Nucleic Acids and Proteins

Question 1

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

Answer 1

DNA is made up of two strands. At one end of each strand there is a phosphate group attached to the carbon atom number 5 of the deoxyribose (this indicates the 5’ terminal) and at the other end of each strand is a hydroxyl group attached to the carbon atom number 3 of the deoxyribose (this indicates the 3’ terminal). The strands run in opposite directions and so we say that they are antiparallel. One strand runs in a 5’-3’ direction and the other runs in a 3’-5’ direction. Adjacent nucleotides are attached together via a bond between the phosphate group of one nucleotide and the carbon atom number 3 of the deoxyribose of the other nucleotide.

The bases of each strand link together via hydrogen bonds. Adenine and Guanine are purines as they have two rings in their molecular structure. Thymine and Cytosine are pyrimidines as they only have one ring in their molecular structure. A purine will link with a pyrimidine. Adenine and thymine link together by forming two hydrogen bonds while Guanine and cytosine link together by forming 3 hydrogen bonds.

Question 2

Outline the structure of nucleosomes.

Answer 2

Nucleosomes consiste of DNA wrapped around eight histone proteins and held together by another histone protein.

Question 3

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

Answer 3

Nucleosomes help to supercoil chromosomes and help regulate transcription.

Question 4

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

Answer 4

Not all of the base sequences in DNA are translated. Highly repetitive base sequences are not translated. They consist of sequences of between 5 and 300 bases that may be repeated up to 10 000 times. They constitute 5-45% of eukaryotic DNA. Single-copy genes or unique genes are translated and constitute a surprisingly small proportion of eukaryotic DNA.

Question 5

State that eukaryotic genes can contain exons and introns.

Answer 5

Eukaryotic genes can contain exons and introns.

Question 6

Which direction does DNA replication occur in?

Answer 6

5? → 3? direction

Question 7

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.

Answer 7

The first stage of DNA replication in prokaryotes is the uncoiling of the DNA double helix by the enzyme helicase. Helicase separates the DNA into two template strands. RNA primase then adds a short sequence of RNA to the template strands. This short sequence of RNA is a primer which allows DNA polymerase III to bind to the strands and start the replication process. Once this is done, DNA polymerase III adds nucleotides to each template strand in a 5’→3’ direction. The nucleotides have 3 phosphate groups and are called deoxyribonucleoside triphosphates. Two of these phosphate groups break off during the replication process to release energy. Since the strands are anti-parallel (the two strands have their 5’ end and 3’ end in opposite sides) and replication can only occur in a 5’→3’ direction, one of the strands will be replicated in the same direction as the replication fork and the other will be replicated in the opposite direction of the replication fork. This means that one of the strands is synthesised in a continuous manner (named the leading strand) while the other one is synthesised in fragments (named the lagging strand). The leading strand only needs one primer while the lagging strand needs quite a few as it is formed in fragments. These fragments are called Okazaki fragments. DNA polymerase I will remove the RNA primers and replace these with DNA. The enzyme DNA ligase then joins the Okazaki fragments together to form a continuous strand.

Question 8

When is DNA replication initiated?

Answer 8

DNA replication is initiated at many points in eukaryotic chromosomes.

Question 9

Which direction is transcription carried out in?

Answer 9

Transcription is carried out in a 5’→3’ direction.

Question 10

Distinguish between the sense and antisense strands of DNA.

Answer 10

The antisense strand is the template DNA strand which is transcribed. The sense strand on the other hand is the DNA strand which has the same base sequence as the mRNA with thymine instead or uracil.

Question 11

Distinguish between the sense and antisense strands of DNA.

Answer 11

The antisense strand is the template DNA strand which is transcribed. The sense strand on the other hand is the DNA strand which has the same base sequence as the mRNA with thymine instead or uracil.

Question 12

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

Answer 12

mRNA is produced during transcription. In prokaryotes, RNA polymerase recognises a specific sequence of DNA called the promoter. The promoter basically “tells” the RNA polymerase where to start the transcription process. Transcription is initiated with the binding of RNA polymerase to the promoter site. The RNA polymerase then uncoils the DNA and separates the two strands. One of the strands is used as the template strand for transcription. The RNA polymerase will then use free nucleoside triphosphates to build the mRNA in a 5’→3’ direction. These nucleoside triphosphates bond to their complementary base pairs on the template strand. As they bind they become nucleotides by losing two phosphate groups to release energy. Since RNA does not contain thymine, uracil pairs up with adenine instead. RNA polymerase forms covalent bonds between these nucleotides. It moves along the DNA to keep elongating the sequence of mRNA until it reaches a sequence of DNA called the terminator. This sequence of DNA “tells” the RNA polymerase to stop transcription. The RNA polymerase is then released from the DNA and the newly created mRNA separates from the template DNA strand. Finally, the DNA rewinds back to its original double helical structure.

Question 13

Eukaryotic RNA needs the removal of _____ to form mature mRNA.

Answer 13

Eukaryotic RNA needs the removal of introns to form mature mRNA.

Question 14

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.

Answer 14

There are many different types of tRNA and each tRNA is recognised by a tRNA-activating enzyme. This enzyme binds a specific amino acid to the tRNA by using ATP as an energy source. The tRNA molecule has a specific structure. It contains double stranded sections (due to base pairing via hydrogen bonds) and loops. It has an anticodon loop which contains the anticodon and two other loops. The nucleotide sequence CCA is found at the 3’ end of the tRNA and allows attachment for an amino acid. Each type of tRNA has slightly different chemical properties and three dimensional structure which allows the tRNA-activating enzyme to attach the correct amino acid to the 3’ end of the tRNA. There are 20 different tRNA-activating enzymes as there are 20 different amino acids. Each enzyme will attach a specific amino acid to the tRNA which has the matching anticodon for that amino acid. When the amino acid binds to the tRNA molecule a high energy bond is created. The energy from this bond is used later on to bind the amino acids to the growing polypeptide chain during translation.

Question 15

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

Answer 15

Ribosomes have a particular structure. They are made up of proteins and ribosomal RNA. They have two subunits, one large the other small. On the surface of the ribosome there are three sites to which tRNA can bind to. However not more than two tRNA molecules can bind to the ribosome at one time. Also there is a site on the surface of the ribosome to which mRNA can bind to.

Question 16

What does translation consist of?

Answer 16

Translation consists of initiation, elongation, translocation and termination.

Question 17

What direction does translation go in?

Answer 17

Translation occurs in a 5’→3’ direction.

Question 18

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

Answer 18

Translation occurs in the cytoplasm. It starts off with the tRNA containing the matching anticodon for the start codon AUG binding to the small subunit of the ribosome. This tRNA carries the amino acid methionine and is always the first tRNA to bind to the P site. The small subunit of the ribosome then binds to the 5’ end of the mRNA. This is because translation occurs in a 5’→3’ direction. The small subunit will move along the mRNA until it reaches the start codon AUG. The large subunit of the ribosome can then binds to the small subunit. The next tRNA with the matching anticodon to the second codon on the mRNA binds to the A site of small subunit of the ribosome. The amino acids on the two tRNA molecules then form a peptide bond. Once this is done, the large subunit of the ribosome moves forward over the smaller one.The smaller subunit moves forward to join the larger subunit and as it does so the ribosome moves 3 nucleotides along the mRNA and the first tRNA is moved to the E site to be released. The second tRNA is now at the P site so that another tRNA with the matching anticodon can then bind to the A site. As this process continues the polypeptide is elongated. Once the ribosome reaches the stop codon on the mRNA translation will end as no tRNA will have a matching anticodon to the stop codon. The polypeptide is then released. Many ribosomes can translate the same mRNA at the same time. They will all move along the mRNA in a 5’→3’ direction. These groups of ribosomes on a single mRNA are called polysomes.

Question 19

State that free ribosomes synthesize proteins for use primarily within ____ and that bound ribosomes synthesize proteins primarily for ________________

Answer 19

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

Question 20

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

Answer 20

Primary Structure: The primary structure of a protein is its amino acid sequence. This amino acid sequence is determined by the base sequence of the gene which codes for the protein.
Secondary Structure: Secondary structures have α-helices and β-pleated sheets. These form as a result of hydrogen bonds between the peptide groups of the main chain. Therefore, proteins that contain secondary structures will have regions that are cylindrical (α-helices) and/or regions that are planar (β-pleated sheets).

Tertiary Structure: The tertiary structure of a protein is its three-dimensional conformation which occurs as a result of the protein folding. This folding is stabilised by hydrogen bonds, hydrophobic interactions, ionic bonds and disulphide bridges. These intramolecular bonds form between the R groups of different amino acids.

Quaternary Structure: A quaternary structure is formed when two or more polypeptide chains associate to form a single protein. An example is haemoglobin which consists of four polypeptide chains. In some cases, some proteins can have a non-polypeptide structure called a prosthetic group. These proteins are called conjugated proteins. The haem group in haemoglobin is a prosthetic group.

Question 21

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

Answer 21

Protein shape can be categorised as either fibrous or globular. Fibrous proteins tend to be elongated, physically tough and insoluble in water. Collagen found in the skin and keratin found in hair are examples of fibrous proteins. Globular proteins tend to be compact, rounded and water soluble. Haemoglobin and enzymes are examples of globular proteins.

Question 22

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

Answer 22

Amino acids have different R groups. Some of these R groups will be hydrophilic, making the amino acid polar, while others will be hydrophobic, making the amino acid non-polar. The distribution of the polar and non-polar amino acids in a protein influences the function and location of the protein within the body. Non-polar amino acids are found in the centre of water soluble proteins while the polar amino acids are found at the surface.

Examples of how the distribution of non-polar and polar amino acids affect protein function and location:

Controlling the position of proteins in membranes: The non-polar amino acids cause proteins to be embedded in membranes while polar amino acids cause portions of the proteins to protrude from the membrane.

Creating hydrophilic channels through membranes: Polar amino acids are found inside membrane proteins and create a channel through which hydrophilic molecules can pass through.

Specificity of active site in enzymes: If the amino acids in the active site of an enzyme are non-polar then it makes this active site specific to a non-polar substance. On the other hand, if the active site is made up of polar amino acids then the active site is specific to a polar substance.

Question 23

What is an example of the structural function of proteins?

Answer 23

Collagen strengthens bones, skin and tendons.

Question 24

What is an example of the movement function of proteins?

Answer 24

Myosin found in muscle fibers causes contraction of the muscle which results in movement.

Question 25

What is an example of the transport function of proteins?

Answer 25

Haemoglobin transports oxygen from the lungs to other tissues in the body.

Question 26

What is an example of the defense function of proteins?

Answer 26

Immunoglobulin acts as an antibody.

Question 27

What to metabolic pathways consist of?

Answer 27

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

Question 28

Describe the induced-fit model.

Answer 28

Initially the substrate does not fit perfectly into the active site of the enzyme. When the substrate binds to the active site, this changes the shape of the active site and only then does it perfectly fit the substrate. As the substrate binds it changes the shape of the active site and this weakens the bonds in the substrate and therefore reduces the activation energy. This model is a more precise version of the lock and key one. The reason for this is that it explains why some enzymes can bind to many different substrates. If the shape of the active site changes when a substrate binds, this allows many different but similar substrates to bind to the one enzyme.

Question 29

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

Answer 29

Reactants of a chemical reaction need to gain energy before they can undergo the reaction. This required energy is called the activation energy of the reaction and it is needed to break bonds within the reactants. At a later stage in the reaction energy will be released as new bonds form. The majority of biological reactions are exothermic. In exothermic reactions the energy released by the new bonds formed is greater than the activation energy. In other words, the reaction releases energy. Enzymes make it easier for reactions to occur by decreasing the activation energy required in the reactions that they catalyse.

Question 30

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

Answer 30

Enzyme inhibitors are substances which inhibit enzyme activity. There are competitive enzyme inhibitors and non-competitive inhibitors.

Competitive Inhibitors

These are structurally similar to the substrate of the enzyme and bind to the active site. This means that when a competitive inhibitor binds to the active site of an enzyme, it prevents the substrate from binding to the active site. Only once the inhibitor has been released from the active site can the substrate bind. The inhibitor is called a competitive inhibitor as it competes with the substrate for the active site.

The effects of a competitive inhibitor can be reduced by increasing the substrate concentration. More substrate would successfully bind to the active site than inhibitor and therefore reducing the effect of the inhibition. The maximum rate of reaction or a level very close to the maximum rate of reaction can be reached.

An example of a competitive inhibitor is malonate. Malonate is structurally similar to the substrate succinate. Succinate is found in the Krebs cycle of aerobic respiration and binds to the active site of the dehydrogenase enzyme. Malonate can compete with succinate for the active site and in doing so it can prevent succinate from binding.

Non-Competitive Inhibition

These are not similar to the substrate and they do not bind to the active site of the enzyme. Instead they bind to a different site on the enzyme and change the conformation of the active site. The substrate may still be able to bind to the active site however the enzyme is not able to catalyse the reaction or can only do so at a slower rate.

In the presence of a non-competitive inhibitor, increasing the substrate concentration cannot prevent the inhibitor from binding to the enzyme as the two bind to different sites. Therefore, no matter how high the concentration of substrate is, some of the enzymes will still be inhibited. The maximum rate of reaction will always be lower in the presence of a non-competitive inhibitor.

An example of a non-competitive inhibitor is ATP. When ATP accumulates it binds to a site other than the active site on the enzyme phosphofructokinase. In doing so it changes the enzyme conformation and lowers the rate of reaction so that less ATP is produced.

Question 31

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

Answer 31

Metabolic pathways are made up of many chemical reactions and these reactions are catalysed by enzymes. Often, the product of the last reaction in the pathway inhibits the enzyme that catalyses the first reaction of the pathway. This is called end-product inhibition and it involves non-competitive inhibitors.

The product of the last reaction of the metabolic pathway will bind to a site other than the active site of the enzyme that catalyses the first reaction. This site is called the allosteric site. When it binds to the allosteric site it acts as non-competitive inhibitor and changes the conformation of the active site. Therefore, it makes the binding of the substrate to the enzyme unlikely. Once the inhibitor is released from the allosteric site, the active site returns to its original conformation and the substrate is able to bind again.

There is a clear advantage in using end-product inhibition for controlling metabolic pathways. When there is an excess of end-product, the whole metabolic pathway is shut down as the end product inhibits the first enzyme of the pathway. Therefore less of the end product gets produced and by inhibiting the first enzyme it also prevents the formation of intermediates. When the levels of the end product decrease, the enzymes start to work again and the metabolic pathway is switched on.