7. Nucleic acids and proteins

Question 1

Describe the structure of DNA, including the antiparal- lel 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 consist 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 in- trons.

Answer 5

Eukaryotic genes can contain exons and introns.

Question 6

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

Answer 6

DNA replication occurs in a 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.

1. Helicase uncoils the DNA
2. RNA primase adds short sequences of RNA to both strands (the primer)
3. The primer allows DNA polymerase III to bind and start replication
4. DNA polymerase III adds nucleotides to each template strand in a 5’3’ direction
5. These nucleotides are initially deoxyribonucleoside triphosphates but they lose two phosphate groups during the replication process to release energy
6. One strand is replicated in a continuous manner in the same direction as the replication fork (leading strand)
7. The other strand is replicated in fragments (Okazaki fragments) in the opposite direction (lagging strand)
8. DNA polymerase I removes the RNA primers and replaces them with DNA
9. DNA ligase then joins the Okazaki fragments together to form a continuous strand

Question 8

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

Answer 8

DNA replication is initiated at many points in eukaryotic chromosomes.

Question 9

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

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

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

Answer 11

Summary:

1. RNA polymerase binds to the promoter region
2. This initiates transcription
3. RNA polymerase uncoils the DNA
4. Only one strand is used, the template strand
5. Free nucleoside triphosphates bond to their complementary bases on the template strand
6. Adenine binds to uracil instead of thymine
7. As the nucleoside triphosphates bind they become nucleotides and release energy by losing two phosphate groups
8. The mRNA is built in a 5’3’ direction
9. RNA polymerase forms covalent bonds between the nucleotides and keeps moving along the DNA until it reaches the terminator
10. The terminator signals the RNA polymerase to stop transcription
11. RNA polymerase is released and mRNA separates from the DNA
12. The DNA rewinds

Question 12

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

Answer 12

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

Question 13

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 13

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.

Summary:

• Each tRNA activating enzyme recognises a specific tRNA molecule
• The tRNA molecule is made up of double stranded sections and loops
• At the 3’ end of the tRNA there is the nucleotide sequence CCA to which the amino acid attaches to
• The different chemical properties and three dimensional structure of each tRNA allows the tRNA-activating enzymes to recognise their specific tRNA
• Each tRNA enzyme binds a specific amino acid to the tRNA molecule
• The tRNA-activating enzyme will bind the amino acid to the tRNA with the matching anticodon
• Energy from ATP is needed during this process

Question 14

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

Answer 14

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 15

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

Answer 15

Translation consists of initiation, elongation, translocation and termination.

Question 16

State that translation occurs in a 5’ 3’ direction.

Answer 16

Translation occurs in a 5’3’ direction.

Question 17

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

Answer 17

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 contain- ing 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.

Summary:

1. The tRNA containing the matching anticodon to the start codon binds to P site of the small subunit of the ribosome
2. The small subunit binds to the 5’ end of the mRNA and moves along in a 5’3’ direction until it reaches the start codon
3. The large subunit then binds to the smaller one
4. The next tRNA with the matching anticodon to the next codon on the mRNA binds to the A site
5. The amino acids on the two tRNA molecules form a peptide bond
6. The larger subunit moves forward over the smaller one
7. The smaller subunit rejoins the larger one, this moves the ribosome 3 nucleotides along the mRNA and moves the first tRNA to the E site to be released
8. The second tRNA is now at the P site so that another tRNA with the matching anticodon to the codon on the mRNA can bind to the A site
9. As this process continues, the polypeptide is elongated
10. Once the ribosome reaches the stop codon on the mRNA translation ends and the polypeptide is released
11. Many ribosomes can translate a single mRNA at the same time, these groups of ribosomes are called polysomes

Question 19

State that free ribosomes synthesize proteins for use pri- marily within the cell, and that bound ribosomes syn- thesize proteins primarily for secretion or for lysosomes.

Answer 19

Free ribosomes synthesise proteins for use primarily within the cell while bound ribosomes synthesise proteins primarily for secretion or for lysosomes.

Question 20

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

Answer 20

There are four levels of protein structure:

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 -helices and -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 (-helices) and/or re- gions that are planar (-pleated sheets).

Tertiary Structure : The tertiary structure of a protein is its three- dimensional conformation which occurs as a result of the protein fold- ing. This folding is stabilised by hydrogen bonds, hydrophobic interac- tions, 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 exam- ple 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 pro- teins, 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. Col- lagen 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 hy- drophilic, 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

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

Answer 23

Function

Example

Structural

Collagen strengthens bones, skin and tendons.

Movement

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

Transport

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

Defense

Immunoglobulin acts as an antibody.

Question 24

State that metabolic pathways consist of chains and

cycles of enzyme-catalysed reactions.

Answer 24

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

Question 25

## Footnote Describe the induced-fit model.

Answer 25

## Footnote 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 sub- strates. 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 26

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

Answer 26

## Footnote Reactants of a chemical reaction need to gain energy before they can un- dergo 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 en- ergy. 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 27

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

Answer 27

## Footnote 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 con- centration. 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 sub- strate 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.