mutations
o Base pair substitution – Swap 1 base for another
 Missense – Change amino acid by changing one letter
 Nonsense – Causes an early stop codon
 Silent – No change in Amino Acid
o Insertion/Deletion – adding or subtracting bases
 Frameshift – Everything “downstream” gets shifted
Microbial Genetics
10.17 Compare the lytic and lysogenic reproductive cycles of a phage.
10.18 Compare the structures and reproductive cycles of the mumps virus and a
herpesvirus.
10.19 Describe three processes that contribute to the emergence of viral disease and note examples of
each.
10.19 Explain why RNA viruses tend to have an unusually high rate of mutation.
10.20 Explain how the AIDS virus enters a host cell and reproduces.
10.21 Describe the structure of viroids and prions and explain how they cause disease.
10.22 Define and compare the processes of transformation, transduction, and
conjugation.
10.23 Describe the roles of bacterial F factors. Define a plasmid and explain why R plasmids pose
serious human health problems.
The molecule of Heredity
Griffith’s Experiments
Chargaff
Watson/Crick
Found the double helix structure of the DNA via putting together work from other scientists
 Mutation: any change in the nucleotide sequence of DNA
 Mutations within a gene can be divided into two general categories:
o Nucleotide substitutions
o Nucleotide insertions or deletions
 Nucleotide substitutions: the replacement of one nucleotide and its base-pairing partner with another pair of
nucleotides.
o Example: If A replaces G in the fourth codon of the mRNA.
o Effects:
 Because the genetic code is redundant, some substitution mutations have no effect at all.
This is called a silent mutation
 Missense mutations: does change amino acid coding
 Example: if a mutation causes an mRNA codon to change from GGC to AGC, the
resulting protein will have a serine (Ser) instead of a glycine (Gly) at this position.
 Some have little to no effect on the shape or function of the resulting protein, but
others, as in the case of sickle-cell disease, prevent the protein from performing its
normal function.
 Nucleotide substitution: leads to an improved protein that enhances the success of the
mutant organism and its descendants.
 Introns: The things that need to be removed from pre-mRNA
Big ideas:
DNA replication
DNA structure
Transcription
Translation