Exam Questions

On Figure 1, draw a circle around the part of the fatty acid molecule and the part of the glycerol molecule that is removed to form a bond in a triglyceride molecule.

Name the bond formed between a fatty acid and glycerol in a triglyceride molecule.

Name the reaction involved in forming a bond between a fatty acid and glycerol in a triglyceride molecule.

• Bond =
• Reaction =

Name the fatty acid shown in Table 1 that is a saturated fatty acid.

The melting point is the temperature at which a solid changes state to be a liquid.

Use Table 1 to describe the relationship between fatty acid structure and fatty acid melting point

The ratio of saturated to unsaturated fatty acids in a cell-surface membrane determines the extent of the membrane’s fluidity.

Scientists provided a cell culture of mouse phagocytes with liquid broth rich in unsaturated fatty acids.

The scientists observed:

• an increase in the proportion of phospholipids in the phagocytes containing unsaturated fatty acids
• more phagocytosis.

Suggest and explain why there was more phagocytosis.

Describe the processes of facilitated diffusion and active transport.

• Facilitated diffusion

• Active transport

What are microvilli?

Vitamin A is a fat-soluble substance.

Micelles are involved in the process of vitamin A absorption.

Describe the process of vitamin A absorption into cells lining the ileum

Describe how we breathe in.

A scientist prepared alveolar tissue to view using an optical microscope. The scientist cut very thin slices of the alveolar tissue.

Explain why the scientist used very thin slices of alveolar tissue with the optical microscope.

Identify the tube labelled A.

The scientist used a ruler to measure the diameter of some of the alveoli.

Table 2 shows the scientist’s results.

The magnification of the image in Figure 2 is × 40.

Use this information and Table 2 to calculate the mean diameter, in μm, of the alveoli.

Show your working (Answer in μm)

Give the uncertainty associated with taking a measurement using a ruler with 1 mm graduations.

Calculate the percentage error for a measurement using the ruler of 4 mm

Give the:

• Uncertainty ± (mm)
• Percentage error

Molasses is a solution obtained from sugar beet plants.

The sugars present in molasses are sucrose, glucose and fructose.

Give the number of different types of monosaccharides present in molasses.

A student used the biochemical test for reducing sugars on a clear sample of molasses.

Describe the biochemical test for a reducing sugar.

Explain the result expected from the test on the sample of molasses

Include:

• Description of biochemical test
• Explanation of expected result

‘Free sugar’ is the sugar in food and drinks released when food is crushed or when sugar is added to food at home or by manufacturers.

Scientists recommend that no more than 5% of the energy consumed per day should come from ‘free sugar’.

The mean daily energy requirement for a 10-year-old child is 8100 kJ. The ‘free sugar’ in one tablespoon of molasses contains 250 kJ of energy.

Calculate the number of tablespoons of molasses required for a 10-year-old child to reach the recommendation for energy consumed in ‘free sugar’ per day.

Answer in number of tablespoons per day

A scientist used the apparatus in Figure 3 to investigate osmosis.

Use your understanding of osmosis to explain why the air pressure in the tube increased.

The scientist repeated the investigation, but made one change to the molasses. The scientist did not change the volume of molasses at the start of the investigation.

The scientist observed that the air pressure inside the tube increased by 160 kPa compared with 800 kPa in the first investigation.

Suggest the change the scientist made to the molasses to cause this smaller increase in air pressure.

Use the air pressure figures in a calculation to support your answer.

Describe the appearance and behaviour of chromosomes during prophase and during anaphase of mitosis.

Prophase

Anaphase

Give three structural differences between an mRNA molecule and a tRNA molecule

Table 3 shows mRNA codons and the amino acid coded by each codon.

Figure 5 shows the mRNA base sequence produced when part of a gene coding for an enzyme is transcribed.

Use Table 3 to give the amino acids coded by the mRNA base sequence in Figure 5

A mutation occurred in the part of the gene transcribed in Question 06.2. Figure 6 shows the mRNA base sequence produced when the identical part in the mutated gene is transcribed.

The amino acids coded by this mRNA base sequence form part of the enzyme’s active site.

Use all the information in this question to:

• name the type of mutation that occurred to produce the mutated gene
• give the change in DNA caused by this mutation
• explain the effect this mutation will have on the function of the enzyme.

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).

Method

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

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.

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.

Include:

• Change to method
• Explanation

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?

Give two structural features of an aorta wall and explain how they are related to the function of an aorta.

Small tears may occur in the layers of tissue in an aorta wall. These tears weaken the wall without bursting the aorta.

Scientists:

• measured the aorta diameter (d) in a large population of people over 60 years of age
• calculated the risk of an aorta wall developing a tear.

Table 4 shows their results

Blood may push through the tears in the aorta wall. This produces a balloon-like swelling called an aneurysm and increases the aorta diameter. Aneurysms can cause the aorta to burst.

Using all the information, what can you conclude about aorta diameter and the risk of developing an aneurysm?

• measured the aorta diameter (d) in a large population of people over 60 years of age
• calculated the risk of an aorta wall developing a tear.

A scientist investigated changes in a diseased heart and changes in a healthy heart during cardiac cycles.

For each heart, the scientist obtained a value for: • the mean blood volume in a full ventricle just before the ventricle contracts (BVB) • the mean ejection fraction (EF).

• The EF is the proportion of blood pumped out of a full ventricle in one heartbeat.
• The EF is calculated using this formula:

EF = Blood volume pumped out of a full ventricle in one heartbeat (stroke volume) / BVB

Table 5 shows the scientist’s results.

Using Table 5, a student calculated that the percentage change in the stroke volume of the diseased heart compared with the stroke volume of a healthy heart is –30%.

The student’s answer is wrong because the final step of the calculation was performed incorrectly. Using the equation and Table 5, calculate the correct percentage change in the stroke volume of the diseased heart compared with the stroke volume of the healthy heart.

Identify the mathematical error in the final step of the student’s calculation

Include

• Correct answer (%)
• Mathematical error

Courtship behaviour in the frog species, Xenopus laevis box, involves male frogs calling to:

• attract sexually active females - these are advertisement calls
• start and continue mating – these are mating calls
• signal when a male is not sexually active – these are rasping calls.

Scientists investigated frog courtship behaviour by feeding a population of sexually active male frogs a diet containing the hormone EE2. The scientists also fed a separate control population of sexually active male frogs a diet without EE2.

They determined the percentage of males making advertisement calls or rasping calls in each population.

Table 6 shows their results

There were 800 males in the control population.

Each male made one type of call.

Use this information and Table 6 to calculate the number of males making mating calls in the control population.

Suggest one change the scientists could make to both frog populations to increase the number of mating calls.

The scientists also investigated the effect on female frog courtship behaviour of feeding EE2 to male frogs.

Table 7 shows their results.

EE2 is contained in human contraceptive pills. Some EE2 is released in human urine and collects in sewage. Untreated sewage pollutes the water in frog habitats.

Suggest and explain the effect EE2 pollution in frog habitats will have on frog breeding.

Use information from Table 6 and Table 7 in your answer.

Include:

• Effect on frog breeding
• Explanation

Describe and explain how you would use cell fractionation and ultracentrifugation to obtain a sample of nuclei from muscle tissue.

Describe the role of organelles in the production and release of enzymes by animal cells. Do not include details of transcription in your answer.

Describe the structure of ATP.

Outline how named enzymes break down and resynthesise ATP.

Give the three structural features found in all virus particles and describe the function of one of these features.

Explain why viruses are described as acellular and non-living.

Give one reason why antibiotics are not effective against viruses.

Chitin is a polysaccharide. The chitin monomer is a β-glucose molecule with one OH group replaced by an NHCOCH3 group. NHCOCH3 can be represented by N(Ac). Figure 1 shows the monomer that forms chitin and the chitin polymer

Chitin has a similar structure to cellulose.

Use Figure 1 to describe three ways the structure of chitin is similar to the structure of cellulose.

Chitin keeps the tracheae open in the tracheal system of gas exchange in an insect. Gas exchange does not occur in the tracheae.

Explain the importance of one adaptation of the gas exchange surface in the tracheal system of an insect.

Lignin is a polymer found in the walls of xylem vessels in plants. Lignin keeps the xylem vessel open as a continuous tube.

Explain the importance of the xylem being kept open as a continuous tube.

The human disease, malaria, is caused by infection with a single-celled eukaryotic organism.

Figure 2 shows a diagram of Plasmodium vivax, one of the species that can cause malaria.

Other than the Golgi apparatus, name one structure in Figure 2 which shows that P. vivax is a eukaryote.

Describe two functions of the Golgi apparatus in a eukaryotic cell.

P. vivax box evolved from a common ancestor in Africa. As humans migrated around the world, new strains of P. vivax evolved.

P. vivax is now extremely rare in Africa but there are several different strains of P. vivax in other parts of the world.

Figure 3 shows a phylogenetic diagram of the evolution of these different strains

What does Figure 3 suggest is the order of human migration out of Africa?

Tick () one box.

There are an estimated 229 million cases of human malaria worldwide per year.

94% of these cases are found in Africa, but are not caused by P. vivax.

P. vivax does cause 61% of the cases of human malaria outside Africa.

Use this information to calculate the number of cases worldwide caused by P. vivax each year.

Answer in cases of malaria

In Africa today, most of the human population are resistant to malaria caused by P. vivax.

Use your knowledge of natural selection to explain why this resistance is so common in Africa.

Some hospital patients suffer from diarrhoea caused by infection with the bacterium Clostridium difficile. The C. difficile bacteria release toxins. These toxins cause the diarrhoea.

The toxins damage the cells lining the ileum, causing them to lose their microvilli. The damage to the cells reduces the absorption of the products of digestion and reduces the absorption of water, resulting in diarrhoea.

Explain why the damage to the cells lining the ileum reduces absorption of the products of digestion and why this reduces absorption of water.

Not all patients in hospital with C. difficile develop diarrhoea. box Scientists measured the anti-toxin antibody concentration in hospital patients with and without C. difficile infection. They measured the anti-toxin antibody concentration four times:

• on admission to hospital (day 0)
• on day 3
• on day 6
• on the day the patient left the hospital.

Figure 4 shows the scientists’ results.

The scientists suggest that the anti-toxin antibody could be given to some patients as a form of passive immunity.

Use Figure 4 to suggest how this passive immunity would work and which patients should be offered this anti-toxin antibody.

To be used as passive immunity treatment, the anti-toxin antibody would be injected. If it was given by mouth, it would be digested.

Describe how the anti-toxin antibody would be digested.

A student investigated the use of cinnamon oil as box an antimicrobial substance. She investigated the effect of cinnamon oil on the growth of five different bacterial cultures grown on agar plates.

The student added 100 mm³ of each bacterial culture from its glass bottle onto a separate agar plate. She spread each bacterial culture evenly over the agar using a spreader.

Describe the aseptic techniques the student should use

On each agar plate, the student cut a well (a hole) in the agar. The well had a diameter of 6 mm. The student added 50 mm³ of cinnamon oil into the well.

Calculate the minimum depth of the well to allow the addition of 50 mm³ of cinnamon oil.

Use the following equation in your calculation: Volume of a cylinder = πr² × l Use 3.14 as the value for π. Show your working.

The student kept the plates at 25 °C for 24 hours.

Figure 5 shows what one of her plates looked like after 24 hours.

Suggest exactly what the student added to the wells to get the positive control and negative control results.

Complete Table 1 to show the median and mean diameters.

The mean ± 2 standard deviations includes over 95% of the data. Use this information to consider whether the standard deviations suggest the differences in means are likely to be due to chance.

Explain your answer, including at least one calculation.

Define genome and proteome.

The classification system used in the early 20th century grouped different species of bacteria according to the position and shape of flagella on bacterial cells and by the number of flagella per cell. These were observed using an optical microscope.

Each species of bacterium has a characteristic cell shape and arrangement of flagella.

These characteristics may be shared with other species within a genus.

Flagella are fragile, difficult to stain and may extend from the cell at any angle.

Consider the accuracy and limitations of the early classification of bacteria using the arrangement of flagella.

Suggest why several bacterial species have been renamed in recent years.

Figure 6 shows an image from an optical microscope of a single bacterial cell.

This bacterial cell is 2.3 μm long (excluding the flagellum).

Calculate the magnification of this image. Show your working.

Carbon monoxide (CO) is released during incomplete combustion of fossil fuels. Figure 7 shows the dissociation curve for oxyhaemoglobin when:

• not exposed to CO
• exposed to CO such that 50% of the oxygen binding sites are occupied by CO (50% COHb)

Using Figure 7, what can you conclude about how exposure to CO affects the loading and unloading of oxygen by haemoglobin?

Explain your answer

The World Health Organisation (WHO) suggests that to avoid long-term health effects, COHb concentrations should be kept below 2.5%. WHO recommends that people should not be exposed to:

• air with > 10 mg m⁻³ CO for more than 8 hours
• air with > 30 mg m⁻³ CO for more than 1 hour.

Scientists have used a mathematical model to calculate the exposure to carbon monoxide that would result in 2.5% COHb in both adults and children. Table 2 shows the scientists’ results. The scientists suggest that the WHO recommendations for carbon monoxide concentrations resulting in 2.5% COHb should be reduced.

Evaluate the scientists’ conclusion

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

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 Answer (μg)

Dengue box fever is a human disease caused by the dengue virus. Scientists designed an ELISA test to detect antibodies to the dengue virus in a patient’s blood sample. Figure 10 shows a diagram of this test and some information about how it works.

Suggest what is on the test at line T and explain what causes the line to appear in a positive test.

A line at C shows that the test has worked.

Suggest one reason why a line at C shows the test has worked.

Figure 12 shows a flowchart of how the anti-human antibodies with enzyme attached are produced. Suggest why the fused cells allow continuous production of monoclonal antibodies.

Evaluate the ethics of the production process shown in Figure 12.

Early identification of dengue fever can be difficult as many other diseases produce box the same symptoms. Early identification is important because people suffering with dengue fever can become ill very quickly and may need hospital treatment.

Scientists compared the effectiveness of three diagnostic tests for dengue fever.

• Laboratory-based test – a patient’s blood sample is sent from the doctor’s clinic to a laboratory for testing.
• Current test used in the doctor’s clinic.
• New test to be used in the doctor’s clinic – the ELISA test shown in Figures 10 and 11 (on page 24).

The scientists’ results are shown in Table 3.

A blood sample from each patient with confirmed dengue fever at each time after onset of symptoms was tested with all three diagnostic tests.

The scientists recommend that the new test is used for the identification of dengue fever in all countries around the world.

Discuss this recommendation. Use all the information given

The dengue virus causes damage to capillaries so that blood proteins move out of the capillaries into the tissue fluid.

Explain how damage to capillaries would affect the return of tissue fluid into the capillaries.

Describe how a quaternary protein is formed from its monomers. Do not include translation.

Describe the structure of DNA and the structure of a chromosome.

Mutation can result in an increase in genetic variation within a species. Describe and explain the other processes (apart from mutation) that result in increases in genetic variation.

Describe the structure and function of the nucleus.

Name the main polymer that forms the following cell walls.

• Plant cell wall
• Fungal cell wall

Scientists investigated the effect of the number of fungal species in soil on the diversity of plant species.

Suggest one reason the scientists used biomass instead of the number of individuals of each plant species when collecting data to measure diversity.

The scientists used this equation to calculate the plant species index of diversity.

d = 1 − Σ(n/N)²

where n = shoot biomass of each plant species and N = total shoot biomass of all plant species

Use this equation to calculate the index of diversity for the data in Table 1.

Clostridium difficile is a bacterial species that causes disease in humans. Antibiotic-resistant strains of C. difficile have become a common cause of infection acquired when in hospital.

Explain how the use of antibiotics has led to antibiotic-resistant strains of bacteria becoming a common cause of infection acquired when in hospital.

Scientists suggested that factors, other than antibiotic use, led to the increase in antibiotic-resistant C. difficile infections. One suggested factor is people eating more trehalose in their diet. Trehalose is a disaccharide formed from two glucose molecules.

Name another disaccharide formed from two glucose molecules.

Scientists investigated the effect of trehalose on the growth rate of C. difficile. They grew populations of non-resistant and antibiotic-resistant C. difficile on separate agar plates with:

• no carbohydrate added
• trehalose added.

They measured the growth rate of the C. difficile.

Describe how the scientists could use aseptic techniques to transfer 0.3 cm³ of C. difficile in liquid culture from a bottle onto an agar plate.

Use Figure 2 to evaluate whether more trehalose in the diet could be a factor in the increased number of antibiotic-resistant C. difficile infections.

Give two features of all prokaryotic cells that are not features of eukaryotic cells.

This AP has a secondary structure in a helical shape.

Tick the box to show which type of bond maintains the helical structure of the polypeptide.

The amino acids on one side of each AP helix have hydrophobic properties. The amino acids on the opposite side of each helix have hydrophilic properties. Figure 4 shows this.

Suggest how these properties of the APs allow them to become positioned across the membrane (as shown in Figure 3) and make a channel through which ions can pass.

Figure 5 shows further information about a channel formed in the cell-surface membrane by the APs.

Use Figure 5 to calculate the cross-sectional area of the channel through which ions can pass. Assume the cross-sectional area is circular. Use π = 3.14 in your calculation.

Give your answer in nm² and to 1 decimal place

Answer in nm²

The APs damage prokaryotic cells but do not damage the eukaryotic cells in the organisms that produce them. Prokaryotic cell membranes do not contain cholesterol.

Assess why the APs do not damage the eukaryotic cells of the organisms that produce them.

Scientists observed these APs on prokaryotes using a transmission electron microscope. They stained the APs using a monoclonal antibody with gold attached to it.

Suggest how these techniques allowed observation of APs on prokaryotes.

Describe viral replication.

Complete Table 2 by putting a tick () where the feature is part of a cell cycle involving mitosis or a cell cycle involving binary fission.

Complete Figure 6 to show the chromosomes inside the daughter cells formed after the second meiotic division.

A student concluded that there were more mothers of age ˃37 with MM2 errors than with MM1 errors. Using Figure 7 and suitable calculations show why this conclusion is not valid.

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.

Mangrove trees grow near the sea. Sea water surrounds the lower parts of the trees at high tide. Scientists investigated the rate of transpiration in a mangrove tree. Figure 9 shows the scientists’ results.

Explain the rate of transpiration between 5 am and midday shown in Figure 9.

Use Figure 9 to calculate the percentage increase in the rate of transpiration from 1 pm to 2 pm.

Percentage increase in rate of transpiration % =

The higher rate of transpiration at high tide shows that the mangrove tree is absorbing water from the sea water surrounding its roots.

Describe an experiment that you could do to investigate whether the mangrove root cells have a lower water potential than sea water. You are given:

• a piece of fresh mangrove root
• sea water
• access to laboratory equipment.

Complete Table 3 to give three differences between DNA molecules and tRNA molecules.

The scientists broke open the cells to produce a suspension of cell contents.

Describe how the scientists would remove large organelles from this suspension of cell contents.

Explain the position of the bands of ribosomes in tubes A and B in Figure 10.

To observe the fish gills with the optical microscope, the scientists used two different stains. The first stain binds to DNA; the second stain binds to the red blood cells.

Explain why a second stain would be needed to stain the red blood cells.

Suggest which molecule the stain could bind to in the red blood cells.

Using Figure 11, the scientists calculated the surface area to volume ratios for each gill filament in these two fish. Some of their results are shown in Table 4.

Complete Table 4. State your calculated volume and surface area:volume ratio to 2 significant figures.

The damage to the gills causes uncontrolled cell division in the cells around the capillaries in the gill filaments. Other than surface area:volume ratio, describe one way this uncontrolled cell division changes the gills, as shown in Figure 11.

Explain how this difference would affect gas exchange.

Include:

• Difference
• Explanation

Use Figure 12 to complete Table 5 to show two differences between the circulation of blood in fish and the circulation of blood in a mammal.

Describe the transport of carbohydrate in plants.

Compare and contrast the structure of starch and the structure of cellulose.

Describe the complete digestion of starch by a mammal.

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:

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.

Explain the advantage for larger animals of having a specialised system that facilitates oxygen uptake.

Suggest how the environmental conditions have resulted in adaptations of systems using Model A rather than Model B.

Figure 3 shows changes in concentration of oxygen in two gas exchange systems.

A student studied Figure 3 and concluded that the fish gas exchange system is more efficient than the human gas exchange system.

Use Figure 3 to justify this conclusion.

Explain how the counter-current principle allows efficient oxygen uptake in the fish gas exchange system.

Table 1 shows features of two mammals. Bats are flying mammals; shrews are ground-living mammals.

Calculate how many times the lung volume per unit of body mass of the bat is greater than that of the shrew.

Give your answer to an appropriate number of significant figures. Give one suggestion to explain this difference.

Describe how one amino acid is added to a polypeptide that is being formed at a ribosome during translation.

Table 2 shows:

• mRNA codons and the amino acid coded for by each codon
• the type of bond formed by the R group of some of the amino acids.

Crystallin is a structural protein found in the human eye. An inherited disease that leads to blindness is caused by changes in properties of crystallin. The replacement of the amino acid Arg with the amino acid Gly causes these changes.

Use information in Table 2 to suggest why this amino acid replacement changes the properties of crystallin.

The amino acid replacement of Arg with Gly is caused by a single base substitution mutation in the DNA. The non-mutant DNA triplet is TCC. Complete Table 3.

Give:

• the mRNA codon complementary to the non-mutant DNA triplet
• the mutated mRNA codon that could cause the change from Arg to Gly in the crystallin protein
• the DNA triplet complementary to this mutated mRNA codon.

A student box dissected an organ from a mammal to observe blood vessels. He dissected a slice of the organ and identified two blood vessels.

Suggest two ways the student could improve the quality of his scientific drawing of the blood vessels in this dissection.

Identify the type of blood vessel labelled as X and the type of blood vessel labelled as Y in Figure 4.

Describe one feature that allowed you to identify the blood vessels. Blood vessel X = Blood vessel Y = Feature =

Describe two precautions the student should take when clearing away after the dissection.

Describe how a sample of chloroplasts could be isolated from leaves.

Scientists grew two groups of plants:

• control plants with all the inorganic ions needed
• iron-deficient plants with all the inorganic ions needed but without iron ions.

After 1 week, the scientists measured the mass of protein and the mass of chlorophyll in the chloroplasts isolated from samples of leaves of these two groups of plants. Table 4 shows the scientists’ results.

Some proteins found inside the chloroplast are synthesised inside the chloroplast.

Give one feature of the chloroplast that allows protein to be synthesised inside the chloroplast and describe one difference between this feature in the chloroplast and similar features in the rest of the cell.

Feature = Structural difference =

The ratio of protein to chlorophyll in control plants is 9:1 Use the information in Table 4 to calculate the ratio of protein to chlorophyll in iron-deficient plants.

Use Figure 6 to suggest why iron-deficient plants have a reduced growth rate

Figure 7 shows the mean distance between centromeres and the poles (ends) of the spindle during mitosis.

Calculate the rate of movement of the centromeres during phase E.

Give your answer in μm minute–1 and to 3 decimal places

Name the three phases of mitosis shown by C, D and E on Figure 7.

Describe the role of the spindle fibres and the behaviour of the chromosomes during each of these phases.

ADCs are molecules made of a monoclonal antibody linked to a cancer drug. Figure 8 shows how an ADC enters and kills a tumour cell. The process of entering the cell and the breakdown of the antibody to release the drug is very similar to phagocytosis.

Use your knowledge of phagocytosis to describe how an ADC enters and kills the tumour cell.

Some of the antigens found on the surface of tumour cells are also found on the surface of healthy human cells.

Use this information to explain why treatment with an ADC often causes side effects

Calculate the volume of antibody solution that the scientists would have injected into a 23 g mouse.

Give your answer in dm³ and in standard form.

Suggest one reason why there are no data for Group G and Group H after day 8

Suggest and explain two further investigations that should be done before this ADC is tested on human breast cancer patients.

Describe how a triglyceride molecule is formed.

Put a tick () in one box that contains correct information about one of these fatty acids.

The percentage of saturated fatty acids compared with unsaturated fatty acids found in lipid stores in seeds differs in different populations. Scientists investigated two populations of the plant, Helianthus annuus. The scientists grew young plants from seeds collected from each population. They placed the seeds on wet tissue paper so that the root growth was visible.

They grew seeds from each population at two temperatures:

• warm temperature of 24 °C
• cool temperature of 10 °C

After 10 days, the scientists measured the length of each root.

Table 6 shows some of the properties of the two populations and the scientists’ results.

The mean ±2 × standard deviation includes 95% of the data. The scientists used a data logger to measure the length of the root rather than a ruler.

Suggest one reason why they used a data logger and explain why this was important in this investigation.

It is known that:

Use this information and Table 6 to show how each population is better adapted for its natural environment when compared with the other population.

• during respiration saturated fatty acids yield more energy than unsaturated fatty acids
• saturated fatty acids have higher melting points than unsaturated fatty acids
• lipases in seeds act more rapidly on liquid substrates.

Although these two populations are completely separate and show genetic variation, they are both called Helianthus annuus.

Explain why they are both given this name.

Complete Table 7 with ticks () to show which elements are found in the following biological molecules.

After Watson and Crick proposed the model of DNA structure, scientists investigated box the possible mechanisms for DNA replication. Two scientists grew a bacterial population, providing them with a nitrogen source containing only the heavy isotope of nitrogen, 15 N. As soon as all the DNA in this population contained 15 N, the scientists changed the nitrogen source to one containing only the lighter isotope of nitrogen, 14 N. They changed the nitrogen source at 0 hours. During the investigation, the scientists measured the size of the population of bacterial cells. Figure 11 shows the scientists’ results.

The generation time for a population of bacteria is the time taken for all the bacteria to divide once by binary fission.

Use Figure 11 and the following equation to calculate the generation time for this population of bacteria.

Give your answer in hours.

At intervals during this investigation, the scientists removed samples of the bacterial population, isolated the DNA and measured the density of the DNA. DNA made using 15 N has a higher density than DNA made using 14 N. Figure 12 shows the scientists’ results.

Which of these models, P, Q or R, is supported by the results shown in Figure 12?

Give the letter and name of the model supported and explain why the results do not support the other models.

Describe the structure of DNA.

Name and describe five ways substances can move across the cell-surface membrane into a cell.

Contrast the structure of the two cells visible in the electron micrographs shown in Figure 14.

Figure 1 shows a cell from the lining of the ileum specialised for absorption of products of digestion. SGLT1 is a carrier protein found in the cell-surface membrane of this cell, it transports glucose and sodium ions (Na⁺) into the cell.

The action of the carrier protein X in Figure 1 is linked to a membrane-bound ATP hydrolase enzyme.

Explain the function of this ATP hydrolase.

The movement of Na⁺ out of the cell allows the absorption of glucose into the cell lining the ileum. Explain how

Describe and explain two features you would expect to find in a cell specialised for absorption.

Draw phospholipids on Figure 2 to show how the carrier protein, SGLT1, would fit into the cell-surface membrane. Do not draw more than eight phospholipids.

Figure 2 shows the SGLT1 polypeptide with NH₂ at one end and COOH at the other end.

Describe how amino acids join to form a polypeptide so there is always NH₂ at one end and COOH at the other end. You may use a diagram in your answer.

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

At P on Figure 3, the pressure in the left ventricle is increasing. At this time, the rate of blood flow has not yet started to increase in the aorta.

Use evidence from Figure 3 to explain why.

At Q on Figure 3 there is a small increase in pressure and in rate of blood flow in the aorta.

Explain how this happens and its importance

A student correctly plotted the right ventricle pressure on the same grid as the left ventricle pressure in Figure 3.

Describe one way in which the student’s curve would be similar to and one way it would be different from the curve shown in Figure 3.

Use information from Figure 3 to calculate the heart rate of this dog.

Anthocyanins are coloured pigments found in the cell vacuole of some plant cells. Anthocyanins cannot move across undamaged cell membranes. A student investigated how to extract anthocyanins from blueberries. She mixed 10 g of crushed, fresh blueberries with 100 cm³ of extraction solvent for 1 hour. She investigated three different extraction solvents:

When making up extraction solvent E, the student used a volume ratio of 70:30:1 ethanol:water:acid.

Tick () one box that shows the most appropriate volumes she would use to make up 100 cm³ of extraction solvent E.

A) 63.6cm³ ethanol, 27.3cm³ water, 9.1cm³ acid B) 69.3cm³ ethanol, 29.7cm³ water, 1.0cm³ acid C) 70.0cm³ ethanol, 30.0cm³ water, 1.0cm³ acid D) 70.7cm³ ethanol, 30.3cm³ water, 1.0cm³ acid

The student kept constant:

Name two other variables the student should have kept constant during this investigation.

• the mass of fresh blueberries
• the volume of extraction solvent
• the time for the mixture to stand.

After 1 hour, the student filtered the samples.

She placed the filtrate in a colorimeter and measured the light absorbance.

Her results are shown in Figure 4.

Use your knowledge of membrane structure to explain the results in Figure 4

A different student did this investigation. He did not have a colorimeter.

Describe a method this student could use to prepare colour standards and use them to give data for the total anthocyanin extracted.

Describe the role of DNA polymerase in the semi-conservative replication of DNA

Figure 5 shows the percentage of rat cells undergoing DNA replication. Some cells contained a protein called cyclin D and some cells did not contain cyclin D. All cells were in early interphase at time 0

It took less time for 25% of cells with cyclin D to be undergoing DNA replication than for 25% of cells without cyclin D.

Use Figure 5 to calculate this time difference as a percentage decrease. Show your working.

Cyclin D stimulates the phosphorylation of DNA polymerase, which activates the DNA polymerase.

Describe how an enzyme can be phosphorylated.

Some tumour cells contain higher than normal concentrations of cyclin D.

Use Figure 5 to suggest why higher than normal concentrations of cyclin D could result in a tumour.

Particulate matter is solid particles and liquid particles suspended in air. Polluted air contains more particulate matter than clean air.

A high concentration of particulate matter results in the death of some alveolar epithelium cells. If alveolar epithelium cells die inside the human body they are replaced by non-specialised, thickened tissue.

Explain why death of alveolar epithelium cells reduces gas exchange in human lungs

Scientists grew alveolar epithelium cells and exposed the epithelium cells to different box concentrations of particulate matter. They calculated the percentage of these alveolar epithelium cells that died after 24 hours of exposure to particulate matter. Their results are shown in Figure 6.

Do the data in Figure 6 show a linear relationship between concentration of particulate matter and percentage of dead cells?

Use suitable calculations to justify your answer.

Alpha-gal is made of two galactose molecules. Galactose has the chemical formula C₆H₁₂O₆.

Give the chemical formula for the disaccharide, alpha-gal, and describe how it is formed from two galactose molecules.

Some people eat red meat for many years without having any reaction, then have an allergic reaction to the alpha-gal in red meat. An allergic reaction is caused by an immune response.

Draw a labelled diagram of an antibody and identify the specific alpha-gal binding site.

A tick is a small animal that bites humans and feeds on their blood. This results in proteins from the tick saliva entering the human body.

Scientists have suggested one hypothesis for the allergic reaction to alpha-gal in red meat. They think that an earlier immune response to a tick bite can cause a person to have an allergic reaction to alpha-gal in red meat.

Suggest how one antibody can be specific to tick protein and to alpha-gal.

Complete Table 2 to show three differences between DNA in the nucleus of a plant cell and DNA in a prokaryotic cell.

Scientists investigated the genetic diversity between several species of sweet potato. They studied non-coding multiple repeats of base sequences.

Define ‘non-coding base sequences’ and describe where the non-coding multiple repeats are positioned in the genome.

Use the information in Table 3 to complete the phylogenetic tree shown in Figure 8.

Write the letter that represents the correct species into each box.

The scientists studied five individuals from each species. Within the five individuals of species T they found a percentage similarity of 66%.

Use Table 3 to evaluate how this information affects the validity of the phylogenetic tree.

Scientists investigated stomatal density on leaves of one species of tree. Figure 9 shows three examples of the square fields of view the scientists used to calculate a mean stomatal density.

Calculate the mean stomatal density in the three fields of view in Figure 9.

Give your answer as number of stomata per mm² Show your working.

The scientists used leaves from individual trees that had grown in different areas of box the world in different years. Each tree had grown in an area and year with known carbon dioxide concentration. Their results are shown in Figure 10.

Give a null hypothesis for this investigation and name a statistical test that would be appropriate to test your null hypothesis.

From 1910 to 2000, the carbon dioxide concentration in the atmosphere increased from 300 parts per million to 365 parts per million.

Use Figure 10 to calculate the mean rate of change in stomatal density from 1910 to 2000.

Give your answer as number of stomata per mm² per 10-year period. Show your working.

Give as number of stomata per mm² per 10-year period

A journalist saw Figure 10 and suggested that future increases in atmospheric carbon dioxide concentration could result in less transpiration.

Evaluate his suggestion.

Describe how mRNA is formed by transcription in eukaryotes.

Describe how a polypeptide is formed by translation of mRNA

Define ‘gene mutation’ and explain how a gene mutation can have:

• no effect on an individual
• a positive effect on an individual.

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 box . 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.

Table 1 shows cell wall components in plants, algae, fungi and prokaryotes.

Complete Table 1 by putting a tick () where a cell wall component is present.

Cell walls make up much of the fibre that people eat. Scientists investigated the relationship between the mass of fibre people ate each day and their risk of cardiovascular disease (CVD). They gathered data from a large sample of people and used this to calculate a relative risk.

• A relative risk of 1 means there is no difference in risk between the sample and the whole population.
• A relative risk of < 1 means CVD is less likely to occur in the sample than in the whole population.
• A relative risk of > 1 means CVD is more likely to occur in the sample than in the whole population.

Their results are shown in Figure 3. A value of ± 2 standard deviations from the mean includes over 95% of the data.

A student concluded from Figure 3 that eating an extra 10 g of fibre per day would significantly lower his risk of cardiovascular disease.

Evaluate his conclusion

The scientists estimated the mean mass of fibre eaten per day using a food frequency questionnaire (FFQ). The FFQ asks each person how often they have eaten many types of food over the past year. An alternative method to calculate fibre eaten is for a nurse to ask each person detailed questions about what they have eaten in the last 24 hours.

Suggest one advantage of using the FFQ method and one disadvantage of using the FFQ method compared with the alternative method.

A group of students investigated biodiversity of different areas of farmland. They collected data in each of these habitats:

• the centre of a field
• the edge of a field
• a hedge between fields.

Their results are shown in Figure 4.

What data would the students need to collect to calculate their index of diversity in each habitat? Do not include apparatus used for species sampling in your answer.

Give two ways the students would have ensured their index of diversity was representative of each habitat.

Modern farming techniques have led to larger fields and the removal of hedges between fields.

Use Figure 4 to suggest why biodiversity decreases when farmers use larger fields.

Farmers are now being encouraged to replant hedges on their land.

Suggest and explain one advantage and one disadvantage to a farmer of replanting hedges on her farmland.

Scientists collected data on 800 000 human births. The data showed the mass of each baby at birth and whether the baby needed to be transferred to a special care unit for very ill babies. Their results are shown in Figure 5.

Use Figure 5 to explain how human mass at birth is affected by stabilising selection

The scientists studied the effect of one form, KIR2DS1, of the human KIR gene on mass at birth. In the following passage the numbered spaces can be filled with biological terms.

Write the correct biological term beside each number below, that matches the space in the passage.

The scientists studied 1500 more births. They recorded the mass at birth of each baby and the nature of the KIR gene in the mother’s genome. Some of their results are shown in Table 2

The scientists used a statistical test to test the following null hypothesis: ‘The presence of KIR2DS1 in the mother’s genome does not affect the frequency of births above 4500g’

Tick () one box that gives the name of the statistical test that the scientists should use with the data in Table 2 to test this null hypothesis.

The scientists calculated a P value of 0.03 when testing their null hypothesis. What can you conclude from this result?

Explain your answer.

Describe the structure of the human immunodeficiency virus (HIV).

Some people infected with HIV do not develop AIDS. These people are called box HIV controllers. Scientists measured the number of HIV particles (the viral load) and the number of one type of T helper cell (CD4 cells) in the blood of a group of HIV controllers and also in a group of HIV positive patients who had symptoms of AIDS. The median values and the range of their results are shown in Table 3.

A test sample of 500mm³ of blood is taken from an HIV controller to determine the viral load.

Tick () one box that shows the number of virus particles that would be present in a test sample of blood taken from an HIV controller with the median viral load.

Use the data in Table 3 and your knowledge of the immune response to suggest why HIV controllers do not develop symptoms of AIDS.

Scientists investigated the cell cycle in heart cells taken from mice 6 days before their birth and then at 4, 14 and 21 days after their birth. Their results are shown in Table 4. Age 0 days = day of birth.

Describe and explain the data in Table 4

The scientists determined the percentage of heart cells undergoing DNA replication by using a chemical called BrdU. Cells use BrdU instead of nucleotides containing thymine during DNA replication.

Describe how BrdU would be incorporated into new DNA during semi-conservative replication

Cells with BrdU in their DNA are detected using an anti-BrdU antibody with an enzyme attached.

Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA.

Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by seawater. Figure 6 shows a diagram ofUnlike plants, Ulva lactuca does not have xylem tissue.

Suggest how Ulva lactuca is able to survive without xylem tissue one Ulva lactuca alga.

On Figure 7 complete each box with an appropriate letter to show the type of cell division happening between each stage in the life cycle.

Use ‘T’ to represent mitosis and ‘E’ to represent meiosis.

Ulva prolifera also produces haploid, mobile single cells that can fuse to form a zygote.

Suggest and explain one reason why successful reproduction between Ulva prolifera and Ulva lactuca does not happen.

The water potential of leaf cells is affected by the water content of the soil. Scientists grew sunflower plants. They supplied different plants with different volumes of water.

After two days, they determined the water potential in the leaf cells by using an instrument that gave a voltage reading. The scientists generated a calibration curve to convert the voltage readings to water potential.

Figure 8 shows their calibration curve

The scientists needed solutions of known water potential to generate their calibration curve.

Table 5 shows how to make a sodium chloride solution with a water potential of −1.95 MPa

Complete Table 5 by giving all headings, units and volumes required to make 20 cm³ of this sodium chloride solution.

There is a linear relationship between the water potential and the concentration of sodium chloride solution.

Use the data in Table 6 to calculate the concentration of sodium chloride solution with a water potential of −3.41MPa

In addition to determining the water potential in the leaf cells, the scientists measured the growth of the leaves. They recorded leaf growth as a percentage increase of the original leaf area. Their results are shown in Figure 9.

One leaf with an original area of 60cm² gave a voltage reading of −7µV Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area.

Give your answer in cm²

Sunflowers are not xerophytic plants. The scientists repeated the experiment with xerophytic plants.

Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9.

Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly

A scientist investigated the affinity for oxygen of horse haemoglobin and mouse haemoglobin. Some of their results are shown in Table 7.

Plot the haemoglobin saturation data from Table 7 and use these points to sketch the full oxyhaemoglobin dissociation curves for a horse and a mouse.

The following equation can be used to estimate the metabolic rate of an animal. Metabolic rate = 63 × BM–0.27 BM = body mass in grams

Use this equation to calculate how many times faster the metabolic rate of a mouse is than the metabolic rate of a horse

The data in Table 7 show differences between the oxyhaemoglobin dissociation curve for a mouse and the oxyhaemoglobin dissociation curve for a horse.

Suggest how these differences allow the mouse to have a higher metabolic rate than the horse.

Mammals such as a mouse and a horse are able to maintain a constant body temperature.

Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse.

Explain five properties that make water important for organisms.

Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample.

Describe the chemical reactions involved in the conversion of polymers to monomers and monomers to polymers.

Give two named examples of polymers and their associated monomers to illustrate your answer.

Figure 1 shows all the chromosomes present in one human cell during mitosis. A scientist stained and photographed the chromosomes. In Figure 2, the scientist has arranged the images of these chromosomes in homologous pairs.

Give two pieces of evidence from Figure 1 that this cell was undergoing mitosis.

Explain your answers.

Tick one box that gives the name of the stage of mitosis shown in Figure 1.

When preparing the cells for observation the scientist placed them in a solution that had a slightly higher (less negative) water potential than the cytoplasm. This did not cause the cells to burst but moved the chromosomes further apart in order to reduce the overlapping of the chromosomes when observed with an optical microscope.

Suggest how this procedure moved the chromosomes apart.

The dark stain used on the chromosomes binds more to some areas of the chromosomes than others, giving the chromosomes a striped appearance.

Suggest one way the structure of the chromosome could differ along its length to result in the stain binding more in some areas.

In Figure 2 the chromosomes are arranged in homologous pairs.

What is a homologous pair of chromosomes?

Give two ways in which the arrangement of prokaryotic DNA is different from the arrangement of the human DNA in Figure 1.

A student investigated the effect of surface area on osmosis in cubes of potato.

• He cut two cubes of potato tissue, each with sides of 35 mm in length.
• He put one cube into a concentrated sucrose solution.
• He cut the other cube into eight equal-sized smaller cubes and put them into a sucrose solution of the same concentration as the solution used for the large cube.
• He recorded the masses of the cubes at intervals.

His results are shown in Figure 3.

Describe the method the student would have used to obtain the results in Figure 3. Start after all of the cubes of potato have been cut. Also consider variables he should have controlled.

The loss in mass shown in Figure 3 is due to osmosis. The rate of osmosis between 0 and 40 minutes is faster in B (the eight small cubes) than in A (single large cube). Is the rate of osmosis per mm² per minute different between A and B during this time?

Use appropriate calculations to support your answer.

Bees are flying insects that feed on nectar made in box flowers. There are many different species of bee.

Scientists investigated how biodiversity of bees varied in three different habitats during a year. They collected bees from eight sites of each habitat four times per year for three years.

The scientists’ results are shown in Figure 4 in the form they presented them.

What is meant by ‘species richness’?

From the data in Figure 4, a student made the following conclusions. 1. The natural habitat is most favourable for bees. 2. The town is the least favourable for bees.

Does the data in Figure 4 support these conclusions?

Explain your answer.

Explain the following: 1. The natural habitat is most favourable for bees. 2. The town is the least favourable for bees

The scientists collected bees using a method that was ethical and allowed them to identify accurately the species to which each belonged.

In each case, suggest one consideration the scientists had taken into account to make sure their method 1. was ethical. 2. allowed them to identify accurately the species to which each belonged.

Suggest and explain two ways in which the scientists could have improved the method used for data collection in this investigation.

Three of the bee species collected in the farmland areas were Peponapis pruinosa, Andrena chlorogaster and Andrena piperi.

What do these names suggest about the evolutionary relationships between these bee species?

Explain your answer.

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 × 10−9 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. ATP → ADP + Pi 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 dm⁻³ 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.

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 box 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?

Use information from Table 1 to tick ( ) one box that shows a base substitution mutation in DNA that would result in a change from Val to Ala at amino acid number 203

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.

Complete Figure 7 to show the chromosome content of the cells that would result from a normal meiotic division of the diploid parent cell shown in Figure 6

If two diploid (2n) gametes fuse at fertilisation, it can result in the growth of a tetraploid plant which has 4 copies of each chromosome. Red clover is a plant grown to produce cattle feed. Tetraploid red clover plants produce a higher yield than diploid red clover plants. Whether a red clover plant produces 2n gametes is genetically controlled. Scientists investigated the possibility of breeding red clover plants that only produced 2n gametes.

The scientists used the following null hypothesis. ‘The proportion of plants that produce 2n gametes will not change from one breeding cycle to the next.’

Complete Table 3 to show the expected number of plants that did not produce 2n gametes and the expected number of plants that did produce 2n gametes after 1 cycle.

• In breeding cycle 0, they grew red clover plants and identified plants that produced 2n gametes.
• In breeding cycle 1, they used the plants producing 2n gametes to produce offspring.
• In breeding cycles 2 and 3, they identified plants producing 2n gametes and used these to produce offspring.

Their results are shown in Table 3.

Give each answer to the nearest whole number.

The scientists tested their null hypothesis using the chi-squared statistical test. After 1 cycle their calculated chi-squared value was 350 The critical value at P=0.05 is 3.841

What does this result suggest about the difference between the observed and expected results and what can the scientists therefore conclude?

Use your knowledge of directional selection to explain the results shown in Table 3.

When a person is bitten by a venomous snake, the snake injects a toxin into the person. Antivenom is injected as treatment. Antivenom contains antibodies against the snake toxin. This treatment is an example of passive immunity.

Explain how the treatment with antivenom works and why it is essential to use passive immunity, rather than active immunity.

A mixture of venoms from several snakes of the same species is used. Suggest why.

Horses or rabbits can be used to produce antivenoms. When taking blood to extract antibody, 13 cm³ of blood is collected per kg of the animal’s body mass. The mean mass of the horses used is 350 kg and the mean mass of the rabbits used is 2 kg

Using only this information, suggest which animal would be better for the production of antivenoms.

Use a calculation to support your answer.

During the procedure shown in Figure 8 the animals are under ongoing observation by a vet.

Suggest one reason why.

During vaccination, each animal is initially injected with a small volume of venom. Two weeks later, it is injected with a larger volume of venom.

Use your knowledge of the humoral immune response to explain this vaccination programme.

Scientists investigated the effect of a heat treatment on mass transport in barley plants.

• They applied steam to one short section of a leaf of the heat-treated plants. This area is shown by the arrows in Figure 9.
• They did not apply steam to the leaves of control plants.
• They then supplied carbon dioxide containing radioactively-labelled carbon to each plant in the area shown by the rectangular boxes in Figure 9.
• After 4 hours, they:
• found the position of the radioactively-labelled carbon in each plant. These results are shown in Figure 9.
• recorded the water content of the parts of the leaf that were supplied with radioactively-labelled carbon dioxide. These results are shown in Table 4.

The scientists concluded that this heat treatment damaged the phloem.

Explain how the results in Figure 9 support this conclusion.

The scientists also concluded that this heat treatment did not affect the xylem.

Explain how the results in Table 4 support this conclusion

The scientists then investigated the movement of iron ions (Fe3+) from the soil to old and young leaves of heat-treated barley plants and to leaves of plants that were not heat treated. Heat treatment was applied half way up the leaves. The scientists determined the concentration of Fe3+ in the top and lower halves of the leaves of each plant. Their results are shown in Figure 10.

What can you conclude about the movement of Fe3+ box in barley plants?

Use all the information provided.

Describe the role of two named enzymes in the process of semi-conservative replication of DNA.

Scientists investigated the function of a eukaryotic cell protein called cyclin A. This protein is thought to be involved with the binding of one of the enzymes required at the start of DNA replication. The scientists treated cultures of cells in the following ways.

• C – Control cells, untreated
• D – Added antibody that binds specifically to cyclin A
• E – Added RNA that prevents translation of cyclin A
• F – Added RNA that prevents translation of cyclin A and added cyclin A protein

They then determined the percentage of cells in each culture in which DNA was replicating. Their results are shown in Table 5.

Suggest explanations for the results in Table 5

Describe the gross structure of the human gas exchange system and how we breathe in and out.

Mucus produced by epithelial cells in the human gas exchange system contains triglycerides and phospholipids.

Compare and contrast the structure and properties of triglycerides and phospholipids.

Mucus also contains glycoproteins. One of these glycoproteins is a polypeptide with the sugar, lactose, attached.

Describe how lactose is formed and where in the cell it would be attached to a polypeptide to form a glycoprotein.

Give the two types of molecule from which a ribosome is made.

Describe the role of a ribosome in the production of a polypeptide. Do not include transcription in your answer.

Table 1 shows the base sequence of part of a pre-mRNA molecule from a eukaryotic cell.

Complete the table with the base sequence of the DNA strand from which this pre-mRNA was transcribed.

In a eukaryotic cell, the base sequence of the mRNA might be different from the sequence of the pre-mRNA. Explain why

In mammals, in the early stages of pregnancy, a developing embryo exchanges substances with its mother via cells in the lining of the uterus. At this stage, there is a high concentration of glycogen in cells lining the uterus.

Describe the structure of glycogen.

During early pregnancy, the glycogen in the cells lining the uterus is an important energy source for the embryo.

Suggest how glycogen acts as a source of energy. Do not include transport across membranes in your answer.

Suggest and explain two ways the cell-surface membranes of the cells lining the uterus may be adapted to allow rapid transport of nutrients.

In humans, after the gametes join at fertilisation, every cell of the developing embryo undergoes mitotic divisions before the embryo attaches to the uterus lining.

What is the mean volume of each cell after 3 days? Express your answer in standard form. Show your working.

• The first cell division takes 24 hours.
• The subsequent divisions each take 8 hours. After 3 days, the embryo has a total volume of 4.2 × 10⁻³ mm³ .