What is an ionic bond? What is the final product of an ionic bond?

 

Molecular Biology & Biochemistry ­ Part I

Time allowed: 1 hour and 30 minutes

Total marks available for this paper: 50

  • Answer all questions in the spaces provided on the examination paper
  • The marks available for each question are indicated on the paper
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For marker use only:

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Answer all questions in the spaces provided

  1. a) The molecular weight of one deoxyribose nucleotide is approximately 320g/mol. Calculate the number of moles in 1 microgram (µg) of a single stranded DNA of 200 bases length. (3 marks)

200 base strand has the Mw of 320 g/mol x 200 = 64,000 g/mol (1 mark)

1 µg = 0.001 mg = 0.000001 g (1 mark)

Therefore there are 0.000001/64000 moles = 1.56 x 10­11 = 1.56 x 10­8 mMoles = 0.00001563 µMoles = 0.01563 nMoles = 15.625 pMoles (1 mark)

  1. What is an ionic bond? What is the final product of an ionic bond? (2 marks)

Answer: Ionic bonds are formed as a result of electrostatic attraction of two atoms and transfer of electrons from one atom to another (1 mark) resulting in the production of a salt (1 mark).

  1. Propionic acid has a pKa of 4.87, calculate the ratio of conjugate base to acid at a pH of 5.0. (3 marks)

Using the Henderson­Hasselbalch equation,

­

pH = pKa + log ([A ]/[HA]) (1 mark) ­

5.0 = 4.87 + log ([A ]/[HA]) (1 mark) log ([A ]/[HA])= 0.13

0.13 = 1.35 (1 mark)

[A ]/[HA]= 10

2          In a double stranded conformation, which of these DNA sequences has a lower melting temperature? Circle the correct answer (1 mark).

  1. GGAATCGGCGTCCGG
  2. GTTTATAACATGAATC

Justify your answer briefly (2 marks).

  1. ii) (1 mark)

ii has a lower melting temperature because it has a higher AT content. AT forms two H­bonds per base whereas GC forms 3 H­bonds which are more difficult to break apart. High GC content requires energy to melt. (2 marks)

  1. Describe one difference between DNA and RNA? (1 mark)

Any of below:

  • RNA uses Uracil instead of Thymine
  • RNA is single stranded though it does fold to make secondary structures, but DNA is double stranded (strands in anti­parallel) ● Different sugar: RNA has 2’ OH instead of 2’H (in DNA)
  1. What are histones? (2 marks)

Positively charged proteins involved in DNA packaging. DNA wraps around histones to form nucleosomes.

  1. Name the two enzymes responsible for polymerisation of DNA and state which strand each enzyme synthesises. (2 marks)
    1. DNA Polymerase III (leading and lagging strands)
    2. DNA Polymerase I (lagging strand)
    3. a) In the diagram below, pK1 = 2.19, pKR = 4.25 and pK2 = 9.67. Calculate the isoelectric point of glutamate. (1 mark).

(pK1 + pKR)/2 = 3.22

  1. In a protein, why might this calculated isoelectric point not match the experimental isoelectric point? (1 mark)

The ionisation of a sidechain in a folded protein is affected by its environment.

  1. Why are charged amino acid side chains important in the formation of protein structures? (1 mark)

Side chains of opposite charge can interact to form a salt bridge/ion pair.

  1. Which type of bond plays a critical role in the formation of α helices and between which groups within an α helix do these bonds form? (2 marks)

Hydrogen bonds are important (1 mark) and form between backbone NH and CO groups four residues apart (1 mark).

  1. What are London dispersion forces and why are they important in stabilising protein structure? (3 marks)

They are charge­charge interactions between two induced dipoles (1 mark), formed as a result of the charge distribution around an atom at any instant not being perfectly symmetric (1 mark). These forces are important because there are many of them within a protein (1 mark).

 

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4          The table below provides data on the purification of an enzyme. Calculate the missing value. (1 mark)

Fraction volume (ml) Total protein (mg) Activity (units) Specific activity

(units/mg)

Step 1 1500 15000 150000 10
Step 2 200 3500 140000 40
Step 3 70 470 41
Step 4 40 120 18000 150
Step 5 3 1 15000 15000
Fraction volume (ml) Total protein (mg) Activity (units) Specific activity

(units/mg)

Step 1 1500 15000 150000 10
Step 2 200 3500 140000 40
Step 3 70 470 19270 41
Step 4 40 120 18000 150
Step 5 3 1 15000 15000
  1. If you were required to omit one of the five purification steps above, which step would you omit? Explain your reasoning. (2 marks)

Step 3 should be omitted (1 mark). This is because there is very little increase in specific activity (1 mark).

  1. Describe an experiment which demonstrated that the primary structure of a protein specifies the tertiary structure. (6 marks).

Anfinsen used an enzyme that could be assayed easily for activity, with activity indicating the native fold (1 mark). He purified ribonuclease and BIO00004C

measured its enzymatic activity (1 mark). He then disrupted the structure by adding a denaturant and and a reducing agent that reduced all disulphide bonds in the enzyme (1 mark). The protein unfolded and enzymatic activity was lost (1 mark). He then removed the denaturant and the reducing agent by dialysis (1 mark). Enzymatic activity was recovered, indicating reformation of the native fold (1 mark).

  1. a) Describe what is meant by the term “transition state” of an enzyme­catalysed reaction. (2 marks)

A non­stable intermediate in the reaction pathway. The least energetically favourable point in the reaction.

  1. b) Draw a fully labelled diagram that illustrates the change in energy as a reaction proceeds, both with and without enzyme catalysis. (6 marks)

Marks for: showing where transition state is (1), indicating activation energy (1), axes with labels (1) , showing enzyme catalysed reaction has lower activation energy (1), showing Substrate and product have same energy regardless of whether enzyme catalysed (1), showing that Enzyme substrate and enzyme product complexes can have lower energy than substrate / product alone (1). Marks could also be given for students who give the TS stabilisation energy (difference between non­catalysed and catalysed TS energy) (1)

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6          What is an enzyme active site? (2 marks)

The portion of an enzyme that forms the binding site for the substrate(s). Catalysis occurs here.

  1. b) With reference to any particular example enzyme, outline the key structural and functional properties of its active site. (7 marks)
  2. i) Structural

e.g. chymotrypsin. Contains catalytic triad of Ser, His, Asp. Substrate binding cleft includes hydrophobic pocket.

  1. ii) Functional

Binds polypeptides. Specific for peptides including aromatic side chains (which fit into the hydrophobic pocket). Catalytic triad stabilizes Ser anion, which forms covalent bond with substrate. Transition state stabilized by oxyanion hole formed by peptide backbone amides.

[Could develop other examples, based on key enzymes discussed in the lectures, particularly carbonic anhydrase, aspartate transcarbamylase, possibly lysozyme]

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