• Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

Nucleic Acids and Proteins Exam Prep. Notes

Extracts from this document...


Topic 7 – Nucleic Acids and Proteins 7.1 – DNA Structure 7.1.1 Describe the structure of DNA, including the antiparallel strands, 3’–5’ linkages and hydrogen bonding between purines and pyrimidines. ï At one end of each strand is a phosphate linked to carbon atom 5 of deoxyribose. This is the 5’ terminal. ï At the other end of the strand is a hydroxyl group attached to carbon atom 3 of deoxyribose. This is therefore the 3’ terminal. ï There are two hydrogen bonds between adenine and thymine and three hydrogen bonds between guanine and cytosine. Pyrimidine = thymine and cytosine; Purines = guanine and adenine. 7.1.2 Outline the structure of nucleosomes. Nucleosomes ï In Eukaryotes, the DNA is associated with proteins to form nucleosomes – globular structures that contain eight histone proteins, with DNA wrapped around. Another histone protein bonds the structure together. In an interphase nucleus in eukaryotes the DNA resembles a string of beads. ï Prokaryotes have naked DNA because there are no histone proteins. 7.1.3 State that nucleosomes help to supercoil chromosomes and help to regulate transcription. Nucleosomes have two functions: They help to package up the DNA during mitosis and meiosis by the process of supercoiling. They can be used to mark particular genes, either to promote gene expression by transcription and translation, or to cause silencing of a gene by preventing transcription. ...read more.


7.2.2 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. Leading strand: DNA replication takes place towards the replication fork. Lagging strand: DNA replication takes place away from the replication fork. 7.2.3 State that DNA replication is initiated at many points in eukaryotic chromosomes. The rate of replication in fruit flies (Drosophila) is 2600 nucleotides/minute. The largest chromosome of Drosophila is 6.5 x 107 nucleotides. If DNA replication started at both ends of the chromosome, it would take 8.5 days to replicate the chromosome. In fact, it only takes 3- 4 minutes. In order to explain this discrepancy, scientists have determined that replication must start at many points along the same DNA helix at the same time. This process is needed in eukaryotic cells because they have very large amounts of DNA, which would take much longer to replicate if replication was done from one end of the chromosomes to the other. 7.3 Transcription 7.3.1 State that transcription is carried out in a 5? ? 3? direction. Transcription is the enzyme-controlled process of synthesizing RNA from a DNA template. It is carried out in a 5? to 3? direction (of the new RNA strand). This means that new RNA nucleotides are added to the 3? end of the growing RNA strand. ...read more.


Termination Initiation mRNA binds to the small sub-unit of the ribosome ribosome slides along mRNA to start codon anticodon of tRNA pair with codon on mRNA complementary base pairing between codon and anti-codon anticodon of tRNA with methionine pairs with start codon/AUG is start codon second tRNA pairs with next codon Elongation Peptide bond forms between amino acids Ribosome moves along the mRNA by one codon Movement in 5? to 3? direction tRNA that has lost its amino acid detaches Another tRNA pairs with the next codon/moves to A site tRNA activating enzymes Link amino acids to specific tRNA 3. Termination 7.4.4 State that translation occurs in a 5? ? 3? direction. During translation, the ribosome moves along the mRNA towards the 3? end. The start codon is nearer to the 5? end. There are 3 stop codons. 7.4.5 Draw and label a diagram showing the structure of a peptide bond between two amino acids. 7.4.6 Explain the process of translation, including ribosomes, polysomes, start codons and stop codons. Codon: three consecutive bases in DNA (or RNA), which specify an amino acid. Anticodon: three consecutive bases in tRNA, complementary to a codon on RNA. 7.4.7 State that free ribosomes synthesize proteins for use primarily within the cell, and that bound ribosomes (in RER) synthesize proteins primarily for secretion or for lysosomes. 7.5 Proteins 7.5.1 Explain the four levels of protein structure, indicating the significance of each level. The bonds in a tertiary structure form between R groups. ...read more.

The above preview is unformatted text

This student written piece of work is one of many that can be found in our International Baccalaureate Biology section.

Found what you're looking for?

  • Start learning 29% faster today
  • 150,000+ documents available
  • Just £6.99 a month

Not the one? Search for your essay title...
  • Join over 1.2 million students every month
  • Accelerate your learning by 29%
  • Unlimited access from just £6.99 per month

See related essaysSee related essays

Related International Baccalaureate Biology essays

  1. Biology - Observing the Process of Guttation

    The green bean sprouts emerged from beneath the soil on Day 4, one day before those grown in the exposed containers (shown in Figure 12 below). Figure 12. The growth of green bean sprouts in exposed containers D, E, and F is shown in this graph.

  2. What is the time-course of enzyme to catalyze the breakdown of a protein, into ...

    For example, if the samples give an absorbance (AU) of 0.0478, then I will compare this result to the graphical representation of the calibration results and plot to see the concentration of the sample. Data Collection and Processing: Table 1: Calibration of Colorimeter Protein Concentration (%) Absorbance AU) 1 0.607 0.8 0.565 0.6 0.489 0.4 0.380 0.2 0.234 0.0

  1. DNA Structure questions. Outline DNA nucleotide structure in terms of sugar (deoxyribose), base and ...

    According to the base pair rule, the opposite complementary bases attract to one another which is why DNA nucleotides can be linked together into a single strand (Bowen 1).

  2. IB Biology notes on infection and the body's responses.

    The reason for vaccination III. Active versus Passive Immunity A. active immunity: immunity that results from the body producing its own antibodies after exposure to a particular antigen 1. can be induced by vaccination with a microbe that has been treated so that it still has antigen but is not able to cause disease B.

  1. Genetics NOTES

    a particular characteristic - Capital letters are used to indicate dominance - Co-dominance is the expression of both alleles in a heterozygous individual - Incomplete dominance is a situation where neither gene dominates the other nor both exercise an influence on the individual.

  2. IB Genetic Unit Notes

    They occur in a diploid cell, contain the same sequence of genes, but have come from different parents. 4.3.4: Outline the process of meiosis, including pairing of homologous chromosomes and crossing over, followed by two divisions, which results in four haploid cells.

  1. Experimental Design Notes

    Poverty level of the world o At least of 80 percent of the world live off of 10 used * Descriptive Stats o numerical summaries of data o Mean: average of data points o Range: measure of the spread of data.

  2. Penicillin - its discovery, properties and uses.

    of the bacteria staphylococcus and streptococcus, which helped in the development of penicillin. Everything changed in 1928 by a British bacteriologist Alexander Fleming17. In September, 1928, Fleming returned from holiday and began to sort through the mess in his laboratory.

  • Over 160,000 pieces
    of student written work
  • Annotated by
    experienced teachers
  • Ideas and feedback to
    improve your own work