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Topic 4 Assessment Booklet

Marks = 164 Time Allowed 207 minutes

Q1.Patau syndrome is a condition caused by a mutation affecting chromosome number. All the cells of the body will have this mutation.

Figure 1 shows the chromosomes from one of the cells of a female who has Patau syndrome.

(a) What is the effect of Patau syndrome on the chromosomes of this female?

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(b) Describe how the change in chromosome number in Patau syndrome was produced.

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(c) Explain why all the cells of the body will have this mutation.

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(d) Most children born with Patau syndrome die in the first 12 months, often due to defects of circulation of blood.

One of these defects is patent ductus arteriosus (PDA). This can result in some of the blood flowing between the aorta and the pulmonary artery.Figure 2 shows a healthy child’s heart and the heart of a child with PDA.

Suggest how the flow of some of the blood between the aorta and pulmonary artery

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could cause children to die in the first 12 months.

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(Total 8 marks)

Q2.(a) There are many different species of field mouse in Europe. Using a phylogenetic

classification, all of these species have names that start with Apodemus.

What information does this give about field mice?

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The long-tailed field mouse, Apodemus sylvaticus, is a small mammal common in mainland Britain.

(b) Complete Table 1 to show the classification of the long-tailed field mouse.

Table 1

Taxon Name of Taxon

Eukarya

Kingdom Animalia

Chordata

Mammalia

Order Rodentia

Family Muridae

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The St. Kilda field mouse lives only on one island off the coast of Scotland. It is very similar in appearance to the long-tailed field mouse but is larger and has lighter coloured fur.

Biologists wanted to find out if the St. Kilda field mouse and the long-tailed field mouse populations belonged to different species. They measured the length of the same features of a large number of individuals from the two populations.

The results are shown in Table 2.

Table 2

PopulationMean length (±SD) / mm

Head and body Tail

St. Kilda field mouse 112.3 (±9.3) 105.5 (±8.4)

Long-tailed field mouse 95.2 (±8.2) 90.2 (±7.3)

(c) Do the data in Table 2 provide evidence that the two populations belong to different species? Use calculations of ratios to support your answer.

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(d) Describe how breeding experiments could determine whether the two populations are from the same species.

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(Total 9 marks)

Q3.

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(a) The genetic code is degenerate and non-overlapping.

Explain the meaning of:

Degenerate _________________________________________________________

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Non-overlapping _____________________________________________________

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The table shows a short section of a messenger RNA (mRNA) molecule and the section of a polypeptide for which it codes.

mRNA G G G G C U U C A C C G G C A A C G

Polypeptide glycine alanine serine proline alanine threonine

(b) Name the bases represented in the table by:

A _________________________________

C _________________________________

G _________________________________

U _________________________________(2)

(c) Use information in the table to give the sequence of bases in DNA that codes for serine.

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(Total 5 marks)

Q4.Figure 1 represents the phylogenetic classification of four different species of fruit fly.

Figure 1

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(a) Figure 1 shows a hierarchy. Explain how.

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(b) Name the taxon to which Drosophilidae belongs.

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Drosophila fruit flies display courtship behaviour. One of the stages of courtship is singing by males. Normally a male will produce a ‘sine song’, in which continual noise is made, and a ‘pulse song’, in which there is continual noise with some louder peaks.

Scientists showed fruit flies a visual stimulus that made them sing. They made recordings of these songs.

Figure 2 shows the recordings of the songs of three flies over the same time period.

Figure 2

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(c) D. erecta and D. willistoni are closely related species but different species.

Describe evidence from Figure 2 that supports this statement.

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(d) The scientists repeated their experiments, using female fruit flies as the visual stimulus. When a male and female D. willistoni were together, their songs led to mating.

When two female D. willistoni were together, their songs did not lead to any attempt to mate.

Use information from Figure 2 to suggest why the two females did not attempt to mate.

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(Total 7 marks)

Q5.To reduce the damage caused by insect pests, some farmers spray their fields of crop plants with pesticide. Many of these pesticides have been shown to cause environmental damage.

Bt plants have been genetically modified to produce a toxin that kills insect pests. The use of Bt crop plants has led to a reduction in the use of pesticides.

Scientists have found that some species of insect pest have become resistant to the toxin produced by the Bt crop plants.

The figure below shows information about the use of Bt crops and the number of species of insect pest resistant to the Bt toxin in one country.

Year

(a) Can you conclude that the insect pest resistant to Bt toxin found in the years 2002 to 2005 was the same insect species? Explain your answer.

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(b) One farmer stated that the increase in the use of Bt crop plants had caused a mutation in one of the insect species and that this mutation had spread to other species of insect. Was he correct? Explain your answer.

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(c) There was a time lag between the introduction of Bt crops and the appearance of the first insect species that was resistant to the Bt toxin.Explain why there was a time lag.

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(Total 8 marks)

Q6.(a) Name the process by which bacterial cells divide.

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A microbiologist investigated the ability of different plant oils to kill the bacterium Listeria monocytogenes. She cultured the bacteria on agar plates. She obtained the bacteria from a broth culture.

(b) Describe two aseptic techniques she would have used when transferring a sample of broth culture on to an agar plate.Explain why each was important.

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The microbiologist tested five different plant oils at two different temperatures and determined the minimum concentration of plant oil that killed the L. monocytogenes.

The table below shows her results.

Plant oil

Minimum concentration of plant oil that killed Listeria

monocytogenes / percentage

4 °C 35 °C

Bay 0.10 0.04

Cinnamon 0.08 0.08

Clove 0.05 0.05

Nutmeg >1.00 0.05

Thyme 0.02 0.03

(c) Which plant oil is least effective at killing L. monocytogenes at 35 °C?

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L. monocytogenes is a pathogen of great concern to the food industry, especially in foods stored in refrigeration conditions (4 °C) where, unlike most food-borne pathogens, it is able to multiply. It has been suggested that plant oils, together with refrigeration may help to reduce the growth of L. monocytogenes.

(d) What conclusions can be drawn about the effectiveness of using plant oils with refrigeration to reduce food-borne infections caused by L. monocytogenes?

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(e) Plant oils are hydrophobic and can cross the cell-surface membrane of the bacterium. The low temperature of 4 °C can slow the rate of entry of plant oils into the cells.

Suggest how the low temperature slows the rate of entry.

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(Total 10 marks)

Q7.A student investigated the species richness and index of diversity of insects in three different habitats, a barley field, a wheat field and a hedge.

Her results are shown in the table below.

Number of individuals of each insect species in each habitat

Insect species Barley field Wheat field Hedge

a 32 4 34

b 78 0 12

c 0 126 22

d 0 5 12

e 0 0 8

f 0 0 42

g 0 25 13

h 0 10 12

i 0 0 12

j 42 41 0

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Species richness

Total number of insects (N)

(a) Complete the table for species richness and the total number of insects of each habitat.

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(b) Calculate the index of diversity of the wheat field.

Use the following formula:

where N = total number of organisms

and n = total number of organisms of each species.

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(c) The index of diversity of the insects was higher in the hedge than in the barley field.Suggest why.

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(Total 7 marks)

Q8.The diagram below represents one process that occurs during protein synthesis.

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(a) Name the process shown.

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(b) Identify the molecule labelled Q.

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(c) In the diagram above, the first codon is AUG. Give the base sequence of:

the complementary DNA base sequence __________________________________

the missing anticodon _________________________________________________(2)

The table below shows the base triplets that code for two amino acids.

Amino acid Encoding base triplet

Aspartic acid GAC, GAU

Proline CCA, CCG, CCC, CCU

(d) Aspartic acid and proline are both amino acids. Describe how two amino acids differ from one another. You may use a diagram to help your description.

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(e) Deletion of the sixth base (G) in the sequence shown in the diagram above would

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change the nature of the protein produced but substitution of the same base would not. Use the information in the table and your own knowledge to explain why.

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(Total 8 marks)

Q9.(a) Give the two types of molecule from which a ribosome is made.

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

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(c) The table below 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.

DNA

A C G C A U U A U pre-mRNA

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(d) In a eukaryotic cell, the base sequence of the mRNA might be different from the sequence of the pre-mRNA.

Explain why.

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(Total 7 marks)

Q10.Scientists investigated treatment of a human bladder infection caused by a species of bacterium. This species of bacterium is often resistant to the antibiotics currently used for treatment.

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

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

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

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

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

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(c) The scientists tested the new antibiotic on mice with the same bladder infection.

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They divided these mice into three groups, C, R and A.

• Group C was the control (untreated).• Group R was treated with an antibiotic currently used against this bladder

infection.• Group A was treated with the new antibiotic.

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

Their results are shown in the graph.

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

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

Show your working.

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Answer = ____________________ mg(2)

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

Show your working.

Answer = ____________________ %(2)

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

Explain why the scientists made this suggestion.

Use information from the graph in part (c) and your knowledge of evolution of antibiotic resistance in bacteria in your answer.

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(Total 9 marks)

Q11.The diagram shows two different ways of classifying the same three species of snake.

• Classification X is based on the frequency of observable characteristics• Classification Y is based on other comparisons of genetic characteristics.

All three species of snake belong to the Python family.

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(a) What do these classifications suggest about the evolutionary relationships between these species of snake?

Classification X ______________________________________________________

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Classification Y ______________________________________________________

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(b) Complete the table below to show the missing names of the taxa when classifying these snakes.

Taxon (hierarchical order) Name

Eukaryote

Animal

Chordata

Reptilia

Squamata

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Family Python

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(c) There is a debate about the name of one of these species of snake. Some scientists name it Liasis papuana and other scientists name it Apodora papuana.

Give the name of the taxon about which the scientists disagree.

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(d) State three comparisons of genetic diversity that the scientists used in order to generate Classification Y.

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(Total 7 marks)

Q12.(a) Define each of the following terms.

Species ____________________________________________________________

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Species richness _____________________________________________________

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Scientists investigated the species richness of fish caught at various depths in the Pacific Ocean close to the western coast of Chile.

The graph shows the scientists’ results. 68% of all the fish caught in this investigation came from sample A.

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(b) What is the modal value of species richness?

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(c) 68% of all the fish caught in this investigation came from sample A.A student thought this showed that sample A had a greater index of diversity than any of the other samples.

It is not possible to draw this conclusion from the given data. Give reasons why.

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(Total 6 marks)

Q13.Figure 1 shows a faulty form of meiosis that can occur in some plants.

Figure 1

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(a) Complete Figure 2 to show the chromosome content of the cells that would result from a normal meiotic division of the diploid parent cell shown in Figure 1.

Figure 2

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(b) 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.

• 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 the table.

Observed Expected

Breeding cycle

Number of plants that did not produce 2n gametes

Number of plants that did

produce 2n gametes

Number of plants that did not produce 2n gametes

Number of plants that did

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0 50 4 50 4

1 14 42

2 2 44

3 0 56

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 the table 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.Give each answer to the nearest whole number.

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(c) The scientists tested their null hypothesis using the chi-squared statistical test.After 1 cycle their calculated chi-squared value was 350The 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?

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(d) Use your knowledge of directional selection to explain the results shown in the table.

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(3)(Total 9 marks)

Q14.(a) Most human cells contain two copies of each gene. However, there might be up to

15 copies of the gene for amylase (AMY1). Scientists investigated the number of copies of the AMY1 gene in individual people in two populations. One population had a high-starch diet and the other population had a low-starch diet.

The graph below shows their results.

Describe what their results show.

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(b) Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Use your knowledge of protein synthesis and enzyme action to explain the

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advantage of this adaptation.

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(c) Multiple copies of the AMY1 gene is an adaptation to a high-starch diet.

Suggest how this evolved through natural selection.

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(Total 9 marks)

Q15.Figure 1 shows the life cycle of a moss plant. In this life cycle, only the stalk and spore capsule are diploid. All the cells in all the other stages of the life cycle of the moss are haploid.

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(a) Which letter, A, B, C or D, in Figure 1, shows where meiosis occurs in the life cycle of the moss? Write the appropriate letter in the box provided.

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(b) Explain how the chromosome number is halved during meiosis.

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(c) Figure 2 shows a cell from the moss plant.

The cell is in the second meiotic division.

What is the haploid number of chromosomes for this species of moss?

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(d) Crossing over greatly increases genetic diversity in this species of moss.

Describe the process of crossing over and explain how it increases genetic diversity.

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(Total 8 marks)

Q16.In women, the first division of meiosis produces one daughter cell that has almost all of the cytoplasm. The other daughter cell consists of a nucleus surrounded by a very small amount of cytoplasm and a cell-surface membrane. This very small daughter cell is called a polar body. Polar bodies do not usually divide. The same process occurs in the second division of meiosis, resulting in one egg cell and two polar bodies.

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The diagram shows the formation of an egg cell and two polar bodies during meiosis. It also shows what happens to one pair of homologous chromosomes. This pair carries two alleles of gene A.

(a) Complete the diagram by putting A or a in the boxes. One box has been completed for you with A.

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(b) Put a tick (✓) in the box next to the name of the process that produced the combination of alleles on the chromosome in the first polar body in the diagram.

Anaphase

Crossing over

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Independent assortment

Semi-conservative replication

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(c) A scientist measured the diameter of a polar body and the diameter of the nucleus inside it. The diameter of the polar body was 10.4 μm and the diameter of the nucleus was 7.0 μm. The density of mitochondria in the cytoplasm of the polar body (outside of the nucleus) was 0.08 mitochondria per μm3.

Calculate the number of mitochondria in the polar body. You should assume polar bodies and nuclei are spherical.

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

Show your working.

Number of mitochondria =____________________ (2)

(d) Mitochondrial diseases are caused by faulty mitochondria. All of a person’s mitochondria are inherited from their mother via the egg cell. An egg cell contains approximately 3 × 105 mitochondria.

One proposed treatment to prevent passing on faulty mitochondria involves

• removing the nucleus from an egg cell donated by a woman with healthy mitochondria

• replacing this nucleus with the contents of the polar body from a woman whose egg cells are affected by mitochondrial disease.

Suggest how this treatment prevents inheritance of mitochondrial diseases.

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(e) If most of the mitochondria in a cell are faulty, this prevents many important enzyme-catalysed reactions taking place or slows them down.

Suggest and explain one reason why.

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(Total 8 marks)

Q17.Bees are flying insects that feed on nectar made in 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 below in the graphs in the form they presented them.

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(a) What is meant by ‘species richness’?

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(b) From the data in the graphs, a student made the following conclusions.

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

Do the data in the graphs support these conclusions? Explain your answer.

1. The natural habitat is most favourable for bees.

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2. The town is the least favourable for bees.

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(c) 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.

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2. allowed them to identify accurately the species to which each belonged.

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

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(e) 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.

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(Total 11 marks)

Q18.(a) What is the proteome of a cell?

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(b) Give two structural differences between a molecule of messenger RNA (mRNA) and a molecule of transfer RNA (tRNA).

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2. _________________________________________________________________

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(c) Starting with mRNA in the cytoplasm, describe how translation leads to the production of a polypeptide.

Do not include descriptions of transcription and splicing in your answer.

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(Total 8 marks)

Q19.(a) A student used a dilution series to investigate the number of cells present in a liquid

culture of bacteria.

Describe how he made a 1 in 10 dilution and then used this to make a 1 in 1000 dilution of the original liquid culture of bacteria.

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(b) Using an optical microscope, the student determined there were 15 cells in 0.004 mm3 of the 1 in 1000 dilution of the culture.

Calculate the number of cells in 1 cm3 of undiluted liquid culture.

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Answer = ____________________ Number of cells(2)

(c) The student looked at cells in the 1 in 10 dilution during his preliminary work. He decided not to use this dilution to determine the number of cells in the undiluted liquid culture.

Suggest an explanation for the student’s decision.

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(d) On some farms, animals are routinely given antibiotics in their food.

Scientists investigated whether these farm animals had antibiotic-resistant bacteria in their intestines. They tested the bacteria for resistance to two antibiotics, tetracycline and streptomycin.

Their results are shown in the table.

AntibioticPercentage of

antibiotic-resistant bacteria

Tetracycline 29

Streptomycin 13

Suggest and explain one reason why bacteria resistant to tetracycline are more common than bacteria resistant to streptomycin in these farm animals.

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(e) In recent years, these farm animals have not been given tetracycline in their food. Despite this, the percentage of bacteria resistant to tetracycline has remained constant.

Suggest one reason why.

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(Total 10 marks)

Q20.(a) Compare and contrast the DNA in eukaryotic cells with the DNA in prokaryotic cells.

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(b) Haemoglobins are chemically similar molecules found in many different species.

Differences in the primary structure of haemoglobin molecules can provide evidence

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of phylogenetic (evolutionary) relationships between species.

Explain how.

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(Total 10 marks)

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Mark schemes

Q1.(a) Three of chromosome 13 / an extra chromosome 13;

Accept trisomy 13Accept circle around three chromosomes or any other correct indication on Figure 1Do not allow references to any other chromosomes.Do not accept chromatids for chromosomes.

1

(b) 1. In meiosis;2. Homologous chromosomes / sister chromatids do not separate;

2. Accept non-disjunction2 max

(c) 1. Mutation / extra chromosome in gamete / egg / sperm (that formed zygote);

2. All cells derived (from a single cell / zygote) by mitosis;OR3. All cells derived from a single cell / zygote by mitosis;4. Mitosis produces genetically identical cells / a clone;

Mark points 1 and 2 OR 3 and 44. Accept: have same DNA / same alleles

2

(d) 1. (Some) oxygenated blood (from the aorta) flows into pulmonary artery;ORLess oxygenated blood flows out through aorta;ORLower blood pressure in aorta;

2. Less oxygen delivered to cells / tissues / organs / named organ / via named blood vessel;

3. So less / not enough oxygen for aerobic respiration (in cell / tissue / organ);

4. Tissue / organ doesn’t grow / develop properly (causing death);ORTissue dies / organ stops working (causing death);

1. Accept mixing of deoxygenated with oxygenated blood in pulmonary artery

2. Do not accept “no oxygen”3. Do not accept “produce energy”

3 max[8]

Q2.(a) 1. Same genus;

2. Same evolutionary origin / common ancestor.2

(b)

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Taxon Name of Taxon

Domain Eukarya

Kingdom Animalia

Phylum Chordata

Class Mammalia

Order Rodentia

Family Muridae

3 correct = 2 marks2 correct = 1 mark1 or 0 correct = 0 marks

2

(c) 1. (No) SDs of means of body sizes / sizes of parts of bodies overlap;

2. Calculation of correct head and body: tail ratios;

3. Almost identical, so same body shape / proportions;3

(d) 1. Breed the two mice together;

2. (Same species) produce fertile offspring.2

[9]

Q3.(a) 1. Degenerate: more than one (base) triplet for each amino acid;

2. Non-overlapping: each base is part of only one triplet.Accept codon (as would be applicable to mRNA code)

2

(b) A = adenine

C = cytosine

G = guanine

U = uracilAll four correct = 2One error = 1Two or more errors = 0

2 max

(c) AGT;1

[5]

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Q4.(a) 1. (It shows) smaller groups within larger groups / larger groups containing

smaller groups;Accept groups within groups

2. With no overlap (between groups);2

(b) Family; Accept phonetic spellings

1

(c) 1. Sine song is (very) similar / same length (for both, so closely related).

2. (But) have different peaks / pulses (in pulse song);Must give a difference, not just state they are differentAccept suitable differences eg number / length / amplitude / interval

2

(d) 1. (Three) peaks (in pulse song) occur at the same time (since both female) / songs identical / male peaks are different;

Accept suitable differences in male peaks eg number / length / amplitude / interval

2. (Therefore) no male (song) to stimulate / cause mating;ORNothing to stimulate / cause mating;

2[7]

Q5.(a) (No – no mark)

Graph / bar chart only shows number of species, not the name of the species.1

(b) (No – no mark)1. Mutations are spontaneous / random;2. Only the rate of mutation is affected by environment;3. Different species do not interbreed / do not produce fertile offspring;4. So mutation / gene / allele cannot be passed from one species to

another.Ignore references to correlation does not prove causation

4

(c) 1. Initially one / few insects with favourable mutation / allele;2. Individuals with (favourable) mutation / allele will have more offspring;3. Takes many generations for (favourable) mutation / allele to become the

most common allele (of this gene).3

[8]

Q6.(a) Binary fission;

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Reject mitosis1

(b) 1. Keep lid on Petri dishOROpen lid of Petri dish as little as possible.

2. To prevent unwanted bacteria contaminating the dish.ORL. monocytogenes may be dangerous / may get out.

OR

3. Wear glovesORWear maskORWash hands;

4. To prevent contamination from bacteria on hands / mouthORPrevent spread of bacteria outside the lab;

OR

5. Use sterile pipetteORFlame the loopORFlame the neck of the container of the culture;

6. To maintain a pure culture of bacteria4 max

(c) Cinnamon;1

(d) 1. Thyme is the most effective / best (at 4 °C);

2. Clove and cinnamon same effectiveness at 4 °C as 35 °C (so suitable);

3. Bay and nutmeg are less effective at 4 °C than 35 °C (so unsuitable).3

(e) Less kinetic energyORLess movement of oil molecules / of phospholipid molecules

1 max[10]

Q7.(a)

3 6 9

152 211 1672

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(b) 2.45Use of the correct denominator = 1 mark

2

(c) 1. More plant species;

2. More food sources / variety of food;

3. More habitats / niches;Allow converse for barley fieldMore food = neutral

3[7]

Q8.(a) Translation.

1(b) Transfer RNA / tRNA.

1(c) TAC;

UAC.2

(d) Have different R group.Accept in diagram

1

(e) 1. Substitution would result in CCA / CCC / CCU;2. (All) code for same amino acid / proline;3. Deletion would cause frame shift / change in all following codons /

change next codon from UAC to ACC.3

[8]

Q9.(a) 1. One of RNA / ribonucleic acid(s) / nucleotide(s)/nucleic acid(s) / rRNA /

ribosomal RNA / ribosomal ribonucleic acidandone of protein(s) / polypeptide(s) / amino acid(s) / peptide(s) / ribosomal protein;

Reject DNA, deoxyribonucleic acid, tRNA, transfer RNA, transfer ribonucleic acid, mRNA, messenger RNA, messenger ribonucleic acid.Ignore enzyme(s), base(s).

1

(b) 1. mRNA binds to ribosome;2. Idea of two codons / binding sites;3. (Allows) tRNA with anticodons to bind / associate;4. (Catalyses) formation of peptide bond between amino acids (held by

tRNA molecules);5. Moves along (mRNA to the next codon) / translocation described;

Assume ‘it’ refers to ribosome.3 max

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(c) TGCGTAATA;Any errors = 0 marks

1

(d) 1. Introns (in pre-mRNA);2. Removal of sections of (pre-mRNA) / splicing;

Introns removed’ scores 2 marks.Reference to ‘introns present in mRNA’ disqualifies mp1 but allow ECF for mp2.Accept for 1 mark mRNA contains only exons.

2[7]

Q10.(a)

ATP ⟶ ADP + Pi + H2O

ATP + H2O ⟶ ADP + Pi

ADP + Pi ⟶ ATP + H2O

ADP + Pi + H2O ⟶ ATP

1

(b) 1. Human ATP synthase has a different tertiary structure to bacterial ATP synthaseORHuman ATP synthase has a different shape active site to bacterial ATP synthaseORAntibiotic cannot enter human cells/mitochondriaORAntibiotic not complementary (to human ATP synthase);

1

(c) 0.75;One mark for showing 30 g = 0.03 kg;One mark for showing 0.025 mg g −1

2

(d) Answer in range 97.0 − 97.8%;ORAnswer in range 3288 − 4368%;

2. 1 mark for correct log10 readings from graph converted to actual numbers

(16.98 − 19.50 and 660.7 − 758.6)2

(e) 1. (From the graph in part c) New / old antibiotic does not kill all bacteria;OR

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(From the graph in part c) Some bacteria are resistant to the new / old antibiotic;

2. Resistant bacteria will reproduce to produce (more) resistant bacteria;3. (Use of both) one antibiotic will kill bacteria resistant to the other

antibiotic;ORUnlikely that bacteria are resistant to both the new and the old antibiotic;ORUse of both antibiotics (likely to) kill all / most bacteria;

Accept use of ‘A’ for ‘new antibiotic’ and ‘R’ for ‘old antibiotic’.1. Must relate to the bacteria that are still present –

‘some bacteria are killed’ or ‘the bacteria number is reduced ’ is insufficient.

2. Accept ‘resistant bacteria reproduce to pass on resistance gene / allele’

3. ‘Use of both antibiotics will be more effective’ is insufficient.

3[9]

Q11.(a) 1. (Without genetic analysis / X) mackloti and olivaceus have a more recent

common ancestor with each other (than with papuana);2. (Genetic analysis indicates / Y) papuana and mackloti have a more

recent common ancestor with one another (than with olivaceus);Accept ‘more closely related to’ for ‘more recent common

ancestor’2

(b)

Domain Eukaryote

Kingdom Animal

Phylum Chordata

Class Reptilia

Order Squamata

Family Python ;

All 5 correct = 1 markAny errors = 0 marks

1

(c) Genus / genera;If the response has two answers no mark is awarded.

1

(d) 1. The (base) sequence of DNA;Accept ‘DNA hybridisation’

2. The (base) sequence of mRNA;3. The amino acid sequence (of proteins);

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3[7]

Q12.(a) 1. Species = (A group of) organisms that are able to produce fertile offspring;

2. Species richness = the number of (different) species in a community;2. Accept in a habitat / ecosystem / area2. Reject in a population2. Ignore ‘types’ unqualified

2

(b) 5;1

(c) 1. Number of individuals of each species not known;2. Almost all (of sample A / the 68%) could be of the same species;3. Two / other samples have a higher number of species / higher species

richness but a lower number of individuals / fish;4. Other samples may have more individuals of each species;

2. If not stated otherwise, assume MP2 relates to sample A / 68%

3 max[6]

Q13.(a) 1. 1 long and 1 short chromosome, each made up of 2 chromatids held (by

centromere), in each cell of 1st division;

2. 1 long and 1 short (separate) chromosome in each cell of 2nd division;Allow ECF for correct chromosomes shown in each cell from candidate’s 1st division cells.Ignore drawing of centromere.

2

(b)

52 4

;;

Allow 1 mark for numbers totalling 56 except 14/42 - repetition of observed values.

If table is blank, award 1 mark for evidence of 56.Both 52 and 4 required in table for two marks, do not credit 52 or 4 for one mark.Award 1 max for answers not given as whole numbers.

2

(c) 1. There is a less than 0.05/5% probability that the difference(s) (between observed and expected) occurred by chance;

Reject ‘results (without reference to difference) occurring by chance’. Overall max 1 with this statement.Accept ‘there is a greater than 0.95/95% probability that the difference did not occur by chance’.

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Ignore ‘difference is significant’

2. Calculated value is greater than critical value so the null hypothesis can be rejected;

Ignore ‘difference is significant’Do not accept ‘P value’ for ‘critical value’.

3. (The scientists can conclude that) the proportion of plants that produce 2n gametes does change from one breeding cycle to the next;

2 max

(d) 1. The scientists selected/used for breeding plants that produced 2n gametes;Answer must be in context of the scientists selecting plants to breed. Accept ‘artificial selection’ or ‘selectively bred’.

2. (So these plants) passed on their alleles (for production of 2n gametes to the next generation);

Both mark points can be awarded if one correct reference is made to alleles (in either context).

3. The frequency of alleles for production of 2n gametes increased (in the population).

Both mark points can be awarded if one correct reference is made to alleles (in either context).For ‘production of 2n gametes’ accept ‘abnormal meiosis’.Do not accept ‘number’ for frequency.Accept converse answers linked to plants that produce n gametes.

3[9]

Q14.(a) 1. Low starch, fewer copies;

2. Ranges overlap almost completely;ORRanges overlap from 2 − 13 copies;

3. (surprisingly) very few / 2 or 3% have only 2 copies / are diploid;4. the mode / highest percentage for low starch is 4 copies and for high

starch is 6;5. the range / spread is greater with high starch;

4. “most people” is not equivalent to mode3 max

(b) 1. More mRNA / more transcription;2. More translation / enzyme;3. So reaction faster;

The idea of “more” must be stated at least once.2. Accept ‘amylase’ for enzyme3. “More starch digested” is insufficient

3

(c) 1. Mutation(s) produce extra copies of (AMY1) gene;2. Those with more copies / this adaptation/mutation reproduce / survive

better on high starch diet;

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2. And pass on multiple copies / this adaptation/mutation (to offspring);Ignore ref. to single allele/gene

3[9]

Q15.(a) D;

1

(b) 1. Homologous chromosomes (pair);2. One of each (pair) goes to each (daughter) cell / to opposite poles;

Ignore descriptions of the second division of meiosis.2

(c) 6;1

(d) 1. Homologous pairs of chromosomes associate / form a bivalent;2. Chiasma(ta) form;3. (Equal) lengths of (non-sister) chromatids / alleles are exchanged;4. Producing new combinations of alleles;

1. Accept descriptions of homologous pairs2. Accept descriptions of chiasma(ta) e.g. chromatids /

chromosomes entangle / twist2. Neutral Crossing / cross over3. Reject genes are exchanged3. Accept lengths of DNA are exchanged4. Do not accept references to new combinations of

genes unless qualified by alleles4

[8]

Q16.(a) Lowercase a in both boxes

1

(b) Tick in box next to ‘Crossing over’;1

(c) 32.73 / 32.7 / 32 / 33;;

Award 1 max for either

409 (409.2) for difference in volume (but incorrect number of mitochondria);

OR

Answer of 262 (261.9) (using diameter, rather than radius);2

(d) 1. Egg (created) has nucleus / DNA / genes of (affected) woman / mother;Accept ref. to zygote / embryo / child for eggAccept genetic informationIgnore references to alleles

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Reject if nucleus from wrong egg / woman

2. It has mostly / many / lots of normal mitochondria (of unaffected woman)ORThere are few faulty mitochondria;

Reject ref. to production of healthy mitochondria as result of treatment

2

(e) 1. Not enough / little ATP produced;One reason asked for, so list rule appliesIgnore ref. to no ATP produced

2. ATP provides energy for (enzyme) reactionsORATP phosphorylates substrates / enzymes, so making them (more) reactive;

Accept (leads to) lower activation energy for reactionReject if mention energy produced

2 max[8]

Q17.(a) (A measure of) the number of (different) species in a community;

For ‘community’ accept ‘habitat/ecosystem/one area/environment’Reject ‘in a population’.

1

(b) Yes, natural best, because

1. Peak of (mean) bee numbers in natural habitat is highest;For accept description for ‘peak’.

2. The (mean) number of bees was higher in the natural habitat until day 200;2. For ‘day 200’ accept any day between 190 and 210.2. For ‘until day 200’ accept ‘for 200 days’.

3. (Mean) species richness in natural habitat higher at all times;

No, natural not best, because

4. Lowest (mean) number of bees after day 220;4. For ‘day 220’ accept any day between 210 and 230.

Yes, town worst, because

5. Peak of species richness higher in both natural and farmland

OR

Species richness lowest in town from day 125;For ‘day 125’ accept any day between 115 and 135.

No, town not worst, because

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(Mean) species richness is lower in farmland until day 125;For ‘day 125’ accept any day between 115 and 135.For ‘until day 125’ accept ‘for 125 days’.

7. Similar (mean) number of bees to farmland;

OR

(Mean) number of bees lower in farmland until day 140;For ‘day 140’ accept any day between day 130 and 150.For ‘until day 140’ accept ‘for 140 days’

General, no, because

8. Index of diversity of bees not measured

OR

The number of bees of each species is not known;4 max

(c) 1. Must not harm the beesORMust allow the bee to be released unchanged;

2. Must allow close examinationORUse a key (to identify the species);

Accept method that allows close examinationIgnore references to DNA sequencingAccept ‘use photographs/specimens (to identify species)'

2

(d) 1. Collect at more times of the year so more points on graph/better line (of best fit) on graph;

Both suggestion and explanation is required for each mark point.The explanation must relate to the graph.

2. Counted number of individuals in each species so that they could calculate index of diversity;

3. Collected from more sites/more years to increase accuracy of (mean) data;For ‘accuracy’ accept ‘representative’.

2 max

(e) 1. A. chlorogaster and A. piperi are more closely related (to each other than to P. pruinosa);

Must be a comparative statement.Accept A. chlorogaster and A. piperi share a more recent/closer common ancestor (than they do with P. pruinosa);Ignore references to A. chlorogaster and A. piperi not being related to P. pruinosa or not having a common ancestor with P. pruinosa.

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2. Because they are in the same genus;2

[11]

Q18.(a) (The proteome is the full) range of / number of different proteins that a cell is able to

produce (at a given time);

OR

(The proteome is the full) range of / number of different proteins the genome / DNA is able to code for;

Do not accept number of proteins unqualified1

(b) 1. mRNA does not have hydrogen bonds / base pairing, tRNA does;ORmRNA is linear / straight chain, tRNA is cloverleaf;

2. mRNA does not have an amino acid binding site, tRNA does;Accept mRNA cannot carry an amino acid, tRNA can

3. mRNA has more nucleotides;Accept mRNA is longer or converse

4. (Different) mRNAs have different lengths, all tRNAs are similar / same length;

5. mRNA has codons, tRNA has an anticodon;Statements must be comparative

2 max

(c) 1. mRNA associates with a ribosome / ribosome attaches to mRNA;Idea of association is required

2. Ribosome moves to / finds the start codon / AUG;

3. tRNA brings / carries (appropriate / specific) amino acid;Must be explicitly stated and not inferred.

4. Anticodon (on tRNA complementary) to codon (on mRNA);

5. Ribosome moves along to next codon;ORRibosome ‘fits’ around two codons / can fit two tRNAs;

Must be explicitly stated and not inferred.

6. (Process repeated and) amino acids join by peptide bonds / condensation reaction (to form polypeptide);OR(Process repeated and) amino acids joined using (energy from) ATP (to form polypeptide);

5[8]

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Q19.(a) 1. Add 1 part (bacteria) culture to 9 parts (sterile) liquid (to make 10–1 dilution);

Accept water / nutrient / broth for liquid

2. Mix (well);Accept stir

3. Repeat using 9 parts fresh (sterile) liquid and 1 part of 10–1 and 10–2 dilutions to make 10–3 dilution;ORAdd 1 part 10–1 (suspension) to 99 parts (sterile) liquid (to make 10–3 dilution);

Accept water / nutrient / broth for liquidReject 1 part (undiluted) culture added to 999 parts liquid

3

(b) 3.75 × 109 / 3 750 000 000;;Accept for 1 mark: 3750 000 / 3.75 × 106 (cells per mm3)OR3.75 × 1012 (wrong volume conversion)OR3750 (cells per mm3 of diluted culture)OREvidence of using correct dilution conversion and correct volume conversion, i.e., × 1000 and × 1000

2

(c) 1. Count unlikely to be accurate / repeatable / reproducible / reliable;

2. Because too many cells;ORBecause cells overlapping / not spread out;

2

(d) 1. Tetracycline used more often / in higher doses;

2. Resistant bacteria more likely to (survive and reproduce and) pass on allele/gene for (tetracycline) resistance;OR

3. More / higher frequency of mutations (for tetracycline resistance);Reject reference to mutation being caused by use of antibiotic

4. (so) gene passed on to more bacteria;OR

5. Tetracycline used over longer time period;

6. More time for (chance) mutation to occur / for selection to occur;Ignore reference to resistant animalsIgnore reference to immunity

2

(e) No selection against resistant bacteria / resistance gene/allele;OR

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Bacteria pass on (resistance) gene / allele when they reproduce;ORBacteria resistant to tetracycline are passed on from one generation of farm animals to the next (probably via faeces);OREnvironment does not change, so stabilising selection occurs;

Accept no selection to get rid of itReject reference to mitosis or immunity

1[10]

Q20.(a) Comparisons

1. Nucleotide structure is identical;Accept labelled diagram or description of nucleotide as phosphate, deoxyribose and base

2. Nucleotides joined by phosphodiester bond;

OR

Deoxyribose joined to phosphate (in sugar, phosphate backbone);

3. DNA in mitochondria / chloroplasts same / similar (structure) to DNA in prokaryotes;

Accept shorter than nuclear DNA/is circular not linear/is not associated with protein/histones unlike nuclear DNA;

Contrasts

4. Eukaryotic DNA is longer;

5. Eukaryotic DNA contain introns, prokaryotic DNA does not;

6. Eukaryotic DNA is linear, prokaryotic DNA is circular;

7. Eukaryotic DNA is associated with / bound to protein / histones, prokaryotic DNA is not;

5 max

(b) 1. Mutations change base / nucleotide (sequence);Reject if mutation in amino acid

2. (Causing) change in amino acid sequence;

3. Mutations build up over time;

4. More mutations / more differences (in amino acid / base / nucleotide sequence / primary structure) between distantly related species;

OR

Few(er) mutations / differences (in amino acid / base / nucleotide sequence / primary structure) in closely related species;

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5. Distantly related species have earlier common ancestor;

OR

Closely related species have recent common ancestor;Accept “order” for “sequence”If neither MP4 or MP5 accept for 1 mark, idea of more mutations /differences as evidence of earlier common ancestor OR converse

5[10]

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