IB Biology Review Test 1 Part 2

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IB Biology Review Test 1 Part 2
2011-03-24 00:09:04
IB Biology Review Test Part transcription translation DNA Replication

IB Biology Review Test 1 Part 2
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  1. Explain the consequence of a base substitution mutation in relation to the processes of transcription and translation, using the example of sickle-cell anemia
    Sickle cell anaemia is a genetic disease that affects red blood cells in the body. It is due to a mutation on the Hb gene which codes for a polypeptide of 146 amino acids which is part of haemoglobin (haemoglobin is an important protein component in red blood cells). In sickle cell anaemia the codon GAG found in the normal Hb gene is mutated to GTG. This is called a base substitution mutation as adenine (A) is replaced by thymine (T). This means that when the mutated gene is transcribed, a codon in the messenger RNA will be different. Instead of the normal codon GAG, the messenger RNA will contain the codon GUG. This in turn will result in a mistake during translation. In a healthy individual the codon GAG on the messenger RNA matches with the anticodon CUC on the transfer RNA carrying the amino acid glutamic acid. However, if the mutated gene is present then GUG on the messenger RNA matches with the anticodon CAC on the transfer RNA which carries the amino acid valine. So the base substitution mutation has caused glutamic acid to be replaced by valine on the sixth position on the polypeptide. This results in haemoglobin S being present in red blood cells instead of the normal haemoglobin A. This has an effect on the phenotype as instead of normal donut shaped red blood cells being produced some of the red blood cells will be sickle shaped. As a result these sickle shaped red blood cells cannot carry oxygen as efficiently as normal red blood cells would. However, there is an advantage to sickle cell anemia. The sickle cell red blood cells give resistance to malaria and so the allele Hbs on the Hb gene which causes sickle cell anemia is quite common in parts of the world where malaria is found as it provides an advantage over the disease.
  2. Explain the process of DNA Replication in prokaryotes.
    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
  3. Explain the process of transcription in prokaryotes
    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.
  4. RNA needs the removal of ____ to form mature mRNA
  5. Explain the process of translation in eukaryotes
    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.
  6. List examples of proteins and their functions
    • Enzymatic Proteins: selective acceleration of chemical reactions Ex) Digestive enzymes
    • Structural Proteins: Support ex) collagen
    • Storage Proteins: storage of amino acids ex) Ovalbumin
    • Transport Proteins: transport substance ex)Hemoglobin, transports oxygen
    • Hormonal Proteins: coordination of activites ex) insulin
    • Receptor Proteins: response of cell to chemical stimuli ex) receptors on nerve cells
    • Motor Proteins: movement Ex) actin and myosin
    • Defensive Proteins: protection against disease ex) antibodies
  7. Draw and label a diagram showing the structure of a peptide bond between two amino acids.