Unit 3: Molecular Biology

Molecular Biology is the study of what genes are, and how they work, at the molecular level.

Central Dogma of Genetics:

Genes (DNA) mRNA protein

transcription translation

replication

Genes and Chromosomes in the late 1800s and early

1900s

By early 1900s, it was demonstrated that whatever the genes were, they are located on chromosomes in the nucleus of the cell.

Chromosomes were known to consist of DNA and protein. Which of these was the substance of the genes?

In the 1860s Mendel described the behavior of hereditary factors (now known as genes).

Genes have three properties

they are able to:

•store and retrieve information

•be copied

•change over time

Evidence for DNA as the Genetic Material

DNA transfer in bacteria and viruses

The introduction of pure DNA to bacterial cells can change the characteristics of bacteria (transformation, Griffith 1928, Avery 1944).

Work on bacteriophages (aka, phages; viruses that infect bacteria) by Hershey and Chase, 1952.

Fig. 6-15 Klug, Cummings, and Spencer, 8th ed.

The replication cycle of a typical virus

When viruses infect E. coli, DNA, but not protein, enters the bacterial cell.

Thus DNA has the information to produce new phage particles.

Solving the Structure of DNA

James Watson & Francis Crick

1953

Rosalind Franklin’s X-ray

crystallogram

of DNA

Nucleic Acids

DNA – deoxyribonucleic acid

RNA – ribonucleic acid

Both are linear polymers of nucleotide subunits.

Both function to store and transmit hereditary information.

5 end

Nucleoside

Nitrogenousbase

Phosphategroup Sugar

(pentose)

(b) Nucleotide

(a) Polynucleotide, or nucleic acid

3 end

3C

3C

5C

5C

Nitrogenous bases

Pyrimidines

Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA)

Purines

Adenine (A) Guanine (G)

Sugars

Deoxyribose (in DNA) Ribose (in RNA)

(c) Nucleoside components

Ribose (in RNA)Deoxyribose (in DNA)

Sugars

(c) Pentose sugars in DNA and RNA

(c) Nucleoside components: nitrogenous bases

Purines

Guanine (G)Adenine (A)

Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA)

Nitrogenous bases

Pyrimidines

5 end

Nucleoside

Nitrogenousbase

Phosphategroup Sugar

(pentose)

(b) Nucleotide

(a) Polynucleotide, or nucleic acid

3 end

3C

3C

5C

5C

Nitrogenous bases

Pyrimidines

Cytosine (C) Thymine (T, in DNA) Uracil (U, in RNA)

Purines

Adenine (A) Guanine (G)

Sugars

Deoxyribose (in DNA) Ribose (in RNA)

(c) Nucleoside components

Nitrogenous base connected to 1’ carbon of sugar – nucleoside

Phosphate group added to 5’ carbon of sugar - nucleotide

Making a nucleotide

5’ C

3’ OH (free)

1’ C

5’ PO4 (free)

DNA is a linear polymer of nucleotide subunits joined together by phosphodiester bonds - covalent bonds between phosphate group at 5’ carbon and 3’ carbon of next nucleotide – uses oxygens as bridges.

Chain of nucleotides has alternating sugar and phosphate components, called the “sugar-phosphate backbone.” Nitrogenous bases stick off backbone at regular intervals.

Any linear chain of nucleotides has a free 5’ PO4 on one end, and a free 3’ OH on the other. A chain of DNA thus has POLARITY

Different DNA molecules differ only in the identities of the nitrogenous bases at any given position – they have different DNA sequences. A simple way to represent this strand of DNA is:

5’-TACG-3’

Segments of this sequence, which can be 1000s of nucleotides long, are the genes that code for single, specific proteins.

3’ C

RNA is single stranded

DNA – almost always exists as a double helix.

RNA

Sugar is ribose

Nitrogenous bases are A G C U

DNA

Sugar is deoxyribose

Nitrogenous bases are A G C T

5’

3’

In a double helix, 2 strands of DNA wrapped around each other in shape of helix

Strands are held together by hydrogen bonding between nitrogenous bases.

Only pairings that work are A with T and G with C. Strands held at constant distance from one another because of the similar geometry of A-T and G-C base pairs

Also, only way pairings will work is if strands have opposite polarity 5’ to 3’ “Antiparallel”

Complementary base pairing

A bonds with T (2 H-bonds)

G bonds with C (3 H-bonds).

Note that A-T and G-C base pairs both contain a purine and a pyrimidine – similar geometry, same overall diameter.

Sugar-phosphatebackbones

3' end

3' end

3' end

5' end

5' end

5' end

Nitrogenous base pair (joined byhydrogen bonding)

Note again that the two chains of double helix have opposite polarity.

If two chains can form proper base pairs when oriented with opposite polarity, the chains are said to be complementary to each other.

Complementary

3' end

3' end

3' end

3' end

5' end

5' end

5' end

5' end

Old strands

Newstrands

For example, if you have a double helix of DNA, and one of strands is:

5’ ATCCGCT 3’

3’ TAGGCGA 5’

You know its partner has to be this.

Because the two strands are complementary, if you know the sequence of one strand, you can easily determine the sequence of the other.

Complementary

The cell replicates (duplicates) DNA by simply synthesizing new strands of DNA complementary to old strands

You have a single strand of DNA with the following nucleotide sequence:

5’ ATTGCCGACGTACCG 3’

You have four additional strands of DNA as follows:

1) 5’ TAACGGCTGCATGGC 3’

2) 3’ ATTGCCGACGTACCG 5’

3) 5’ CGGTACGTCGGCAAT 3’

4) 3’ GCCATGCAGCCGTTA 5’

For each of these four strands, indicate if they are identical, complementary, or unrelated to the original strand.

A. 1,4-identical; 2-unrelated; 3-complementary

B. 1- unrelated; 2,3-complementary; 4-identical

C. 1,3,4- identical; 2-unreleated

D. 1,4-identical; 2-unrelated; 3-complementary

E. 1,2-unrelated; 3-complementary; 4-identical

3’ TAACGGCTGCATGGC 5’5’ ATTGCCGACGTACCG 3’

2. A brief, 5 minute incubation at a near boiling temperature (95oC) has the effect of disrupting (breaking) hydrogen bonds, but has no effect on covalent bonds. If double stranded DNA in solution is incubated for 5 minutes at 95oC, what will happen to the DNA?

A. The DNA will be unaffected by the treatment.

B. The DNA will separate into 2, intact, single strands of DNA.

C. The nitrogenous bases will be released from the sugar phosphate backbones which will remain intact.

D. The DNA will be broken down into individual nucleotides.

E. The DNA will be broken down into its sugar, phosphate, and nitrogenous base components.

When the temperature is lowered back to room or body temperature, the complimentary strands will find each other and reform a double helix. No enzymes or energy are needed for the process – it is truly spontaneous.