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Summary IB Biology Topic 7: Nucleic Acids

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Detailed objective-by-objective summary notes for Topic 7: Nuclear acids for IB Biology SL/HL. Contains information on everything you need to know from 7.1 to 7.3, according to each understanding, application or skill. Written by a IB HL Biology student who graduated with a 45/45.

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IB TOPIC 7 | NUCLEIC ACIDS
2016 | SYJ0014


Topic 7.1: Nucleic acids – DNA Structure and replication
The evolution of multicellular organisms allowed cell specialization and cell replacement.

• Understanding: Nucleosomes help to supercoil the DNA.

 Nucleosomes: base units of DNA packaging in eukaryotes that consists of a segment of
DNA wound around a histone octamer
 Nucleosome and constituents: forms the repeating units of a chromosome
 Function: pacts the DNA in nucleus while ensuring appropriate access

 Histone octamer: consists of four different type of histones
 H1 histone: one histone outside each of the nucleosomes (that includes
histone octamer) to organize and hold the nucleosomes together

 Supercoiling of chromosome the DNA:
 Supercoiling allows a great length of DNA to be packed into a smaller space
 2m linear DNA can be packed into a nucleus of 10 µm diameter

• Understanding: DNA structure suggested a mechanism for DNA replication.

 DNA structures that suggested mechanism for DNA replication:
 Complimentary base pairing: the stability of complimentary base pairs led to semi conservation mechanism
 3’ and 5’ ends: 3’ ends terminate with sugar molecules, 5’ ends terminate with phosphate molecules, that acts as the basic
backbone of DNA that can be ‘read’ and replicated in certain direction (3’ to 5’)

• Understanding: DNA polymerases can only add nucleotides to the 3’ end of a primer.

 3’ to 5’ direction (antisense strand): DNA is read in this direction
 5’ to 3’ direction (replicating DNA): DNA is replicated in this direction
 DNA polymerase add 5’ ends of nucleotides to the growing 3’ end of the replicating chain

• Understanding: DNA replication is continuous on the leading strand and discontinuous on the lagging strand.

 Leading strand: strand replicated continuously; polymerase replicates
towards the replication fork
 Direction of helicase propagation: 3’ → 5’ direction
 Direction of DNA being read: 3’ → 5’ direction
 Polymerase replication: continuous

 Lagging strand: strand replicated discontinuously; polymerase replicates
away from the replication fork
 Direction of helicase propagation: 5’ → 3’ direction
 Direction of DNA being read: 3’ → 5’ direction
 Polymerase replication: needs to disassociate, move further up
the molecule then reattach

 Okazaki fragment: DNA fragments that are formed on the
lagging template strand

• Understanding: DNA replication is carried out by a complex system of enzymes.

Organism type Enzymes (proteins) involved Process Description
General DNA gyrase (topoisomerase) relieves torsional strain relieves strain while the double stranded DNA is
being unwound by helicase
DNA helicase unwinds and separates double unwinds the helix structure and separate double
helix stranded DNA to single strands by cutting H-
bonds between them
single stranded stabilizing protein stabilizes single strand binds to the DNA when it has be separated by
(replication protein A) helicase to stabilize the strands, preventing them
from recoiling or form secondary structures
DNA primase produces/lays RNA primer on synthesizes and lays short RNA strand (RNA
DNA primer) complementary to the DNA strand
DNA ligase joins Okazaki fragments closes nicks in the phosphodiester backbone of
DNA, especially between Okazaki fragments in
the lagging strand
Prokaryotes DNA polymerase III replicative elongation binds to RNA primer and joins free DNA
nucleotides in a continuous fashion (5’ → 3’)
DNA polymerase I RNA primer removal replaces the initial RNA primers (laid by DNA
primase) with DNA nucleotides




LAST EDITED 2017-03-16 | 1

, IB TOPIC 7 | NUCLEIC ACIDS
2016 | SYJ0014


• Understanding: Some regions of DNA do not code for proteins but have other important functions.

 Noncoding DNA: components of an organism's DNA that do not encode protein sequences (~70% of genome)

 Introns: intervening sequences within a gene that is removed by splicing during the maturation of mRNA
 Function: encode function RNAs, allows alternative splicing to generate multiple proteins from a single gene
 Post transcriptional modification: exon skipping, mutually exclusive exons, intron retention, alternative binding sites, intron
splicing to remove introns so they do not get translated

 Telomere: region of repetitive nucleotide sequences (TTAGGG)n at each end of a chromosome
 Function: protects the end of the chromosome from deterioration or from fusion with neighbouring chromosomes
 End replication problem: the very 3’ end of the DNA cannot be synthesized, and every time the cell replicates, 3’ end of
telomere sequence is lost and eventually the cells become unable to replicate and die (~4kb)

 Genes for tRNA: sections of DNA that are transcribed to create tRNA
 Function: generation of tRNA used for amino acid transfer during translation
 Location: located in both nuclear and mitochondrial DNA

 Regulatory sections: sections of nucleotides that are involved in regulating the gene expression
 Function: allows or prevents or RNA polymerase from transcribing the gene sequence
 Regulation in prokaryotes: activator or repressor proteins bind to operators or promoters

• Application: Rosalind Franklin’s and Maurice Wilkins’ investigation of DNA structure by X-ray diffraction.

 Rosalind Franklin and Maurice Wilkin’s investigation of DNA:
 Contribution to discovery of structure of DNA: Franklin and Wilkins were also credited for the discovery of the DNA structure
 Franklin’s crystallography: obtained the sharpest images of diffraction patterns of the DNA

 Process of X-ray diffraction:
 Beam of X-ray is directed at a sample
 X-ray detector is placed close to the same to collect the diffraction

 Deductions made from the X-ray diffraction patterns:
 Cross in the centre: indicated the molecule with a helix
 Gaps in the cross: indicates the distance between adjacent base pairs
 Outer diamond shape: suggests the helix is continuous and of constant dimension
 Distance between the bands: indicated the length of one full turn of the helix

• Application: Use of nucleotides containing dideoxyribonucleic acid to stop DNA replication in preparation of samples for base sequencing.

 Dideoxyribonucleic acids: DNA group that lack the hydroxyl groups at the 3’ position
 Usage: addition of dideoxyribonucleic acid terminates synthesis because no attachable 3’ hydroxyl
 Production: there are four different dideoxyribonucleic acids corresponding to the 4 base pairs

 Process of DNA sequencing:
 DNA replication: copies of unknown DNA is placed into test tubes with materials necessary to carry out replication
 Addition of dideoxyribonucleotides: dideoxyribonucleotides labelled with different fluorescent markers are added
 Termination of replication: when they are added, the replication stops at the point where they were added
 Gel electrophoresis: the fragments are separated by length using electrophoresis
 Electropherogram: fluorescence markers are detected and the sequence of DNA is then analysed

• Application: Tandem repeats are used in DNA profiling.

 Variable tandem repeats (VNTR): short nucleotide sequence that show variation between individuals
 VNTR inheritance: variation of VNTR can be inherited in forms of alleles
 VNTR coding sequence: some tandem repeats form part of coding sequences (e.g. CAG in individuals with Huntington’s)

 VNTR are used in DNA profiling:
 A unique DNA profile can be collected by amplifying a sample of DNA that contains one of these repeat sections
 The repeating sequence of tandem repeats can be used to deduce variation in hypervariable regions

 Maternal lineage: determined by analysing mitochondrial DNA
 Paternal lineage: deduced by analysing tandem repeats in Y chromosomes




LAST EDITED 2017-03-16 | 2

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IB Diploma New Zealand graduate. Graduated with 45/45 with A in both Extended Essay and TOK. Took HL Chemistry, HL Biology, HL Economics, SL English Literature A, SL Korean B and SL mathematics.

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