UNIT II Molecular Biology

DNA and RNA : The Molecular Basis       

DNA and the Importance of Proteins DNA Structure and Function DNA Structure Base pairing in DNA Replication The repair of genetic information The DNA code RNA Structure and Function

Q 1.

State the names of the four bases in DNA.

Adenine, Guanine, Cytosine and Thymine. Q 2. Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and phosphate. A nucleotide is made of the sugar deoxyribose, a base (which can be either adenine, guanine, cytosine or thymine) and a phosphate group. Below is a representation of a nucleotide. Q 3. Explain how a DNA double helix is formed using complementary base pairing and hydrogen bonds. DNA is made up of two nucleotide strands. The nucleotides are connected together by covalent bonds within each strand. The sugar of one nucleotide forms a covalent bond with the phosphate group of another. The two strands themselves are connected by hydrogen bonds. The hydrogen bonds are found between the bases of the two strands of nucleotides. Adenine forms hydrogen bonds with thymine whereas guanine forms hydrogen bonds with cytosine. This is called complementary base pairing. Below is a digram showing the molecular structure and bonds within DNA. Q 4. Explain DNA replication in terms of unwinding the double helix and separation of the strands by helicase, followed by formation of the new complementary strands by DNA polymerase. DNA replication is semi-conservative as both of the DNA molecules produced are formed from an old strand and a new one. The first stage of DNA replication involves the unwinding of the double strand of DNA (DNA double helix) and separating them by breaking the hydrogen bonds between the bases. This is done by the enzyme helicase. Each separated strand now is a template for the new strands. There are many free nucleotides around the replication fork which then bond to the template strands. The free nucleotides form hydrogen bonds with their complimentary base pairs on the template strand. Adenine will pair up with thymine and guanine will pair up with cytosine. DNA polymerase is the enzyme responsible

for this. The new DNA strands then rewind to form a double helix. The replication process has produced a new DNA molecule which is identical to the initial one.

Q 5.

Draw and label a simple diagram of the molecular structure of DNA.

Q 6. Explain the significance of complementary base pairing in the conservation of the base sequence of DNA. Complementary base pairing is very important in the conservation of the base sequence of DNA. This is because adenine always pairs up with thymine and guanine always pairs up with cytosine. As DNA replication is semi-conservative (one old strand an d one new strand make up the new DNA molecules), this complementary base pairing allows the two DNA molecules to be identical to each other as they have the same base sequence. The new strands formed are complementary to their template strands but also identical to the other template. Therefore, complementary base pairing has a big role in the conservation of the base sequence of DNA.

Q 7.

Compare the structure of RNA and DNA.

DNA and RNA both consist of nucleotides which contain a sugar, a base and a phosphate group. However there are a few differences. Firstly, DNA is composed of a double strand forming a helix whereas RNA is only composed of one strand. Also the sugar in DNA is deoxyribose whereas in RNA it is ribose. Finally, both DNA and RNA have the bases adenine, guanine and cytosine. However DNA also contains thymine which is replaced by uracil in RNA. Q 8.

State that DNA replication is semi- conservative. (Assignment Question 2)

Q 9. State that DNA replication occurs in a 5? → 3? direction. (Assignment Question 3)

Q 10.

Describe the genetic code in terms of codons composed of triplets of bases.

A triplet of bases (3 bases) forms a codon. Each codon codes for a particular amino acid. Amino acids in turn link to form proteins. Therefore DNA and RNA regulate protein synthesis. The genetic code is the codons within DNA and RNA, composed of triplets of bases which eventually lead to protein synthesis. Q 11. Outline DNA transcription in terms of the formation of an RNA strand complementary to the DNA strand by RNA polymerase. DNA transcription is the formation of an RNA strand which is complementary to the DNA strand. The first stage of transcription is the uncoiling of the DNA double helix. Then, the free RNA nucleotides start to form an RNA strand by using one of the DNA strands as a template. This is done through complementary base pairing, however in the RNA chain, the base thymine is replaced by uracil. RNA polymerase is the enzyme involved in the formation of the RNA strand and the uncoiling of the double helix. The RNA strand then elongates and then separates from the DNA template. The DNA strands then reform a double helix. The strand of RNA formed is called messenger RNA. Q 12. 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 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. Summary:           

Helicase uncoils the DNA RNA primase adds short sequences of RNA to both strands (the primer) The primer allows DNA polymerase III to bind and start replication DNA polymerase III adds nucleotides to each template strand in a 5'→3' direction These nucleotides are initially deoxyribonucleoside triphosphates but they lose two phosphate groups during the replication process to release energy One strand is replicated in a continuous manner in the same direction as the replication fork (leading strand) The other strand is replicated in fragments (Okazaki fragments) in the opposite direction (lagging strand) DNA polymerase I removes the RNA primers and replaces them with DNA DNA ligase then joins the Okazaki fragments together to form a continuous strand State that DNA replication is initiated at many points in eukaryotic chromosomes. DNA replication is initiated at many points in eukaryotic chromosomes. Q 13.

Discuss the relationship between one gene and one polypeptide.

A polypeptide is formed by amino acids liking together through peptide bonds. There are 20 different amino acids so a wide range of polypeptides are possible. Genes store the information required for making polypeptides. The information is stored in a coded form by the use of triplets of bases which form codons. The sequence of bases in a gene codes for the sequence of amino acids in a polypeptide. The information in the genes is decoded during transcription and translation leading to protein synthesis.

Synthetic Biology Protein Synthesis : Central dogma Step-1 : Transcription --- Making RNA Step -2 : Translation----Making Protein The Control of Protein Synthesis Controlling Protein Quantity Controlling Protein Quality

Transcription Q 14. State that transcription is carried out in a 5’ → 3’ direction. (Assignment Question 4) Q 15.

Distinguish between the sense and antisense strands of DNA.

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. Q 16. Explain the process of transcription in prokaryotes, including the role of the promoter region, RNA polymerase, nucleoside triphosphates and the terminator. 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. Summary:              

RNA polymerase binds to the promoter region This initiates transcription RNA polymerase uncoils the DNA Only one strand is used, the template strand Free nucleoside triphosphates bond to their complementary bases on the template strand Adenine binds to uracil instead of thymine As the nucleoside triphosphates bind they become nucleotides and release energy by losing two phosphate groups The mRNA is built in a 5'→3' direction RNA polymerase forms covalent bonds between the nucleotides and keeps moving along the DNA until it reaches the terminator The terminator signals the RNA polymerase to stop transcription RNA polymerase is released and mRNA separates from the DNA The DNA rewinds State that eukaryotic RNA needs the removal of introns to form mature mRNA. Eukaryotic RNA needs the removal of introns to form mature mRNA.

Translation Q 17. 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. 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 Q 18. Outline the structure of ribosomes, including protein and RNA composition, large and small subunits, three tRNA binding sites and mRNA binding sites.

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. Q 19. State that translation consists of initiation, elongation, translocation and termination. Translation consists of initiation, elongation, translocation and termination.

Q 20. State that translation occurs in a 5? → 3? direction. (Assignment Question 5) Q 21.

Explain the process of translation, leading to polypeptide formation.

Translation is the process through which proteins are synthesized. It uses ribosomes, messenger RNA which is composed of codons and transfer RNA which has a triplet of bases called the anticodon. The first stage of translation is the binding of messenger RNA to the small subunit of the ribosome. The transfer RNA’s have a specific amino acid attached to them which corresponds to their anticodons. A transfer RNA molecule will bind to the ribosome however it’s anticodon must match the codon on the messenger RNA. This is done through complementary base pairing. These two form a hydrogen bond together. Another transfer RNA molecule then bonds. Two transfer RNA molecules can bind at once. Then the two amino acids on the two transfer RNA molecules form a peptide bond. The first transfer RNA then detaches from the ribosome and the second one takes it’s place.The ribosome moves along the messenger RNA to the next codon so that another transfer RNA can bind. Again, a peptide bond is formed between the amino acids and this process continues. This forms a polypeptide chain and is the basis of protein synthesis. Q 22. Draw and label a diagram showing the structure of a peptide bond between two amino acids. Q 23. Explain the process of translation, including ribosomes, polysomes, start codons and stop codons. 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. Summary:   

The tRNA containing the matching anticodon to the start codon binds to P site of the small subunit of the ribosome 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 The large subunit then binds to the smaller one

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The next tRNA with the matching anticodon to the next codon on the mRNA binds to the A site The amino acids on the two tRNA molecules form a peptide bond The larger subunit moves forward over the smaller one 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 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 As this process continues, the polypeptide is elongated Once the ribosome reaches the stop codon on the mRNA translation ends and the polypeptide is released Many ribosomes can translate a single mRNA at the same time, these groups of ribosomes are called polysomes Q 24. Explain the four levels of protein structure, indicating the significance of each level.

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 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. Q 25. Outline the difference between fibrous and globular proteins, with reference to two examples of each protein type. 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. Q 26.

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

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. Q 27. 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. Q 28.

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

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.