Chapter 16:

The Molecular Basis of Inheritance

• Mendel discovered the existence of heritable factors…

• Just about 60 years ago biologists had a dilemma: Where were the genes located?

• T.H. Morgan’s group showed that genes are located on chromosomes.

• This led to the next question: Which one, protein or DNA, is the “genetic material”?

“Little was known abut nucleic acids, whose physical and chemical properties seemed far too uniform to account for the multitude of specific inherited traits exhibited by every organism.”

Appropriate experimental organisms:Viruses and Bacteria

Scientists Who Contributed to the Discovery of DNA as The Unit of Heredity:

1) Frederick Griffith

2) Avery, MacLeod & McCarty

3) Hershey and Chase

4) Erwin Chargaff

5) Rosiland Franklin

6) James Watson and Francis Crick

1) Frederick Griffith 1928 Transformation, the change in a genotype and phenotype due to the assimilation of external material by a cell, is possible.

• Studied Streptococcus pneumoniae, bacterium• 2 varieties: pathogenic S (smooth) or harmless R (rough)• Experiment:

1) kill pathogenic strain with heat (inject in mouse)2) mix dead pathogenic bacteria with harmless live bacteria.

(then inject in mouse)

Figure 16.1 Transformation of bacteria

•Results: some bacteria became pathogenic(injected mice with the bacterial strain- they died)•Transformation, the change in a genotype and phenotype due to the assimilation of external material by a cell, is possible.

2) Avery, MacLeod and McCarty The transforming agent was DNA.

Attempted successfully to pinpoint the transforming agent.

Experiment:1. Purified various chemicals from

the heat-killed pathogenic bacteria.

2. tried to transform live “harmless” or non-pathogenic bacteria with each chemical.

• Result: only DNA worked.• The transforming agent was

DNA.

3) Alfred Hershey & Martha ChaseDNA is a virus’ genetic material

• 1952 used bacteria infecting viruses• bacteriophages “bacteria eaters” (virus) = DNA + protein• Experiment to show that it was DNA not protein that actually entered

the bacterium- DNA was virus’ genetic material (viral DNA can program cells).

1. Tag DNA and protein with different radioactive isotopes then centrifuge cells- DNA is in the pellet, the protein coat is in the supernatant (liquid above the pellet.)

• Infect E. coli w/ radioactive-sulfur (35S protein) grown viruses. Centrifuge cells. Radioactivity in supernatant.

• Infect E. coli w/ radioactive-phosphorus (32P DNA) grown viruses. Centrifuge cells. Radioactivity in pellet.

• Pellet = Cells , supernatant = outside of cells

Bacteriophage reproduction

• DNA enters cells

• Protein remains outside.

• Phages act like a hypodermic needle- inject DNA into their host cell.

Figure 16.2a The Hershey-Chase experiment: phages

Conclusion:DNA (not

protein) functions as

the T2 phage’s genetic

material.

4) Erwin Chargaffin DNA, the nucleotides A & T exist in equal proportions; as do C and G

• 1947 biochemist who analyzed the base composition of DNA from different organisms.

• DNA composition varies from one species to another:• In any one species, the amounts of the four nitrogenous

bases are not all equal but are present in a characteristic ratio.

• A = T (approximately) and G = C Chargaff’s rules

Chargaff’s rules: A=TG=C

This is how we knowThe base pair rules.

5) Rosiland Franklinshape of DNA is twisted

• X-ray diffraction/ X-ray crystallography• Basically took x-ray picture of DNA- based on the

“shadow” it cast.• Worked down the hall from Watson & Crick, who

were also studying the structure of DNA.• Victim of sexism… pioneer of her time.• Nobel Prize can only go to three people… the

structure of DNA is credited to Watson, Crick, & Wilkins.

Figure 16.4 Rosalind Franklin and her X-ray diffraction photo of DNA

6) James Watson and Francis CrickDNA is double helix

• American (Watson) and Englishman (Crick)• Interpret Franklin’s pictures of DNA• DNA helical in shape• Able to calculate width of helix and spacing of

nitrogenous bases along it.• Double helix• Explained Chargaff’s rules.

Figure 16.0 Watson and Crick

Figure 16.0x James Watson

The Molecular Structure of

DNA• Backbone: sugar-phosphate

• In a 3’ to 5’ and 5’ to 3’ direction

• Note the numbering of the sugar!

• 2 types of Nitrogenous Bases:

• Purines:

Adenine & Guanine

• Pyrimindines:

Thymine & Cytosine

• Base Pairing: A-T & G-C

G & C look like eachother

Unnumbered Figure (page 292) Purine and pyridimine

1/2 of the DNA Double Helix1/2 of the Twisted Ladder

1)What are the uprights of the ladder?2)What are the rungs of the ladder?3)What would the other side look like?4)What kinds of bonds hold sugars to phosphates?5)What kinds of bonds hold nitrogenous bases together?6)Which side runs 3’ to 5’?

1/2 of the DNA Double Helix1/2 of the Twisted Ladder

1) Uprights = sugar-phosphates2) Rungs = bases3) A-T-G-C (inverted sugar-phosphates)4) Covalent bonds5) Hydrogen bonds (2 AT & 3 CG)6) Right side- 5’ = phosphate group

Strands are “anti-parallel”

Figure 16.5 The double helix

Quiz Time

• What holds the two strands of DNA together?

• HYDROGEN BONDS (2 AT or 3 GC)• What shape is the DNA molecule?• DOUBLE HELIX• When did Watson and Crick complete their

model?• FEBRUARY 28th 1953 @ 7:07am

Figure 16.8 Three alternative models of DNA replication

Semiconservative Model- two strands of the parental molecule separate,and each functions as a template for synthesis of a new complementary strand.

DNA REPLICATION• What is a template? 1/2 DNA Double Helix… a single strand of

DNA.• What was Watson and Crick’s template theory? DNA is a pair of

templates complementary to each other. Implies a copying mechanism: each new strand is 1/2 original template and 1/2 new complementary strand.

• What is DNA polymerase? The enzyme that connects a nucleoside-tri-phosphate to a DNA template nucleotide according to base-pair rules.

• Each step of DNA replication is managed by a specific enzyme.• What fuels this process? Phosphates from the nucleoside-tri-

phosphate.

Figure 16.11 Incorporation of a nucleotide into a DNA strand

Figure 16.7 A model for DNA replication: the basic concept (Layer 1)

Figure 16.7 A model for DNA replication: the basic concept (Layer 2)

HELICASE ENZYME splits the DNA double helix

Figure 16.7 A model for DNA replication: the basic concept (Layer 3)

DNA POLYMERASE attracts matching nucleotides and covalentlybonds them to the free 3’ end of the existing strand.Also “proof-reads” for mistakes.

Figure 16.7 A model for DNA replication: the basic concept (Layer 4)

PROKARYOTIC REPLICATION• Prokaryotic chromosomes are single, circular,

and about 5 million base pairs.• Replication begins at the origin of replication.• The direction of replication is 3’ to 5’ along the

template strand… new DNA is created 5’ to 3’.• Replication can be as fast as one hour- some

quicker than 20 minutes.• Ex. A staph infection is deadly because

doubling occurs every 20 minutes.

EUKARYOTIC REPLICATION• Eukaryotic chromosomes are LINEAR• Replication begins at the originS of

replication (multiple spots)• The direction of replication is 3’ to 5’ along

the template strand (so new strand is “elongated” 5’ to 3’)

• Replication can be as fast as a few hours in humans- even though our 6 billion base pairs is over a 1000x more DNA than in bacteria.

• Because eukaryotic chromosomes are larger, replication begins at many sites along the giant DNA molecule of each chromosome.

Figure 16.10 Origins of replication in eukaryotes

1. Helicase- unzips the DNA at an “origin of replication”2. Single Stranded Binding Proteins- keep strands apart3. Primase attracts RNA nucleotides to form RNA primer4. DNA Polymerase attracts DNA nucleosides to build the

new strand and replaces the primer with DNA.5. DNA ligase joins Okazaki fragments on the lagging

strand

LEADING STRAND is built by the continuous addition of nucleotides along the 3’-5’template toward the replication fork. Elongation occurs5’ to 3’ because DNA polymerase can only bondNucleotides to the 3’ end of a nucleotide.

LAGGING STRAND is built by the fragmented addition of nucleotides along the 5’-3’ template strand away from the replication fork. Sincepolymerase attaches to the template on the 3’ end and moves towardthe 5’ end, elongation occurs in “spurts” & creates Okazaki fragments.

• DNA strands are anti-parallel

5’------------------------------------------3’

3’------------------------------------------5’

• 5’ end is the phosphate end

• 3’ end is the hydroxyl end.

• DNA POLYMERASE attaches to the template on the 3’ end and moves toward the 5’ end.

• ELONGATION of the new strand occurs in the 5’ to 3’ direction because DNA polymerase can only add nucleotides to the free 3’ end of an existing nucleotide… this is why PRIMER must be laid down first.

1) getting started: origins of replication• At the origin(s) of replication, proteins

attach to the DNA separating the two strands with HELICASE and holding them apart with SINGLE STRANDED BINDING PROTEINS opening up one or many “replication bubble (s)”.

• At each end of a replication bubble is a replication fork a Y-shaped region where the new strands of DNA are elongating.

• Replication proceeds in both directions along the template from it’s 3’ to 5’ end.

Figure 16.14 Priming DNA synthesis with RNA

2)Elongating a new DNA strand:• PROBLEM!!! DNA polymerase can only add

to a started chain… so how do we start the new chain?

• PRIMASE adds RNA nucleotides to create the beginning of the complementary new strand = PRIMER (a stretch of about 10 nucleotides in Eukaryotes)

• Next, elongation of new DNA at a replication fork is catalyzed by enzymes called DNA POLYMERASES.

• DNA POLYMERASE covalently bonds the phosphate of the new nucleotide to the 3’ sugar of the next nucleotide.

• A different DNA polymerase switches DNA nucleotides for the primer.

HOW LONG DOES IT TAKE?

• 500 nucleotides /second in bacterial cells

• 50/second in human cells

• SOURCE OF ENERGY??? They are actually A,T, G, C nucleoside tri-phosphates not nucleotides.

Figure 16.13 Synthesis of leading and lagging strands during DNA replication

REPLICATION PROBLEM: telomeres

Telomeres are the ends of linearChromosomes.

There is a problem with DNAreplication where thedaughter molecules of DNA get shorter and shorter each time the DNA is copied.

Some cells are “immortal” andcarry an enzyme called TELOMERASE which solvesthe “end replication problem”.

Figure 16.19b Telomeres and telomerase

The 5’ end of the Dna has a problemBeing replicated.

DNA can then getShorter and shorterWith each reproduction.

Enzyme- telomeraseExtends the 3’ end of the DNA and then the otherEnd is extended with primaseDNA polymerase, and ligase.

Figure 16.19a Telomeres and telomerase: Telomeres of mouse chromosomes

ORGANELLE REPLICATION

• The two organelles that have DNA are the MITOCHONDRION & CHLOROPLAST

• Their chromosomes are CIRCULAR, JUST LIKE A PROKARYOTE.

• These organelles are inherited from an organism’s MOTHER (mitochondrial disorders pose a problem)

* Relate this to our family tree… Eve.

Mutations are changes in the DNA code.1) When chromosomes replicate, the error rate is: 1/10,000 base pairs or 1/1billion

To correct these errors: DNA polymerase itself proofreads each nucleotide against its template (like “delete” key) & mismatch repair enzymes fix missed errors (NUCLEASE).

2) Additional factors that cause mutation: mutagens, carcinogens

a. Chemicalsb. X and UV radiation- ex. thymine dimer (bond to e/o not the A’s fixed by excision repair)c. Bases undergo spontaneous changes

Figure 16.17 Nucleotide excision repair of DNA damage

DNA REPLICATION

EUKARYOTIC DNA REPLICATION

• Origin of Replication• Helicase• Replication fork• Single strand binding

proteins• Primase• Primer• DNA polymerase• Direction 3’ --> 5’• Elongation 5’ --> 3’ • Leading strand• Lagging strand• Okazaki fragments• DNA ligase

Leading Strand (narrated)

DNA REPLICATION narrated