Regulation of transcription & translation

Cleft lip and palate (CLP) is a split in the upper lip and/or the roof of the mouth. CLP is present from birth.

CLP can be caused by various genetic and environmental factors.

CDH1 is a gene involved in lip and mouth development in a fetus before a baby is born.

Scientists hypothesised that CLP is caused by a ‘two-hit’ model:

• a ‘genetic hit’ through loss of CDH1 function

• an ‘environmental hit’ leading to CDH1 methylation caused when inflammation affects a fetus.

Inflammation can result from the mother having an infection or diabetes.

Scientists investigated the ‘two-hit’ model using human cells grown in vitro.

They:

• grew normal cells and cells with a loss of CDH1 function

• treated some of each type of cell with a drug to cause inflammation and left some of each type of cell untreated

• measured the percentage CDH1 methylation of the cells.

Figure 4 shows the scientists’ results.

The frequency of live births with CLP is 1 in 700

Despite knowing this frequency, the Hardy–Weinberg equation cannot be used to estimate any allele frequencies for CLP.

Suggest two reasons why.

Describe one effect methylation could have on CDH1

Suggest and explain three reasons why the data in Figure 4 might not provide evidence for the ‘two-hit’ model.

Following a body injury, bone marrow stem cells move to the site of damage and undergo cell differentiation.

Figure 5 shows how this differentiation occurs.

Suggest how SCFR is destroyed by a lysosome.

After a heart attack, cardiomyocytes (cardiac muscle cells) die, and become infarcted tissue. Infarcted tissue cannot contract.

Stem cells in bone marrow cannot move to the infarcted tissue and differentiate into cardiomyocytes. Scientists used laboratory rats to investigate if bone marrow stem cell transplants could be used to repair infarcted tissue resulting from a heart attack.

They split the rats into three groups.

• Control group did not get a transplant of bone marrow stem cells.

• c-KIT+ group got a transplant of bone marrow stem cells with a functioning c-KIT gene.

• c-KIT– group got a transplant of bone marrow stem cells with no functioning c-KIT gene.

After 9 days, the scientists measured the mean ventricular blood pressure of each of the three groups.

Figure 6 shows their results.

The differences between the groups were all statistically significant.

Using all of the information, suggest explanations for the results for the Control group and the c-KIT– group shown in Figure 6.

Nine days after transplantation, the c-KIT+ group showed that 68% of infarcted tissue was made up of new cardiomyocytes. The control group had no new cardiomyocytes.

Assuming that mean ventricular blood pressure is directly proportional to the number of cardiomyocytes, calculate the percentage of infarcted tissue that was made up of new cardiomyocytes in the c-KIT– group.

All new cardiomyocytes produced:

• Connexin-43, a channel protein that allows electrical impulses to pass between cardiomyocytes

• GATA-4, a transcriptional factor that stimulates the expression of genes for actin and myosin.

Suggest how production of Connexin-43 and GATA-4 could give the result seen in the c-KIT+ group in Figure 6 (on page 13).

Do not include details of transcription or translation in your answer.

BRCA1 and BRCA2 are human genes that code for tumour suppressor proteins. Mutations in BRCA1 and BRCA2 can cause cancer (lines 1–2). Explain how.

Genetic testing, using DNA from saliva, can screen for all known harmful mutations in both genes (lines 5–6). Describe how this DNA could be screened for all known harmful mutations in both genes.

Effective treatment of ER-positive breast cancers often involves the use of drugs which have a similar structure to oestrogen (lines 9–10).

Suggest and explain how these drugs are an effective treatment of ER-positive breast cancers.

Blood tests can be used to test for cancers (line 11). However, the results of blood tests may not be conclusive when testing for prostate cancer. Explain why.

Treatment with drugs might be able to reverse the epigenetic changes that cause cancers (lines 16–17). Suggest and explain how.

Hepatitis B is a life-threatening liver infection caused by the hepatitis B virus (HBV).

Figure 7 shows the structure of HBV.

HBV infects a liver cell. The liver cell is 25 µm in diameter.

Use Figure 7 to calculate how many times larger in diameter this cell is than HBV. You should use the lipid layer to measure the diameter of HBV.

Scientists investigated the effectiveness of two types of RNA interference (RNAi) molecules on reducing HBV replication. These molecules were:

• short hairpin RNA (shRNA)

• long hairpin RNA (IhRNA).

The scientists infected mouse liver cells with HBV and transferred either shRNA or lhRNA into these cells. Then they determined the concentration of the attachment proteins, HBsAg, in these cells.

The concentration of HBsAg is a measure of HBV replication.

Figure 8 shows the scientists’ results.

The error bars represent ±2 standard deviations from the mean, which includes over 95% of the data.

One method of transferring RNAi molecules into cells involves combining these molecules with a lipid. Suggest why this increases uptake of RNAi molecules into cells.

Using all the information provided, evaluate the use of the two types of RNAi in treating hepatitis B in humans.

Do not refer in your answer to how RNAi reduces HBV replication.

Plants transport sucrose from leaves to other tissues for growth and storage. SUT1 is a sucrose co-transporter protein.

Scientists investigated whether the cells of tobacco plant leaves used SUT1 to transport sucrose to other tissues.

The scientists used a radioactively labelled DNA probe to show that the cells of tobacco plant leaves contained the SUT1 gene.

Describe how they would do this.

Do not include PCR in your answer.

To study the role of SUT1 in tobacco plants, scientists reduced the expression of the SUT1 gene.

When the SUT1 gene is transcribed, the SUT1 mRNA produced is called ‘sense’ SUT1 mRNA. The scientists genetically modified plants by inserting an extra gene so that this also allowed the production of ‘antisense’ SUT1 mRNA.

The scientists had two types of tobacco plants:

• type A – plants that were genetically modified

• type B – plants that were not genetically modified.

Suggest how the production of ‘antisense’ SUT1 mRNA in type A plants would reduce the expression of the SUT1 gene.

The scientists hypothesised that lower rates of sucrose transport from leaves would cause reduced growth.

To test this hypothesis, the scientists provided leaves of type A and type B plants with labelled carbon dioxide (¹⁴CO₂). To estimate sucrose transport out of leaves, they measured the percentage of¹⁴C remaining in the leaves for 16 hours.

Figure 5 shows their results.

Calculate the ratio of percentage of¹⁴C remaining in leaves of type B to type A plants 16 hours after providing¹⁴CO₂

In type B plants, the percentage of¹⁴C remaining in the leaves does not reach zero per cent, as shown in Figure 5.

Suggest two reasons why.

1

2

The scientists measured physiological differences between type A plants and

type B plants.

Table 3 shows the scientists’ results as they presented them.

Sucrose is able to inhibit the production and activity of rubisco in leaves of a plant. Type A plants have decreased dry mass compared with type B plants.

Use all the information to suggest and explain how the physiological factors in Table 3 would contribute to the decreased dry mass observed in type A plants.

Turn over for Section B

Section B

one question.

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

Testosterone is a steroid hormone that belongs to a group of male sex hormones called androgens.

Steroid hormones are hydrophobic.

Explain why steroid hormones can rapidly enter a cell by passing through its cell-surface membrane.

In the cytoplasm, testosterone binds to a specific androgen receptor (AR). An AR is a protein.

Suggest and explain why testosterone binds to a specific AR.

The binding of testosterone to an AR changes the shape of the AR. This AR molecule now enters the nucleus and stimulates gene expression.

Suggest how the AR could stimulate gene expression.

The gene that codes for the AR has a variable number of CAG repeats. Some studies have shown an association between the number of CAG repeats and the risk of developing prostate cancer.

Table 1 shows the results of a statistical test from one study.

What can you conclude from the data in Table 1 ?

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.

Damage to the myelin sheath of neurones can cause muscular paralysis (lines 2–4).

Explain how.

Sometimes Guillain–Barré syndrome causes heart rate irregularities (lines 4–5).

Suggest and explain why.

The first successful drug trial to reduce concentrations of huntingtin in the brain used single-stranded DNA molecules (lines 13–14).

Suggest and explain how this drug could cause a reduction in the concentration of the protein huntingtin.

Scientists from the first successful drug trial to reduce concentrations of huntingtin (lines 9–11) reported that the drug is not a cure for Huntington’s disease.

Suggest two reasons why the drug should not be considered a cure. Do not include repeats of the drug trial in your answer.

Suggest two reasons why people had the drug injected into the cerebrospinal fluid (lines 12–13) rather than taking a pill containing the drug.

What is meant by a genome?

Chromatin immunoprecipitation is one method to determine where a transcription factor binds to DNA. The principle behind this procedure is shown in Figure 8.

Figure 8

Explain why the antibody binds to the transcription factor.

Use Figure 8 to explain what ‘precipitated DNA’ consists of.

Soybeans are used in a number of processed foods. However, soybeans contain a protein known as P34 that causes an allergic response in some people. Scientists have created transgenic soybeans that produce single-stranded cDNA, which prevents transcription of the P34 gene. They used recombinant plasmids as vectors to transform soybean cells. After they had screened these cells for production of the P34 protein, they cultured the transformed cells to form soybean plants.

Suggest how single-stranded cDNA could prevent transcription of the P34 gene.

Describe the roles of two named types of enzymes used to insert DNA fragments into plasmids.

The soybean cells were screened for the presence of the P34 protein. This process involved the use of gel electrophoresis to separate proteins extracted from soybean cells.

Suggest two features of the structure of different proteins that enable them to be separated by gel electrophoresis.

Lactose is the main sugar in milk and is hydrolysed by the enzyme lactase. Lactase is essential to newborn mammals as milk is their only source of food. Most mammals stop producing lactase when they start feeding on other food sources. Humans are an exception to this because some continue to produce lactase as adults. The ability to continue producing lactase is known as lactase persistence (LP) and is controlled by a dominant allele. A number of hypotheses based on different selection pressures have been put forward to explain LP in humans.

One hypothesis for LP in humans suggests that the selective pressure was related to some human populations farming cattle as a source of milk.

Describe how farming cattle as a source of milk could have led to an increase in LP.

Use the information provided to explain why the number of people showing LP would rapidly increase once selection for this condition had been established.

Lactase persistence is caused by a mutation in DNA. This mutation does not occur in the gene coding for lactase.

Suggest and explain how this mutation causes LP.

Chronic lymphocytic leukaemia (CLL) is a cancer that affects some B cells of a person’s immune system.

Rituximab is a drug used to treat CLL. It binds to a protein called CD20 on the surface of B cells. If enough Rituximab binds to a B cell, it can kill the cell. Rituximab kills both healthy and cancerous B cells. The body then produces new B cells.

The amount of CD20 on the surface of B cells varies from one person to another. Doctors investigated the relationship between the amount of CD20 on the B cells of a patient and the percentage of B cells destroyed by Rituximab.

Figure 2 shows the doctors’ results. Each cross is the result for one patient.

What statistical test could the scientists have used to determine whether there was a significant relationship between the amount of CD20 on the surface of B cells and the percentage of B cells destroyed by Rituximab? Give a reason for your answer.

From these data, what can you conclude about the effectiveness of Rituximab in treating patients with CLL?

Do not include considerations of statistical analyses in your answer.

Approximately 11 percent of people with CLL also have mutations of a gene called NOTCH1. This leads to production of a non-functional transcription factor associated with CD20 production.

The doctors determined the median percentage of B cells destroyed by Rituximab in people with CLL who had the NOTCH1 mutation and those who did not.

The doctors’ results are shown in Table 1.

Human blood contains (approximately) 1.0 10⁹ B cells per dm³.

Use the median values in Table 1 to calculate the difference between the number of B cells per dm³ in the blood of people treated with Rituximab with the NOTCH1 mutation and people without the NOTCH1 mutation.

Express your answer in standard form. Show your working.

Use all of the information to suggest how the mutation of NOTCH1 led to the difference in the percentage of B cells destroyed by Rituximab.

Scientists investigated the role of a protein called CENP-W in mitosis. Their method involved cell fractionation and ultracentrifugation.

The scientists began by lysing (breaking open) cells and organelles using a detergent that dissolves lipids in water.

Suggest how the detergent releases CENP-W from cells.

Explain how ultracentrifugation separates CENP-W from other molecules.

CENP-W is involved in the formation of spindle fibres in mitosis. Spindle fibres are made of molecules of a protein called tubulin.

The scientists treated cells in a culture with small interfering RNA (siRNA). This siRNA causes RNA interference of expression of the CENP-W gene. The scientists took samples of cells at 0, 48 and 72 hours after adding the siRNA. They then used gel electrophoresis to separate CENP-W and tubulin from these samples.

Figure 5 shows the results of the electrophoresis. The size of each band is proportional to the amount of CENP-W or tubulin present.

Suggest how the siRNA produced these results.

Ammonia in soil is oxidised to nitrites and nitrates by species of nitrifying bacteria.

Scientists investigated whether two soils with a different pH contained different communities of nitrifying bacteria. These communities consist of all the nitrifying bacteria of different species in each soil. They took samples of soil from two sites, A and B.

They measured the pH of the samples and found that

• the soil from site A had a pH of 6.9

• the soil from site B had a pH of 4.3

The scientists measured the concentration of ammonia in soil samples over 20 days. Each sample contained the same concentration of ammonia at the start and had the same mass. They recorded the concentration of ammonia in

• soil A with a pH of 6.9

• soil B with a pH of 4.3

• a mixture of equal masses of soils A and B with its pH adjusted to 6.9

Their results are shown in Figure 6.

The scientists used units of µg g¹ for the concentration of ammonia in soil.

Suggest why, in this investigation, the scientists used these units.

g¹

Calculate the difference in the rate of breakdown of ammonia per day between day 0 and day 2 in soil A and soil B.

Show your working and the units for your answer.

The scientists concluded that the soil mixture experiment showed there were different communities of bacteria in soils A and B.

What evidence from Figure 6 supports their conclusions? Give reasons for your

answer.

The oxidation of ammonia by nitrifying bacteria involves the enzyme ammonia monooxygenase. Each species of nitrifying bacteria has its own specific amoA gene that codes for production of ammonia monooxygenase.

In a second investigation, the scientists determined the expression of the amoA gene in two species of bacteria, S and T. Species S was from acid soil and species T was from soil with a neutral pH.

The scientists grew cultures of each species separately in soils of different pH. They determined the amount of mRNA from the amoA gene in each culture.

Their results are shown in Figure 7.

In which species was the number of copies of mRNA more affected by changes in soil pH from 4.9 to 7.5? Use a calculation to support your answer.

This method allowed the scientists to estimate the expression of the amoA gene in each culture but not the growth of the bacterial population in each culture.

Explain why.

The scientists set up their cultures in sterile glass bottles.

Suggest one suitable method for sterilising the bottles and explain why it was necessary to sterilise them.

Turn over for Section B

Section B

one question.

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

Each year, a few people with type I diabetes are given a pancreas transplant. Pancreas transplants are not used to treat people with type II diabetes.

Give two reasons why pancreas transplants are not used for the treatment of type II diabetes.

The pancreas produces the hormone insulin.

Put a tick (✓) in the box next to the statement which describes incorrectly the action of insulin.

Scientists investigated the use of induced pluripotent stem cells (iPS cells) to treat type I diabetes in mice. The scientists used four transcription factors to reprogramme skin cells to form iPS cells. The scientists then stimulated the in vitro differentiation of iPS cells into pancreatic cells.

The scientists set up three experimental groups:

Group A – 30 mice with type I diabetes received pancreatic cell transplants derived from iPS cells.

Group B – 30 mice with type I diabetes were left untreated.

Group C – 30 mice without diabetes were left untreated.

The scientists measured the blood glucose concentration of all the mice on a weekly basis for 12 weeks.

The results the scientists obtained are shown in Figure 6.

Suggest how transcription factors can reprogramme cells to form iPS cells.

Using all the information provided, evaluate the use of iPS cells to treat type I diabetes in humans.

Alzheimer’s disease (AD) is a non-reversible brain disorder that develops over a number of years. At the start of 2014 the number of Americans with AD was estimated to be 5.4 million. Every 30 seconds another person in America develops AD.

In the brain of a person with AD there is a lower concentration of acetylcholine. 5 This affects communication between nerve cells and initially results in memory loss and confusion. Some of the symptoms of AD that are associated with communication between nerve cells are reduced by taking the drug donepezil. Donepezil inhibits the enzyme acetylcholinesterase.

A gene mutation called E280A found on chromosome 14 causes early-onset AD 10 at a mean age of 49 years. The age at which the E280A mutation is expressed to cause AD varies.

Yaramul is a town in a historically isolated region of the Andes Mountains. The population of this town has the highest frequency of the E280A mutation in the world. The origin of the E280A mutation in this population has been traced back 15 to a common ancestor in the 17th century. Natural selection has not reduced the frequency of the E280A mutation in the population.

This autosomal dominant mutation involves a change in triplet 280 from GAA to GCA. Scientists analysed chromosome 14 from 102 individuals from Yaramul. They recorded a sample size of 204 and detected 75 E280A mutations but only 20 74 potential AD cases. The scientists identified individuals with the mutation by whole genome sequencing. They had decided that a DNA probe would not be a suitable method to detect the E280A mutation.

Assuming no one with AD died in 2014, calculate the annual percentage increase in AD cases in America for 2014 (lines 2–4).

Explain how donepezil could improve communication between nerve cells (lines 7–9).

Suggest and explain two reasons why there is a high frequency of the E280A mutation in Yaramul (lines 13–15).

Explain why natural selection has not reduced the frequency of the E280A mutation in the population (lines 16–17).

The age at which the E280A mutation is expressed to cause AD can vary (lines 11 – 12).

Suggest and explain one reason for this.

One scientific study which analysed chromosome 14 involved 102 individuals. The scientists recorded a sample size of 204. In this sample they detected 75 E280A mutations but only 74 potential AD cases (lines 19 – 21).

Suggest explanations for the figures the scientists recorded.

Suggest why a DNA probe for the mutated triplet was not considered a suitable method for detection of the E280A mutation (lines 22–23).

A student used a potometer to measure the movement of water through the shoot of a plant. The potometer is shown in Figure 5. As water is lost from the shoot, it is replaced by water from the capillary tube.

In one experiment, the air bubble moved 7.5 mm in 15 minutes. The diameter of the capillary tube was 1.0 mm.

Calculate the rate of water uptake by the shoot in this experiment.

Give your answer in mm³ per hour. Show your working. (The area of a circle is found using the formula, area = πr²)

The student wanted to determine the rate of water loss per mm² of surface area of the leaves of the shoot in Figure 5.

Outline a method she could have used to find this rate. You should assume that all water loss from the shoot is from the leaves.

The rate of water movement through a shoot in a potometer may not be the same as the rate of water movement through the shoot of a whole plant.

Suggest one reason why.

Aquaporins are channel proteins that allow the diffusion of water across membranes. One type of aquaporin, called PIP1, can also transport carbon dioxide molecules across membranes.

Figure 6 shows the structure of a water molecule and of a carbon dioxide molecule. They are drawn to the same scale.

Suggest two reasons why water molecules and carbon dioxide molecules can both pass through PIP1.

The scientists first produced transgenic poplar trees. These trees all had a length of foreign DNA inserted into them. This DNA led to the production of single-stranded RNA that specifically inhibited expression of the gene for PIP1.

The scientists then measured the difference in the amount of PIP1 in leaves of transgenic poplars and in leaves of wild type poplars without the foreign DNA. The amount of PIP1 in the transgenic poplars was approximately 15% of that in the wild type poplars.

Using this information, what can you conclude about the effect of the foreign DNA in the transgenic poplar trees?

The transgenic poplars still produced some PIP1. Suggest why.

The scientists investigated the importance of PIP1 in the movement of water and carbon dioxide through the tissues of leaves of poplar trees.

They measured the mean rates of movement of carbon dioxide and water through the tissues of leaves of transgenic poplars and through the tissues of leaves of wild type poplars.

Their results are shown in Figure 7.

Figure 7

Using only Figure 7, evaluate the importance of PIP1 in the movement of carbon dioxide and water through leaves of poplar trees.

Section B

one question.