Enzyme action

Describe how mRNA is obtained from pre-mRNA.

Describe how RNA polymerase joins RNA nucleotides.

Figure 1 shows the energy contained in molecules during an enzyme-controlled reaction.

Which expression represents the activation energy (E_a) of this reaction?

Tick (✓) one box.

In 1914, a group of scientists investigated enzyme action.

After studying their results, the scientists reached the following conclusions:

Our current understanding of enzyme action is based on the induced-fit model.

Describe similarities and differences between the conclusions from 1914 and our current understanding of enzyme action.

Complete Table 1 by adding a tick (✓) if the feature is present in the biological molecules.

Lactase catalyses the hydrolysis of lactose.

A student investigated the effect of galactose on lactase activity.

The student used the following method.

1. Mix together (in tube X) solutions of lactose, galactose and lactase in a 15: 3: 1 volume ratio so that the total volume is 90 cm³

2. Leave the mixture at 30 °C for 10 minutes.

3. Measure the glucose concentration in the mixture.

4. Repeat steps 1 to 3 with two different mixtures.

Calculate the volume of galactose solution used in tube X.

Table 2 shows the student’s results.

Galactose is a competitive inhibitor of lactase.

Use this information and your knowledge of enzymes to explain the results obtained in Table 2.

Give:

• one change the student could make in tube X to confirm that galactose is a competitive inhibitor

• the result you would expect.

A student investigated the activity of the enzyme polyphenol oxidase (PPO) in apple tissue.

When apple tissue is exposed to air, PPO catalyses a reaction between colourless phenol compounds in apple tissue and oxygen.

Figure 7 shows the reaction.

Figure 7

PPO phenol compounds + oxygen → brown pigment

The student measured the time taken for the brown pigment to appear in two apple varieties (D and E).

1. Cut a 1 cm cube of apple tissue from variety D.

2. Put the cube on a plate and leave the plate at 30 °C

3. Measure the time for the brown pigment to appear.

4. Repeat steps 1 to 3 two more times.

5. Repeat steps 1 to 4 with apple tissue from variety E.

The student obtained the results shown below, but did not record the data in a results table.

Variety D = 15 min 50 s, 18 min, 14 min 30 s

Variety E = 6 min 30 s, 8 min, 7 min

In the box labelled Figure 8, design a suitable results table.

• Enter the student’s results into the table.

• Calculate the mean results and include these in the table.

• Use 1 decimal place for both mean results.

Figure 8

Suggest and explain why the results for variety D are different from the results for variety E.

The student repeated the investigation but made one change to the method used to prepare the apple tissue.

The student then observed shorter times for the brown pigment to appear in both apple varieties.

Suggest the change the student made to the method of preparing the apple tissue.

Explain why the brown pigment appeared in a shorter time.

Do not suggest using a different volume or mass of apple.

The student wants to change the procedure to obtain a measure of PPO activity either in terms of how much substrate is used or how much product is produced.

Which change in procedure will provide a successful measure of PPO activity for the student?

Tick (✓) one box.

Scientists investigated a drug called MiTMAB as a treatment for cancer. MiTMAB inhibits cytokinesis.

Figure 8 shows drawings of cancer cells seen with an optical microscope from a:

• sample treated with MiTMAB

• control sample.

Figure 8

The cells in drawing A can be identified as those treated with MiTMAB.

Explain why.

MiTMAB acts as a non-competitive inhibitor of an enzyme called dynamin.

Suggest how MiTMAB can cause dynamin to become inactive.

When active, dynamin has two functions:

it stimulates cytokinesis

it inhibits cell death.

The scientists treated actively growing cultures of cancer cells with MiTMAB.

They incubated:

• one sample of 2500 cells without MiTMAB as a control

• eight samples, each with 2500 cells and a different concentration of MiTMAB.

After 72 hours, the scientists measured the number of cells in each sample.

Figure 9 shows the scientists’ results.

A negative value for proportion of control growth means that fewer than 2500 cells were counted after 72 hours.

Use all the information given to explain the results shown in Figure 9.

0.01 dm³ of MiTMAB solution was added to the treated cells.

Calculate the increase in mass of MiTMAB (in μg) added to the cells to reduce the cell growth from equal to the control to 0.0 of the control.

Show your working.

Two enzymes, P and Q, are proteins with quaternary structure which catalyse the same reaction, but they have different amino acid sequences.

Define the quaternary structure of a protein.

Explain how two enzymes with different amino acid sequences can catalyse the same reaction.

Scientists investigated the effect of pH 8.4 and pH 7.5 on the activity of enzymes P and Q.

Figure 8 shows their results.

Describe what the scientists should place in the control tubes in this investigation.

Give three conclusions you can make from Figure 8.

Describe the induced-fit model of enzyme action and how an enzyme acts as a catalyst.

Scientists investigated the action of the enzyme ATP synthase. They made reaction mixtures each containing:

• ATP synthase

• buffer (to control pH)

• substrates.

One of the substrates required in these reaction mixtures is inorganic phosphate (Pi).

Tick () one box to show which other substrate the scientists must add to the reaction mixtures to produce ATP.

The scientists investigated the effect of concentration of inorganic phosphate (Pi) on ATP synthase activity.

After 2 minutes, they stopped each reaction and then measured the concentration of ATP.

Figure 1 shows the scientists’ results.

Suggest and explain a procedure the scientists could have used to stop each reaction.

Explain the change in ATP concentration with increasing inorganic phosphate concentration.

Write an essay on one of the topics below.

Either

The importance of complementary shapes of molecules in organisms

Or

The importance of ions in metabolic processes

To study lipid digestion, a scientist placed a tube into the gut of a healthy 20-year-old man. The end of the tube passed through the stomach but did not reach as far as the ileum.

The scientist fed the man a meal containing triglycerides through the tube. The scientist also used the tube to remove samples from the man’s gut at intervals after the meal.

The scientist measured the type of lipid found in the samples. Some of her results are shown in Table 1.

Use your knowledge of lipid digestion to explain the differences in the results for samples A and B shown in Table 1.

You should assume that no absorption had occurred.

After collecting the samples, the scientist immediately heated them to 70 °C for 10 minutes.

Explain why.

Describe the role of micelles in the absorption of fats into the cells lining the ileum.

In the UK in 2016, there were 525 048 deaths. Cancer caused 30.4% of all deaths. Throat cancer caused 5% of all deaths from cancer.

Calculate the mean number of people who died of throat cancer per month in 2016.

Show your working.

Increased methylation of the promoter region of a tumour suppressor gene causes one type of human throat cancer.

In this type of throat cancer, cancer cells are able to pass on the increased methylation to daughter cells. The methylation is caused by an enzyme called DNMT.

Scientists have found that a chemical in green tea, called EGCG, is a competitive inhibitor of DNMT. EGCG enables daughter cells to produce messenger RNA (mRNA) from the tumour suppressor gene.

Suggest how EGCG allows the production of mRNA in daughter cells.

The scientists investigated the effect of different amounts of EGCG on the growth rate of the throat cancer cells grown in vitro. Their results are shown in Figure 2.

Figure 2

This figure has been removed due to third-party copyright restrictions.

A reporter who reviewed all of this work concluded that drinking green tea could be a cure for cancer.

Suggest three reasons why his conclusion might not be valid.

Write an essay on one of the topics below.

Either

The functions of enzymes and their importance in organisms.

or

The causes and importance of variation and diversity in organisms.

Describe how a non-competitive inhibitor can reduce the rate of an enzyme-controlled reaction.

Pectin is a substance found in some fruit and vegetables.

A scientist investigated the effect of pectin on the hydrolysis of lipids by a lipase enzyme.

His results are shown in Figure 1.

The scientist concluded that pectin is a non-competitive inhibitor of the lipase enzyme.

Use Figure 1 to explain why the scientist concluded that pectin is a non-competitive inhibitor.

The scientist also found that pectin stops the action of bile salts. He prepared two suspensions:

• suspension A – lipid and bile salts

• suspension B – lipid, bile salts and pectin.

He did not add lipase to either suspension.

He observed samples from the suspensions using an optical microscope. Figure 2 shows what he saw in a typical sample from each suspension.

Calculate the maximum length of the large lipid droplet marked X in Figure 2.

Using a ruler with millimetre intervals always includes an uncertainty in the measurement. Use the uncertainty in your measurement to determine the uncertainty of your calculated maximum length.

You can assume there is no uncertainty in the magnification.

No large lipid droplets are visible with the optical microscope in the samples from suspension A.

Explain why.

Uncontrolled cell division can cause tumours to form.

Figure 7 shows the growth pattern followed by a type of tumour.

Use Figure 7 to calculate the percentage of maximum growth this type of tumour reaches before it can be detected.

You will need to use the 10^x button on your calculator.

Figure 7 can also be used to calculate the age of this type of tumour.

At diagnosis, a patient had a tumour of 3.98 10¹¹ cells. Calculate the age of the tumour.

You will need to use the log₁₀ button on your calculator.

Trexall is a drug that can be used to slow the development of various forms of cancer.

Trexall slows cell division by interacting with an enzyme called dihydrofolate reductase (DR).

DR is involved in making nucleotides; the substrate for DR is folic acid.

Figure 8 shows the chemical structure of Trexall.

Figure 9 shows the chemical structure of folic acid.

Use the information provided to suggest how Trexall slows cell division.

Doctors investigated how the concentration of Trexall given to patients affected the growth of lung tumours. The doctors measured the volume of tumours at the beginning of the study and after 8 months.

Figure 10 shows the results of this investigation. The bars represent ± 2 standard deviations. A value of ± 2 standard deviations from the mean includes over 95% of the data.

The scientists measured the percentage change in tumour volume.

Suggest why they recorded both percentage change and tumour volume.

Percentage change

A lung cancer patient received 15 mg of Trexall per week. After treatment, the diameter of his lung tumour was 35.8 mm

Assuming the tumour was spherical, use the mean percentage change in tumour volume shown in Figure 10 to calculate the volume of the patient’s tumour before treatment with Trexall.

The formula for the volume of a sphere is πr³ where π = 3.14

To reduce the size of tumours, would it be better to use 30 mg of Trexall per week, or 20 mg of Trexall per week?

Explain your answer.

Trexall can also be used to slow the development of rheumatoid arthritis (a pain-causing joint disease).

Scientists investigated the effectiveness of Trexall as a pain relief treatment in 12 rheumatoid arthritis patients. All of the patients were female. They randomly divided the patients into two groups:

• Group R received Trexall tablets for 35 days

• Group S was a control group.

They asked both groups to rate their pain on a scale of 0–10 (0 being no pain and 10 being the worst pain possible) at the start and then every 7 days for 35 days. They calculated mean scores for each group.

Their results can be seen in Table 3.

Apart from age and general health, give two important factors when choosing patients for this investigation.

A student analysed Table 3 and concluded that Trexall was effective in reducing pain in arthritis patients.

Evaluate the student’s conclusion.

Turn over for Section B

Section B

one question.

You are advised to spend no more than 45 minutes on this section.

Formation of an enzyme-substrate complex increases the rate of reaction.

Explain how.

A scientist measured the rate of removal of amino acids from a polypeptide with and without an enzyme present. With the enzyme present, 578 amino acids were released per second. Without the enzyme, 3.0 10⁹ amino acids were released per second.

Calculate by how many times the rate of reaction is greater with the enzyme present. Give your answer in standard form.

Another scientist investigated an enzyme that catalyses the following reaction.

The scientists set up two experiments, C and L.

Experiment C used

• the enzyme

• different concentrations of ATP.

Experiment L used

• the enzyme

• different concentrations of ATP

• a sugar called lyxose.

The scientists measured the rate of reaction in each experiment. Their results are shown in Figure 5.

Calculate the rate of reaction of the enzyme activity with no lyxose at 2.5 mmol dm³ of ATP as a percentage of the maximum rate shown with lyxose.

Lyxose binds to the enzyme.

Suggest a reason for the difference in the results shown in Figure 5 with and without lyxose.

Draw the general structure of an amino acid.

Table 1 shows mRNA codons and the amino acids coded for by each codon. It also shows some properties of the R group of each amino acid.

The genetic code is described as degenerate.

What is meant by this? Use an example from Table 1 to illustrate your answer.

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.

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

A change from Glu to Lys at amino acid 300 had no effect on the rate of reaction catalysed by the enzyme. The same change at amino acid 279 significantly reduced the rate of reaction catalysed by the enzyme.

Use all the information and your knowledge of protein structure to suggest reasons for the differences between the effects of these two changes.

Bacteria are often used in industry as a source of enzymes. One reason is because bacteria divide rapidly, producing a large number of them in a short time.

Describe how bacteria divide.

Washing powders often contain enzymes from bacteria. These enzymes include proteases that hydrolyse proteins in clothing stains.

Figure 1 shows the effect of temperature on a protease that could be used in washing powder.

Explain the shape of the curves at 50 C and 60 C.

Some proteases are secreted as extracellular enzymes by bacteria.

Suggest one advantage to a bacterium of secreting an extracellular protease in its natural environment.

Explain your answer.

Mammals have some cells that produce extracellular proteases. They also have cells with membrane-bound dipeptidases.

Describe the action of these membrane-bound dipeptidases and explain their importance.

Scientists investigated treatment of a human bladder infection caused by a species of bacterium. This species of bacterium is often resistant to the antibiotics currently used for treatment.

They investigated the use of a new antibiotic to treat the bladder infection. The new antibiotic inhibits the bacterial ATP synthase enzyme.

Place a tick () in the appropriate box next to the equation which represents the reaction catalysed by ATP synthase.

The new antibiotic is safe to use in humans because it does not inhibit the ATP synthase found in human cells.

Suggest why human ATP synthase is not inhibited and bacterial synthase is inhibited.

The scientists tested the new antibiotic on mice with the same bladder infection. They divided these mice into three groups, C, R and A.

Group C was the control (untreated).

Group R was treated with an antibiotic currently used against this bladder infection.

Group A was treated with the new antibiotic.

They removed samples from the bladder of these mice after treatment and estimated the total number of bacteria in the bladder.

Their results are shown in Figure 2.

The antibiotics were given to the mice at a dose of 25 mg kg^–1 per day.

Calculate how much antibiotic would be given to a 30 g mouse each day.

Show your working.

Calculate the percentage difference in actual numbers of bacteria in group A compared with group R. The actual number of bacteria can be calculated from the log₁₀ value by using the 10^× function on a calculator.

Show your working.

The scientists suggested that people newly diagnosed with this bladder infection should be treated with both the current antibiotic and the new antibiotic.

Explain why the scientists made this suggestion.

Use information from Figure 2 and your knowledge of evolution of antibiotic resistance in bacteria in your answer.

Name two enzymes involved in the semi-conservative replication of DNA.

Sometimes, damage occurs during DNA replication. One enzyme involved in repairing damage to DNA is called ATR.

ATR works as follows.

ATR phosphorylates other enzymes involved in repairing DNA.

ATR also phosphorylates substrates required to repair DNA.

When ATR phosphorylates other enzymes, these enzymes become able to bind to their substrates.

Use your knowledge of enzyme structure to suggest why.

The enzyme-catalysed reactions activated by ATR only occur if the substrates have been phosphorylated.

Use your knowledge of energy changes in enzyme-catalysed reactions to suggest why.

Sometimes, a mutagenic agent causes DNA to break. A different enzyme called ATM binds to the broken DNA. This leads to the activation of a protein coded for by a tumour suppressor gene. The effect of ATM binding is to stop cell division until DNA is repaired.

A mutation could result in a person having non-functional forms of the gene that produces ATM.

What can you predict about the possible effects of having a non-functional form of ATM?