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Worked solutions to student book questions Chapter 13 DNA Heinemann Chemistry 2 4th edition Enhanced Copyright © Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1 Q1. Copy the structural formula of deoxyribose sugar. Now indicate where covalent bonds form to link this group to: a a phosphate group b a base group A1. Q2. Give the sequence of bases that will pair with a sequence of 5´-AAACCTGAACGA-3´. A2. TTTGGACTTGCT Q3. a What structural similarities do the purine bases adenine and guanine share? b What structural similarities do the pyrimidine bases cytosine and thymine share? c In terms of the structure of these bases, explain how the two polymer chains of phosphate and deoxyribose sugar in the double helix bring about a constant separation along the length of the DNA molecule. A3. a Adenine and guanine are derivatives of purine (Figure 13.2a). b Cytosine and thymine are derivatives of pyrimidine (Figure 13.2b). c The width of the double helix (the ‘ladder’ when the double helix untwists) is constant because one purine base is hydrogen bonded to one pyrimidine base.

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  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 1

    Q1. Copy the structural formula of deoxyribose sugar. Now indicate where covalent bonds form to link this group to: a a phosphate group b a base group

    A1.

    Q2. Give the sequence of bases that will pair with a sequence of 5-AAACCTGAACGA-3.

    A2. TTTGGACTTGCT

    Q3. a What structural similarities do the purine bases adenine and guanine share? b What structural similarities do the pyrimidine bases cytosine and thymine share? c In terms of the structure of these bases, explain how the two polymer chains of

    phosphate and deoxyribose sugar in the double helix bring about a constant separation along the length of the DNA molecule.

    A3. a Adenine and guanine are derivatives of purine (Figure 13.2a). b Cytosine and thymine are derivatives of pyrimidine (Figure 13.2b). c The width of the double helix (the ladder when the double helix untwists) is

    constant because one purine base is hydrogen bonded to one pyrimidine base.

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 2

    Q4. a i What structural change occurs in the DNA of a person who suffers from

    sickle-cell anaemia? ii How does this change affect the polypeptide chains of their haemoglobin

    molecules? b i What structural change occurs in the DNA of a person who suffers from

    cystic fibrosis? ii How does this change affect the polypeptide chains of their transmembrane

    chloride-transport molecules?

    A4. a i Thymine is replaced by adenine at one point in the DNA sequence on

    chromosome number 11. ii The sixth amino acid in the -globin polypeptide chains is valine, instead of

    glutamic acid. b i A sequence of just three nucleotides is missing on chromosome 7.

    ii One phenylalanine molecule is omitted from sequence of amino acids used to make the chloride transport polypeptide chain.

    E1. Why would the hair sample found on the murdered child not be suitable for producing a DNA profile similar to the ones shown in Figures 13.18on page 218 of the student book?

    AE1. The DNA profiles were prepared using DNA found in the nuclei of cells. Hair cells do not have a nucleus so mitochondrial DNA is used for profiling.

    E2. Suggest the ways in which mitochondrial DNA might become contaminated from an external source before analysis.

    AE2. Mitochondrial DNA may be contaminated from cells:

    discarded by other people or animals (e.g. skin, hair) who have been to the crime scene before collection of samples for forensic analysis

    discarded by police or the analyst if correct procedures (e.g. gloves for handling samples) are not followed

    E3.

    Discuss the similarities and differences in mtDNA of Tsar Nicholas, his wife, their daughters and the imposter.

    AE3. The Tsar and his daughters have same mtDNA profile; the impostor will have a different mtDNA profile from the Tsar. They will have different profiles unless she is related to the Tsar via maternal links.

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 3

    Q5. Make a flow chart of the steps involved in producing DNA profiles such the one shown in Figure 13.15 or 13.16.

    A5.

    Fragments labelled with radioactive DNA or fluorescent dyes

    Position of bands determined by UV light or computer-controlled lasers

    Tissue is collected

    DNA is extracted

    Restriction enzymes cut DNA in segments of varying length

    DNA segments are duplicated through a sequence of denaturation, annealing, and elongation

    Fragments separated by size using electrophoresis

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 4

    Chapter review

    Q6. a Draw structural formulas for the bases adenine and thymine, showing clearly how

    hydrogen bonds form during complementary base pairing. b Draw structural formulas for the bases cytosine and guanine, showing clearly

    how hydrogen bonds form during complementary base pairing.

    A6.

    a, b

    Q7. Sketch a single nucleotide that includes the base thymine.

    A7.

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 5

    Q8. A segment of DNA includes the base sequence GATTATCAA. List the base sequence on the complementary strand of the molecule.

    A8. CTAATAGTTA

    Q9. a List three diseases that result from changes in the base sequence on a gene. b By referring to the inherited disease sickle-cell anaemia, explain how a small

    change in the nucleotide sequence in a protein-coding region of a DNA molecule can have serious biochemical consequences.

    A9. a cystic fibrosis, PKU, sickle-cell anaemia b Just one missing nucleotide in the DNA sequence that codes for haemoglobin

    production causes a defective molecule in red blood cells, so that they change shape where oxygen concentration is low. The red blood cells have a shorter lifespan and this causes anaemia.

    Q10. a What is a repeating base sequence? b Give an example of a repeating base sequence, including the number of the

    chromosome on which it is found. c Explain the significance of repeating base sequences to forensic analysis.

    A10. a Repeating base sequences are found in the non-coding sections of DNA. In these

    regions a sequence of bases may be repeated several times. The number of times these base sequences are repeated varies between individuals.

    b The AGAT base sequence found on chromosome number 5 may be repeated between seven and 15 times.

    c The number of times a base sequences repeats varies between individuals. The number of repeating base sequences at ten locations across various chromosome pairs is considered sufficiently accurate to identify an individual.

    Q11.

    Describe the ways in which those sections of DNA used for forensic analysis can differ from individual to individual.

    A11. The non-coding part of DNA is used in forensic analysis. In these regions a sequence of bases may be repeated. The number of times a sequence of bases is repeated varies from individual to individual. In DNA profiling, the repeating base sequences are cut using special restriction enzymes and thousands of these segments are duplicated. They are separated on the basis of their size by gel electrophoresis. This pattern of bands of DNA fragments is unique to an individual.

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 6

    Q12. Explain why a DNA molecule is able to produce a replica of itself.

    A12. Because nitrogen base pairing ensures that the opposite polymer strand is produced from each half of the double helix

    Q13. How are DNA databases useful for forensic analysis?

    A13. To eliminate individual from suspicion; to identify the culprit of a crime, to identify victims of a natural disaster or terrorist actions

    Q14. Why might the reliability of DNA fingerprinting be questioned and withdrawn as evidence in a court case?

    A14. Contamination of the DNA sample by DNA from another individual.

    Q15. A database of DNA fingerprints of everyone in the country should be established. Discuss the merits and disadvantages of this proposal with your class members. You might consider the following questions in your discussion. a Should such a database contain both coding and non-coding DNA sequences? b Who should have access to a DNA database? Police and security services?

    Private organisations such as insurance companies?

    A15.

    Issues that could be considered: Non coding sequences vary between individuals while coding sequences are

    similar. DNA profiles uses to non coding sequences to identify individuals Coding sequence could be used by insurance companies to determine whether

    or not an individual suffers from a disease and deny them insurance cover.

    Q16. a How does mitochondrial DNA (mtDNA) differ from nuclear DNA? b Give an example to illustrate the usefulness of mtDNA in forensic analysis.

    A16. a Nuclear DNA is only found in the 46 chromosomes in the nucleus of the cell and

    contain the genetic code that controls the protein synthesis. MtDNA is found outside the nucleus and controls energy releasing reactions. Because cells contain many thousands of copies of mtDNA, there is in some circumstances a better chance of extracting this than extracting nuclear DNA

    b Identification of the bodies of the Tsar and his family 60 years after their execution.

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 7

    Q17. Distinguish between the primary, secondary and tertiary structures of DNA. A17. The sequence of nitrogen bases along the sugarphosphate backbone that forms the primary structure; covalent bonds are responsible for producing the primary structure of DNA. The right-handed double helix formed a pair of DNA polynucleotide strands is the secondary structure; hydrogen bonding is responsible for maintaining the secondary structure. The tertiary structure refers to the way DNA molecules wrap around histones and become supercoiled. Q18. a Outline the main steps in the polymerase chain reaction (PCR). b Why is PCR such an important tool in forensic analysis? A18. a Each cycle involves three stages:

    1. Denaturation: The DNA fragments are heated so that hydrogen bonding between the DNA strands are broken.

    2. Annealing: The mixture is cooled so that primers bond to the start and end of the sequence to be copied.

    3. Elongation: The mixture is heated so that complementary base pairs are added to the single DNA strands to form a double strand of DNA.

    b to increase the amount of DNA available for testing Q19.

    Figure 13.19 Figure 13.19 shows a small segment of a strand of DNA.

    a Identify that part of the strand that represents a nucleotide. b Name the species that have combined to form the nucleotide. c What type of reactions are involved in the formation of nucleotides?

  • Worked solutions to student book questions Chapter 13 DNA

    Heinemann Chemistry 2 4th edition Enhanced Copyright Pearson Australia 2010 (a division of Pearson Australia Group Pty Ltd) 8

    A19. a Adenosine monophosphate nucleotide (top), cytosine monophosphate nucleotide

    (lower) b Phosphate, deoxyribose, adenine (top). Phosphate, deoxyribose, cytosine (lower) c Condensation reactions, as shown in Figure 13.6 on page 207 of the student book.