Bio Chap 8: DNA&genomics 1

Structure of a nucleotide

Nitrogenous bases

DNA: GCTA

RNA:GCUA

Purines: Guanine, Adenine (2 rings)

Pyramidines: Cytosine, Thymine, Uracil (1 ring)

CBP: C≡G, A=T, A=U

A:T = 1:1 AND C:G= 1:1

(A+G) = (C+T)

Structure of DNA

• Constant width between sugar phosphate backbone = 2nm

• 2 strands are anti-parallel, one runs in 5’ to 3’ while other runs in 3’ to 5’ (DNA said to have directionality)

• 1 complete turn of double helix has 10 base pairs and spans a distance of 3.4nm

• 5’ phosphate, 3’ OH

Semi-conservative replication

• Both strands separate by the breaking of H bonds and each strand acts as a template for the synthesis of a new strand through CBP

• Thus each DNA molecule formed is a hybrid consisting of one original strand and 1 newly synthesized strand

Gene mutations

Alteration in the sequence of nucleotides which may change the sequence of amino acids in a polypeptide chain; may change the 3D shape and hence function of protein subsequently changing phenotype of organism (mutation may result in new alleles)

Substitution mutation

• Replacement of one nucleotide by another

• 1 codon changed

• Minor/major depending on whether a new AA is synthesized, its properties, and its position in the protein

Inversion mutation

• Segment of nucleotides separates from the allele and rejoins at the original position but it is inverted

• 1 or more codons changed

• Minor/major depending on the properties of the new AA and their positions in the protein

Insertion/deletion mutation

• One or several nucleotides are inserted/removed into sequence

• shifts reading frame from point of mutation

• usually major, unless the number of nucleotides inserted or deleted are a multiple of 3. Then there will be a change in the primary sequence but a frame shift will not result

Frame shift mutation

Due to the insertion or deletion of several nucleotides that are not in a multiple of 3, reading frame will be disrupted and a completely different, non-functional polypeptide will result as the ribosomes read the incorrect triplets from point of I or D

Silent mutation

• Point mutation that does not change AA sequence in. a polypeptide

• due to degeneracy of the genetic code, more than one codon can code for the same AA, hence same polypeptide will be synthesized

Missense mutation

• Point mutation in which, a single nucleotide change results in a codon that codes for a different AA

• If new AA has different biochemical properties to the one replaced, mutation said to be conservative; If different, mutation said to be non-conservative

Nonsense mutation

Point mutation which results in premature stop codon, causing polypeptide to be truncated and non-fucntional

Sickle cell anaemia

• Beta-globin chain of haemoglobin affected

• Change in DNA: CTC to CAC; Change in mRNA: GAG to GUG; change in AA: glutamate to valine

• Charged and hydrophilic glutamate replaced by non-polar, hydrophobic valine in HbS

• At low O2 conc, HbS will lose the O2 and undergo a conformation change, which will cause the hydrophobic patches of HbS to stick out

• Hydrophobic areas of the different HbS molecules will stick together and this polymerisation of HbS results in the formation of abnormal, rigid, rod-like fibres, which will distort the shape of the biconcave RBC and make it sickle-shaped

• When O2 binds to HbS again, it will return to its original conformation and the RBC will return to their original biconcave shape(hence reversible)

• Sickle-shaped RBC are more fragile and break more easily, resulting in a shortage of RBC and poor O2 transport

• Sickle-shaped RBC may also lodge in small blood vessels and interfere with blood circulation, leading to organ damage

Chromosomal abberations in structure

1. Deletion removes a chromosomal segment

2. duplication repeats a chromosomal segment

3. Inversion reverses the segment within a chromosome

4. Translocation moves a segment from one chromosome to another, non-homologous one

Deletions and duplications can result in phenotypic abnormalities due to reduced or additional genes respectively

Inversions and reciprocal translocations can result in disease, as although the amount of genetic material remains the same, the expression of a gene can be influenced by its new location

Chromosomal abberations in chromosomal number

1. Aneuploidy is a condition where the cell does not have a chromosome number that is a multiple of the haploid number (could be extra or few chromosomes)

2. Aneuploidy is a result of non-disjunction where:

a) Homologous chromosomes do not move properly to opp poles during Meiosis I OR

b)When sister chromatids fail to separate properly to opp poles in Meiosis II

Role of DNA

To store info and pass it down from one gen to the next; suitable store of info as:

a) It can be replicated accurately - daughter cells have identical copies of DNA as parent cell (weak Hbonds between 2 strands allow them to separate and act as template for new strand synthesis)

b) Stable molecule - Can be passed on to the next gen without loss of coded info (numerous Hbonds hold the 2 strands of DNA together and adjacent nucleotides in each strand joined by covalent phosphodiester bonds)

c) there is a backup of code - DNA is double stranded and one strand can serve as template for the repair of the other if a mutation occurs on either one

d) coded info can be readily utilised/accessed (weak Hbond allows template strand to separate from non-template strand allowing transcription to take place; CBP allows faithful transfer of info from DNA to RNA in transcription

Role of mRNA

1. Serves as messenger that in eu, takes the info out of the nucleus via nuclear pore to the cytoplasm where translation takes place

2. Acts as template for translation

3. As each codon within the coding region of the mRNA represents an AA in polypeptide, sequence of codons will determine polypeptide sequence

Role of tRNA

tRNA can facilitate translation by:

1. Binding to a specific single AA during AA activation and carries it to the ribosome

2. Anticodon of tRNA CBPs with the complementary codon mRNA during translation

3. Thus tRNA brings specific AA in a sequence corresponding to the sequence of codons in mRNA to the growing polypeptide

Role of rRNA

1. Associates with a set of proteins to form ribosomes

2. Main constituent of the interface between large and small subunits; thus small ribosomal subunit can bind to mRNA, as CBP can occur between the rRNA and mRNA binding site of the small subunit and mRNA

3. Main constituent of the P site and A site on the large subunit, hence it allows the binding of aminoacyl-tRNA to the P site and A site via CBP

4. An rRNA molecule, peptidyl transferase, on the large subunit catalyses the formation of the peptide bond between the amino group of a new AA in the A site and the carboxyl end of the growing polypeptide in the P site

Features of genetic code

1. It is a triplet code

2. It is a universal code

3. It is a degenerate code

4. It is a non-overlapping code

5. It is a continuous code

6. It includes start (AUG) and stop (UAG, UAA, UGA) codons

What happens before DNA replication

1. Free deoxyribonucleoside triphosphates are manufactured in the cytoplasm and transported into the nucleoplasm via nuclear pores

2. DNA replication occurs at S phase of interphase

Unzipping of parental strand

1. Replication begins at specific points of the DNA molecule, each of which is known as the origin of replication (Ori)

2. Helicase binds to ori, disrupting the Hbonds between CBPs, causing parental strands to unzip and separate (process requires ATP)

3. Single-strand binding proteins keep the strands apart so that they can serve as templates for the synthesis of new strands

4. Topoisomerase relieves the ‘overwinding’ strain ahead of replication forks by breaking, swivelling and rejoining the DNA strands

Addition of Primer

1. RNA primer is added to each template strand by the enzyme primase

2. RNA primeer provides a free 3’ OH end for DNA polymerase to recognise and start DNA synthesis of the complementary daughter strand

3. DNA polymerase can only add deribooxynucleotides to a pre-existing 3’ OH end of a nucleotide

Synthesis of daughter strands

1. DNA polymerase uses the parental strand as a template and aligns the free activated deoxyribonucleoside triphosphates (dNTP) in a sequence complementary to that of the parental strand

2. A pairs with T, and C pairs with G

3. DNA polymerase catalyses the formation of phosphodiester bonds between adjacent daughter DNA nucleotides of the newly synthesised strand

4. As DNA polymerase moves along the template, it proofreads the previous region for proper base pairing. Any incorrect deoxyribonucleotide is removed and replaced by the correct one

5. The leading strand is synthesised continuously in the 5’ to 3’ direction

6. The lagging strand is synthesised discontinuously(3’ to 5’), being synthesised in Okazaki fragments, being initiated by an RNA primer before the addition of DNA nucleotides

7. A different DNA polymerase then removes the RNA primer and replaces it with deoxyribonucleotides

8. DNA ligase seals the nicks by forming phosphodiester bonds between adjacent nucleotides of each of the DNA fragments on the new strand

End of replication

1. Complementary parental and daughter strands rewind into double helix

2. Each resultant helix consists of one parental strand and one daughter strand. Thus semi-conservative