IB Biology/Nucleic Acids and Proteins

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Topic 6: Nucleic Acids and Proteins[edit]

DNA Structure[edit]

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

  • DNA has a double stranded helix which has uniform diameter along its entire length
  • Both helices are right handed which allows it to fit within a defined three dimensional space.
  • Two polynucleotide chains are 'anti-parallel', running in opposite directions.
  • The polynucleotides are form around the outside of the helix with the bases extending into the centre.
  • Polynucleotide chains are held together by the bases (in centre) hydrogen bonding with bases on the opposite polynucleotide.
  • The hydrogen bonding is specific and known as complementary base pairing.

Outline the structure of nucleosomes.

  • A nucleosome is a section of DNA wrapped around eight smaller protein molecules called histones. Another histone seals the octamer of histones to the DNA double strand molecule.
  • DNA is bonded to proteins called Histones.
  • Nucleosomes enable the DNA molecule to be super coiled to make chromosomes about 8000 times smaller in length.
  • A nucleosome is a basic unit of DNA packaging. It consists of 8 histones (histone is a kind of protein) with the DNA double helix wrapped around it. This "bead' is fastened onto the "string" of DNA by another histone.
  • The combination of DNA and histones is secured by the 'H1 linker'

State that only a small proportion of the DNA in the nucleus constitutes genes and that the majority of DNA consists of repetitive sequences.

  • Only a small proportion of the DNA in the nucleus constitutes genes and that the majority of DNA consists of repetitive sequences.

Describe the structure of DNA including the antiparallel strands, 3'-5' linkages and hydrogen bonding between purines and pyrimidines.

  • The sides of the ladder of DNA consist of alternating phosphate groups and deoxyribose (a sugar). The two sides are antiparallel, meaning that the sugar and phosphates are running in opposite directions (one looks "upside down"). Each side has a 5' end and a 3' end. If a strand is structured from 3' to 5', that means that the sugar-phosphate backbone runs from sugar to phosphate. Since the sides are antiparallel, one side goes in the 3' to 5' direction, and the other goes in the 5' to 3' direction. There are two types of nucleotides, pyrimidines and purines. Pyrimidines hydrogen bond to purines to create the rungs of the DNA ladder.
Thymine and cytosine are pyrimidines, adenine and guanine are purines.

DNA Replication[edit]

State that DNA replication occurs in a 5' to 3' direction.

  • DNA replication occurs in a 5' to 3' direction.

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

  • The process of replication begins at specific nucleotide sequences called the origins of replication on the DNA strand. It is at these points that helicase splits the DNA into its two antiparallel strands. On the strand running in the 5'--->3' direction, DNA polymerase III latches on at one end of the opening, called the replication bubble, and begins to continuously lay a new DNA strand from free nucleotides in the nucleus. As always, an exact copy of the now-detached strand is formed from this template due to base-pairing rules. At the same time DNA polymerase III is laying new DNA, helicase is continuing to split the strands, thus allowing replication to continue uninterrupted. On the opposite strand running 3'--->5', replication is not so simple. Because new strands have to be laid in the 5'---3' direction, DNA polymerase III cannot lay continuously as it can on the other strand. Instead, RNA primase lays short segments of RNA primer nucleotides at many points along the strand. Polymerase III fills in between the RNA primers with DNA nucleotides, until it comes into contact with another RNA primer. DNA polymerase I replaces the RNA primer with DNA. Although the lagging strand is completely made up of DNA nucleotides, between the RNA primer points that are now replaced by DNA nucleotides there are small gaps; these segments of DNA are called Okazaki fragments. Once these fragments have been laid, they are joined by yet another enzyme known as DNA ligase, which attaches DNA into the gaps between fragments and completes the new strand. The 3'--->5' strand with Okazaki fragments is called the lagging strand, while the leading strand is the continuously replicating one.

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

  • In eukaryotic chromosomes, replication is initiated at many points.


Define Transcription

  • Transcription - is the synthesis of a strand RNA using a strand of the DNA as a template. An enzyme (RNA polymerase) moves along the DNA strand, temporarily unwinding the double strands. Nucleotides attach to the template strand and elongate the forming RNA strand. The building RNA strand is separated from the DNA as the enzyme moves along the DNA, and the DNA rewinds. RNA differs from DNA in composition, where all instances of thymine are replaced by uracil.

State that transcription is carried out in a 5' to 3' direction.

  • Transcription is carried out in a 5' to 3' direction (from phosphate to sugar). The 5' end (phosphate) of the RNA nucleotide is added to the 3' end (sugar) of the part of the RNA molecule which has already been synthesized.

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

  • RNA polymerase separates the two strands of the DNA and bonds the RNA nucleotides when the base-pair joins to the DNA template.
  • RNA polymerase binds to parts of the DNA called promoters in order for separation of DNA strands to occur.
  • Transcription proceeds as nucleoside triphosphates (type of nucleotide) bind to the DNA template and are joined by RNA polymerase in the 5' to 3' direction.
  • Transcription ends when RNA polymerase reaches a termination site on the DNA. When it reaches the terminator, the RNA polymerase releases the RNA strand.

Distinguish between the sense and antisense strands of DNA

  • The sense strand is the coding strand and has the same base sequence as the mRNA (with uracil instead of thymine). The antisense strand is transcribed and has the same base sequence as tRNA.
  • The sense strand is also called the non-template strand, and the antisense strand is also called the template strand.

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

  • Eukaryotic RNA needs the removal of introns to form mature mRNA. This is because the introns are Highly Repetitive Sequences, or non-coding sequences.
  • Introns contain nucleotides that do not code for amino acids.
  • After introns are removed after transcription and before translation, the exons are joined. This process is called splicing.

State that reverse transcriptase catalyses the production of DNA from RNA.

  • Reverse transcriptase catalyses the production of DNA from RNA. This helps to show the aspects of the DNA viral life cycle to that of the AIDS virus (an RNA virus).

Explain how reverse transcriptase is used in molecular biology.

  • This enzyme can make DNA from mature mRNA (eg insulin) which can then be spliced into host's (eg bacteria) DNA without the introns. Then when the host's DNA is transcripted, proteins like insulin are made.
  • It is important that the DNA created by reverse transcriptase have no introns, because the host does not have the genes (and therefore proteins) necessary to remove the introns.


Explain how the structure of tRNA allows recognition by a tRNA-activating enzyme that binds a specific amino acid to tRNA, using ATP for energy.

  • Each amino acid has a specific tRNA-activating enzyme that help to tRNA to combine with its complementary mRNA codon. The enzyme has a 3-part active site that recognizes three things: a specific amino acid, ATP, and a specific tRNA. The enzyme attaches the amino acid to the 3' end of the tRNA. The amino acid attachment site is always the base triple CCA.
  • It is important to note that each tRNA molecule can attach to one specific amino acid, but an amino acid can have a few tRNA molecules with which is can combine.

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

  • A ribosome consists of two subunits: small and large. These two subunits separate when they are not in use for protein synthesis.
  • In eukaryotes, the large subunit consists of three different molecules of rRNA (ribosomal RNA) and about 45 different protein molecules. A small subunit consists of one rRNA molecule and 33 different protein molecules.
  • A ribosome can produce all kinds of proteins. It has a mRNA binding site, and two tRNA binding sites where the tRNA are in contact with the mRNA.

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

  • Translation consists of initiation, elongation, and termination.

State that translation occurs in a 5' to 3' direction.

  • Translation occurs in a 5' to 3' direction, the ribosome moves along the mRNA toward the 3' end.
  • The start codon is nearer to the 5' end than the stop codon.

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

  • Initiation: Once the RNA reaches the cytoplasm, it attaches its 5' end to the small subunit of the ribosome. AUG is called the start codon (remember: codon is a base triple on the mRNA) because it initiates the translation process. The anticodon on one end of a tRNA molecule is complementary to a specific codon on the mRNA, meaning that the anitcodon and the codon bond by hydrogen bonds. The codon AUG hydrogen bonds with a tRNA molecule holding the amino acid methionine (the initiator tRNA). The large ribosomal subunit has two tRNA binding sites: the P site and the A site. tRNA molecules in these sites attach to the mRNA "conveyor belt". tRNA molecules move from the A site to the P site (because the mRNA moves that way), so when initiation is done the initiator tRNA is in the P site.
  • Elongation: Another tRNA (let's call it tRNA "X") carrying a specific amino acid attaches itself to the next codon at the A site of the larger subunit. These two amino acids (methionine and the amino acid on X) now make a peptide bond with each other. The initiator tRNA breaks off and tRNA X moves from the A site to the P site. Then another tRNA molecule (tRNA "Y") attaches to the codon in the A site. The amino acid that is attached to tRNA Y makes a peptide bond with the amino acid from tRNA X. The codon keeps moving through the ribosome in a 5' to 3' direction (from A to P). (In actuality, it is the ribosome that is moving across the mRNA chain.) This makes the A site vacant for another tRNA to attach a new amino acid etc. etc.
  • Termination: The stop codon is one that does not code for an amino acid and that terminates the translation process. The polypeptide is released and the mRNA fragments return to the nucleus. These nucleotides are recycled and used for RNA and DNA synthesis. tRNA also is returned to its free state and attaches to its specific amino acid so as to be ready for the translation process when needed.

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

  • Free ribosomes synthesize proteins for use primarily within the cell and bound ribosomes synthesize proteins primarily for secretion or for lysosomes.


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

  • The primary structure is the basic order of amino acids in the polypeptide protein chain, before any folding or bonding between amino acids has occurred. Proteins are usually not functional on the primary level, and all proteins have a primary structure.
  • The secondary structure is the repeated, regular structure protein chains take due to hydrogen bonding between amino acids. The secondary structure is usually in the form of an alpha helix (similar to a DNA chromosome) or a beta- pleated sheet (similar in form to the corrugations of cardboard).

Alpha helix is due to hydrogen bonding within one polypeptide chain, while the beta pleated sheet is due to hydrogen bonding between separate polypeptide chains.

  • The third structure is the tertiary structure. It is the complex, three-dimensional protein shape resulting from the folding of the polypeptide due to different types of bonds between the amino acids. These bonds include hydrogen bonds, disulphide linkages and electrovalent bonds. A disulphide linkage is where one of the amino acids, cysteine, contains sulfur. Two of these can have a bond between their sulfur atoms which results in a disulphide linkage which leads to a folding in the chain. An electrovalent bond is between the negative and positive molecules or amino acid groups along the chain.
  • The quaternary structure is the most complicated form of a protein. It encompasses the primary, secondary and tertiary, and then adds another level: the quaternary is one or more peptide chains bonded together. This is the functional form of many proteins, but again, just as not all proteins have secondary or tertiary structure, not all proteins have a quaternary structure. For a common example of a quaternary-structured protein, see hemoglobin.

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

  • Fibrous proteins are in their secondary structure, which could be in the alpha helix or beta pleated forms. They are made of a repeated sequence of amino acids that can be coiled tightly around in a pattern that makes it a very strong structure. Two examples are keratin (in hair and skin) and collagen (in tendons, cartilage, and bones).
  • Globular proteins are in their tertiary or quaternary structure, which is folded, creating a globular, three-dimensional shape. Two examples are all enzymes and microtubules (form centrioles, cilia, flagella, and cytoskeleton).

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

  • Non-polar amino acids have non-polar (neutrally charged) R groups. Polar amino acids have R chains with polar groups (charged either positive or negative). Proteins with a lot of polar amino acids make the proteins hydrophyllic and therefore able to dissolve in water. Proteins with many non-polar amino acids are more hydrophobic and are less soluble in water. With these abilities, proteins fold themselves so that the hydrophilic ones are on the inner side and allow hydrophilic molecules and ions to pass in and out of the cells through the channels they form. These channels are vital passages for many substances in and out of the cell.

State six functions of proteins, giving a named example of each.

  • One function is transport. An example is hemoglobin, which transports oxygen around the body within a blood cell.
  • Another function is the contraction of muscles. An example is actin and myosin which are involved in the contraction of muscles.
  • Enzymes are all proteins and catalyze reactions. Some examples are trypsin and amylase.
  • Some hormones are proteins. An example is insulin, a hormone secreted by the pancreas which is used to regulate blood sugar levels.
  • Another function is antibodies that fight against disease are made of proteins.
  • The sixth function is support through structural proteins. Examples include nails (keratin) or tendons (collagen).
Note that membrane proteins should not be included.


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

  • Metabolic pathways consist of chains and cycles of enzyme catalysed reactions.

Describe the induced fit model.

  • This is an extension of the lock-and-key model. It accounts for the broad specificity of some enzymes. The induced fit model of enzymes states that the active site of an enzyme does not fit perfectly with the substrate, as would a lock and key. Instead, the fit is not quite perfect; thus when a substrate collides with an enzyme and enters the active site, the substrate is stressed in a way that allows its bonds to break easier. When another collision occurs with the substrate in the active site, the bonds are then broken and the substrate released. Enzymes can process reactions such as this very rapidly.

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

  • All reactions, either with or without enzymes, need collisions between molecules in order to occur. Many molecules have strong bonds holding them together, and as such require powerful collisions at high speed in order to break these bonds. However, increasing the rate of collision to a rate at which these reactions would occur would require prohibitive amounts of energy, usually in the form of heat. Enzymes, by stressing substrate bonds in such a way that a weaker collision is required to break them, reduce the amount of energy needed to cause these reactions to occur, or the activation energy.

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

  • Competitive inhibition occurs when an inhibiting molecule structurally similar to the substrate molecule binds to the active site, preventing substrate binding. Examples are the inhibition of butanedioic acid (succinate) dehydrogenase by propanedioic acid (malonate) in the Krebs cycle, and inhibition of folic acid synthesis in bacteria by the sulfonamide prontosil (an antibiotic).
  • Non-competitive are an inhibitor molecule binding to an enzyme (not to its active site) that causes a conformational change in its active site, resulting in a decrease in activity. Examples include Hg, Ag, Cu and CN inhibition of many enzymes (eg cytochrome oxidase) by binding to SH groups, thereby breaking -S-S- linkages; and nerve gases like Sarin and DFP (diisopropyl fluorophosphate) inhibiting ethanyl (acetyl) cholinesterase.

Explain the role of allostery in the control of metabolic pathways by end product inhibition.

  • Allostery is a form of non-competitive inhibition. The shape of allosteric enzymes can be altered by the binding of end products of an enzyme to an allosteric site (an area of the enzyme seperate from the active site), thereby decreasing its activity. Metabolites, a type of allosteric inhibitor, can act as allosteric inhibitors of enzymes earlier in a metabolic pathway and regulate metabolism according to the requirements of organisms; they are a form of negative feedback. Examples include ATP inhibition of phosphofructokinase in glycolysis and inhibition of aspartate carbamoyltransferase (ATCase) which catalyses the first step in pyrimidine synthesis.