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Access to HE Diploma Applied Biochemistry
Access to HE Diploma Applied Biochemistry
2026-01-05T09:07:31.486+00:00 Applied Biochemistry Assignment open

Working folder Med_Biochem_TAQ_questions.pdf

TAQ 1 - 500 words

TAQ 1.1 – 150 words

  1. Define the term ‘anabolic reaction’ and explain why the formation of glycogen is a good example of this type of reaction.

Anabolic reactions combine molecules into larger more complex ones, consuming energy during the reaction.

The formation of glycogen is an example of an anabolic reaction. The polysaccharide glycogen is created by bonding of the monosaccharide glucose into a chain through successive condensation reactions. Each condensation reaction consumes energy--often provided in the form of ATP--and creates a glycosidic bond between two glucose molecules, and one free water molecule.

  1. Define the term ‘catabolic reaction’ and explain why the formation of glucose molecules from a polysaccharide such as glycogen is a good example of this type of reaction.

Catabolic reactions break down molecules into smaller simpler ones, releasing energy during the reaction.

The formation of a smaller glucose molecule from a larger glycogen macromolecule is an example of a catabolic reaction. The glycosidic chemical bonds that join glucose into chains to form glycogen, can be broken via a hydrolysis reaction, during which a water molecule is utilised and the stored chemical energy in the bond released. It is the reverse of the condensation reaction and is usually facilitated by an enzyme.

TAQ 1.2 – 200 words

TAQ 1.2.a

  1. Reaction A indicates hydrolysis as the molecule is split in two by the consumption of a water molecule during the reaction, breaking a glycosidic bond.

  2. Reaction A shows the splitting of a dimer into its two constituent monomers. This creation of two smaller molecules from a larger one is an example of a catabolic reaction.

  3. Reaction B shows the production of a water molecule during a reaction that chemically bonds two molecules together, this indicates a condensation reaction.

  4. Reaction B shows the bonding of two monomers into a dimer, this joining of two smaller molecules into a larger one is an example of an anabolic reaction.

TAQ 1.2.b

  1. Benedict's reagent. The sample moved from Blue to brick red, this is consistent with the Benedict reagent colour range, moving from no indication to a very strong indication.
  2. The results indicate a 'large amount' of reducing sugars following the procedure.
  3. Lactose
  4. Colours after heating:
    • Without enzyme = Brick red
    • With enzyme = Blue
  5. The prediction was correct, all the reducing sugars glucose and fructose (assuming an initial mixture of equal parts) will have been converted the non-reducing sugar, sucrose.

TAQ 1.3 – 150 words

TAQ 1.3.a:

  1. Dipeptide formation from two free amino acids

!2026-01-22 2253 dipeptide condensation reaction.excalidraw

  1. This is a condensation reaction due to the production of a free water molecule during a reaction that bonds the two amino acids into a larger dipeptide.

TAQ 1.3.b

  1. Both solutions contain proteins.
  2. Hydrolysis
  3. No, it should show a negative result. The violet colour is created when the copper ions in the Biuret solution react with a protein's peptide bonds, by their very nature free amino acids have no peptide bonds for the copper ions to react with.
  4. The albumin may have been completely converted into it constituent amino acids by the protease, however protease enzyme itself is a protein, therefore the Biuret solution will react with the peptide bonds within the enzyme, producing a positive result.

TAQ 1 Bibliography

TAQ 2 - 450 words

TAQ 2.1 - 250 words

TAQ 2.1.a - 138 words

Part 1a: The catalytic function of an enzyme is based on its structural architecture.

Explain how the structural features of an enzyme allow it to carry out this function. You should focus on the following points in your account:

  • tertiary structure (and shape),
  • stabilising bonds,
  • active site,
  • substrate specificity,
  • activation energy,
  • rate of a reaction.

Through the vast combinations of amino acid sequences in a polypeptide, the possible folding patterns, and the stabilising bonds that hold those folds, proteins exist in a nearly endless variety of shapes.

Enzymes are specialised proteins, each with unique tertiary structures that enable them to accelerate reactions of a specific substrate. They do this via a feature on their tertiary structure called the active site. It is this active site which dictates an enzyme's substrate specificity; only specific substrates with a complementary shape and chemical profile will bond with an enzyme's active site.

Once an enzyme-substrate complex is formed, the enzyme increases the reaction rate within the substrate by lowering the required activation energy. Once an enzyme-product complex has been formed the substrate product no longer matches the active site and is released from the enzyme.

TAQ 2.1.b - 138 words

The enzymes maltase and sucrase have similar structural features, and they are both involved in the chemical digestion of disaccharides. With reference to the induced fit model of catalysis, explain why maltase is able to catalyse the breakdown of the disaccharide maltose but not sucrose.

The induced fit theory shows us that the initial tertiary structure of the maltase enzyme is not a perfect fit for the maltose or sucrose disaccharides. However, the precise arrangement of R-groups within the active site of the maltase align with chemical groups of an opposite charge in the maltose. These opposite charges create attractive forces which elastically deforms the maltase enzyme, changing the shape of the active site, enabling an enzyme-substrate complex to form. Through this bonding and elastic deformation, the enzyme induces tension into the maltose substrate, lowering the reaction activation energy and breaking its glycosidic bond.

However, as the chemical groups of sucrose differ to maltose, they are also not complementary to the arrangement of R-groups within the maltase active site, and so its proximity does not activate the change in shape required for binding.

TAQ 2.2 - 200 words

  1. Type 1 = Non-competitive inhibition Type 2 = Competitive inhibition

  2. A & F = Active site B & E = Substrate C = Non-competitive inhibitor D = Competitive inhibitor

  3. A non-competitive inhibitor embeds into an allosteric site. Whilst the allosteric site is distinct from the active site, this does cause some deformation to the active site. This deformation can completely restrict a substrate from binding at the active site, or the substrate can bind but enzyme activity is inhibited. So such, non-competitive inhibitors can range in their effects on enzyme activity, from slowing down the catalytic reaction, to stopping them altogether.

  4. A competitive inhibitor is almost always similar in shape and chemical properties to an enzymes substrate, this allows it to bind strongly via the active site, preventing an enzyme substrate from binding, ending the enzyme's ability to catalyses a substrate reaction.

  5. The effects of competitive inhibitors can be reduced by increasing the concentration of the enzymes substrate. Not only does this increased substrate to inhibitor ratio increase the probability that a substrate will have chance to interact with an active enzyme before the inhibitor disables it, but the presence of a substrate in the active site also blocks inhibitor interactions, further reducing the probability of an inhibitor binding with the enzyme.

TAQ 2 Bibliography

TAQ 3 - 700 words

TAQ 3.1

TAQ 3.1.a - 300 words

With direct reference to data in Table 4, and the trend shown in the plotted graph, explain the effect that a range of temperatures has had on the catalytic activity of the enzyme used in the investigation.

You are expected to refer to specific data values and support your explanation with scientific theory.

You need to explain (using the tabulated data values and line graph trend) the effect that different temperatures have had on the rate at which an enzyme can catalyse a chemical reaction.

To provide an effective discussion, the data values documented in Table 4, should be referred to directly (covering rate of reaction at low, optimal, and high temperatures as a minimum).

You need to explain the rate of reaction values stipulated, using level 3 Scientific theory on the effects that temperature has on enzyme activity. Consideration should be given to;

  • kinetic energy
  • collision theory
  • enzyme-substrate complex formation
  • enzyme denaturation In addition, you should refer to the trend evident in the plotted graph. What does the direction and gradient of the line indicate has happened as temperature is increased at 10°C intervals?

Temperature is a measure of the average thermal energy within a system, or more specifically, it is the total kinetic energy of the atoms and molecules. This feature has a significant effect on enzyme led catabolic reactions inside the human body. A reduction in temperature signifies a reduction in the kinetic energy, or movement, of the enzymes and their substrates. As their movement slows, so too does the rate of random interactions between molecules, and it's these interactions that drive enzyme activity.

The experiment results support this effect; as the test temperatures increase from 10°C to the enzymes optimum result at 40°C, so to does the rate of reaction. It should be noted that the graph shows an exponential relationship of reaction rate against temperature; with each 10°C increment from 10°C the reaction rate roughly doubles, \ce{8 x 10^{-4}} \ce{17 x 10^{-4}} \ce{33 x 10^{-4}} \ce{65 x 10^{-4}}.

Whilst the increased kinetic energy within the system increases the frequency of the interactions that drive enzymes activity, as the temperature increases past the enzymes optimum of 37°C the increased force of those interactions starts to break the chemical stabilising bonds that hold the enzymes shape, which can render the enzyme denatured. The experiment results support this with the reaction rate dropping off fairly linearly from the optimum rate at 40°C to \ce{27 x 10^{-4}} @ 50°C and then finally \ce{0} @ 60°C as all catalytic activity stops due to complete enzyme denaturing.

Most human biological enzymes operate optimally at a temperature of approximately 37°C. This temperature provides an ideal balance of kinetic energy and structural stability. Whilst the resolution of the test temperatures did not include 37°C directly, the closest test temperature of 40°C shows the highest reaction rate of \ce{65 x 10^{-4}} out of the tests, with the a drop off in the rate when the temperature increases or decreases from this point.

TAQ 3.1.b

  1. Q_{10}=\frac{17\times10^{-4}}{8\times10^{-4}}=2.12
  2. Q_{10}=\frac{65\times10^{-4}}{33\times10^{-4}}=1.97
  3. Q_{10}=\frac{27\times10^{-4}}{65\times10^{-4}}=0.41

TAQ 3.1.c - 100 words

The temperature co-efficient shows us by what factor the rate of reaction will rise or fall by given a 10c increase, this case the rate of reaction catalysed by the enzyme.

The temperature co-efficient of 1.97, from the second equation shows us that a 10C rise from 30C to 40C, the rate of reaction will nearly double;

The temperature co-efficient of 0.41, from the second equation shows us that a 10C rise in temperature from 40C to 50C, the rate of reaction will decrease by over half or 59%.

TAQ 3.2

TAQ 3.2.a - 250 words

With direct reference to data in Table 5, and the trend shown in the plotted graph, explain the effect that pH has had on the catalytic activity of the two enzymes used in the investigation.

You are expected to refer to specific data values and support your explanation with scientific theory.

you need to take a similar approach but in this section, the focus is on the effect of pH on enzyme activity. Again, as this question asks you to discuss, you must refer to data values and use theory to explain this data.

Enzymes have a optimum pH at which their activity rate peaks, this optimum pH does differ between different enzymes. As the pH deviates from this optimum, the increased acidity or alkalinity starts to break the hydrogen and ionic bonds between the R-groups of the amino acids in a proteins structure, without these stabilising bonds, the protein begins to unfold, with the enzyme becoming increasing denatured, making it no longer complementary to the substrate, the further the pH is from its optimum.

The enzyme pepsin breaks down protein in the acidic environment, pH 1.5 to 2.5, of the stomach. The tests results confirm pepsin's optimum pH is within this range with the highest rate of reaction of \ce{60 x 10^{-4}} at a pH 2. Either side of this optimum pH the rate of reaction roughly halves for the first pH point deviation from pH 2; \ce{30 x 10^{-4}} @ pH 1 and \ce{25 x 10^{-4}} @ pH 3. Enzyme activity completely stops at 2 pH points from the optimum (pH 0 and pH 4).

Trypsin breaks down proteins in the much more alkali environment of the small intestines, where the ambient pH ranges from pH 6 to 8.5. The test results confirm trypsin's optimum pH falls within this range with a reaction rate of \ce{61 x 10^{-4}} at a pH 8. Either side of this optimum pH the rate of reaction roughly halves for the first pH point deviation from pH 2; \ce{35 x 10^{-4}} @ pH 7 and \ce{30 x 10^{-4}} @ pH 9. Trypsin enzyme activity completely stops at 2 pH points from it optimum (pH 6 and pH 10), signifying that a pH ≤ 6 or ≥ 10 renders the enzyme completely denatured.

For both enzymes the enzymes have fairly narrow pH ranges that they are active within. The graph shows a sharp activity drop off, with enzyme activity reducing to zero over 2 pH points away from their respective optimum pH.

TAQ 3.2.b - 50 words

Based on the two graphs, explain why pepsin would not be able to catalyse the hydrolysis of polypeptides once it has exited the stomach (mixed with partially digested food) and enters the duodenum (first section of the small intestine).

As the pepsin, mixed with partially digested food, moves from the acidic pH ~2 of the stomach to much more alkali pH of ~8 of small intestine, the graph shows that at any pH ≥ 4, enzyme activity drop to zero, signifying that the pH level has denatured the enzyme.

TAQ 3 Bibliography

  • DistanceLearningCentre.com (2022). Applied Biochemistry Learning Materials by DistanceLearningCentre.com Ltd. Available at: DistanceLearningCentre.com (Accessed: 18 December 2025).
  • Wikipedia Contributors (2019). Enzyme. [online] Wikipedia. Available at: https://en.wikipedia.org/wiki/Enzyme (Accessed: 28 January 2026).
  • Wikipedia Contributors (2019). Trypsin. [online] Wikipedia. Available at: https://en.wikipedia.org/wiki/Trypsin (Accessed: 28 January 2026).
  • Wikipedia Contributors (2019). Pepsin. [online] Wikipedia. Available at: https://en.wikipedia.org/wiki/Pepsin (Accessed: 28 January 2026).

TAQ 4 - 850 words

TAQ 4.1

TAQ 4.1.a - 300 words

Complete the Table below by summarising the role of different enzymes in the process of producing synthetic insulin using recombinant DNA technology

Due to the constraints of the word limit for this section of TAQ.4, you are only required to describe the specific function of the tabulated enzymes which are involved in the roduction of synthetic insulin rather than describing the process of recombinant DNA technology as a whole.

You need to remember that one of the listed enzymes is used in isolating the required gene (from DNA in a human cell) and in preparing a plasmid to create recombinant DNA. Therefore, reference to the role of this enzyme in both processes is necessary. One of the enzymes in the Table is not covered in the learning materials so to summarise the function of this enzyme, you will need to do some wider reading.

Restriction Endonuclease

Restriction Endonuclease, also called restriction enzymes, are used to cut sections of DNA. This enzyme is used in the process of splicing the cDNA into a bacteria plasmid, before replication. Initially, Restriction Endonuclease cuts the ends of the cDNA to make them 'sticky', so the the sticky ends match, the same type of restriction endonuclease is also used to cut into the bacteria plasmid (its circular DNA), before the cDNA in spliced into the plasmid.

Reverse Transcriptase

Whilst it is possible to isolate the gene for human insulin directly from DNA, it is far easier to acquire via the appropriate mRNA. From this mRNA, the enzyme reverse transcriptase constructs a single chain of cDNA, the mRNA is removed before the enzyme DNA Polymerase completes the cDNA double helix structure, which is then spliced into a bacterium plasmid to create the rDNA. Reverse transcriptase role in enabling the use of mRNA, greatly improves the process.

DNA Ligase

DNA Ligase is an enzyme that has the ability to join two pieces of DNA. It has two functions during the production of synthetic insulin; in the first instance it is used to join the Okazaki fragments (disconnections in the DNA backbone) following DNA polymerases work on the lagging strand. Later it is used during the annealing stage, to splice the cDNA into the a bacterium plasmid to create the recombinant plasmid.

Alkaline Phosphatase

When a the enzyme restriction endonuclease cuts the vector plasmid open, it creates two "sticky ends" that are complimentary to each other, this and the circular form of the plasmid creates a tendency for the plasmid to reseal itself. The enzyme alkaline phosphatase can remove phosphates from the 5' end, ensuring the plasmid remains open until the enzyme DNA ligase splices in the cDNA, creating the recombinant plasmid.

TAQ 4.1.b - 200 words

pBR322 is a plasmid that is commonly used in the process of creating synthetic proteins because of its antibiotic resistance genes.

With reference to this plasmid specifically, explain how the process of replica plating can be used to isolate only transformed bacteria that have taken up recombinant plasmids that contain the insulin gene.

By using a feature of the pBR322 plasmid--its two antibiotic resistance genes; tetracycline and ampicillin--and combining this with the technique of replica plating, bacteria that have taken up the recombinant plasmid can be identified and isolated.

A BamHI restriction enzyme is used to cut the plasmid in the centre of the tetracycline gene, a successful insertion splits the gene rendering it inactive. However, if the plasmid reconnected itself before insertion, the tetracycline gene remains active. The result, any bacteria that has taken up a recombinant plasmid will have resistance to ampicillin, but not tetracycline.

Using the replica plating procedure, a 'master plate' is prepared containing agar and ampicillin and a sample of the bacteria introduced, only bacteria resistant to Ampicillin will cultivate. With the use of a sterile cloth to preserve the relative locations of the colonies, a sample pressing is transferred to a 'replica plate' containing tetracycline. Following cultivation, the master and replica plates are compared to identify the colonies that contain the recombinant plasmids. Bacteria that have taken up the recombinant plasmid will fail to cultivate on the replica plate, due to the inactive tetracycline resistant gene, and are harvested from the master, ampicillin-containing, plate.

TAQ 4.1.c - 50 words

Briefly explain how a commercial scale fermenter is used to yield synthetic insulin from transformed bacteria that have taken up the recombinant plasmid, once they have been isolated.

A commercial scale fermenter provides the optimal environment for bacterial growth and protein production through the precise control of the pH, temperature, oxygen and nutrient levels. Agitation ensures that condition and nutrients are uniform throughout the vessel. Insulin is harvested from the fermenter once the required synthesis has been achieved.

TAQ 4.2 - 300 words

TAQ 4.2.a

Explain how PCR would have been used to ensure that the trace sample of DNA discovered at a crime scene was able to be used to create the clear bands evident in the Southern blot.

As part of your answer, you should describe the steps involved in PCR to include the role of named enzymes and changes in temperature.

Polymerase chain reaction can be used to amplify trace DNA for forensic profiling. A mixture containing the required nucleotides and enzyme DNA polymerase, is added to the sample DNA and a the mixture is cycled through three thermal stages:

  1. The mixture is then heated to 95°C for 30 seconds, which splits the complementary base pairs into two separate chains.
  2. The mixture is cooled to 37°C allowing the DNA primers to bond to their complementary sequence.
  3. The mixture is heated to 72°C, the optimum temperature for DNA polymerase enzyme to build the new strands, ending with a completed replicated DNA.

A single strand of DNA can be duplicated rapidly, with the number of copies growing exponentially with each cycle, providing the DNA to create clear bands in the Southern blot.

TAQ 4.2.b

The results of the PCR reaction, along with DNA samples taken from three suspects were loaded onto agarose gel to undergo Gel Electrophoresis. Describe the principles of this technique.

As part of your answer, you should outline the method used to produce bands of DNA, as well as to create a Southern blot x-ray film.

Gel electrophoresis separates molecules, and therefore also DNA, by size and charge. The sample DNA is loaded into wells in agarose gel and then an electric current is passed through the gel, causing the negatively charged DNA to migrate through the gel towards the positive electrode. The gel acts as a sieve; smaller fragments travel faster and further than larger ones. By the end of the test the DNA fragments have separating in to bands. For analysis, the DNA bands are stained to make them visible, then transferred from the gel to a nitrocellulose filter through blotting. Following this, Radioactive probes hybridise with specific DNA sequences, allowing the pattern to be visualised on X-ray film.

TAQ 4.2.c

By analysing the Southern blot x-ray film (Figure.1) which of the three suspects would be charged with committing the crime?

Briefly explain your answer.

The banding produced by the test is unique to each DNA profile. When comparing the test of sample recovered from the crime scene to that of Suspect 2, the pattern, position and thickness of all the bands are a match. This is not the case with suspects 1 and 3, they can be excluded.

Therefore, suspect 2 would be charged with committing the crime.

TAQ 4 Bibliography

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