Topic 7 -Nucleic acids and proteins

7.1 DNA Structure

7.1.1 Describe the structure of DNA, including the antiparallel strands, 3’→5’ linkages and hydrogen bonding between purines and pyrimidines.

Major and minor grooves, direction of the “twist”, alternative B and Z forms, and details of the dimensions are not required.

• DNA consists of two anti-parallel strands of nucleotides, forming a double helix.
• Nucleotides consist of a nitrogenous base, deoxyribose and a phosphate group.
• Nitrogenous bases can be adenine, guanine, cytosine or thymine.
• Nucleotides are linked in a 3’ to 5’ direction.
• Hydrogen bonding between purines and pyrimidines on opposite chains holds the DNA molecule together. Complementary base pairing forms between A-T and G-C.
• Two hydrogen bonds exist between adenine and thymine and three hydrogen bonds exist between guanine and cytosine.

7.1.2 Outline the structure of nucleosomes.

Limit this to the fact that a nucleosome consists of DNA wrapped around eight histone proteins and held together by another histone protein.

• A nucleosome consists of a core of eight histone proteins, which DNA wraps around twice. Another histone proteins holds the structure together.

7.1.3 State that nucleosomes help to supercoil chromosomes and help to regulate transcription.

7.1.4 Distinguish between unique or single-copy genes and highly repetitive sequences in nuclear DNA.

Highly repetitive sequences (satellite DNA) constitutes 5–45% of the genome. The sequences are typically between 5 and 300 base pairs per repeat, and may be duplicated as many as 105 times per genome.

Highly repetitive genes

• Known as satellite DNA and constitutes 5-45% of the genome.
• Sequences are usually between 5 to 300 base pairs per repeat, and may duplicate as many as 105 times per genome.
• Satellite DNA is not translated and its function is not yet clear.

Unique/single-copy genes

• This is the gene coding region (exons) that codes for mRNA, which codes for polypeptides.

7.1.5 State that eukaryotic genes can contain exons and introns.

7.2 DNA Replication

7.2.1 State that DNA replication occurs in a  5 ‘→ 3’ direction.

The 5’ end of the free DNA nucleotide is added to the 3’ end of the chain of nucleotides that is already synthesized.

7.2.2 Explain the process of DNA replication in prokaryotes, including the role of enzymes (helicase, DNA polymerase, RNA primase and DNA ligase), Okazaki fragments and deoxynucleoside triphosphates.

The explanation of Okazaki fragments in relation to the direction of DNA polymerase III action is required. DNA polymerase III adds nucleotides in the  5 ‘→ 3’ direction. DNA polymerase I excises the RNA primers and replaces them with DNA.

helicase uncoils/splits DNA double helix;
RNA primase;
(creates a primer for) DNA polymerase III to bind/begin replication;
deoxyribonucleoside triphosphates (free in cell);
two phosphates removed to release energy;
DNA polymerase III adds nucleotides in 5′  to 3′  direction;
A paired with T / C paired with G;
discontinuous copying / Okazaki fragments / short lengths of DNA formed (between
RNA primers) on lagging strand;
continuous on leading strand;
DNA polymerase I removes RNA primers/replaces them with DNA;
DNA ligase joins the fragments;
initiated at many points in the (eukaryotic) chromosomes;  [8 max]

• Replication is initiated at multiple points simultaneously.
• Helicase unwinds DNA double helix [by breaking hydrogen bonds between complementary base pairs.
• RNA primase synthesizes the short RNA primers to initiate the replication process, because DNA polymerase III is only able to add DNA nucleotides to a free 3’ end on an existing DNA strand. Therefore replication occurs in a 5’ to 3’ direction.
• DNA pol. III catalyzes the condensation reaction of freefloating deoxynucleoside triphosphates, creating the daughter strand of DNA. Two phosphate groups are removed to release energy.
• Adenine is paired with thymine and guanine is paired with cytosine.
• For the leading strand, unwinding of the DNA and replication occurs in the same direction, so replication occurs continuously (primer is only laid down once).
• For the lagging strand, unwinding and replication occur in opposite directions so replication occurs discontinuously. DNA poly. III synthesizes the new daughter strand in ‘Okazaki fragments’. Once fragments are assembled, DNA polymerase I removes RNA primers and replaces them with DNA.
• DNA ligase then joins the sugar-phosphate backbones of the fragments.

7.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomes.

7.3 Transcription

7.3.1 State that transcription is carried out in a  5 ‘→ 3’ direction.

The 5’ end of the free RNA nucleotide is added to the 3’ end of the RNA molecule that is already synthesized.

7.3.2 Distinguish between the sense and antisense strands of DNA.

The sense strand (coding strand) has the same base sequence as mRNA with uracil instead of thymine. The antisense (template) strand is transcribed.

Sense

• The sense strand has the same base sequence as mRNA (except U for T).
• It carries promoter site, where RNA polymerase binds and begins transcription.
• It carries the terminator site, where RNA polymerase stops transcription.

Antisense

• The antisense strand is the template and it is transcribed to build mRNA.
• It has the same base sequence as the complementary tRNA molecule.

7.3.3 Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and the terminator.

The following details are not required: there is more than one type of RNA polymerase; features of the promoter region; the need for transcription protein factors for RNA polymerase binding;
TATA boxes (and other repetitive sequences); and the exact sequence of the bases that act as terminators.

• RNA polymerase binds to the promoter site at the gene that will be transcribed. This is located on the antisense strand, which acts as a template for the mRNA. This initiates transcription.
• RNA polymerase unwinds the DNA double helix by breaking the hydrogen bonds between base pairs.
• At the same time, RNA polymerase catalyzes the synthesis of the mRNA strand. It does so by removing two phosphate groups from the freefloating nucleoside triphosphates in the nucleoplasm, and links the 5’ end of the resulting nucleotide with the 3’ end of the growing mRNA molecule.
• RNA polymerase reads 3’ to 5’ but builds 5’ to 3’.
• The helix zips back together as the RNA polymerase continues along.
• The mRNA molecule elongates as the RNA polymerase continues along the antisense strand.
• RNA polymerase releases at the terminator site. and the mRNA is completed.

7.3.4 State that eukaryotic RNA needs the removal of introns to form mature mRNA.

Further details of the process of post-transcriptional modification of RNA are not required.

• Post-translational modifications of mRNA involves the removal of introns through enzyme-catalyzed reactions in the cytoplasm. Mature mRNA results.

7.4 Translation

7.4.1 Explain that each tRNA molecule is recognized by a tRNA-activating enzyme that binds a specific amino acid to the tRNA, using ATP for energy.

Each amino acid has a specific tRNA-activating enzyme (the name aminoacyl-tRNA synthetase is not required). The shape of tRNA and CCA at the 3’ end should be included.

tRNA activation

• Each tRNA molecule is specific to a certain amino acid and a certain tRNA-activating enzyme.

• The tRNA- activating enzyme recognizes a corresponding uncharged tRNA molecule and, using energy from ATP, catalyzes the binding of an amino acid to the tRNA.
• The amino acid becomes covalently bonded to the amino acid binding site, and the resulting charged tRNA is released from the enzyme.

Structure of tRNA

• tRNA is composed of one chain of (RNA) nucleotides;
• tRNA has a position/end/site attaching an amino acid; (Reject tRNA contains an amino acid.) at the 3' terminal / consisting of CCA / ACC;
• tRNA has an anticodon; anticodon of three bases which are not base paired / single stranded / forming part of a loop;
• tRNA has double stranded sections formed by base pairing;
• double stranded sections can be helical;
• tRNA has (three) loops (sometimes with an extra small loop);
• tRNA has a distinctive three dimensional / clover leaf shape;

7.4.2 Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites and mRNA binding sites.

• The ribosome is comprised of proteins and rRNA. It has a small subunit where the mRNA binds, and it has a large subunit.
• There are three tRNA binding sites on the small subunit.
• The A site is the activation site, where charged tRNA enters.
• The P site is the peptidyl site, where the growing polypeptide chain is held.
• The E site is the exit site, where uncharged tRNA is released.

7.4.3 State that translation consists of initiation, elongation, translocation and termination.

7.4.4 State that translation occurs in a   5 ‘→ 3’ direction.

During translation, the ribosome moves along the mRNA towards the 3’ end. The start codon is nearer to the 5’ end.

7.4.5 Draw and label a diagram showing the structure of a peptide bond between two amino acids.


7.4.6 Explain the process of translation, including ribosomes, polysomes, start codons and stop codons.

Use of methionine for initiation, details of the T factor and recall of actual stop codons are not required.

• Initiation:
• First, the small subunit of ribosome binds to mRNA; and moves along mRNA until it reaches the start codon.
• Then the initial tRNA carrying methionine binds to the ribosome / mRNA;
• and the large subunit binds to small subunit;The small ribosomal unit, large ribosomal unit, and initiator tRNA (with the first amino acid, methionine)binds to the promoter site (codon ‘AUG’) on the mRNA.
• Elongation:
• Each tRNA carries an amino acid.
• As the ribosome moves on along the mRNA in a 5' to 3', tRNAs with the corresponding anticodon to the codon on the mRNA enter into the A site.
• A peptide linkage forms between the amino acids on the tRNAs of the A and P site, and the attached amino acids are added to the growing polypeptide chain at the P site.
• Unactivated/uncharged tRNA which have just given their amino acid leave the ribosome from the E site.
• During elongation, there are two tRNAs bound to ribosome at any given time;
• Termination:
• Eventually the stop codon is reached;
• The polypeptide/protein is released / tRNA and mRNA detached from ribosome;
• ribosome splits into (large and small) subunits;

7.4.7 State that free ribosomes synthesize proteins for use primarily within the cell, and that bound ribosomes synthesize proteins primarily for secretion or for lysosomes.

7.5 Proteins

7.5.1 Explain the four levels of protein structure, indicating the significance of each level.

Quaternary structure may involve the binding of a prosthetic group to form a conjugated protein.

Primary structure

• This describes the specific sequence of amino acids in a polypeptide.
• Amino acids are linked by peptide bonds.
• The primary structure determines the subsequent configuration of the secondary and tertiary structure (and the overall type and function of the protein.)

Secondary structure

• This describes the regular repeating structures, including beta-pleated sheets and alpha-helices, between groups in the main chain of the polypeptide.
• This is stabilized by hydrogen bonding.

Tertiary structure

• This describes the 3-dimensional conformation of a polypeptide/protein.
• It is as a result of folding of the polypeptide, which is held by various bonds including: ionic bonds, hydrogen bonds, disulphide bridges and hydrophobic bonds.
• It determines the overall shape of a polypeptide, such as the enzyme lysozyme --its active site is shaped by the tertiary structure.

Quaternary structure

• This describes the linking together of two or more polypeptides to form a single protein.
• For example, collagen consists of three polypeptides and insulin consists of four.
• Quaternary structure may involve the binding of a prosthetic group, and this occurs in conjugated proteins. For example, in hemoglobin, each polypeptide in hemoglobin is bonded to a heam group.

7.5.2 Outline the difference between fibrous and globular proteins, with reference to twoexamples of each protein type.

Fibrous

• repetitive amino-acid sequences, usually dominated by secondary structure
• long and narrow / long strands
• repetitive amino-acid sequence
• support/structural functions
• (mostly) insoluble in waters
• E.g. collagen or myosin

Globular

• irregular amino acid sequences, usually dominated by tertiary structure
• rounded / spherical/ball shaped
• metabolic / other functions;
• (mostly) soluble in water;
• E.g. any enzyme like lysozyme; hemoglobin or insulin

7.5.3 Explain the significance of polar and non-polar amino acids.

Limit this to controlling the position of proteins in membranes, creating hydrophilic channels through membranes, and the specificity of active sites in enzymes.

• Chemically, proteins can be divided into two groups: polar, which have hydrophilic R groups; and non-polar, which have hydrophobic R groups.

Polar amino acids:

• hydrophilic;
• can make hydrogen bonds;
• found in hydrophilic channels/parts of proteins projecting from membranes;
• found on surface of water-soluble proteins, allowing them to be water soluble

Non-polar amino acids:

• hydrophobic;
• forms van der Waals/hydrophobic interactions with other hydrophobic amino acids;
• found in proteins and parts of proteins embedded in interior of membranes;
• found in interior of water-soluble proteins;

7.5.4 State four functions of proteins, giving a named example of each.

Membrane proteins should not be included.

7.6 Enzymes

7.6.1 State that metabolic pathways consist of chains and cycles of enzyme-catalysed reactions.

7.6.2 Describe the induced-fit model.

This is an extension of the lock-and-key model. Its importance in accounting for the ability of some enzymes to bind to several substrates should be mentioned.

• The induced fit model is an extension of the lock-and-key model because it accounts for why some enzymes can catalyze a range of chemically-similar substrates, e.g. proteases.
• The substrate makes contact with the active site of its enzyme, which induces a critical change of shape/conformational change in the enzyme that complements the shape of the substrate.
• This allows the substrate to bind and achieve the enzyme-substrate complex. Bonds of the substrate are broken by amino acids in the active site during this transition state, and new bonds are formed.
• This enzyme releases the product once it is formed and then returns to its original shape.

7.6.3 Explain that enzymes lower the activation energy of the chemical reactions that they catalyse.

Only exothermic reactions should be considered. Specific energy values do not need to be recalled.

• In order for a chemical reaction to occur, there must be sufficient energy in order to break the bonds of the reactants and form products.
• The activation energy of a reaction is the minimum amount of energy required in order for a reaction to proceed successfully.
• Enzymes work to reduce the activation energy by providing an alternate pathway for a reaction to occur.  
• An enzyme’s active site interacts with substrate, altering stability of substrate bonds, allowing substrate molecules to form a transition state which is different than would be formed without the enzyme.
• Since less energy is required (such as, lwoer temperature), reactions become faster and more likely to occur.

7.6.4 Explain the difference between competitive and non-competitive inhibition, with reference to one example of each.

Competitive inhibition is the situation when an inhibiting molecule that is structurally similar to the substrate molecule binds to the active site, preventing substrate binding.

Limit non-competitive inhibition to an inhibitor binding to an enzyme (not to its active site) that causes a conformational change in its active site, resulting in a decrease in activity.

Reversible inhibition, as compared to irreversible inhibition, is not required.

competitive:

• A molecule structurally and chemically similar to the substrate binds to the active site; which prevents the substrate from binding;
• e.g. inhibition of butanedioic acid (succinate) dehydrogenase by propanedioic acid  (malonate) in the Krebs cycle / other valid example;
• competitive inhibition is reversible;

non-competitive:

• An inhibitor molecule binds to an enzyme; at a location not at the active site;
• This causes a conformational change in the active site;
• preventing substrate binding; and decreasing the activity of the enzyme.
• e.g. CN inhibition of cytochrome oxidase (of the electron transport chain) by binding to SH groups /

7.6.5 Explain the control of metabolic pathways by end-product inhibition, including the role of allosteric sites.

• End-product inhibition is a form of non-competitive inhibition.
• An inhibitor binds to enzyme; at an allosteric site.
• This causes a conformational change, changing the shape of the active site and preventing the true substrate from binding.
• End-product inhibition uses a feedback mechanism, in that the final product of a pathway is used as the allosteric inhibitor for the first enzyme in the pathway.
• Feedback is used the regulate the amount of product produced -- the higher the end product concentration, the higher the inhibition.
• This avoids a build up of all the intermediates.
• End-product inhibition is reversible.
• The pathway can be restarted when there is a shortage of the end product.
• Example: in glycolysis, when ATP is in excess, ATP will act as an allosteric inhibitor to the enzyme phosphofructokinase. As a result, glycolysis is inhibited until ATP concentration falls again.