Biochemistry and Molecular Biology solutions

Question 1

Which of the following is true of RNA synthesis (transcription)?

Your answer:

1. a) RNA synthesis is always in the 5 - 3 direction.

Feedback:

In transcription, like DNA replication, nucleotides are added on to the 3' OH of the growing chain so RNA synthesis is always in the 5' - 3' direction. Unlike DNA synthesis, a primer is not needed for initiating a new strand. RNA contains U instead of T, and U base pairs with A and is inserted opposite A (not T) in the template.
Page reference: 353

Question 2

In bacterial promoters, which of the following describes the 'Pribnow box'?

You did not answer the question.

Correct answer:

1. b) The -10 box

Feedback:

The Pribnow box is named after David Pribnow, who discovered it, and it is found around 10 base pairs upstream of the transcriptional start site, i.e. at -10. It is often referred to as the -10 box.
Page reference: 356

Question 3

The deadly 'death cap' mushroom, Amanita palloides, produces a toxin called α-amanitin. Which cellular process is inhibited by this toxin?

You did not answer the question.

Correct answer:

1. c) RNA synthesis

Feedback:

α-amanitin specifically inhibits eukaryotic RNA polymerases II and III so it affects RNA synthesis. It is very potent and as little as 10 mg can kill a human. The main effects are on the liver and kidneys and it can take up to a week for death to occur after ingesting the mushrooms.
Page reference: 360

Question 4

Which of the following can be described as 'a sequence that can be several thousand base pairs upstream or downstream of a eukaryotic promoter and which increases gene expression as much as 200-fold.'

You did not answer the question.

Correct answer:

1. d) Enhancer

Feedback:

Eukaryotic genes often have control elements, called enhancers, which increase gene expression and which can be thousands of base pairs away from the promoter. These can be tissue-specific, enhancing transcription in only certain tissues. The enhancer is brought into proximity with the promoter by looping of the DNA that lies between them.
Page reference: 364-365

Question 5

Which of the following does the abbreviation TBP stand for?

You did not answer the question.

Correct answer:

1. a) TATA-box binding protein

Feedback:

TBP is a protein that attaches the transcription factor TFIID to the TATA box, in eukaryotes, and its full name is 'TATA-box binding protein'.
Page reference: 368

 

 

 

Question 1

Which of the following statements about the active site of an enzyme is correct?

Your answer:

1. a) The active site of an enzyme binds the substrate of the reaction it catalyses more tightly than it does the transition state intermediate.

Correct answer:

1. b) The active site of an enzyme binds the substrate of the reaction it catalyses less tightly than it does the transition state intermediate.

Feedback:

The active or catalytic site of an enzyme is a very small part of the overall molecule. It is a three-dimensional pocket or cleft where the substrate(s) are aligned and bind reversibly via several weak, noncovalent bonds. These bonds are short range, directional, and confer specificity. The substrate molecule(s) form a transition state intermediate which rapidly converts to products because it is unstable. The amino acid groups in the active site help to stabilise the electron distribution of the transition state. The latter binds to the enzyme (active site) more tightly than the substrate in its ground state liberating energy. The active site of the enzyme is therefore complementary to the transition state intermediate. This lowers the energy of activation for the reaction thus increasing the reaction rate. The active site may also position a metal group that may facilitate the reaction.
Page reference: Page 87

Question 2

Which of the following statements about the nature of enzyme catalysis is correct?

You did not answer the question.

Correct answer:

1. d) An enzyme cannot change the equilibrium position of the reaction it catalyses but it lowers the energy of activation of that reaction.

Feedback:

For a biochemical reaction to occur, an energy barrier has to be overcome even if the reaction has a strong negative ΔG (the free energy difference between the reactant(s) and product(s)). A substrate or reactant is first activated to form a higher energy transition state intermediate that is very unstable and rapidly converts to product(s). The energy required for this is the activation energy. In an uncatalysed reaction, this energy is supplied by molecular collisions. In a catalysed reaction, the enzyme positions the substrate molecule(s) in the most favourable relative orientation to form the transition state intermediate. The latter binds to the enzyme more tightly than the substrate (in its ground state) and stabilises it. An enzyme can therefore increase the rate of the reaction it catalyses by lowering the energy of activation but it cannot affect the equilibrium of the reaction.
Page reference: Page 88

Question 3

Which of the following statements about Michaelis-Menten kinetics is correct?

You did not answer the question.

Correct answer:

1. d) Km, the Michaelis constant, is a measure of the affinity the enzyme has for its substrate.

Feedback:

An enzyme that displays hyperbolic kinetics (single substrate enzyme) is referred to as a Michaelis-Menten enzyme. For such an enzyme, the plot of velocity of the reaction against substrate concentration is hyperbolic. The Michaelis-Menten equation describes the kinetics of such an enzyme at initial rates before any product is formed. The Michaelis-Menten equation can be used to calculate the Michaelis constant (Km). This is defined as that concentration of substrate at which the enzyme is working at half maximum velocity. Km is expressed in units of molar concentration and is independent of the enzyme concentration. Km is also a measure of the affinity that the enzyme has for its substrate. The higher the Km, the lower the affinity of the enzyme for its substrate.
Page reference: page 91

Question 4

Which of the following statements about the competitive inhibition of an enzyme-catalyzed reaction is correct?

You did not answer the question.

Correct answer:

1. c) The Vmax for a reaction remains unchanged in the presence of a competitive inhibitor.

Feedback:

Competitive inhibitors have a similar structure to the substrate and compete with the latter for binding to the active site of an enzyme. A competitive inhibitor binds reversibly to the active site. It will have no effect on the reaction at infinite substrate concentration since the substrate will win in the competition to bind to the enzyme active site. Consequently a competitive inhibitor and substrate cannot bind simultaneously to the enzyme. A comparison of the double reciprocal plots of an enzyme reaction with and without competitive inhibitor shows that:
1) Vmax, the maximum velocity for the reaction, remains unchanged in the presence of a competitive inhibitor
2) Km, the Michaelis constant for the reaction, increases in the presence of a competitive inhibitor. This is because the affinity that the enzyme has for its substrate is reduced in the presence of a competitive inhibitor.
Page reference: Page 93

Question 5

Which of the following statements about the mechanism of the catalytic triad of chymotrypsin is correct?

You did not answer the question.

Correct answer:

1. c) A proton moves from the serine to the histidine side chain in the catalytic triad of chymotrypsin.

Feedback:

The active site of chymotrypsin consists of a catalytic triad of serine, histidine and aspartate amino acid residues. The histidine acts as a general base by accepting a proton from the serine. The serine oxygen now makes a nucleophilic attack on the carbonyl carbon of the susceptible peptide bond forming a tetrahedral intermediate. The histidine proton is then transferred to the tetrahedral intermediate produced causing it to break down. An acylated enzyme is formed and the first product is liberated. The histidine side chain again acts as a base accepting a proton from a water molecule, activating it and forming a second tetrahedral intermediate. As before, the protonated histidine (now an acid), donates its acquired proton back to the tetrahedral intermediate, completing the reaction. The aspartate holds the histidine side chain in the correct orientation to accept the proton from the serine.
Page reference: Page 96

 

 

 

Question 1

Which of the following statements about the reactions of glycolysis is correct?

Your answer:

1. a) In glycolysis glucose-6-phosphate is split into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.

Correct answer:

1. b) In glycolysis fructose-1:6-bisphosphate is split into glyceraldehyde-3-phosphate and dihydroxyacetone phosphate.

Feedback:

In glycolysis since glucose-6-phosphate, a six-carbon molecule, does not have the correct aldol structure to be split into two three-carbon molecules, it is isomerized to fructose-6-phosphate first and then phosphorylated to fructose-1:6-bisphosphate by the enzyme phosphofructokinase. It can now be split by aldolase into two three-carbon phosphorylated products. Although the ΔGo' for the aldolase reaction is 24.3kJ mol-1, this refers to 1M concentrations. In the cell these are one thousand times lower, and one molecule of reactant produces two molecules of product. Under cellular conditions, the ΔG is small and the reaction freely reversible.
Page reference: Page 190

 

Question 2

Which of the following statements about the glycolytic intermediate, fructose-6-phosphate is correct?

You did not answer the question.

Correct answer:

1. d) In glycolysis fructose-6-phosphate is an aldol but is not itself split by the aldol reaction until phosphorylated to fructose-1:6-bisphosphate.

Feedback:

Fructose-6-phosphate has, unlike glucose-6-phosphate, the aldol structure. Glucose-6-phosphate is first isomerised to fructose-6-phosphate by phosphohexose isomerase. It is then phosphorylated to fructose-1:6-bisphosphate. This aldol is then split into an aldehyde, glyceraldehyde-3-phosphate and dihydroxyacetone phosphate, a ketone, by aldolase.
Page reference: Page 189

 

Question 3

Which of the following statements about the citric acid cycle is correct?

You did not answer the question.

Correct answer:

1. b) Three molecules of NADH and one molecule of FADH2 are produced in one turn of the citric acid cycle.

Feedback:

The citric acid cycle produces NADH and FADH2 from the cycle metabolites. These are subsequently reoxidised in the electron transport chain and provide energy in the form of ATP synthesized from ADP and Pi (inorganic phosphate). No oxygen is involved in the cycle. The oxygen atoms needed to form CO2 come from water. If there is no oxygen available, the cycle will not be able to proceed because NADH and FADH2 will not be reoxidised by the electron transport system. There is only a limited amount of NAD+ and FAD in cells. One molecule of GTP is produced by substrate-level phosphorylation in the cycle.
Page reference: Page 195

 

Question 4

Which of the following statements about the electron transport chain is correct?

You did not answer the question.

Correct answer:

1. b) The electron transport chain is made up of a chain of electron carriers with increasing redox potential.

Feedback:

The electron transport chain is the final stage in the oxidation of carbohydrates and occurs at the inner mitochondrial membrane. It is made up of a hierarchy of electron carriers of increasing oxidising power (increasing electron affinity or redox potential). The electrons in NADH and FADH2 that are produced in the citric acid cycle are transferred from electron carrier to electron carrier in the electron transport chain and finally to oxygen reducing it to water. NAD+ and FAD are now oxidised (hence the term oxidative). As the electrons are transferred manageable amounts of energy are released and harnessed by the cell to translocate protons across the inner mitochondrial membrane. This proton gradient results in the synthesis of ATP from ADP and Pi (inorganic phosphate).
Page reference: Page 200

 

Question 5

Which of the following statements about the generation of ATP in the electron transport chains is correct?

You did not answer the question.

Correct answer:

1. b) The F1 subunit of the ATP synthase contains the catalytic centre that synthesizes ATP.

Feedback:

ATP is synthesised in mitochondria by the chemiosmotic mechanism. The energy released by the electron flow is harnessed to translocate protons across the inner mitochondrial membrane and establish a proton gradient. This gradient drives the proton flow back into the mitochondrial matrix via ATP synthase. The energy of this is harnessed to rotating the F0 subunit, a circular structure in the membrane. This in turn drives a shaft to rotate in the F1 subunit, the enzyme catalytic centre on which ATP is generated. Each of the F0 and F1 subunits are made up of multiple subunits. It is not exactly known how the energy of the rotating shaft is harnessed to ATP synthesis. The synthesis of ATP from ADP and Pi tightly bound to the ATP synthase involves very little free energy change. It is the release of ATP which requires the energy supplied by the rotating shaft.
Page reference: Page 209

 

 

 

Which of the following statements about column chromatography is not correct?

Your answer:

1. a) Affinity chromatography involves the attachment to the column matrix of groups or molecules known that specifically bind to the wanted protein.

Correct answer:

1. d) Gel-filtration chromatography separates proteins on their ability to bind to specific groups on the column matrix.

Feedback:

Proteins can be separated by column chromatography. The protein mixture is passed through a column containing a commercially available solid matrix. Different proteins are retarded by different amounts as they flow through the column according to their interaction with the solid matrix. The proteins are separately collected and analysed as they emerge from the column. Proteins can be separated according to: size (gel-filtration chromatography), charge (ion-exchange chromatography), ability to bind to specific molecules (affinity chromatography), or hydrophobicity (reverse-phase chromatography). High performance liquid chromatography can also be used. It greatly increases the speed and efficiency of column chromatography. Here the column material is packed in a steel tube and the liquids are forced through at high pressure. The immobile phase method is more finely divided increasing the surface area and efficiency of the separations.
Page reference: Page 72

Question 2

Which of the following statements about SDS polyacrylamide gel electrophoresis is correct?

You did not answer the question.

Correct answer:

1. d) SDS polyacrylamide gel electrophoresis separates proteins on the basis of size.

Feedback:

This method can be used when only micrograms of protein are available. The protein mixture is dissolved in a solution of sodium dodecylsulphate (SDS), a detergent. The SDS molecules insert into the proteins their hydrophobic tails denaturing them, imparting a negative charge to the proteins as well as making them water-soluble. The (chosen) gel porosity allows all the SDS-proteins to migrate in the electric field towards the anode (the positive electrode), the smallest proteins moving the fastest. The proteins are mainly separated by molecular sieving since they are all negatively charged. The gels are stained with Coomassie blue, and the proteins appear as bands. A mixture of pure proteins or markers of known molecular weight run on one of the tracks allows calibration so the molecular weight of the proteins in the samples can be roughly estimated. These proteins however cannot be tested for biological activity as they have been denatured.
Page reference: Page 73

Question 3

Which of the following statements about isoeletric focusing is correct?

You did not answer the question.

Correct answer:

1. b) Proteins separated by isoelectric focusing can be tested for biological activity.

Feedback:

Isoelectric focusing involves the separation of proteins using non-denaturing gels so that the separated proteins can be tested for biological activity. The separation is based partly on net charge and partly on size. The gels used are commercially available with a pH gradient established. These polymers are called ampholines. When subjected to an electric field, the proteins migrate to the oppositely charged electrode but they become stationary when they reach that point in the gel where the pH is the isoelectric point of the protein. At low pH the carboxylic acid groups of proteins tend to be uncharged and the amino groups positively charged giving the proteins an overall positive charge. At high pH the carboxylic acid groups of proteins tend to be negatively charged and the amino groups uncharged giving them an overall positive charge. The isoelectric point of a protein is the pH at which the protein has no overall or net charge.
Page reference: Page 74

Question 4

Which of the following statements about the use of mass spectrometry in protein investigation is not correct?

You did not answer the question.

Correct answer:

1. d) Mass spectrometry involves the separation of ionic fragments on a gel.

Feedback:

Mass spectrometry can be used along with computer searches of databases to match amino acid sequences of protein samples being analysed with catalogued proteins. The most commonly used form of the technique is the MALDI-TOF (matrix-assisted laser desorption ionization-time-of-flight spectrometry). The proteins are hydrolysed into predictable peptides (by known proteases), ionised, and accelerated in an electric and magnetic field. Heavier ions drift more slowly than lighter ones. They are separated according to their mass-to-charge ratio. The molecular weight of the original protein can then be accurately measured. The amino acid sequence of a protein can also be determined with a tandem mass spectrometer by this method as well any post-translational modifications that may have occurred. This information is then used to search genomic databases.
Page reference: Page 75

Question 5

Which of the following statements about Western Blotting is correct?

You did not answer the question.

Correct answer:

1. a) The detection of a particular protein by Western Blotting relies on the very specific interaction between the protein and its antibody.

Feedback:

Specific proteins can be identified after fractionation on a polyacrylamide gel by exposing them to specific antibodies that have been produced in animals. These antibodies are labelled with radioactivity or a fluorescent compound to enable detection of binding. Proteins with a net charge will migrate in an electric field (to the opposite pole) through a gel, the pore size of which can be adjusted. A membrane is placed over the gel and the separated proteins are transferred onto the membrane by an electric field and soaked in a solution of the labelled antibody to detect the particular protein. This technique is known as Western Blotting.
Page reference: Page 74

 

 

 

Which of the following statements about the active site of an enzyme is correct?

Your answer:

1. d) The active site of an enzyme is complementary to the substrate of the reaction it catalyses.

Correct answer:

1. b) The active site of an enzyme binds the substrate of the reaction it catalyses less tightly than it does the transition state intermediate.

Feedback:

The active or catalytic site of an enzyme is a very small part of the overall molecule. It is a three-dimensional pocket or cleft where the substrate(s) are aligned and bind reversibly via several weak, noncovalent bonds. These bonds are short range, directional, and confer specificity. The substrate molecule(s) form a transition state intermediate which rapidly converts to products because it is unstable. The amino acid groups in the active site help to stabilise the electron distribution of the transition state. The latter binds to the enzyme (active site) more tightly than the substrate in its ground state liberating energy. The active site of the enzyme is therefore complementary to the transition state intermediate. This lowers the energy of activation for the reaction thus increasing the reaction rate. The active site may also position a metal group that may facilitate the reaction.
Page reference: Page 87

Question 2

Which of the following statements about the nature of enzyme catalysis is correct?

You did not answer the question.

Correct answer:

1. d) An enzyme cannot change the equilibrium position of the reaction it catalyses but it lowers the energy of activation of that reaction.

Feedback:

For a biochemical reaction to occur, an energy barrier has to be overcome even if the reaction has a strong negative ΔG (the free energy difference between the reactant(s) and product(s)). A substrate or reactant is first activated to form a higher energy transition state intermediate that is very unstable and rapidly converts to product(s). The energy required for this is the activation energy. In an uncatalysed reaction, this energy is supplied by molecular collisions. In a catalysed reaction, the enzyme positions the substrate molecule(s) in the most favourable relative orientation to form the transition state intermediate. The latter binds to the enzyme more tightly than the substrate (in its ground state) and stabilises it. An enzyme can therefore increase the rate of the reaction it catalyses by lowering the energy of activation but it cannot affect the equilibrium of the reaction.
Page reference: Page 88

Question 3

Which of the following statements about Michaelis-Menten kinetics is correct?

You did not answer the question.

Correct answer:

1. d) Km, the Michaelis constant, is a measure of the affinity the enzyme has for its substrate.

Feedback:

An enzyme that displays hyperbolic kinetics (single substrate enzyme) is referred to as a Michaelis-Menten enzyme. For such an enzyme, the plot of velocity of the reaction against substrate concentration is hyperbolic. The Michaelis-Menten equation describes the kinetics of such an enzyme at initial rates before any product is formed. The Michaelis-Menten equation can be used to calculate the Michaelis constant (Km). This is defined as that concentration of substrate at which the enzyme is working at half maximum velocity. Km is expressed in units of molar concentration and is independent of the enzyme concentration. Km is also a measure of the affinity that the enzyme has for its substrate. The higher the Km, the lower the affinity of the enzyme for its substrate.
Page reference: page 91

Question 4

Which of the following statements about the competitive inhibition of an enzyme-catalyzed reaction is correct?

You did not answer the question.

Correct answer:

1. c) The Vmax for a reaction remains unchanged in the presence of a competitive inhibitor.

Feedback:

Competitive inhibitors have a similar structure to the substrate and compete with the latter for binding to the active site of an enzyme. A competitive inhibitor binds reversibly to the active site. It will have no effect on the reaction at infinite substrate concentration since the substrate will win in the competition to bind to the enzyme active site. Consequently a competitive inhibitor and substrate cannot bind simultaneously to the enzyme. A comparison of the double reciprocal plots of an enzyme reaction with and without competitive inhibitor shows that:
1) Vmax, the maximum velocity for the reaction, remains unchanged in the presence of a competitive inhibitor
2) Km, the Michaelis constant for the reaction, increases in the presence of a competitive inhibitor. This is because the affinity that the enzyme has for its substrate is reduced in the presence of a competitive inhibitor.
Page reference: Page 93

Question 5

Which of the following statements about the mechanism of the catalytic triad of chymotrypsin is correct?

You did not answer the question.

Correct answer:

1. c) A proton moves from the serine to the histidine side chain in the catalytic triad of chymotrypsin.

Feedback:

The active site of chymotrypsin consists of a catalytic triad of serine, histidine and aspartate amino acid residues. The histidine acts as a general base by accepting a proton from the serine. The serine oxygen now makes a nucleophilic attack on the carbonyl carbon of the susceptible peptide bond forming a tetrahedral intermediate. The histidine proton is then transferred to the tetrahedral intermediate produced causing it to break down. An acylated enzyme is formed and the first product is liberated. The histidine side chain again acts as a base accepting a proton from a water molecule, activating it and forming a second tetrahedral intermediate. As before, the protonated histidine (now an acid), donates its acquired proton back to the tetrahedral intermediate, completing the reaction. The aspartate holds the histidine side chain in the correct orientation to accept the proton from the serine.