Corrections

• Describe how a gene is a code for the production of a polypeptide.

  Do not include information about transcription or translation in your answer. (3)

  • Because base sequence
  • In triplets
  • Determines sequence of amino acid in polypeptide
• Describe how a phosphodiester bond is formed between two nucleotides within a DNA molecule. (2)

  • Condensation reaction
  • Between phosphate and deoxyribose
  • Catalysed by DNA polymerase
• Complete Table 1 to show three differences between DNA in the nucleus of a plant cell and DNA in a prokaryotic cell. (3)

  DNA in the nucleus of a plant cell:

  1. Linear
  2. Introns
  3. Longer

  DNA in a prokaryotic cell:

  1. Circular
  2. No introns
  3. Shorter
• Not all mutations in the nucleotide sequence of a gene cause a change in the structure of a polypeptide.

  Give two reasons why. (2)

  • Triplets code for same amino acid
  • Occurs in introns
• Compare and contrast the DNA in eukaryotic cells with the DNA in prokaryotic cells. (5)

  Comparison:

  • Nucleotide structure is identical.
  • Nucleotides joined by phosphodiester bond.

  Contrast:

  • Eukaryotic DNA is longer.
  • Eukaryotic DNA contain introns whereas prokaryotic DNA does not.
  • Eukaryotic DNA is linear whereas prokaryotic DNA is circular.
  • Eukaryotic DNA is associated with histones whereas prokaryotic DNA is not.
• Figure 1 shows all the chromosomes present in one human cell during mitosis.

  A scientist stained and photographed the chromosomes.

  In Figure 2, the scientist has arranged the images of these chromosomes in homologous pairs.

  The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.

  Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas. (1)

  • Differences in base sequence.
• In Figure 2, the chromosomes are arranged in homologous pairs.

  What is a homologous pair of chromosomes? (1)

  • 2 chromosomes that carry the same genes at the same loci.
• Explain five properties that make water important for organisms. (5)

  • A metabolite in condensation reaction.
  • A solvent so metabolic reactions can occur.
  • High heat capacity so buffers changes in temperature.
  • Large latent heat of vaporisation so provides a cooling effect through evaporation.
  • Cohesion between water molecules so produces surface tension supporting small organisms.
• Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample. (5)

  Lipid:

  • Add ethanol and shake.
  • Then add water.
  • White emulsion.

  Non-reducing sugar:

  • Do Benedict's test and stays blue.
  • Boil with HCl.
  • Neutralise with Sodium hydrogencarbonate.
  • Heat with Benedict's and becomes red precipitate.

  Amylase:

  • Add Biuret reagent and becomes purple.
  • Add starch.
  • Leave for a time.
  • Test for reducing sugar.
• Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

  Give two named examples of polymers and their associated monomers to illustrate your answer. (5)

  • A condensation reaction joins monomers together and forms a chemical bond and releases water.
  • A hydrolysis reaction breaks a chemical bond between monomers and uses water.
  • Amino acid and polypeptide - peptide bond.
  • Nucleotide and DNA - phosphodiester bond.
• Compare and contrast the DNA in eukaryotic cells with the DNA in prokaryotic cells. (5)

  Comparison:

  • Nucleotide structure is identical.
  • Nucleotides joined by phosphodiester bond.

  Contrast:

  • Eukaryotic DNA is longer.
  • Eukaryotic DNA contains introns whereas prokaryotic DNA does not.
  • Eukaryotic DNA is linear whereas prokaryotic DNA is circular.
  • Eukaryotic DNA is associated with histones whereas prokaryotic DNA is not.
• Haemoglobins are chemically similar molecules found in many different species.

  Differences in the primary structure of haemoglobin molecules can provide evidence of phylogenetic (evolutionary) relationships between species.

  Explain how. (5)

  • Mutations change base sequence.
  • Causing change in amino acid sequence.
  • Mutations build up over time.
  • More mutations in base sequence between distantly related species.
  • Distantly related species have earlier common ancestor.
• Describe the gross structure of the human gas exchange system and how we breathe in and out. (6)

  • Trachea, bronchi, bronchioles, alveoli.

  Breathing in:

  • Diaphragm contracts and external intercostal muscles contract.
  • Causes volume increase and pressure decrease in thoracic cavity to below atmospheric, resulting in air moving in.

  Breathing out:

  • Diaphragm relaxes and internal intercostal muscles contract.
  • Causes volume decrease and pressure increase in thoracic cavity to above atmospheric, resulting in air moving out.
• Mucus produced by epithelial cells in the human gas exchange system contains triglycerides and phospholipids.

  Compare and contrast the structure and properties of triglycerides and phospholipids. (5)

  Comparison:

  • Both contain ester bonds between glycerol and fatty acid.
  • Fatty acids on both may be saturated or unsaturated.
  • Both are insoluble in water.

  Contrast:

  • Triglyceride has three fatty acids and phospholipid has two fatty acids plus phosphate group.
  • Triglycerides are hydrophobic and phospholipids have hydrophilic and hydrophobic region.
• Mucus also contains glycoproteins.

  One of these glycoproteins is a polypeptide with the sugar, lactose, attached.

  Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein. (4)

  • Glucose and galactose.
  • Joined by condensation reaction.
  • Joined by glycosidic bond.
  • Added to polypeptide in Golgi apparatus.
• Describe two differences between the structure of a tRNA molecule and the structure of an mRNA molecule. (2)

  • tRNA is clover leaf shaped whereas mRNA is linear.
  • tRNA has an amino acid binding site whereas mRNA does not.
• In eukaryotic cell, the structure of the mRNA used in translation is different from the structure of the pre-mRNA produced by transcription.

  Describe and explain a difference in the structure of these mRNA molecules. (2)

  • mRNA has fewer nucleotides.
  • Because of splicing.
• Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation. (3)

  • tRNA brings specific amino acid to ribosome.
  • Anticodon on tRNA binds to codon on mRNA.
  • Amino acids join by condensation reaction, using ATP.
• Give two structural differences between a molecule of messenger RNA (mRNA) and a molecule of transfer RNA (tRNA). (2)

  • mRNA is linear whereas tRNA is clover leaf.
  • mRNA has more nucleotides.
• A scientist investigated changes in the amino acid sequence of a human enzyme, resulting from mutations.

  All these amino acid changes result from single base substitution mutations.

  This enzyme is a polypeptide 465 amino acids long.

  Table 2 shows the result of three of the base substitutions (table not needed)

  What is the minimum number of bases in the gene coding for this polypeptide? (1)

  • 465 x 3 = 1395
• In a eukaryotic cell, the base sequence of the mRNA might be different from the sequence of the pre-mRNA.

  Explain why. (2)

  • Introns in pre-mRNA.
  • Splicing.
• The diagram shows part of a DNA molecule.

  5 base pairs = 1.7nm

  The length of a gene is described as the number of nucleotide base pairs it contains.

  Use information in diagram to calculate the length of a gene containing 4.38 x 103 base pairs. (2)

  5 base pairs = 1.7nm

  4.38 x 103 / 5 = 876

  1.7 x 876 = 1489