A-level: Biology

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A-level BIOLOGY Paper 1 Thursday 6 June 2019

Morning

Time allowed: 2 hours

Materials

For Examiner’s Use

For this paper you must have: • a ruler with millimetre measurements • a scientific calculator.

Question 1

Instructions

2

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • Show all your working. • Do all rough work in this book. Cross through any work you do not want to be marked.

• • • •

Information

3 4 5 6 7 8 9 10

• The marks for the questions are shown in brackets. • The maximum mark for this paper is 91.

*JUN197402101*

Mark

TOTAL

IB/M/Jun19/E16

7402/1

2 Do not write outside the box

Answer all questions in the spaces provided. 0 1 . 1

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

[3 marks]

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3 Do not write outside the box

Pectin is a substance found in some fruit and vegetables. A scientist investigated the effect of pectin on the hydrolysis of lipids by a lipase enzyme. His results are shown in Figure 1. Figure 1

0 1 . 2

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. [1 mark]

Question 1 continues on the next page

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4 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. Figure 2

0 1 . 3

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.

[2 marks]

Maximum length =

µm

Uncertainty of your calculated maximum length =

µm

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5 0 1 . 4

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

Do not write outside the box

[2 marks]

8

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6 0 2 . 1

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. [3 marks] Table 1 Cell wall component

Plants

Algae

Fungi

Prokaryotes

Cellulose Murein Chitin

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. Figure 3

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7 0 2 . 2

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.

[4 marks]

[Extra space]

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8 0 2 . 3

The scientists estimated the mean mass of fibre eaten per day using a food frequency questionnaire (FFQ).

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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. [2 marks] Advantage

Disadvantage

9

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9 Do not write outside the box

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10 0 3

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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. Figure 4

0 3 . 1

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.

[1 mark]

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11 0 3 . 2

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

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[2 marks]

1

2

0 3 . 3

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. [1 mark]

Question 3 continues on the next page

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12 0 3 . 4

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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. [2 marks] Advantage

Disadvantage

6

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13 Do not write outside the box

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14 0 4

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. Figure 5

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15 0 4 . 1

Use Figure 5 to explain how human mass at birth is affected by stabilising selection. [3 marks]

[Extra space]

Question 4 continues on the next page

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16 0 4 . 2

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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. KIR2DS1 is an

(1)

(2)

of the KIR gene, found at a (3)

chromosome 19. KIR2DS1 is 14 021 bases long and is that is 1101 bases long. This mRNA is then

(4)

(6)

into mRNA

into a polypeptide 304

amino acids long. The polypeptide is then modified in the organelle, before forming its functional

on

(5)

,

protein structure.

Write the correct biological term beside each number below, that matches the space in the passage. [3 marks] (1) (2) (3) (4) (5) (6)

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17 0 4 . 3

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. Table 2 Presence or absence of KIR2DS1 in mother’s genome

Number of babies with mass at birth: between 2500 g and 4500 g

above 4500 g

Present

389

148

Absent

606

173

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 4500 g’ 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. [1 mark] Chi-squared Correlation coefficient Student’s t-test

Question 4 continues on the next page

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18 0 4 . 4

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

Do not write outside the box

[3 marks]

10

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19 0 5 . 1

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

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[4 marks]

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20 Some people infected with HIV do not develop AIDS. These people are called 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. Table 3

HIV status of people

HIV controllers HIV positive people with AIDS symptoms

0 5 . 2

Median viral load / virus particles per cm3 of blood (range)

Median number of CD4 cells per mm3 of blood (range)

212 (<50 to 609)

693 (529 to 887)

66 274 (30 206 to 306 163)

248 (107 to 365)

A test sample of 500 mm3 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. [1 mark] 106 000 10 600 1060 106

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21 0 5 . 3

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

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[Extra space]

8

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22 0 6

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. Table 4 Percentage of heart cells undergoing mitosis

Percentage of heart cells undergoing DNA replication

−6

13.9

8.5

4

8.5

2.6

14

1.6

0.2

21

0.6

0.0

Age / days

0 6 . 1

Describe and explain the data in Table 4.

[2 marks]

[Extra space]

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23 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. 0 6 . 2

Describe how BrdU would be incorporated into new DNA during semi-conservative replication. [5 marks]

Question 6 continues on the next page

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24 0 6 . 3

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

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Use your knowledge of the ELISA test to suggest and explain how the scientists identified the cells that have BrdU in their DNA. [3 marks]

[Extra space]

10

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25 Do not write outside the box

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26 0 7

Ulva lactuca is an alga that lives on rocks on the seashore. It is regularly covered by seawater. Figure 6 shows a diagram of one Ulva lactuca alga. Figure 6

0 7 . 1

Unlike plants, Ulva lactuca does not have xylem tissue. Suggest how Ulva lactuca is able to survive without xylem tissue.

[1 mark]

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27 Do not write outside the box

Ulva lactuca has a haploid and a diploid form. Figure 7 shows the life cycle of Ulva lactuca. Figure 7

0 7 . 2

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. [2 marks]

0 7 . 3

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. [2 marks]

5 Turn over ►

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28 0 8

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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. Figure 8

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29 0 8 . 1

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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. [2 marks] Table 5 Water potential / MPa

−1.95

Concentration of sodium chloride solution / mol dm–3

Volume of 1 mol dm–3 sodium chloride solution / /

0.04

Table 6 shows some of the concentrations of sodium chloride solution the scientists used and the water potential of each solution. Table 6 Concentration of sodium chloride solution / mol dm–3 0.04 0.10 0.12

0 8 . 2

Water potential / MPa −1.95 −4.87 −5.84

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 −3.41 MPa [2 marks]

Answer =

mol dm–3

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30 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. Figure 9

0 8 . 3

One leaf with an original area of 60 cm2 gave a voltage reading of −7 µV

Use Figure 8 (on page 28) and Figure 9 to calculate by how much this leaf increased in area. Give your answer in cm2 [2 marks]

Answer =

cm2

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31 0 8 . 4

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

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Suggest and explain one way the leaf growth of xerophytic plants would be different from the leaf growth of sunflowers in Figure 9. [2 marks]

0 8 . 5

Use your knowledge of gas exchange in leaves to explain why plants grown in soil with very little water grow only slowly. [2 marks]

10

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32 0 9

A scientist investigated the affinity for oxygen of horse haemoglobin and mouse haemoglobin. Some of their results are shown in Table 7. Table 7 Partial pressure of oxygen when haemoglobin is 50% saturated / kPa

Partial pressure of oxygen when haemoglobin is 25% saturated / kPa

Body mass of one animal / g

Horse

3.2

1.9

550 000

Mouse

6.5

3.3

23

Animal

0 9 . 1

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. [3 marks]

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33 0 9 . 2

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. [2 marks]

Answer =

0 9 . 3

times faster

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. [2 marks]

[Extra space]

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34 0 9 . 4

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

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Use your knowledge of surface area to volume ratio to explain the higher metabolic rate of a mouse compared to a horse. [3 marks]

[Extra space]

10

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35 1 0 . 1

Explain five properties that make water important for organisms.

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[5 marks]

[Extra space]

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36 1 0 . 2

Describe the biochemical tests you would use to confirm the presence of lipid, non-reducing sugar and amylase in a sample. [5 marks]

[Extra space]

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37 1 0 . 3

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

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Give two named examples of polymers and their associated monomers to illustrate your answer. [5 marks]

[Extra space]

15 END OF QUESTIONS

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38 Do not write outside the box

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39 Do not write outside the box

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Copyright information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material are published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

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*196A7402/1* IB/M/Jun19/7402/1

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A-level BIOLOGY Paper 2 Thursday 13 June 2019

Morning

Time allowed: 2 hours

Materials

For Examiner’s Use

For this paper you must have: • a ruler with millimetre measurements • a scientific calculator.

Question

Mark

1 2

Instructions

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • Show all your working. • Do all rough work in this book. Cross through any work you do not want to be marked.

3

Information

10

• • • •

• The marks for the questions are shown in brackets. • The maximum mark for this paper is 91.

4 5 6 7 8 9

TOTAL

*JUN197402201* IB/M/Jun19/E11

7402/2

2 Do not write outside the box

Answer all questions in the spaces provided. 0 1 . 1

Succession occurs in natural ecosystems. Describe and explain how succession occurs. [4 marks]

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3 Do not write outside the box

Figure 1 shows percentages of energy transferred from sunlight to a zebra in a grassland ecosystem. Figure 1

0 1 . 2

Use Figure 1 to calculate the percentage of sunlight energy that would be transferred into the faeces and urine of a zebra. Give your answer to 3 significant figures. [1 mark]

Answer = 0 1 . 3

%

In this ecosystem the net productivity of the vegetation is 24 525 kJ m−2 year−1 Use this information and Figure 1 to calculate the energy stored in new tissues of the zebra in kJ m−2 year−1 [2 marks]

Answer =

kJ m–2 year–1

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*03* IB/M/Jun19/7402/2

7

4 0 2

Sickle cell disease (SCD) is a group of inherited disorders. People with SCD have sickle-shaped red blood cells. A single base substitution mutation can cause one type of SCD. This mutation causes a change in the structure of the beta polypeptide chains in haemoglobin.

0 2 . 1

Explain how a single base substitution causes a change in the structure of this polypeptide. Do not include details of transcription and translation in your answer.

[3 marks]

Haematopoietic stem cell transplantation (HSCT) is a long-term treatment for SCD. In HSCT, the patient receives stem cells from the bone marrow of a person who does not have SCD. The donor is often the patient’s brother or sister. Before the treatment starts, the patient’s faulty bone marrow cells have to be destroyed. 0 2 . 2

Use this information to explain how HSCT is an effective long-term treatment for SCD. [3 marks]

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5 Do not write outside the box

A new long-term treatment for SCD involves the use of gene therapy. Figure 2 shows some of the stages involved in this treatment in a child with SCD. Figure 2

0 2 . 3

Some scientists have concluded that this method of gene therapy will be a more effective long-term treatment for SCD than HSCT. Use all the information provided to evaluate this conclusion. [3 marks]

9

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6 0 3

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A student investigated the effects of indoleacetic acid (IAA) on the growth of oat seedlings (young plants). The student: removed the shoot tip from each seedling and cut out a 10 mm length of shoot placed 10 lengths of shoot into each of 5 Petri dishes added to each Petri dish an identical volume of 5% glucose solution added to each Petri dish 40 cm3 of a different concentration of IAA solution left the Petri dishes at 20 °C in the dark with their lids on for 5 days removed the shoots after 5 days and measured them • determined the mean change in length of shoot at each concentration of IAA. • • • • • •

Table 1 shows her results. Table 1 IAA concentration added to Petri dish / parts per million Mean change in length of shoot / mm

0 3 . 1

10−5

10−3

10−1

0.0

0.1

1.3

1

2.4

Explain why the student removed the shoot tip from each seedling.

10

3.1

[2 marks]

[Extra space]

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7 Do not write outside the box

0 3 . 2

Explain why the student added glucose solution to each Petri dish.

0 3 . 3

Explain why the lids were kept on the Petri dishes.

0 3 . 4

Describe and explain the results shown in Table 1 and suggest how the results might have differed if lengths of root had been used. [3 marks]

[2 marks]

[2 marks]

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8 0 3 . 5

The student produced the different concentrations of IAA using a stock 1 g dm–3 solution of IAA (1 g dm–3 = 1 part per thousand) and distilled water.

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Complete Table 2 with the volumes of stock IAA solution and distilled water required to produce 40 cm3 of 10 ppm (parts per million) IAA solution. [1 mark] Table 2 Concentration of IAA solution / parts per million

Volume of stock IAA solution / cm3

Volume of distilled water / cm3

10

10

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9 Do not write outside the box

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10 0 4

Scientists investigated the effect of a decrease in pH on muscle contraction. The scientists did the investigation with four different preparations of isolated muscle tissue: A, B, C and D. A - mouse muscle fibres at typical pH of mouse muscle tissue (control 1). B - mouse muscle fibres at 0.5 pH units below typical pH. C - rabbit muscle fibres at typical pH of rabbit muscle tissue (control 2). D - rabbit muscle fibres at 0.5 pH units below typical pH. They measured the force of muscle contraction of the muscle fibres at 12 °C, 22 °C and 32 °C Figure 3 shows the results the scientists obtained for B and D compared with the appropriate control. Figure 3

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11 0 4 . 1

A student looked at the results and concluded that a decrease in pH does cause a decrease in the force of muscle contraction. Use Figure 3 to evaluate this conclusion.

[4 marks]

[Extra space]

Question 4 continues on the next page

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12 0 4 . 2

Another group of scientists suggested that a decrease in the force of muscle contraction is caused by an increase in the concentration of inorganic phosphate, Pi, in muscle tissues.

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Their hypothesis is that an increase in the concentration of Pi prevents the release of calcium ions within muscle tissues. Explain how a decrease in the concentration of calcium ions within muscle tissues could cause a decrease in the force of muscle contraction. [3 marks]

0 4 . 3

In muscles, pyruvate is converted to lactate during prolonged exercise. Explain why converting pyruvate to lactate allows the continued production of ATP by anaerobic respiration. [2 marks]

9

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13 Do not write outside the box

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14 0 5 . 1

Describe the role of glucagon in gluconeogenesis. Do not include in your answer details on the second messenger model of glucagon action.

0 5 . 2

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[2 marks]

The gene that codes for glucagon is 9.531 kilobases in length. The DNA helix makes one complete turn every 10 base pairs. Every complete turn is 3.4 nm in length. Use this information to calculate the length in micrometres (µm) of the gene for glucagon. Give your answer to 3 significant figures. [2 marks]

Answer =

µm

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15 Metformin is a drug commonly used to treat type II diabetes. Metformin’s ability to lower the blood glucose concentration involves a number of mechanisms including:

Do not write outside the box

• increasing a cell’s sensitivity to insulin • inhibiting adenylate cyclase. 0 5 . 3

Explain how increasing a cell’s sensitivity to insulin will lower the blood glucose concentration. [2 marks]

0 5 . 4

Explain how inhibiting adenylate cyclase may help to lower the blood glucose concentration. [3 marks]

9

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16 0 6

In fruit flies, a gene for body colour has a dominant allele for grey body, G, and a recessive allele for black body, g. A gene for eye colour has a dominant allele for red eyes, R, and a recessive allele for white eyes, r, and is located on the X chromosome. Figure 4 shows the phenotypes of fruit flies over four generations. Figure 4

0 6 . 1

Give the full genotype of the fly numbered 6 in Figure 4.

[1 mark]

Genotype = 0 6 . 2

Give one piece of evidence from Figure 4 to show that the allele for grey body colour is dominant. [1 mark]

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17 0 6 . 3

Explain one piece of evidence from Figure 4 to show that the gene for body colour is not on the X chromosome. [2 marks]

0 6 . 4

A heterozygous grey-bodied, white-eyed female fly was crossed with a black-bodied, red-eyed male fly. Complete the genetic diagram below to show all the possible genotypes and the ratio of phenotypes expected in the offspring from this cross. [3 marks] Phenotypes of parents:

Genotypes of parents:

Grey-bodied, white-eyed female

×

Black-bodied, red-eyed male

×

Genotypes of offspring Phenotypes of offspring

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18 0 6 . 5

A population of fruit flies contained 64% grey-bodied flies. Use the Hardy–Weinberg equation to calculate the percentage of flies heterozygous for gene G. [2 marks]

Answer =

%

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9

19 0 7 . 1

In photosynthesis, which chemicals are needed for the light-dependent reaction? Tick () one box. [1 mark] Reduced NADP, ADP, Pi, water and oxygen.

NADP, ATP and water.

Reduced NADP, ATP, water and carbon dioxide.

NADP, ADP, Pi and water.

0 7 . 2

Describe what happens during photoionisation in the light-dependent reaction. [2 marks]

Question 7 continues on the next page

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20 A student obtained a solution of pigments from the leaves of a plant. Then the student used paper chromatography to separate the pigments. Figure 5 shows the chromatogram produced. Figure 5

0 7 . 3

Explain why the student marked the origin using a pencil rather than using ink. [1 mark]

0 7 . 4

Describe the method the student used to separate the pigments after the solution of pigments had been applied to the origin.

[2 marks]

*20* IB/M/Jun19/7402/2

Do not write outside the box

21 0 7 . 5

Calculating the Rf values of the pigments can help to identify each pigment. An Rf value compares the distance the pigment has moved from the origin with the distance the solvent front has moved from the origin. Rf =

Do not write outside the box

distance pigment has moved from the origin distance solvent front has moved from the origin

The distance each pigment has moved is measured from the middle of each spot. Pigment A has an Rf value of 0.95 Use Figure 5 to calculate the Rf value of pigment C.

[1 mark]

Rf value of pigment C = 0 7 . 6

The pigments in leaves are different colours. Suggest and explain the advantage of having different coloured pigments in leaves. [1 mark]

8

Turn over for the next question

Turn over ►

*21* IB/M/Jun19/7402/2

22 0 8 . 1

Do not write outside the box

What is a DNA probe?

[2 marks]

DNA probes are used to detect specific base sequences of DNA. The process is shown in Figure 6. Figure 6

0 8 . 2

Describe how the DNA is broken down into smaller fragments.

[2 marks]

*22* IB/M/Jun19/7402/2

23 0 8 . 3

The DNA on the nylon membrane is treated to form single strands. Explain why. [1 mark]

A scientist used DNA probes and electrophoresis to screen four volunteers for five different viral DNA fragments. Figure 7 shows the results the scientist obtained. The lanes numbered 2 to 5 represent the four volunteers. Figure 7

0 8 . 4

Lane 1 of Figure 7 enabled the size of the different viral fragments to be determined. Suggest and explain how.

[2 marks]

Turn over ►

*23* IB/M/Jun19/7402/2

Do not write outside the box

24 Do not write outside the box

The lengths of the viral DNA fragments were: • • • • • 0 8 . 5

600 base pairs 250 base pairs 535 base pairs 300 base pairs 500 base pairs.

Which volunteers had at least one of the viral DNA fragments with 250 base pairs or 535 base pairs? [1 mark] 8

*24* IB/M/Jun19/7402/2

25 0 9

The sundew is a small flowering plant, growing in wet habitats such as bogs and marshes. The soil in bogs and marshes is acidic and has very low concentrations of some nutrients. The sundew can trap and digest insects.

0 9 . 1

Describe how you could estimate the size of a population of sundews in a small marsh. [5 marks]

0 9 . 2

Suggest and explain how digesting insects helps the sundew to grow in soil with very low concentrations of some nutrients. [2 marks]

Do not write outside the box

7

Turn over ►

*25* IB/M/Jun19/7402/2

26 1 0

Guillain–Barré syndrome is a rare disease in which the immune system damages the myelin sheath of neurones. Myelin sheath damage can cause a range of symptoms, for example numbness, muscular weakness and muscular paralysis. Sometimes, neurones of the autonomic nervous system are affected, causing heart rate irregularities.

Do not write outside the box

5

Huntington’s disease is a disorder caused when a protein called huntingtin damages the brain. Huntingtin is produced because of a dominant, mutant allele. The first successful drug trial to reduce concentrations of huntingtin in the human brain involved 46 patients. The patients received the drug for 4 months. The concentration of huntingtin was reduced in all the patients. The drug was injected at the base of the spine into the cerebrospinal fluid bathing the brain and spinal cord. The drug contains single-stranded DNA molecules. These single-stranded molecules inhibit the mRNA needed to produce huntingtin.

10

15

Symptoms of Huntington’s disease can start at any time, but usually develop between 30 and 50 years of age. The likelihood and age when symptoms start are linked to the number of CAG base sequence repeats in the gene for Huntington’s disease. However, recent studies have suggested that epigenetics may also affect the age when symptoms first start. 1 0 . 1

20

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

[3 marks]

*26* IB/M/Jun19/7402/2

27 1 0 . 2

Sometimes Guillain–Barré syndrome causes heart rate irregularities (lines 4–5). Suggest and explain why.

1 0 . 3

[3 marks]

The first successful drug trial to reduce concentrations of huntingtin in the brain used single-stranded DNA molecules (lines 13–14). Suggest and explain how this drug could cause a reduction in the concentration of the protein huntingtin. [3 marks]

Turn over ►

*27* IB/M/Jun19/7402/2

Do not write outside the box

28 1 0 . 4

Scientists from the first successful drug trial to reduce concentrations of huntingtin (lines 9–11) reported that the drug is not a cure for Huntington’s disease. Suggest two reasons why the drug should not be considered a cure. Do not include repeats of the drug trial in your answer.

Do not write outside the box

[2 marks]

1

2

1 0 . 5

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

[2 marks]

1

2

1 0 . 6

Suggest and explain one way epigenetics may affect the age when symptoms of Huntington’s disease start. [2 marks]

15

END OF QUESTIONS Copyright information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material are published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

*28*

*196A7402/2* IB/M/Jun19/7402/2

Please write clearly in block capitals. Centre number

Candidate number

Surname Forename(s) Candidate signature

A-level CHEMISTRY Paper 1

Inorganic and Physical Chemistry

Tuesday 4 June 2019

Afternoon

Time allowed: 2 hours

Materials

For this paper you must have: • the Periodic Table/Data Sheet, provided as an insert (enclosed) • a ruler with millimetre measurements • a scientific calculator, which you are expected to use where appropriate.

Instructions

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • All working must be shown. • Do all rough work in this book. Cross through any work you do not want to be marked.

• • • •

For Examiner’s Use Question

Mark

1 2 3 4 5 6 7 8 9

TOTAL

Information

• The marks for questions are shown in brackets. • The maximum mark for this paper is 105.

*JUN197405101* IB/G/Jun19/E16

7405/1

2 Do not write outside the box

Answer all questions in the spaces provided. 0 1

Figure 1 shows an incomplete Born–Haber cycle for the formation of caesium iodide. The diagram is not to scale. Figure 1

Table 1 gives values of some standard enthalpy changes. Table 1 Name of enthalpy change

∆Ho / kJ mol–1

Enthalpy of atomisation of caesium

+79

First ionisation energy of caesium

+376

Electron affinity of iodine

–314

Enthalpy of lattice formation of caesium iodide

–585

Enthalpy of formation of caesium iodide

–337

0 1 . 1

Complete Figure 1 by writing the formulas, including state symbols, of the appropriate species on each of the two blank lines. [2 marks]

0 1 . 2

Use Figure 1 and the data in Table 1 to calculate the standard enthalpy of atomisation of iodine.

Standard enthalpy of atomisation of iodine

[2 marks]

kJ mol-1

*02* IB/G/Jun19/7405/1

3 Do not write outside the box

0 1 . 3

The enthalpy of lattice formation for caesium iodide in Table 1 is a value obtained by experiment. The value obtained by calculation using the perfect ionic model is –582 kJ mol–1 Deduce what these values indicate about the bonding in caesium iodide.

0 1 . 4

[1 mark]

Use data from Table 2 to show that this reaction is not feasible at 298 K 1 2

CsI(s) → Cs(s) + I2(s)

∆Ho = +337 kJ mol–1

Table 2

So / J K–1 mol–1

CsI(s)

Cs(s)

I2(s)

130

82.8

117 [4 marks]

9

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*03* IB/G/Jun19/7405/1

4 0 2

Time of flight (TOF) mass spectrometry can be used to analyse large molecules such as the pentapeptide, leucine encephalin (P). P is ionised by electrospray ionisation and its mass spectrum is shown in Figure 2. Figure 2

0 2 . 1

Describe the process of electrospray ionisation. Give an equation to represent the ionisation of P in this process.

[4 marks]

Description

Equation

*04* IB/G/Jun19/7405/1

Do not write outside the box

5 0 2 . 2

What is the relative molecular mass of P? Tick () one box.

555

0 2 . 3

Do not write outside the box

556

[1 mark] 557

A molecule Q is ionised by electron impact in a TOF mass spectrometer. The Q+ ion has a kinetic energy of 2.09 x 10–15 J This ion takes 1.23 x 10–5 s to reach the detector. The length of the flight tube is 1.50 m Calculate the relative molecular mass of Q. 1 2

KE = mv 2

where m = mass (kg) and v = speed (m s –1)

The Avogadro constant, L = 6.022 x 1023 mol–1

[5 marks]

10

Relative molecular mass Turn over ►

*05* IB/G/Jun19/7405/1

6 0 3

This question is about periodicity, the Period 4 elements and their compounds.

0 3 . 1

State the meaning of the term periodicity.

0 3 . 2

Identify the element in Period 4 with the highest electronegativity value.

0 3 . 3

Identify the element in Period 4 with the largest atomic radius. Explain your answer.

Do not write outside the box

[1 mark]

[1 mark]

[3 marks]

Element Explanation

0 3 . 4

The equations for two reactions of arsenic(III) oxide are shown. As2O3 + 6 HCl → 2 AsCl3 + 3 H2O As2O3 + 6 NaOH → 2 Na3AsO3 + 3 H2O Name the property of arsenic(III) oxide that describes its ability to react in these two ways. [1 mark]

0 3 . 5

Complete the equation for the formation of arsenic hydride. As2O3 +

Zn +

HNO3 →

AsH3 +

[1 mark] Zn(NO3)2 +

H2O

*06* IB/G/Jun19/7405/1

7

7 0 4

Do not write outside the box

Figure 3 shows some reactions of aqueous iron ions. Figure 3

0 4 . 1

Give the formula of Precipitate J and state its colour. Give an equation for Reaction 1.

[3 marks]

Formula of J Colour Equation

0 4 . 2

Give the formula of L and an equation for Reaction 2.

[2 marks]

Formula of L Equation

0 4 . 3

Suggest a reagent for Reaction 3.

[1 mark]

Turn over ►

*07* IB/G/Jun19/7405/1

8 0 4 . 4

Give the formula of Precipitate M and state its colour.

Do not write outside the box

[2 marks]

Formula of M Colour 0 4 . 5

Transition metal complexes have different shapes and many show isomerism. Describe the different shapes of complexes and show how they lead to different types of isomerism. Use examples of complexes of cobalt(II) and platinum(II). You should draw the structures of the examples chosen.

[6 marks]

*08* IB/G/Jun19/7405/1

9 Do not write outside the box

14

Turn over ►

*09* IB/G/Jun19/7405/1

10 0 5

This question is about some Group 7 compounds.

0 5 . 1

Solid sodium chloride reacts with concentrated sulfuric acid. Give an equation for this reaction. State the role of the sulfuric acid in this reaction.

Do not write outside the box

[2 marks]

Equation

Role 0 5 . 2

Fumes of sulfur dioxide are formed when sodium bromide reacts with concentrated sulfuric acid. For this reaction • give an equation • give one other observation • state the role of the sulfuric acid.

[3 marks]

Equation

Observation

Role 0 5 . 3

Chlorine reacts with hot aqueous sodium hydroxide as shown in the equation. 3 Cl2 + 6 NaOH → NaClO3 + 5 NaCl + 3 H2O Give the oxidation state of chlorine in NaClO3 and in NaCl

[1 mark]

NaClO3 NaCl

*10* IB/G/Jun19/7405/1

11 0 5 . 4

State, in terms of redox, what happens to chlorine in the reaction in Question 05.3. [1 mark]

0 5 . 5

Solution Y contains two different negative ions.

Do not write outside the box

To a sample of solution Y in a test tube a student adds • silver nitrate solution • then an excess of dilute nitric acid • finally an excess of concentrated ammonia solution. The observations after each addition are recorded in Table 3. Table 3 Reagent added to solution Y

Observation

silver nitrate solution

cream precipitate containing compound D and compound E

excess dilute nitric acid

cream precipitate D and bubbles of gas F

excess concentrated ammonia solution

colourless solution containing complex ion G

Give the formulas of D, E and F. Give an ionic equation to show the formation of E. Give an equation to show the conversion of D into G.

[6 marks]

Formula of D Formula of E Formula of F Ionic equation to form E

Equation to show the conversion of D into G 13

Turn over ►

*11* IB/G/Jun19/7405/1

12 0 6

A student does an experiment to determine the percentage of copper in an alloy. The student • reacts 985 mg of the alloy with concentrated nitric acid to form a solution (all of the copper in the alloy reacts to form aqueous copper(II) ions) • pours the solution into a volumetric flask and makes the volume up to 250 cm3 with distilled water • shakes the flask thoroughly • transfers 25.0 cm3 of the solution into a conical flask and adds an excess of potassium iodide • uses exactly 9.00 cm3 of 0.0800 mol dm–3 sodium thiosulfate (Na2S2O3) solution to react with all the iodine produced. The equations for the reactions are 2 Cu2+ + 4 I– → 2 CuI + I2 2 S2O32– + I2 → 2 I– + S4O62–

0 6 . 1

Calculate the percentage of copper by mass in the alloy. Give your answer to the appropriate number of significant figures.

[6 marks]

% copper

*12* IB/G/Jun19/7405/1

Do not write outside the box

13 0 6 . 2

Suggest two ways that the student could reduce the percentage uncertainty in the measurement of the volume of sodium thiosulfate solution, using the same apparatus as this experiment. [2 marks] 1

2

0 6 . 3

State the role of iodine in the reaction with sodium thiosulfate.

0 6 . 4

Give the full electron configuration of a copper(II) ion.

0 6 . 5

Copper(I) iodide is a white solid. Explain why copper(I) iodide is white.

[1 mark]

[1 mark]

[2 marks]

Question 6 continues on the next page

Turn over ►

*13* IB/G/Jun19/7405/1

Do not write outside the box

14 0 6 . 6

Do not write outside the box

Iodine vaporises easily. Calculate the volume, in cm3, that 5.00 g of iodine vapour occupies at 185 °C and 100 kPa The gas constant R = 8.31 J K–1 mol–1 Give your answer to 3 significant figures.

Volume

[4 marks]

cm3

*14* IB/G/Jun19/7405/1

16

15 0 7

0 7 . 1

Do not write outside the box

Sulfur trioxide decomposes on heating to form an equilibrium mixture containing sulfur dioxide and oxygen. 2 SO3(g) ⇌ 2 SO2(g) + O2(g)

A sample of sulfur trioxide was heated and allowed to reach equilibrium at a given temperature. The equilibrium mixture contained 6.08 g of sulfur dioxide. Calculate the mass, in g, of oxygen gas in the equilibrium mixture.

Mass

[2 marks]

g

Question 7 continues on the next page

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*15* IB/G/Jun19/7405/1

16 0 7 . 2

Do not write outside the box

A different mass of sulfur trioxide was heated and allowed to reach equilibrium at 1050 K 2 SO3(g) ⇌ 2 SO2(g) + O2(g)

The amounts of each substance in the equilibrium mixture are shown in Table 4. Table 4 Substance

Amount at equilibrium / mol

sulfur trioxide

0.320

sulfur dioxide

1.20

oxygen

0.600

For this reaction at 1050 K the equilibrium constant, Kp = 7.62 x 105 Pa Calculate the mole fraction of each substance at equilibrium. Give the expression for the equilibrium constant, Kp Calculate the total pressure, in Pa, of this equilibrium mixture.

[4 marks]

Mole fraction SO3 Mole fraction SO2 Mole fraction O2

Kp

Total pressure

Pa

*16* IB/G/Jun19/7405/1

17 0 7 . 3

Do not write outside the box

For this reaction at 1050 K the equilibrium constant, Kp = 7.62 x 105 Pa For this reaction at 500 K the equilibrium constant, Kp = 3.94 x 104 Pa Explain how this information can be used to deduce that the forward reaction is endothermic. [2 marks]

0 7 . 4

Use data from Question 07.3 to calculate the value of Kp, at 500 K, for the equilibrium represented by this equation. Deduce the units of Kp 1 2

SO3(g) ⇌ SO2(g) + O2 (g)

[2 marks]

Kp 10

Units

Turn over for the next question

Turn over ►

*17* IB/G/Jun19/7405/1

18 Do not write outside the box

0 8

This question is about structure and bonding.

0 8 . 1

Draw a diagram to show the strongest type of interaction between two molecules of ethanol (C2H5OH) in the liquid phase. Include all lone pairs and partial charges in your diagram.

0 8 . 2

[3 marks]

Methoxymethane (CH3OCH3) is an isomer of ethanol. Table 5 shows the boiling points of ethanol and methoxymethane. Table 5 Compound ethanol methoxymethane

Boiling point / °C 78 –24

In terms of the intermolecular forces involved, explain the difference in boiling points. [3 marks]

*18* IB/G/Jun19/7405/1

19 Do not write outside the box

Extra space

0 8 . 3

Draw the shape of the POCl3 molecule and the shape of the ClF4– ion. Include any lone pairs of electrons that influence the shapes. In a POCl3 molecule the oxygen atom is attached to the phosphorus atom by a double bond that uses two electrons from phosphorus. Name each shape. Suggest a value for the bond angle in ClF4 ̄ Shape of POCl3

Shape of ClF4 ̄

[5 marks]

Name of shape of POCl3 Name of shape of ClF4 ̄ 11

Bond angle in ClF4 ̄

Turn over for the next question

Turn over ►

*19* IB/G/Jun19/7405/1

20 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

*20* IB/G/Jun19/7405/1

21 0 9

This question is about different pH values.

0 9 . 1

For pure water at 40 °C, pH = 6.67 A student thought that the water was acidic.

Do not write outside the box

Explain why the student was incorrect. Determine the value of Kw at this temperature.

[4 marks]

Explanation

Kw

mol2 dm–6

Question 9 continues on the next page

Turn over ►

*21* IB/G/Jun19/7405/1

22 0 9 . 2

Sodium hydroxide solution was added gradually from a burette to 25 cm3 of 0.080 mol dm–3 propanoic acid at 25 °C The pH was measured and recorded at regular intervals. The results are shown in Figure 4. Figure 4

*22* IB/G/Jun19/7405/1

Do not write outside the box

23 Do not write outside the box

Use Figure 4 to determine the value of Ka for propanoic acid at 25 °C Show your working.

[3 marks]

mol dm−3

Ka 0 9 . 3

Suggest which indicator is the most appropriate for the reaction in Question 09.2? Tick () one box. [1 mark] Indicator

pH range

methyl orange

3.1 – 4.4

bromothymol blue

6.0 – 7.6

cresolphthalein

8.2 – 9.8

indigo carmine

11.6 – 13.0

Tick () one box

Question 9 continues on the next page

Turn over ►

*23* IB/G/Jun19/7405/1

24 0 9 . 4

A student prepared a buffer solution by adding 0.0136 mol of a salt KX to 100 cm3 of a 0.500 mol dm–3 solution of a weak acid HX and mixing thoroughly. The student then added 3.00 × 10–4 mol of potassium hydroxide to the buffer solution. Calculate the pH of the buffer solution after adding the potassium hydroxide. For the weak acid HX at 25 °C the value of the acid dissociation constant, Ka = 1.41 × 10–5 mol dm–3. Give your answer to two decimal places.

[6 marks]

pH

*24* IB/G/Jun19/7405/1

Do not write outside the box

25 0 9 . 5

Do not write outside the box

A buffer solution has a constant pH even when diluted. Use a mathematical expression to explain this.

[1 mark]

15

END OF QUESTIONS

*25* IB/G/Jun19/7405/1

26 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

*26* IB/G/Jun19/7405/1

27 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

*27* IB/G/Jun19/7405/1

28 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

Copyright information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material will be published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

*28*

*196A7405/1* IB/G/Jun19/7405/1

Please write clearly in block capitals. Centre number

Candidate number

Surname Forename(s) Candidate signature

A-level CHEMISTRY Paper 2

Organic and Physical Chemistry

Tuesday 11 June 2019

Afternoon

Time allowed: 2 hours

Materials

For this paper you must have: • the Periodic Table/Data Sheet, provided as an insert (enclosed) • a ruler with millimetre measurements • a scientific calculator, which you are expected to use where appropriate.

Instructions

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • All working must be shown. • Do all rough work in this book. Cross through any work you do not want to be marked.

• • • •

For Examiner’s Use Question

Mark

1 2 3 4 5 6 7 8 9 10

Information

11

• The marks for questions are shown in brackets. • The maximum mark for this paper is 105.

12 13

TOTAL

*JUN197405201* IB/G/Jun19/E12

7405/2

2 Do not write outside the box

Answer all questions in the spaces provided. 0 1

This question is about amines.

0 1 . 1

Give an equation for the preparation of 1,6-diaminohexane by the reaction of 1,6-dibromohexane with an excess of ammonia. [2 marks]

0 1 . 2

Complete the mechanism for the reaction of ammonia with 6-bromohexylamine to form 1,6-diaminohexane. Suggest the structure of a cyclic secondary amine that can be formed as a by-product in this reaction.

[4 marks]

Mechanism

Cyclic secondary amine

*02* IB/G/Jun19/7405/2

3 0 1 . 3

1,6-Diaminohexane can also be formed in a two-stage synthesis starting from 1,4-dibromobutane. Suggest the reagent and a condition for each stage in this alternative synthesis. [3 marks]

Do not write outside the box

Stage 1 reagent and condition

Stage 2 reagent and condition

0 1 . 4

Explain why 3-aminopentane is a stronger base than ammonia.

0 1 . 5

Justify the statement that there are no chiral centres in 3-aminopentane.

[2 marks]

[1 mark]

12

Turn over for the next question

Turn over ►

*03* IB/G/Jun19/7405/2

4 0 2

A student prepared cyclohexene by heating cyclohexanol with concentrated phosphoric acid. The cyclohexene produced was distilled off from the reaction mixture.

0 2 . 1

Complete the diagram of the apparatus used to distil the cyclohexene from the reaction mixture at 83 °C. [2 marks]

0 2 . 2

The distillate was shaken with saturated sodium chloride solution. The cyclohexene was separated from the aqueous solution using a separating funnel. State why cyclohexene can be separated from the aqueous solution using the separating funnel. [1 mark]

*04* IB/G/Jun19/7405/2

Do not write outside the box

5 0 2 . 3

The cyclohexene separated in Question 02.2 was obtained as a cloudy liquid. The student dried this cyclohexene by adding a few lumps of anhydrous calcium chloride and allowing the mixture to stand. Give one observation that the student made to confirm that the cyclohexene was dry. [1 mark]

0 2 . 4

In this preparation, the student added an excess of concentrated phosphoric acid to 14.4 g of cyclohexanol (Mr = 100.0). The student obtained 4.15 cm3 of cyclohexene (Mr = 82.0). Density of cyclohexene = 0.810 g cm–3 Calculate the percentage yield of cyclohexene obtained. Give your answer to the appropriate number of significant figures.

[5 marks]

% yield

Question 2 continues on the next page

Turn over ►

*05* IB/G/Jun19/7405/2

Do not write outside the box

6 0 2 . 5

Do not write outside the box

Cyclohexene reacts with bromine. Complete the mechanism for this reaction.

[3 marks]

12

*06* IB/G/Jun19/7405/2

7 0 3

The outer layers of some golf balls are made from a polymer called polyisoprene. The isoprene monomer is a non-cyclic branched hydrocarbon that contains 88.2 % carbon by mass. The empirical formula of isoprene is the same as its molecular formula.

0 3 . 1

Deduce the molecular formula of isoprene and suggest a possible structure.

[4 marks]

Molecular formula

Structure

Question 3 continues on the next page

Turn over ►

*07* IB/G/Jun19/7405/2

Do not write outside the box

8 0 3 . 2

The insides of some golf balls are made from a mixture of three other polymers. The repeating unit for one of these polymers is shown.

Draw the skeletal formula of the monomer used to make this polymer. Give the IUPAC name of the monomer.

[2 marks]

Skeletal formula of monomer

IUPAC name

*08* IB/G/Jun19/7405/2

Do not write outside the box

9 0 3 . 3

A second polymer in the mixture has a repeating unit with the structure shown.

Do not write outside the box

The third polymer in the mixture is a stereoisomer of this polymer. Draw the structure of the repeating unit of the third polymer. Give a reason why this type of stereoisomerism arises.

[2 marks]

Repeating unit

Reason

0 3 . 4

Golf balls recovered from lakes and ponds can be used again even after being in water for several years. Explain why these golf balls do not biodegrade.

[1 mark]

9

Turn over for the next question

Turn over ►

*09* IB/G/Jun19/7405/2

10 0 4

Substances P and Q react in solution at a constant temperature. The initial rate of reaction was studied in three experiments by measuring the change in concentration of P over the first five seconds of the reaction. The data obtained are shown in Table 1. Table 1 Experiment

1

2

3

0 4 . 1

Concentration / mol dm

Time after mixing / s

3

P

Q

0

1.00 × 10−2

1.25 × 10−2

5.0

0.92 × 10−2

not measured

0

2.00 × 10−2

1.25 × 10−2

5.0

1.84 × 10−2

not measured

0

0.50 × 10−2

2.50 × 10−2

5.0

0.34 × 10−2

not measured

Complete Table 2 to show the initial rate of reaction of P in each experiment.

[1 mark]

Table 2 Experiment 1

Initial rate / mol dm 3 s

1

1.6 × 10−4

2 3

*10* IB/G/Jun19/7405/2

Do not write outside the box

11 0 4 . 2

Determine the order of reaction with respect to P and the order of reaction with respect to Q.

Do not write outside the box

[2 marks]

Order with respect to P Order with respect to Q 0 4 . 3

A reaction between substances R and S was second order with respect to R and second order with respect to S. At a given temperature, the initial rate of reaction was 1.20 × 10–3 mol dm–3 s–1 when the initial concentration of R was 1.00 × 10–2 mol dm–3 and the initial concentration of S was 2.45 × 10–2 mol dm–3 Calculate a value for the rate constant, k, for the reaction at this temperature. Give the units for k [3 marks]

k

6

Units

Turn over ►

*11* IB/G/Jun19/7405/2

12 0 5

The rate constant, k, for a reaction varies with temperature as shown by the equation

Do not write outside the box

k = Ae–EaIRT For this reaction, at 25 °C, k = 3.46 × 10−8 s−1 The activation energy Ea = 96.2 kJ mol−1 The gas constant R = 8.31 J K−1 mol−1 Calculate a value for the Arrhenius constant, A, for this reaction. Give the units for A.

A

[4 marks]

4

Units

*12* IB/G/Jun19/7405/2

13 Do not write outside the box

0 6

This question is about isomers.

0 6 . 1

Give a reagent and observations for a test-tube reaction to distinguish between 2-methylbutan-1-ol and 2-methylbutan-2-ol. [3 marks] Reagent

Observation with 2-methylbutan-1-ol

Observation with 2-methylbutan-2-ol

0 6 . 2

Compounds A and B both have the molecular formula C4H8Br2 A has a singlet, a triplet and a quartet in its 1H NMR spectrum. B has only two singlets in its 1H NMR spectrum. Draw a structure for each of A and B. A

[2 marks] B

Question 6 continues on the next page

Turn over ►

*13* IB/G/Jun19/7405/2

14 0 6 . 3

Do not write outside the box

Compounds C and D both have the molecular formula C6H3Br3 C has two peaks in its 13C NMR spectrum. D has four peaks in its 13C NMR spectrum. Draw a structure for each of C and D

C

[2 marks] D

*14* IB/G/Jun19/7405/2

15 0 6 . 4

Do not write outside the box

Compounds E, F, and G are isomers.

Figure 1 shows the infrared spectra of these isomers, but not necessarily in the same order. Label each spectrum with the correct letter E, F or G in the box. Figure 1

[1 mark]

8

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*15* IB/G/Jun19/7405/2

16 Do not write outside the box

0 7

Isomers X and Y have the molecular formula C5H8O

0 7 . 1

Give the IUPAC name for isomer X.

0 7 . 2

Explain how and why isomers X and Y can be distinguished by comparing each of their

[1 mark]

• boiling points •

13

C NMR spectra

• infrared spectra. Use data from Tables A and C in the Data Booklet in your answer.

[6 marks]

*16* IB/G/Jun19/7405/2

17 Do not write outside the box

7

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*17* IB/G/Jun19/7405/2

18 0 8

Paracetamol is a medicine commonly used to relieve mild pain. Traditionally, paracetamol has been made industrially in a three-step synthesis from phenol.

0 8 . 1

Name the mechanism of the reaction in Step 1.

0 8 . 2

Complete the equation for the reaction in Step 2.

[1 mark]

[1 mark]

*18* IB/G/Jun19/7405/2

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19 0 8 . 3

In theory, either ethanoyl chloride or ethanoic anhydride could be used in Step 3. Complete the mechanism for the reaction of 4-aminophenol with ethanoyl chloride. RNH2 is used to represent 4-aminophenol in this mechanism. [2 marks]

0 8 . 4

In practice, ethanoic anhydride is used in the industrial synthesis rather than ethanoyl chloride. Give one reason why ethanoyl chloride is not used in the industrial synthesis. [1 mark]

Question 8 continues on the next page

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*19* IB/G/Jun19/7405/2

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20 0 8 . 5

In Step 3 other aromatic products are formed as well as paracetamol. Draw the structure of one of these other aromatic products.

0 8 . 6

Do not write outside the box

[1 mark]

Chemists have recently developed a two-step process to produce paracetamol from phenol. In the first step, phenol is oxidised to hydroquinone.

In the second step, hydroquinone reacts with ammonium ethanoate to form paracetamol. Complete the equation for this second step.

[1 mark]

*20* IB/G/Jun19/7405/2

21 0 8 . 7

Calculate the mass, in kg, of hydroquinone (Mr = 110.0) needed to produce 250 kg of paracetamol. [3 marks]

Mass

kg

Turn over for the next question

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*21* IB/G/Jun19/7405/2

Do not write outside the box

10

22 0 9

Do not write outside the box

This question is about thin-layer chromatography (TLC). • A protein was hydrolysed to form a mixture of amino acids. • A spot of this mixture was added to a TLC plate and the plate placed vertically in a small volume of solvent 1. • When the solvent front reached nearly to the top of the plate, the plate was removed and allowed to dry. • The plate was turned anticlockwise through 90° and placed vertically in a small volume of solvent 2. • When the solvent front reached nearly to the top of the plate, the plate was again removed and allowed to dry. • Figure 2 shows the final TLC plate. Figure 2

0 9 . 1

Suggest a suitable reagent for the hydrolysis of a protein.

[1 mark]

*22* IB/G/Jun19/7405/2

23 0 9 . 2

Suggest how the positions of the amino acids on the TLC plate were located.

0 9 . 3

Deduce the minimum number of amino acids present in the original mixture.

0 9 . 4

Suggest why it was necessary to use two different solvents.

Do not write outside the box

[1 mark]

[1 mark]

[1 mark]

4

Turn over for the next question

Turn over ►

*23* IB/G/Jun19/7405/2

24 1 0

Some compounds with different molecular formulas have the same relative molecular mass to the nearest whole number.

1 0 . 1

A dicarboxylic acid has a relative molecular mass of 118, to the nearest whole number. Deduce the molecular formula of the acid.

[3 marks]

Molecular formula 1 0 . 2

A student dissolved some of the dicarboxylic acid from Question 10.1 in water and made up the solution to 250 cm3 in a volumetric flask. In a titration, a 25.0 cm3 sample of the acid solution needed 21.60 cm3 of 0.109 mol dm−3 sodium hydroxide solution for neutralisation. Calculate the mass, in g, of the dicarboxylic acid used. Give your answer to the appropriate number of significant figures.

Mass

[4 marks]

g

*24* IB/G/Jun19/7405/2

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25 1 0 . 3

Compounds with molecular formula C6H14O2 also have a relative molecular mass of 118 to the nearest whole number. These include the diol shown.

Deduce the number of peaks in the 1H NMR spectrum of this diol.

[1 mark]

1 0 . 4

Draw the structure of a different diol also with molecular formula C6H14O2 that has a 1 H NMR spectrum that consists of two singlet peaks. [1 mark]

1 0 . 5

The dicarboxylic acid in question 10.1 and the isomers of C6H14O2 in Questions 10.3 and 10.4 all have a relative molecular mass of 118 State why the dicarboxylic acid can be distinguished from the two diols by high resolution mass spectrometry using electrospray ionisation.

Do not write outside the box

[1 mark]

10

Turn over for the next question

Turn over ►

*25* IB/G/Jun19/7405/2

26 1 1

This question is about esters including biodiesel.

1 1 . 1

An ester is formed by the reaction of an acid anhydride with CH3CH2OH

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Complete the equation. In your answer show clearly the structure of the ester. Give the IUPAC name of the ester. [3 marks] Equation

Name of ester 1 1 . 2

In a reaction to form biodiesel, one mole of a vegetable oil reacts with an excess of methanol to form two moles of an ester with molecular formula C19H34O2 and one mole of an ester with molecular formula C19H36O2 Draw the structure of the vegetable oil showing clearly the ester links. You should represent the hydrocarbon chains in the form Cx Hy where x and y are the actual numbers of carbon and hydrogen atoms. [2 marks]

*26* IB/G/Jun19/7405/2

27 1 1 . 3

The compound C19H34O2 is the methyl ester of Z,Z-octadeca-9,12-dienoic acid. Part of the structure of the acid is shown. Complete the skeletal formula to show the next part of the hydrocarbon chain to carbon atom number 14. In your answer, show the Z stereochemistry around both C=C double bonds. [2 marks]

1 1 . 4

Give an equation for the complete combustion of the ester C19H34O2

[1 mark]

Question 11 continues on the next page

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*27* IB/G/Jun19/7405/2

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28 1 1 . 5

Combustion of biodiesel produces greenhouse gases such as carbon dioxide that cause global warming. Part of the infrared spectrum of carbon dioxide is shown in Figure 3.

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Figure 3

State how the infrared spectrum of carbon dioxide in Figure 3 is not what you might predict from the data provided in Table A in the Data Booklet. [1 mark]

1 1 . 6

Explain how carbon dioxide causes global warming.

[2 marks]

11

*28* IB/G/Jun19/7405/2

29 1 2

Figure 4 shows two complementary strands in part of a DNA double helix structure.

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Figure 4

1 2 . 1

Draw all the hydrogen bonds between the complementary strands shown in Figure 4. Use dashed lines to show the hydrogen bonds. You do not need to show lone pairs of electrons or partial charges.

[2 marks]

1 2 . 2

Draw a ring around each of the component parts that make up the cytosine nucleotide in the section of DNA shown in Figure 4. [2 marks]

1 2 . 3

State the meaning of the term complementary when it is used to refer to DNA strands.

[1 mark]

5 Turn over ►

*29* IB/G/Jun19/7405/2

30 1 3

Aqueous NaBH4 reduces aldehydes but does not reduce alkenes.

1 3 . 1

Show the first step of the mechanism of the reaction between NaBH4 and 2-methylbutanal. You should include two curly arrows. Explain why NaBH4 reduces 2-methylbutanal but has no reaction with 2-methylbut-1-ene.

Do not write outside the box

[5 marks]

First step of mechanism

Explanation

1 3 . 2

A student attempted to reduce a sample of 2-methylbutanal but added insufficient NaBH4 The student confirmed that the reduction was incomplete by using a chemical test. Give the reagent and observation for the chemical test.

[2 marks]

Reagent

Observation 7

END OF QUESTIONS

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31 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

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32 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

Copyright information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material will be published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

*32*

*196A7405/2* IB/G/Jun19/7405/2

Please write clearly in block capitals. Centre number

Candidate number

Surname Forename(s) Candidate signature

A-level BIOLOGY Paper 3 Monday 17 June 2019

Morning

Time allowed: 2 hours

Materials

For Examiner’s Use

For this paper you must have: • a ruler with millimetre measurements • a scientific calculator.

Question

Mark

1 2

Instructions

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions in Section A. Answer one question from Section B. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • Show all your working. • Do all rough work in this book. Cross through any work you do not want to be marked.

• • • • •

3 4 5 6 7

TOTAL

Information

• The marks for the questions are shown in brackets. • The maximum mark for this paper is 78.

*JUN197402301* IB/M/Jun19/E11

7402/3

2 Do not write outside the box

Section A Answer all questions in this section. You are advised to spend no more than one hour and 15 minutes on this section. 0 1 . 1

Describe how ultrafiltration occurs in a glomerulus.

0 1 . 2

Glucose and water are reabsorbed by the proximal convoluted tubule of a nephron.

[3 marks]

Put a tick () in the box next to the correct ways in which glucose and water are reabsorbed. [1 mark] Glucose by active transport and water against a water potential gradient

Glucose by diffusion and water down a water potential gradient Glucose by facilitated diffusion and active transport and water against a water potential gradient Glucose by facilitated diffusion and active transport and water down a water potential gradient

*02* IB/M/Jun19/7402/3

3 0 1 . 3

Do not write outside the box

The equation shows the relationship between urine concentration in arbitrary units (𝑦𝑦) and mean length of the loop of Henle in mm (𝑥𝑥). 𝑦𝑦 = 0.72𝑥𝑥 + 4

Calculate the mean length of the loop of Henle in an organism that produces urine with a concentration of 16.56 arbitrary units. [1 mark]

Answer =

mm

Question 1 continues on the next page

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*03* IB/M/Jun19/7402/3

4 0 1 . 4

Scientists investigated the relationship between the thickness of the kidney medulla of different species of mammals and the concentration of their urine.

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Figure 1 shows their results. Figure 1

Explain the pattern shown by the results in Figure 1.

[3 marks]

8

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5 Do not write outside the box

0 2 . 1

Describe the role of saprobionts in the nitrogen cycle.

0 2 . 2

One environmental issue arising from the use of fertilisers is eutrophication. Eutrophication can cause water to become cloudy.

[2 marks]

You are given samples of water from three different rivers. Describe how you would obtain a quantitative measurement of their cloudiness. [3 marks]

5

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*05* IB/M/Jun19/7402/3

6 0 3

Do not write outside the box

Figure 2 shows a photograph of a dissected heart. Figure 2

0 3 . 1

Name valve A and chamber B.

[1 mark]

Valve A Chamber B 0 3 . 2

Give two safety precautions that should be followed when dissecting a heart.

[1 mark]

1

2

*06* IB/M/Jun19/7402/3

7 0 3 . 3

Explain how valve A in Figure 2 maintains a unidirectional flow of blood.

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[2 marks]

Question 3 continues on the next page

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8 A research scientist investigated the effect of caffeine on heart rate in human volunteers. The scientist divided volunteers into three groups. Each group was given the same volume of fluid. • Each member of Group I was given a sports drink containing caffeine and sugar. • Each member of Group J was given a sports drink containing caffeine and no sugar. • Each member of Group K was given water. The scientist recorded the volunteers’ heart rate before the drink was given and for 120 minutes after the drink was given. Her results can be seen in Figure 3. Figure 3

*08* IB/M/Jun19/7402/3

Do not write outside the box

9 0 3 . 4

Do not write outside the box

Caffeine affects the autonomic nervous system. Suggest how caffeine could account for the results of Group I in Figure 3 at 60 minutes.

0 3 . 5

[2 marks]

Before taking the drink, the mean heart rate of Group J was 68 beats per minute. Fifteen minutes after taking the drink, the mean volume of blood leaving the hearts of Group J was 4700 cm3 per minute. Calculate the mean volume of blood leaving the heart at each beat fifteen minutes after taking the drink. [1 mark]

Answer =

cm3

Question 3 continues on the next page

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10 0 3 . 6

The increase seen in Group I could be due to the combination of caffeine and sugar.

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Suggest one drink to be given to an additional group that should be investigated to find out if this is true. Give a reason for your answer.

[2 marks]

Group to be given

Reason

9

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11 Do not write outside the box

Turn over for the next question

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

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*11* IB/M/Jun19/7402/3

12 0 4

Mitochondrial DNA (mtDNA) is a small circular DNA molecule located in mitochondria. It is 16 569 nucleotides long and contains 37 genes and a control region. Sports scientists investigated whether a mutation in the control region of mtDNA in human males was related to an ability to exercise for longer. • The males in Group T had thymine at nucleotide position 16 519 • The males in Group C had a mutation resulting in cytosine at nucleotide position 16 519

0 4 . 1

The control regions of Group T and Group C were the same length. Name the type of gene mutation that is most likely to have occurred at nucleotide position 16 519 [1 mark]

Group T and Group C completed the same 8-week training programme. The following measurements were taken at the start of the 8-week programme, and again at the end. 1. VO2 max (a measure of maximal oxygen uptake). 2. Citrate synthase (CS) activity (CS is an enzyme involved in the Krebs cycle). The scientists then calculated the percentage increase in each measurement in both groups. Figure 4 and Figure 5 show their results. Figure 4

Figure 5

*12* IB/M/Jun19/7402/3

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13 0 4 . 2

A student concluded from Figure 4 and Figure 5 that training has a positive effect on VO2 max and CS activity. Evaluate the student’s conclusion.

0 4 . 3

[3 marks]

The mitochondrial DNA (mtDNA) control region is an area of mtDNA that is non-coding. This region stimulates the synthesis of both mtDNA and mitochondrial messenger RNA. Use this information to suggest two reasons why the mutation at nucleotide position 16 519 could lead to the differences seen in Figure 5.

[2 marks]

1

2

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14 The sports scientists investigated whether there was a correlation between the percentage change in VO2 max and percentage change in CS activity in Group T. Figure 6 shows their results. Figure 6

*14* IB/M/Jun19/7402/3

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15 0 4 . 4

‘Having thymine at nucleotide position 16 519 in Group T causes an increase in ability to exercise for longer.’

Do not write outside the box

Evaluate this conclusion. Use all the data in this question.

[3 marks]

[Extra space]

9

Turn over for the next question

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*15* IB/M/Jun19/7402/3

16 0 5

The crown-of-thorns starfish (COTS) is one of the main causes of the decline of the world’s coral reefs. Marine biologists used a choice chamber to investigate the effects of flashing and constant light on the behaviour of COTS. Table 1 shows their results as they presented them. The P values show results from a statistical test. Table 1 Behaviour of COTS

Type of light used in choice chamber Flashing

Constant

COTS moving towards the stimulus

22

12

COTS moving away from the stimulus

28

38

0.69

0.02

P value

0 5 . 1

State a null hypothesis the marine biologists tested in this investigation.

0 5 . 2

The natural habitat of COTS is coral reefs of tropical oceans.

[1 mark]

Suggest two factors that should be kept constant in the choice chambers so that COTS display normal behaviour. [1 mark] 1 2

*16* IB/M/Jun19/7402/3

Do not write outside the box

17 0 5 . 3

A journalist studying Table 1 suggested that either type of light could be used to cause COTS to move away from coral reefs. Evaluate the journalist’s suggestion.

0 5 . 4

Do not write outside the box

[3 marks]

One of the reasons COTS can destroy coral reefs in a short time is because COTS move quickly, allowing them to move from one reef to another. Table 2 shows the maximum speeds recorded of COTS in constant light. Table 2 Maximum speed / mm min–1

Response to light COTS moving towards constant light

259

COTS moving away from constant light

564

Calculate the shortest time one COTS would take to move up a coral reef from 66 m under water to 18 m under water in hours of daylight. Give your answer to the nearest hour.

[2 marks]

Answer =

hours Turn over ►

*17* IB/M/Jun19/7402/3

7

18 0 6

Do not write outside the box

Uncontrolled cell division can cause tumours to form. Figure 7 shows the growth pattern followed by a type of tumour. Figure 7

0 6 . 1

Use Figure 7 to calculate the percentage of maximum growth this type of tumour reaches before it can be detected. You will need to use the 10x button on your calculator.

Answer =

[1 mark]

%

*18* IB/M/Jun19/7402/3

19 0 6 . 2

Do not write outside the box

Figure 7 can also be used to calculate the age of this type of tumour. At diagnosis, a patient had a tumour of 3.98 × 1011 cells. Calculate the age of the tumour. You will need to use the log10 button on your calculator.

Answer =

[1 mark]

years

Question 6 continues on the next page

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20 Do not write outside the box

Trexall is a drug that can be used to slow the development of various forms of cancer. Trexall slows cell division by interacting with an enzyme called dihydrofolate reductase (DR). DR is involved in making nucleotides; the substrate for DR is folic acid. Figure 8 shows the chemical structure of Trexall. Figure 9 shows the chemical structure of folic acid. Figure 8

Figure 9

*20* IB/M/Jun19/7402/3

21 0 6 . 3

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

Do not write outside the box

[3 marks]

Question 6 continues on the next page

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*21* IB/M/Jun19/7402/3

22 Doctors investigated how the concentration of Trexall given to patients affected the growth of lung tumours. The doctors measured the volume of tumours at the beginning of the study and after 8 months. Figure 10 shows the results of this investigation. The bars represent ± 2 standard deviations. A value of ± 2 standard deviations from the mean includes over 95% of the data. Figure 10

0 6 . 4

The scientists measured the percentage change in tumour volume. Suggest why they recorded both percentage change and tumour volume.

[2 marks]

Percentage change

Tumour volume

*22* IB/M/Jun19/7402/3

Do not write outside the box

23 0 6 . 5

Do not write outside the box

A lung cancer patient received 15 mg of Trexall per week. After treatment, the diameter of his lung tumour was 35.8 mm Assuming the tumour was spherical, use the mean percentage change in tumour volume shown in Figure 10 to calculate the volume of the patient’s tumour before treatment with Trexall. 4

The formula for the volume of a sphere is 3 πr3 where π = 3.14

Answer = 0 6 . 6

[2 marks]

mm3

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

[2 marks]

Question 6 continues on the next page

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*23* IB/M/Jun19/7402/3

24 Do not write outside the box

Trexall can also be used to slow the development of rheumatoid arthritis (a pain-causing joint disease). Scientists investigated the effectiveness of Trexall as a pain relief treatment in 12 rheumatoid arthritis patients. All of the patients were female. They randomly divided the patients into two groups: • Group R received Trexall tablets for 35 days • Group S was a control group. They asked both groups to rate their pain on a scale of 0–10 (0 being no pain and 10 being the worst pain possible) at the start and then every 7 days for 35 days. They calculated mean scores for each group. Their results can be seen in Table 3. Table 3 Number of days of treatment

0 6 . 7

Mean score for severity of pain (scale 0–10) Group R

Group S

0

9.7

9.8

7

8.2

9.1

14

8.4

8.6

21

7.6

7.2

28

6.3

7.5

35

5.1

7.8

Apart from age and general health, give two important factors when choosing patients for this investigation. [1 mark] 1 2

*24* IB/M/Jun19/7402/3

25 0 6 . 8

A student analysed Table 3 and concluded that Trexall was effective in reducing pain in arthritis patients. Evaluate the student’s conclusion.

Do not write outside the box

[3 marks]

15

Turn over for Section B

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*25* IB/M/Jun19/7402/3

26 Do not write outside the box

Section B Answer one question. You are advised to spend no more than 45 minutes on this section. 0 7

Write an essay on one of the topics below. EITHER

0 7 . 1

The importance of DNA as an information-carrying molecule and its use in gene technologies. [25 marks] OR

0 7 . 2

The importance of bonds and bonding in organisms.

[25 marks]

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27 Do not write outside the box

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28 Do not write outside the box

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29 Do not write outside the box

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30 Do not write outside the box

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31 Do not write outside the box

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*31* IB/M/Jun19/7402/3

32 Do not write outside the box

*32* IB/M/Jun19/7402/3

33 Do not write outside the box

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34 Do not write outside the box

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35 Do not write outside the box

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36 Do not write outside the box

25

END OF QUESTIONS

Copyright Information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material are published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

*36*

*196a7402/3* IB/M/Jun19/7402/3

Please write clearly in block capitals. Centre number

Candidate number

Surname Forename(s) Candidate signature

A-level CHEMISTRY Paper 3 Wednesday 19 June 2019

Morning

Time allowed: 2 hours

Materials

For this paper you must have: • the Periodic Table/Data Sheet, provided as an insert (enclosed) • a ruler with millimetre measurements • a scientific calculator, which you are expected to use where appropriate.

Instructions

Use black ink or black ball-point pen. Fill in the boxes at the top of this page. Answer all questions. You must answer the questions in the spaces provided. Do not write outside the box around each page or on blank pages. • All working must be shown. • Do all rough work in this book. Cross through any work you do not want to be marked.

• • • •

For Examiner’s Use Question

Mark

1 2 3 4 5 Section B

TOTAL

Information

• The marks for questions are shown in brackets. • The maximum mark for this paper is 90.

Advice

• You are advised to spend about 70 minutes on Section A and 50 minutes on Section B.

*JUN197405301* IB/G/Jun19/E10

7405/3

2 Do not write outside the box

Section A Answer all questions in this section. 0 1

Sodium thiosulfate reacts with dilute hydrochloric acid as shown. Na2S2O3(aq) + 2 HCl(aq) → 2 NaCl(aq) + SO2(g) + S(s) + H2O(l)

0 1 . 1

Give the simplest ionic equation for this reaction.

0 1 . 2

The gas SO2 is a pollutant.

[1 mark]

State the property of SO2 that causes pollution when it enters rivers. Give an equation to show the reaction of SO2 with water.

[2 marks]

Property

Equation

*02* IB/G/Jun19/7405/3

3 0 1 . 3

Do not write outside the box

Draw a diagram to show the shape of a molecule of H2O Include any lone pairs of electrons. State the H−O−H bond angle. Explain this shape and bond angle.

[4 marks]

Diagram

Bond angle Explanation

Question 1 continues on the next page

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*03* IB/G/Jun19/7405/3

4 0 1 . 4

The initial rate of the reaction between sodium thiosulfate and hydrochloric acid can be monitored by measuring the time taken for a fixed amount of sulfur to be produced. Describe an experiment to investigate the effect of temperature on the initial rate of this reaction. Include • • • •

a brief outline of your method how you will measure the time taken for a fixed amount of sulfur to be formed how you will present your results in graphical form a sketch of the graph that you would expect. [6 marks]

*04* IB/G/Jun19/7405/3

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5 Do not write outside the box

13

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6 Do not write outside the box

0 2

This question is about sulfuric acid and its salts.

0 2 . 1

Draw the displayed formula of a molecule of H2SO4

0 2 . 2

In aqueous solution, sulfuric acid acts as a strong acid. The H2SO4 dissociates to form HSO4− ions and H+ ions.

[1 mark]

The HSO4− ions act as a weak acid and dissociate to form SO42− ions and H+ ions. Give an equation to show each stage in the dissociation of sulfuric acid in aqueous solution. Include appropriate arrows in your equations.

[2 marks]

Equation 1

Equation 2

*06* IB/G/Jun19/7405/3

7 0 2 . 3

A student is required to make 250 cm3 of an aqueous solution that contains an accurately measured mass of sodium hydrogensulfate (NaHSO4). Describe the method that the student should use to make this solution.

[4 marks]

Extra space

Question 2 continues on the next page

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*07* IB/G/Jun19/7405/3

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8 0 2 . 4

A solution that contains 605 mg of NaHSO4 in 100 cm3 of solution has a pH of 1.72

Do not write outside the box

Calculate the value of Ka for the hydrogensulfate ion (HSO4−) that is behaving as a weak acid. Give your answer to three significant figures. State the units of Ka

Ka 0 2 . 5

[6 marks]

Units

Some sodium sulfate is dissolved in a sample of the solution from question 02.4. Explain why this increases the pH of the solution.

[2 marks]

15

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9 Do not write outside the box

Turn over for the next question

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

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*09* IB/G/Jun19/7405/3

10 0 3

Figure 1 represents the cell used to measure the standard electrode potential for the Fe3+/Fe2+ electrode. Figure 1

0 3 . 1

Name the piece of apparatus labelled A.

0 3 . 2

State the purpose of A.

0 3 . 3

Name the substance used as electrode B in Figure 1.

[1 mark]

[1 mark]

[1 mark]

*10* IB/G/Jun19/7405/3

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11 0 3 . 4

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Complete Table 1 to identify C, D and E from Figure 1. Include the essential conditions for each.

[4 marks]

Table 1

Identity

Conditions

C

D

E

0 3 . 5

The standard electrode potential, Eo, for the Fe3+/Fe2+ electrode is +0.77 V Give the ionic equation for the overall reaction in the cell in Figure 1. State the change that needs to be made to the apparatus in Figure 1 to allow the cell reaction to go to completion. [2 marks]

Ionic equation Change

Question 3 continues on the next page

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12 0 3 . 6

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A student sets up a cell as shown in the cell representation. Zn(s)|Zn2+(aq)||Cu2+(aq)|Cu(s) The student measures the cell EMF, Ecell, with several different concentrations of Cu2+ ions and Zn2+ ions. The results are shown in Table 2.

Table 2

Complete Table 2 to show the value missing from experiment 4. Plot a graph of Ecell against ln ([Zn2+]/[Cu2+]) on the grid.

[3 marks]

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13 0 3 . 7

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This equation shows how Ecell varies with concentration for this reaction.

This equation is in the form of the equation for a straight line, y = mx + c Calculate the gradient of your plotted line on the graph in question 03.6. You must show your working. Use your gradient to calculate the temperature, T, at which the measurements of Ecell were taken. (If you were unable to calculate a gradient you should use the value −0.016 V This is not the correct value.) [3 marks]

0 3 . 8

Gradient

V

T

K

In experiment 2 in Table 2 the electrode potential of the Cu2+/Cu electrode is +0.33 V Use data from Table 2 in question 03.6 to calculate the electrode potential for the Zn2+/Zn electrode in experiment 2. Give one reason why your calculated value is different from the standard electrode potential for Zn2+/Zn electrode. [2 marks] Electrode potential

V

Reason

17

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0 4

Ethanal reacts with potassium cyanide, followed by dilute acid, to form 2-hydroxypropanenitrile.

0 4 . 1

Name the mechanism for the reaction between potassium cyanide and ethanal. [1 mark]

0 4 . 2

The 2-hydroxypropanenitrile formed by the reaction in question 04.1 is a mixture of equal amounts of two isomers. State the name of this type of mixture. Explain how the structure of ethanal leads to the formation of two isomers. Draw 3D representations of the two isomers to show the relationship between them. [5 marks] Name Explanation

3D representations

*14* IB/G/Jun19/7405/3

15 0 4 . 3

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2-Hydroxypropanenitrile can be used in the synthesis of the monomer, acrylonitrile, CH2=CHCN Suggest a suitable reagent and conditions for the conversion of 2-hydroxypropanenitrile into acrylonitrile.

[2 marks]

Reagent Conditions 0 4 . 4

Draw a section of the polymer polyacrylonitrile, showing three repeating units. [1 mark]

9

Turn over for the next question

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*15* IB/G/Jun19/7405/3

16 0 5

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The percentage by mass of iron in a steel wire is determined by a student. The student • reacts 680 mg of the wire with an excess of sulfuric acid, so that all of the iron in the wire forms Fe2+(aq) • makes up the volume of the Fe2+(aq) solution to exactly 100 cm3 • takes 25.0 cm3 portions of the Fe2+(aq) solution • titrates each portion with 0.0200 mol dm−3 potassium manganate(VII) solution.

0 5 . 1

Give the equation for the reaction between iron and sulfuric acid.

0 5 . 2

The titration results are shown in Table 3.

[1 mark]

Table 3 1

2

3

Final volume / cm3

22.90

45.60

22.60

Initial volume / cm3

0.00

22.90

0.00

22.90

22.70

22.60

Titre / cm3 Calculate the mean titre.

[1 mark]

Mean titre 0 5 . 3

cm3

Give the overall ionic equation for the oxidation of Fe2+ by manganate(VII) ions, in acidic conditions. [1 mark]

*16* IB/G/Jun19/7405/3

17 0 5 . 4

State the colour change seen at the end point of the titration.

0 5 . 5

Name the piece of apparatus used for these stages of the method.

Do not write outside the box

[1 mark]

[1 mark]

Taking the 25.0 cm3 portions Adding the potassium manganate(VII) solution 0 5 . 6

The balance used to weigh the 680 mg of iron wire has an uncertainty of ±0.005 g A container was weighed and its mass was subtracted from the total mass of the container and wire. Calculate the percentage uncertainty in using the balance.

[1 mark]

6

% uncertainty

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18 Do not write outside the box

Section B Answer all questions in this section.

Only one answer per question is allowed. For each answer completely fill in the circle alongside the appropriate answer. CORRECT METHOD

WRONG METHODS

If you want to change your answer you must cross out your original answer as shown. If you wish to return to an answer previously crossed out, ring the answer you now wish to select as shown. You may do your working in the blank space around each question but this will not be marked. Do not use additional sheets for this working.

0 6

Which amount of sodium hydroxide would react exactly with 7.5 g of a diprotic acid, H2A (Mr = 150)? [1 mark] A 50 cm3 of 0.05 mol dm–3 NaOH(aq) B 100 cm3 of 0.50 mol dm–3 NaOH(aq) C 100 cm3 of 1.0 mol dm–3 NaOH(aq) D 100 cm3 of 2.0 mol dm–3 NaOH(aq)

*18* IB/G/Jun19/7405/3

19 0 7

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Lead(II) nitrate and potassium iodide react according to the equation Pb(NO3)2(aq) + 2 KI(aq) → PbI2(s) + 2 KNO3(aq) In an experiment, 25.0 cm3 of a 0.100 mol dm–3 solution of each compound are mixed together. Which amount, in mol, of lead(II) iodide is formed?

[1 mark]

A 1.25 x 10–3 B 2.50 x 10–3 C 1.25 x 10–2 D 2.50 x 10–2

0 8

Nitrogen dioxide is produced from ammonia and air as shown in these equations 4 NH3(g) + 5 O2(g) → 4 NO(g) + 6 H2O(g)

ΔH = –909 kJ mol–1

2 NO(g) + O2(g) → 2 NO2(g)

ΔH = –115 kJ mol–1

What is the enthalpy change (in kJ mol–1) for the following reaction? 4 NH3(g) + 7 O2(g) → 4 NO2(g) + 6 H2O(g) [1 mark] A –679 B –794 C –1024 D –1139

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*19* IB/G/Jun19/7405/3

20 0 9

Which change leads to a higher concentration of SO3 in this equilibrium mixture? ∆H = −188 kJ mol−1

2 SO2(g) + O2(g) ⇌ 2 SO3(g)

[1 mark]

A higher concentration of O2 B higher temperature C lower pressure D use of a catalyst

1 0

The results of an investigation of the reaction between P and Q are shown in this table. Experiment

Initial [P] / mol dm–3

Initial [Q] / mol dm–3

Initial rate / mol dm–3 s–1

1

0.200

0.500

0.400

2

0.600

To be calculated

0.800

The rate equation is:

rate = k [P] [Q]2

What is the initial concentration of Q in experiment 2? [1 mark] A 0.167 B 0.333 C 0.408 D 0.612

*20* IB/G/Jun19/7405/3

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21 1 1

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The equation for the reaction between sulfur dioxide and oxygen is shown. 2 SO2(g) + O2(g) ⇌ 2 SO3(g)

In an experiment, 2.00 mol of sulfur dioxide are mixed with 2.00 mol of oxygen. The total amount of the three gases at equilibrium is 3.40 mol What is the mole fraction of sulfur trioxide in the equilibrium mixture?

[1 mark]

A 0.176 B 0.353 C 0.600 D 1.200

1 2

Nitrogen reacts with hydrogen in this exothermic reaction N2(g) + 3 H2(g) ⇌ 2 NH3(g)

Which change increases the equilibrium yield of ammonia but has no effect on the value of the equilibrium constant Kp? [1 mark] A Add a catalyst B Increase the partial pressure of nitrogen C Decrease the temperature D Decrease the total pressure

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*21* IB/G/Jun19/7405/3

22 1 3

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o

The E values for two electrodes are shown. Fe2+(aq) + 2 e– → Fe(s) Eo= –0.44 V Cu2+(aq) + 2 e– → Cu(s) Eo= +0.34 V What is the EMF of the cell Fe(s)|Fe2+(aq)||Cu2+(aq)|Cu(s)?

[1 mark]

A +0.78 V B +0.10 V C –0.10 V D –0.78 V

1 4

Which atom has the greatest first ionisation energy?

[1 mark]

A H B He C Li D Ne

1 5

What is the correct observation when barium metal is added to an excess of water? [1 mark] A Forms a colourless solution only B Forms a colourless solution and effervesces C Forms a white precipitate only D Forms a white precipitate and effervesces

*22* IB/G/Jun19/7405/3

23 1 6

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An aqueous solution of a salt gives a white precipitate when mixed with aqueous silver nitrate and when mixed with dilute sulfuric acid. Which could be the formula of the salt?

[1 mark]

A BaCl2 B (NH4)2SO4 C KCl D Sr(NO3)2

1 7

Which statement is not correct about the trends in properties of the hydrogen halides from HCl to HI ? [1 mark] A The boiling points decrease. B The bond dissociation energy of H−X decreases. C The polarity of the H−X bond decreases. D They are more easily oxidised in aqueous solutions.

1 8

What is observed when concentrated hydrochloric acid is added to an aqueous solution of CuSO4 until no further change occurs? [1 mark] A A colourless gas is evolved and a precipitate forms. B A colourless gas is evolved and no precipitate forms. C

A precipitate forms that dissolves in an excess of concentrated hydrochloric acid.

D The solution changes colour and no precipitate forms.

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*23* IB/G/Jun19/7405/3

24 1 9

What is the most suitable reagent for detecting the presence of carbonate ions in the presence of an excess of sulfate ions? [1 mark] A dilute NaOH(aq) B dilute H2SO4(aq) C BaCl2(aq) D NaCl(aq)

2 0

Methylbenzene reacts with a mixture of concentrated nitric acid and concentrated sulfuric acid. What is the name of the mechanism for this reaction?

[1 mark]

A Electrophilic addition B Electrophilic substitution C Nucleophilic addition D Nucleophilic substitution

*24* IB/G/Jun19/7405/3

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25 2 1

Do not write outside the box

A possible synthesis of a compound found in jasmine flower oil is shown.

Which mechanism is not used in this synthesis?

[1 mark]

A Electrophilic substitution B Nucleophilic substitution C Free-radical substitution D Nucleophilic addition-elimination

2 2

Which compound is formed when 1-phenylethanol reacts with acidified potassium dichromate(VI)?

[1 mark]

A C6H5CH2CH2OH B C6H5CH2CHO C C6H5COCH3 D C6H5CH2COOH

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*25* IB/G/Jun19/7405/3

26 2 3

Do not write outside the box

Three reagents are added separately to four organic compounds. Which row shows the correct observations?

Sodium hydrogen carbonate

Acidified potassium dichromate(VI)

[1 mark] Tollens’ reagent

A

Propan-1-ol

effervescence

orange solution turns green

B

Propanal

no visible change

orange solution turns green

no visible change silver mirror

C

Propanone

no visible change

no visible change

silver mirror

D

Propanoic acid

effervescence

no visible change

silver mirror

2 4

Which compound is formed by acid hydrolysis of phenylmethyl ethanoate?

[1 mark]

A C6H5CH2OH B C6H5CHO C C6H5COCH3 D C6H5COOH

2 5

A student is required to dry a liquid sample of pentanoic acid. Which drying agent is suitable?

[1 mark]

A Calcium oxide B Calcium sulfate C Potassium hydroxide D Potassium carbonate

*26* IB/G/Jun19/7405/3

27 2 6

The reaction between propanoyl chloride and benzene is an example of acylation. Which is a correct representation of part of the mechanism of this reaction?

[1 mark]

A

B

C

D

2 7

Methylamine reacts with bromoethane by substitution to produce a mixture of products. Which compound is not a possible product of this reaction?

[1 mark]

A C2H5NHCH3 B (C2H5)2NCH3 C [(C2H5)3NCH3]+ Br– D [(C2H5)2N(CH3)2]+ Br–

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*27* IB/G/Jun19/7405/3

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28 2 8

Which polymer has hydrogen bonding between its chains?

Do not write outside the box

[1 mark]

A Kevlar B Polythene C PVC D Terylene

2 9

Which structure shows part of a peptide link in a protein?

[1 mark]

A

B

C

D

*28* IB/G/Jun19/7405/3

29 3 0

3 1

Two strands of DNA are linked together by hydrogen bonding between bases on each strand. Which row shows the number of hydrogen bonds between the pair of bases? Use the Data Booklet to help you answer this question. [1 mark] Base 1

Base 2

Number of hydrogen bonds

A

adenine

guanine

2

B

cytosine

thymine

2

C

guanine

cytosine

3

D

adenine

thymine

3

Which is not responsible for conduction of electricity?

[1 mark]

A The sodium ions in molten sodium chloride B The electrons between layers of carbon atoms in graphite C The bonding electrons in a metal D The lone pair electrons on water molecules

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*29* IB/G/Jun19/7405/3

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30 3 2

In the UK industrial ethanol is now produced by the direct hydration of ethene. This process has largely replaced the fermentation method. Which is a likely reason for this change of method?

[1 mark]

A The direct hydration route produces purer ethanol. B The direct hydration route employs milder conditions. C The direct hydration route does NOT use a catalyst. D The direct hydration route produces ethanol by a slower reaction.

3 3

Which alkene reacts with hydrogen bromide to give 2-bromo-3-methylbutane as the major product? [1 mark] A (CH3)2C=CHCH3 B CH3CH2CH=CHCH3 C CH3CH2C(CH3)=CH2 D (CH3)2CHCH=CH2

3 4

Which compound can be purified by forming a hot aqueous solution that recrystallises on cooling? [1 mark] A Cyclohexene B Ethanoic acid C Phenylamine D Benzoic acid

*30* IB/G/Jun19/7405/3

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31 Do not write outside the box

3 5

Use the Data Booklet to help you answer this question Which is the main aspartic acid species present in an aqueous solution at pH = 14? [1 mark]

A

B

C

D 30

END OF QUESTIONS

*31* IB/G/Jun19/7405/3

32 Do not write outside the box

There are no questions printed on this page

DO NOT WRITE ON THIS PAGE ANSWER IN THE SPACES PROVIDED

Copyright Information For confidentiality purposes, from the November 2015 examination series, acknowledgements of third-party copyright material are published in a separate booklet rather than including them on the examination paper or support materials. This booklet is published after each examination series and is available for free download from www.aqa.org.uk after the live examination series. Permission to reproduce all copyright material has been applied for. In some cases, efforts to contact copyright-holders may have been unsuccessful and AQA will be happy to rectify any omissions of acknowledgements. If you have any queries please contact the Copyright Team, AQA, Stag Hill House, Guildford, GU2 7XJ. Copyright © 2019 AQA and its licensors. All rights reserved.

*32*

*196A7405/3* IB/G/Jun19/7405/3

A-level BIOLOGY 7402/1 Paper 1 Mark scheme June 2019 Version: 1.0 Final

*196A7402/1/MS*

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. 2

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Mark scheme instructions to examiners 1. General The mark scheme for each question shows: • • • •

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent.

2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for the same mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of errors / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the ‘Comments’ column of the mark scheme) are not penalised.

3

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

3.2

Marking procedure for calculations Full marks can be given for a correct numerical answer, without any working shown. However, if the answer is incorrect, mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.4

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the mark scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the mark scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.5

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term.

3.6

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.7

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

4

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

1.1

1. Attaches to the enzyme at a site other than the active site;

Mark 3

Comments 1. Accept ‘attaches to allosteric/inhibitor site’

2. Changes (shape of) the active site OR 3. Accept ‘no longer complementary so less/no enzymesubstrate complexes form’

Changes tertiary structure (of enzyme); 3. (So active site and substrate) no longer complementary so less/no substrate can fit/bind;

3. Accept abbreviations of enzyme-substrate complex. 1.2

(With inhibitor) increase substrate/lipid (concentration) does not increase/affect/change rate of reaction

1

Ignore references to competitive inhibitors.

OR (With inhibitor) increase substrate/lipid (concentration) does not increase/affect/change lipase activity OR High substrate (concentration) does not overcome inhibition OR High substrate (concentration) does not meet maximum rate of reaction/lipase activity; 1.3

(Maximum length) 8-10 (µm);

2

(Uncertainty) (±) 2 (µm); 1.4

1. Emulsification; 2. (Cannot be seen) due to resolution (of optical microscope);

2

1. Ignore ‘micelles’ 2. Ignore reference to magnification. 2. For ‘resolution’ accept ‘wavelength of light’.

5

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

2.1

Mark Comments 3

1st 2 columns correct (Plants and Algae) = 1 mark 3rd column correct (Fungi) = 1 mark

Cell wall Plants Algae Fungi Prokaryotes component   Cellulose  Murein  Chitin ; ; ;

4th column correct (Prokaryotes) = 1 mark Accept alternative symbols that clearly indicate the box but are not ticks eg X. If answer clearly crossed out read box as blank.

6

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question 2.2

Marking Guidance 1. Negative correlation (between fibre eaten per day and risk of cardiovascular disease); 2. Original/current fibre intake (of student) not known; 3. (Idea of) significance linked to (2x) standard deviation overlap (at 10 g day-1 change); 4. If current intake between 5 and 30 (g day-1) then (eating 10g more results in a significant) decrease in risk OR If current intake between 30 and 50 (g day-1) then (eating 10g more results in) no significant decrease in risk; 5. Correlation does not mean causation OR Another named factor may be involved; 6. Little evidence/data for higher mass of fibre per day; 7. Large (2x) standard deviation at high/low mass of fibre makes (mean) less precise OR Large (2x) standard deviation at high/low amounts of fibre means there is a greater uncertainty; 8. No statistical test (to show if differences are significant);

Mark Comments 4 max

1. Accept positive correlation with reduced risk 2. Accept ‘it depends on original/current fibre intake’. 3. This is for the correct concept, ignore stated values. 3. Ignore reference to probability and chance. 4. Accept stated values between 5 and 30 for (significant) decrease in risk. 4. Accept stated values between 30 and 50 for no significant decrease in risk. 4. Ignore stated values less than 5 or more than 50. 5. Examples of named factors smoking, exercise, age, sex, genes, other aspects of diet. 7. For ‘precise’ accept reliable or description of precise/reliable.

7

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

2.3

(Advantage) 1. Over longer period so more representative

Mark Comments 2

Only credit reference to ‘honesty’ once.

OR Diet over 24 hr may not be representative OR Diet may vary during the year/from day to day OR Person more likely to be honest on questionnaire (rather than speaking to nurse) OR More cost effective because fewer people/nurses required; (Disadvantage) 2. Relies on (long term) memory so may not be accurate OR Recall of 24 hr diet likely to be more accurate OR Estimation (from FFQ) may be less accurate (than details of last 24hrs) OR Person may be more honest when being interviewed;

8

2. For ‘accurate’ accept only ‘valid’ or ‘close to true value’.

2. Accept examples of ‘estimation (from FFQ)’ eg frequency of eating may not give mass of fibre, type of food may not give mass of fibre, no information on portion size to give mass of fibre. These must all be accompanied by idea of reduced accuracy.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

3.1

(Number of species and) number of individuals in each species (in each habitat)

3.2

Mark 1

Comments Accept organisms for individuals

OR

Ignore frequency.

(Number of species and) population of each species (in each habitat);

Accept abundance of each species.

1. Random samples;

2

2. Large number (of samples)

Both marks can be awarded on one line. Ignore other answers unless they contradict mark points.

OR (Continue sampling) until stable running mean;

2. Accept many/multiple. Ignore several. 2. If a specified number is given, it must be 10 or more. 2. Accept ‘large sample (size)’. Accept 3.3 organisms for individuals

(Larger fields have relatively)

1

More centre OR Less edge OR

Ignore removal of hedge (as given in stem).

Less hedge OR Fewer species;

9

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

3.4

Advantage 1. Greater (bio)diversity so increase in predators of pests OR Increase in predators of pests so more yield/income/less pesticides/less damage to crops OR Increase in pollinators so more yield/income

Mark 2

Comments Accept description of yield eg crop growth. For ‘crop’ accept ‘plant’.

Accept other valid suggestions with explanation that will affect the farm as a whole.

OR May attract more tourists/subsidies to their farm so more income (from diversification); Disadvantage 2. Reduced land area for crop growth/income OR Greater (bio)diversity so increase pest population OR Increase pest population so less yield/less income/(more) need for pesticides/(more) damage to crops OR Increased (interspecific) competition so less yield/income OR More difficult to farm so less income;

10

Examples of ‘more difficult to farm’ – can’t use large machinery, more difficult to plough/seed/harvest.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

4.1

1. (Most likely to be) transferred to a special care unit are those under 2800 g

Mark 3

OR (Most likely to be) transferred to a special care unit are those over 4200 g;

Comments Accept converse answers linked to those with mass at birth at any value between 2800 and 4200 g.

2. Extreme mass babies least likely to survive (to reproduce) and so less likely to pass on their alleles (for extreme mass at birth);

1. For ‘2800 g’ accept any value between 1400 g and 2800 g.

3. Extreme mass at birth decreases in frequency (in the population)

1. For ‘4200 g’ accept any value between 4200 g and 5200 g.

OR Alleles (for extreme mass at birth) decrease in frequency (in the population);

1. If values for both extremes are given, both must be correct.

If neither 1 or 2 awarded allow correct stated mass less/more likely to survive for 1 mark

1. Reject data quoted below 1400 g or above 5200 g. 3. Accept ‘proportion/percentage’ for ‘frequency’. 3. Do not accept ‘number’ for ‘frequency’.

4.2

1. Allele

3 max

6 correct = 3 marks 4 – 5 correct = 2 marks

2. Locus/loci 3. Transcribed

2 – 3 correct = 1 mark

4. Translated

0 – 1 correct = 0 marks

5. Golgi (apparatus)/Rough endoplasmic reticulum 6. Tertiary;;;

2. Do not accept locust. 3. Accept transcripted. 3. Ignore spliced. 5. Reject smooth endoplasmic reticulum. 5. Ignore RER/ER. 6. Ignore 3D. 6. Accept secondary.

4.3

Automarked q –  Chi-squared

1 11

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

4.4

1. Probability that difference (in frequency of births above 4500 g) is due to chance is less than 0.05

Mark Comments 3

Ignore reference to critical value. 1. Accept 5% for 0.05

OR

1. Accept 3% for 0.03

Probability that difference (in frequency of births above 4500 g) is due to chance is 0.03;

1. Ignore results due to chance.

2. Reject null hypothesis; 3. Presence of KIR2DS1/allele does (significantly) affect the frequency of high birth mass;

1. Accept ‘Probability that difference (in frequency of births above 4500 g) is not due to chance is greater than 0.95’ OR ‘Probability that difference (in frequency of births above 4500 g) is not due to chance is 0.97’ 2. Accept ‘H0’ for null hypothesis. 2. For ‘reject’ accept ‘do not accept’ but not ‘disprove/wrong’. 2. Accept ‘Accept the alternate hypothesis/H1’. 3. Do not accept ‘number’ for ‘frequency’.

12

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

Mark

Comments

5.1

1. RNA (as genetic material);

4 max

Accept a labelled diagram.

2. Reverse transcriptase; 3. (Protein) capsomeres/capsid;

1. Reject nucleus/DNA/plasmids.

4. (Phospho)lipid (viral) envelope

3. Reject capsule. 4. Reject if HIV has a cell membrane or a cell wall.

OR Envelope made of membrane; 5. Attachment proteins;

5. Accept gp41 and/or gp 120. 5. Accept glycoprotein. 5. Accept description of attachment protein. 5. Ignore ‘receptor protein’. Ignore cytoplasm.

5.2

Automarked q –  106

5.3

1. (All) have more T helper/CD4 cells;

1 3 max

2. Lower viral load to infect/destroy helper T/CD4 cells;

1. Accept higher proportion of T helper/CD4 to virus particles.

3. (So more/continued) activation of B cells/cytotoxic T cells/phagocytes;

1. and 2. Statement must be comparative.

4. (With B cells more/continued) production of plasma cells/antibodies

2. For ‘infect’ accept ‘HIV does not reproduce in’.

OR (With cytotoxic T cells more/continued) ability to kill virus infected cells; 5. (More able to) destroy other microbes/pathogens OR

3. Accept ‘stimulation’ for ‘activation’. 4. Ignore reference to B cells acting as phagocytes/antigenpresenting cells.

(More able to) destroy mutated/cancer cells;

13

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

6.1

1. (Trend of) slowing growth from before birth to 21 days

Mark 2

Comments 1. Accept ‘day -6’ for ‘before birth’. 1. For ’21 days’ accept ‘until the end of the investigation’.

OR (Trend of) decreasing percentage undergoing mitosis from before birth to 21 days OR (Trend of) decreasing percentage undergoing DNA replication from before birth to 21 days; 2. DNA replication happens before mitosis

2. Accept ‘Heart growing/developing before birth and becomes (fully) developed’.

OR Heart growth slowing until (fully) developed OR

2. Accept reference to only unipotent cells/cardiomycetes dividing (at 21 days).

These cells lost the ability to divide;

6.2

1. DNA helicase; 2. Breaks hydrogen bonds (between 2 DNA strands);

2. Reject ‘hydrolyses hydrogen bonds’

3. BrdU complementary to adenine (on template strand)

2 and 3. Accept H bonds for hydrogen bonds.

OR BrdU forms hydrogen bonds with adenine (on template strand); 4. DNA polymerase joins (adjacent) nucleotides (to incorporate BrdU into the new DNA strand); 5. Phosphodiester bonds form (between nucleotides);

14

5

4. Reject if DNA polymerase catalyses complementary base pairing or if DNA polymerase catalyses nucleotides joining to template strand.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

6.3

1. Add antibody (anti-BrdU with enzyme attached) to cells/DNA OR Add cells/DNA to antibody (anti-BrdU with enzyme attached); 2. Wash (cells/DNA) to remove excess/unattached antibody OR Wash (immobilised antibody) to remove excess/unattached cells/DNA; 3. Add substrate to cause colour change;

Mark 3

Comments All mark points must relate to procedure. Do not negate any mark point for use of additional antibodies.

2. Allow ECF for absence of cells/DNA. 3. For ‘substrate’ accept description in context of enzyme.

15

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

7.1

Short diffusion pathway (to cells)

Mark 1

OR It has a surface permeable (to water/ions into cells);

Comments Accept the idea of not needing structural support as supported by the water. Ignore pores/stomata

7.2

2

Accept answers written beside the box but clearly intended for that box. 1. Accept ‘meiosis’ for ‘E’ (spelling must be correct) 2. Accept ‘mitosis’ for ‘T’ (spelling must be correct)

1. Reject anything other than ‘E/meiosis’ written in the top right box 1. E in top right box; (1 mark) 2. 3 x T in top and bottom left and bottom right boxes; (1 mark)

2. Reject anything other than ‘T/mitosis’ written in top left, bottom left and bottom right boxes.

If 1 x E and 3 x T but written in incorrect boxes = 0 marks

16

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

7.3

1. They are different species; 2. (So) if fused together they would not produce fertile offspring OR (So) they have named characteristics that means they are reproductively isolated;

Mark Comments 2

2. For ‘fuse’ accept ‘form a zygote’. 2. Accept   

if they fused together meiosis could not occur if they fused together (chromosomes) could not form homologous pairs if they fused production of gametes could not occur.

2. Accept a description of characteristics that would lead to reproductive isolation eg  will not successfully fuse with one another  produce single cells at different times  description of geographical isolation. Accept the description on its own, the phrase ‘reproductive isolation’ is not required.

17

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

8.1

Mark 2

Comments 1 mark for each row. If values do not match the given unit, max 1. Accept dm3 / mm3 for volume unit. Accept 0.0008/8 x 10-4 and 0.0192/1.92 x 10-2 Accept 800 and 19200 Ignore units in 2nd row. Do not accept mm-3/cm-3/dm-3/ ml

8.2

Correct answer of 0.07 (mol dm–3) = 2 marks;;

2

Incorrect answer 1 mark for any evidence of 48.6 to 48.8 OR 0.02 OR 0.7 OR A final answer between 0.04 and 0.10 OR A final answer of minus 0.07/-0.07;

18

Ignore minus signs on other 1 mark options.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

8.3

Correct answer of 9 (cm2) = 2 marks;;

Mark 2

Comments Allow 9.0

Incorrect answer 1 mark for evidence of water potential of between -1.85 and -1.95 (MPa) OR

Accept correct reading labelled on the graph shown on Figure 8 or Figure 9.

growth of 15% OR 69 (cm2) OR A final answer between 8.7 and <9; 8.4

EITHER

2 max

1. Low/slow growth;

Mark as pair – 1 and 2 OR 3 and 4. 2. Reference to stomata must not relate only to water loss.

2. Due to smaller number/area of stomata (for gas exchange); OR 3. Growth may continue at lower water potentials; 4. (Due to) adaptations in enzymes involved in photosynthesis/metabolic reactions; 8.5

1. Stomata close; 2. Less carbon dioxide (uptake) for less photosynthesis/glucose production;

2 2. ‘Less’ only required once. 2. Reject ‘no photosynthesis’ but accept ‘carbon dioxide can’t enter so less photosynthesis’. 2. Ignore oxygen for respiration but reject oxygen for photosynthesis. 2. Ignore less water for photosynthesis. 2. Accept only correct chemical formulae. 2. For ‘glucose’ accept named product of photosynthesis eg triose phosphate, TP, amino acid, lipid.

19

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

9.1

1. y axis 0 – 100 in linear scale and x axis minimum 1 to 8 in linear scale and both axes use at least half size of grid;

Mark 3

2. Correct plots for 50% and 25% for both animals;

Correct answer of 15 (times faster)

1. If tick marks are used on the axis, they must be accurate to within ± half a small square. 2. 25% - 1.9, 3.3 and 50% 3.2 and 6.5

3. Both curves levelling off (at higher partial pressures and at percentage saturations ≤100%); 9.2

Comments

2. Accept plot ± half a small square.

2

= 2marks ;; If ≥3sf given, accept answers in the range 15.0 to 15.4 (times faster) = 2marks;; Incorrect answer 1 mark for evidence of: 23–0.27 divided by 550 000–0.27 OR 0.42888777 OR 0.02819045 OR Between 27 and 27.1 OR Between 1.77599861 and 1.8 OR 0.06

20

Accept any number of significant figures ≥2, if rounding correct.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

9.3

1. Mouse haemoglobin/Hb has a lower affinity for oxygen

Mark 2

OR

Comments 1. For ‘Hb is less saturated’ accept ‘less oxygen will be bound to Hb’.

For the same pO2 the mouse haemoglobin/Hb is less saturated OR At oxygen concentrations found in tissue mouse haemoglobin/Hb is less saturated;

2. Accept ‘oxygen dissociated/released/unloaded more readily/easily/quickly’

2. More oxygen can be dissociated/released/unloaded (for metabolic reactions/respiration);

9.4

Mouse 1. (Smaller so) larger surface area to volume ratio; 2. More/faster heat loss (per gram/in relation to body size); 3. (Faster rate of) respiration/metabolism releases heat;

2. Reject ‘oxygen loaded more readily/easily/quickly’ or ‘more oxygen loaded’ 3

Accept converse answers in relation to the horse. 1. Accept larger SA:V. 1. and 2. must be comparative. 2. Ignore heat lost more easily/readily. 3. Accept respiration/metabolism replaces heat. 3. Reject produce/generate heat/energy.

21

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

Mark

10.1

1. A metabolite in condensation/hydrolysis/ photosynthesis/respiration;

5 max

2. A solvent so (metabolic) reactions can occur OR A solvent so allowing transport of substances; 3. High heat capacity so buffers changes in temperature; 4. Large latent heat of vaporisation so provides a cooling effect (through evaporation); 5. Cohesion (between water molecules) so supports columns of water (in plants); 6. Cohesion (between water molecules) so produces surface tension supporting (small) organisms;

Comments

3. For ‘buffer’ accept ‘resist’. 5. For ‘columns of water’ accept ‘transpiration stream’. Do not credit ‘transpiration’ alone but accept description of ‘stream’. 5. For ‘columns of water’ accept ‘cohesion-tension (theory)’. 5 and 6. For cohesion accept hydrogen bonding Ignore reference to pH. Allow other suitable properties but must have a valid explanation. For example  

22

ice floating so maintaining aquatic habitat beneath water transparent so allowing light penetration for photosynthesis

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

Mark

Comments

10.2

Lipid

5 max

4 max if marks gained from only 2 substance tests.

1. Add ethanol/alcohol then add water and shake/mix OR Add ethanol/alcohol and shake/mix then pour into/add water; 2. White/milky emulsion OR emulsion test turns white/milky; Non-reducing sugar 3. Do Benedict’s test and stays blue/negative; 4. Boil with acid then neutralise with alkali; 5. Heat with Benedict’s and becomes red/orange (precipitate); Amylase 6. Add biuret (reagent) and becomes purple/violet/mauve/lilac; 7. Add starch, (leave for a time), test for reducing sugar/absence of starch;

1. Reject heating emulsion test. 1. Accept ‘Add Sudan III and mix’. 2. Ignore cloudy. 2. Reject precipitate. 2. Accept (for Sudan III) top (layer) red. 3. Ignore details of method for Benedict’s test for this mp. 4. Accept named examples of acids/alkalis. 5. Do not credit mp5 if no attempt at mp4. 5. For ‘heat’ ignore ‘warm’/’heat gently’/’put in a water bath’ but accept stated temperatures ≥60°C. 5. Heat must be stated again, do not accept using residual heat from mp4. 5. Accept ‘do the Benedict’s test’ if full correct method given elsewhere. 5. Accept ‘sodium carbonate, sodium citrate and copper sulfate solution’ for Benedict’s but must have all three if term ‘Benedict’s’ not used. 6. Accept ‘sodium or potassium hydroxide and copper sulfate solution’ for ‘biuret’. 6. Reject heating biuret test.

23

MARK SCHEME – A-LEVEL BIOLOGY – 7402/1 – JUNE 2019

Question

Marking Guidance

10.3

1. A condensation reaction joins monomers together and forms a (chemical) bond and releases water; 2. A hydrolysis reaction breaks a (chemical) bond between monomers and uses water; 3. A suitable example of polymers and the monomers from which they are made; 4. A second suitable example of polymers and the monomers from which they are made; 5. Reference to a correct bond within a named polymer;

Mark Comments 5

Ignore reference to dimers. 3. and 4. Polymers must contain many monomers. 3. and 4: suitable examples include 

  

amino acid and polypeptide, protein, enzyme, antibody or specific example nucleotide and polynucleotide, DNA or RNA Alpha glucose and starch/glycogen Beta glucose and cellulose.

If neither specific carbohydrate example is given, allow monosaccharide/glucose and polysaccharide. 3. and 4. Reject (once) reference to triglycerides. 5. Reject reference to ester bond.

24

A-level CHEMISTRY 7405/1

Paper 1 Inorganic and Physical Chemistry Mark scheme June 2019 Version: 1.0 Final

*196A74051/MS*

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre. 2

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

AS and A-Level Chemistry Mark Scheme Instructions for Examiners 1. General The mark scheme for each question shows: • • • •

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. You should mark according to the contents of the mark scheme. If you are in any doubt about applying the mark scheme to a particular response, consult your Team Leader. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which might confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent. The use of M1, M2, M3 etc in the right-hand column refers to the marking points in the order in which they appear in the mark scheme. So, M1 refers to the first marking point, M2 the second marking point etc. 2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for a mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general ‘List’ principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of error / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the mark scheme) are not penalised.

3

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

For example, in a question requiring 2 answers for 2 marks:

3.2

Correct answers

Incorrect answers (i.e. incorrect rather than neutral)

Mark (2)

1

0

1

1

1

1

They have not exceeded the maximum number of responses so there is no penalty.

1

2

0

They have exceeded the maximum number of responses so the extra incorrect response cancels the correct one.

2

0

2

2

1

1

2

2

0

3

0

2

The maximum mark is 2

3

1

1

The incorrect response cancels out one of the two correct responses that gained credit.

3

2

0

Two incorrect responses cancel out the two marks gained.

3

3

0

Comment

Marking procedure for calculations Full marks should be awarded for a correct numerical answer, without any working shown, unless the question states ‘Show your working’ or ‘justify your answer’. In this case, the mark scheme will clearly indicate what is required to gain full credit. If an answer to a calculation is incorrect and working is shown, process mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the marking scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the marking scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.4

Equations In questions requiring students to write equations, state symbols are generally ignored unless otherwise stated in the ‘Comments’ column. Examiners should also credit correct equations using multiples and fractions unless otherwise stated in the ‘Comments’ column.

4

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

3.5

Oxidation states In general, the sign for an oxidation state will be assumed to be positive unless specifically shown to be negative.

3.6

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.7

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term or if the question requires correct IUPAC nomenclature.

3.8

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.9

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

3.10

Marking crossed out work Crossed out work that has not been replaced should be marked as if it were not crossed out, if possible. Where crossed out work has been replaced, the replacement work and not the crossed out work should be marked.

3.11

Reagents and Observations The command word “Identify”, allows the student to choose to use either the name or the formula of a reagent in their answer. In some circumstances, the list principle may apply when both the name and the formula are used. Specific details will be given in mark schemes. The guiding principle is that a reagent is a chemical which can be taken out of a bottle or container. Failure to identify complete reagents will be penalised, but follow-on marks (e.g. for a subsequent equation or observation) can be scored from an incorrect attempt (possibly an incomplete reagent) at the correct reagent. Specific details will be given in mark schemes. For example, no credit would be given for • the cyanide ion or CN– when the reagent should be potassium cyanide or KCN; • the hydroxide ion or OH– when the reagent should be sodium hydroxide or NaOH; 5

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

• the Ag(NH3)2+ ion when the reagent should be Tollens’ reagent (or ammoniacal silver nitrate). In this example, no credit is given for the ion, but credit could be given for a correct observation following on from the use of the ion. Specific details will be given in mark schemes. In the event that a student provides, for example, both KCN and cyanide ion, it would be usual to ignore the reference to the cyanide ion (because this is not contradictory) and credit the KCN. Specific details will be given in mark schemes. •

•

3.12

Where an observation is required, the answer must state clearly what is seen, heard or detected by smell. Statements such as ‘carbon dioxide is given off’ or ‘barium sulfate is formed’ would not gain marks as observations. Credit would be given for descriptions such as ‘effervescence’ or ‘fizzing’ or for ‘white precipitate or white ppt’. Where relevant, ‘no visible change’ is an acceptable answer, but the statement ‘no observation’ would not gain a mark.

Organic structures Where students are asked to draw organic structures, unless a specific type is required in the question and stated in the mark scheme, these may be given as displayed, structural or skeletal formulas or a combination of all three as long as the result is unambiguous. In general • Displayed formulae must show all of the bonds and all of the atoms in the molecule, but need not show correct bond angles. • Skeletal formulae must show carbon atoms by an angle or suitable intersection in the skeleton chain. Functional groups must be shown and it is essential that all atoms other than C atoms are shown in these (except H atoms in the functional groups of aldehydes, secondary amines and N-substituted amides which do not need to be shown). • Structures must not be ambiguous, e.g. 1-bromopropane should be shown as CH3CH2CH2Br and not as the molecular formula C3H7Br which could also represent the isomeric 2-bromopropane. • Bonds should be drawn correctly between the relevant atoms. This principle applies in all cases where the attached functional group contains a carbon atom, e.g nitrile, carboxylic acid, aldehyde and acid chloride. The carbon-carbon bond should be clearly shown. Wrongly bonded atoms will be penalised on every occasion. (see the examples below) • The same principle should also be applied to the structure of alcohols. For example, if students show the alcohol functional group as C ─ HO, they should be penalised on every occasion. • Latitude should be given to the representation of C ─ C bonds in alkyl groups, given that CH3─ is considered to be interchangeable with H3C─ even though the latter would be preferred. • Similar latitude should be given to the representation of amines where NH2─ C will be allowed, although H2N─ C would be preferred. • Poor presentation of vertical C ─ CH3 bonds or vertical C ─ NH2 bonds should not be penalised. For other functional groups, such as ─ OH and ─ CN, the limit of tolerance is the half-way position between the vertical bond and the relevant atoms in the attached group.

6

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

By way of illustration, the following would apply.

C

CH3 C

C

C

C

CH3CH2

CH3

allowed

OH

allowed

not allowed

OH

not allowed

NH2

not allowed NO2

C

NH2 C

NH2

NH2

allowed

CN

allowed C

C

allowed

CHO

not allowed

C

C

C COOH

not allowed

COCl

C

CHO

CHO

not allowed

C COOH

not allowed

C

not allowed

C

COOH

CN

not allowed

allowed

not allowed

not allowed

not allowed

C COCl

not allowed

not allowed

• Representation of CH2 by C−H2 will be penalised • Some examples are given here of structures for specific compounds that should not gain credit (but, exceptions may be made in the context of balancing equations)

CH3COH

for

ethanal

CH3CH2HO OHCH2CH3 C2H6O

for for for

ethanol ethanol ethanol

CH2CH2 CH2.CH2 CH2:CH2

for for for

ethene ethene ethene

• Each of the following should gain credit as alternatives to correct representations of the structures. CH2 = CH2

for

ethene, H2C=CH2

7

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

CH3CHOHCH3

for

propan-2-ol, CH3CH(OH)CH3

• In most cases, the use of “sticks” to represent C ─ H bonds in a structure should not be penalised. The exceptions to this when “sticks” will be penalised include • when a displayed formula is required • when a skeletal structure is required or has been drawn by the candidate 3.13

Organic names As a general principle, non-IUPAC names or incorrect spelling or incomplete names should not gain credit. Some illustrations are given here. Unnecessary but not wrong numbers will not be penalised such as the number ‘2’ in 2-methylpropane or the number ‘1’ in 2-chlorobutan-1-oic acid. but-2-ol

should be butan-2-ol

2-hydroxybutane

should be butan-2-ol

butane-2-ol

should be butan-2-ol

2-butanol

should be butan-2-ol

ethan-1,2-diol

should be ethane-1,2-diol

2-methpropan-2-ol

should be 2-methylpropan-2-ol

2-methylbutan-3-ol

should be 3-methylbutan-2-ol

3-methylpentan

should be 3-methylpentane

3-mythylpentane

should be 3-methylpentane

3-methypentane

should be 3-methylpentane

propanitrile

should be propanenitrile

aminethane

should be ethylamine (although aminoethane can gain credit)

2-methyl-3-bromobutane

should be 2-bromo-3-methylbutane

3-bromo-2-methylbutane

should be 2-bromo-3-methylbutane

3-methyl-2-bromobutane

should be 2-bromo-3-methylbutane

2-methylbut-3-ene

should be 3-methylbut-1-ene

difluorodichloromethane

should be dichlorodifluoromethane

8

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

3.14

Organic reaction mechanisms Curly arrows should originate either from a lone pair of electrons or from a bond. The following representations should not gain credit and will be penalised each time within a clip.

H3C

Br

H3C

..

H3C

Br

_ : OH

. . Br

_ OH

..

For example, the following would score zero marks H H3C

C

HO

H

Br

When the curly arrow is showing the formation of a bond to an atom, the arrow can go directly to the relevant atom, alongside the relevant atom or more than half-way towards the relevant atom. In free-radical substitution • the absence of a radical dot should be penalised once only within a clip. • the use of half-headed arrows is not required, but the use of double-headed arrows or the incorrect use of half-headed arrows in free-radical mechanisms should be penalised once only within a clip Mechanisms may be drawn using structural, displayed or skeletal formulae. However, if skeletal formulae are used in mechanisms such as elimination reactions (from halogenoalkanes or alcohols) or in electrophilic substitutions, any hydrogen atoms that are essential to a step in the mechanism must be shown. 3.15 Extended responses For questions marked using a ‘Levels of Response’ mark scheme: Level of response mark schemes are broken down into three levels, each of which has a descriptor. Each descriptor contains two statements. The first statement is the Chemistry content statement and the second statement is the communication statement. Determining a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the Chemistry content descriptor for that level. The descriptor for the level indicates the qualities that might be seen in the student’s answer for that level. If it meets the lowest level, then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer.

9

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level. Once the level has been decided, the mark within the level is determined by the communication statement: • •

If the answer completely matches the communication descriptor, award the higher mark within the level. If the answer does not completely match the communication descriptor, award the lower mark within the level.

The exemplar materials used during standardisation will help you to determine the appropriate level. There will be an exemplar in the standardising materials which will correspond with each level of the mark scheme and for each mark within each level. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the exemplar to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the exemplar. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other chemically valid points. Students may not have to cover all of the points mentioned in the indicative content to reach the highest level of the mark scheme. The mark scheme will state how much chemical content is required for the highest level. An answer which contains nothing of relevance to the question must be awarded no marks.

For other extended response answers: Where a mark scheme includes linkage words (such as ‘therefore’, ‘so’, ‘because’ etc), these are optional. However, a student’s marks for the question may be limited if they do not demonstrate the ability to construct and develop a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. In particular answers in the form of bullet pointed lists may not be awarded full marks if there is no indication of logical flow between each point or if points are in an illogical order. The mark schemes for some questions state that the maximum mark available for an extended response answer is limited if the answer is not coherent, relevant, substantiated and logically structured. During the standardisation process, the Lead Examiner will provide marked exemplar material to demonstrate answers which have not met these criteria. You should use these exemplars as a comparison when marking student answers.

10

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

01.1

01.2

Answers

Additional Comments/Guidelines

Cs+(g) + e– + I(g) 1 Cs(s) + 2 I2(s)

1 1

79 + x + 376 –314 = –337 +585

1

Top line Lower line

So enthalpy change = 107 (kJmol-1)

Allow I mark for -107 (kJmol-1) Allow answer to 2sf or more

1

(Almost/Mostly) purely/ perfectly ionic

If ionic not mentioned, allow no/little covalent bonding/character Penalise references to atoms/molecules Ignore electronegativity

1

M1 Correct entropy change value

1

M2 equation or equation with numbers

1

M3 for converting units: ∆S into kJK-1 mol-1 or

1

01.3

1 2

M1 ΔS = [(82.8 + x 117) –130 ] = 11.3 (J K–1 mol–1) M2 ΔG = ΔH – TΔS 01.4

M3 ΔG = 337 - 298× 11.3 x 10–3

OR

337000 – 298 x 11.3

M4 ΔG = (+)334 kJ mol–1 or 334000 J mol–1

11

Mark

∆H into Jmol-1

M4 answer with correct units Any negative answer loses M4

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019 Question

Answers M1: P dissolved or put in/added to a solvent M2: (injected through) a needle or nozzle or capillary and at high voltage/4000 volts or high potential

02.1

M3: Gains a proton / H+ M4: P + H+  PH+

02.2

Additional Comments/Guidelines

Mark

M1: Allow named solvent eg water or methanol M2: Allow needle is positively charged

1 1

M3: Not atoms gain a proton M3: Could be scored from equation

1

Correct equation gains M3 and M4 Ignore state symbols

1

1

555 M1 V = d/t or = 1.22 x 105 ms-1

Recall this equation

1

Rearrangement to give m

1

M3: Calculation of m.

1

M4 = 2.81 x 10-25 x L = 0.169

M4: Allow M3 x L

1

M5

M5: Allow M4 x 1000 169 only scores 5 marks Allow answers to 2 significant figures or more ignore units

M2 m = 2KE v2 or

or

2 x 2.09 x 10–15 (1.22 x 105)2

M2 m = 2KE x t2 or 2 x 2.09 x 10–15 x (1.23 x 10 –5)2 d2 1.502 02.3

M3 m =

2.8(1) x 10-25 (kg)

0.169 x 1000 = 169.(2)

1

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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

13

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019 Question

03.1

03.2

03.3

Answers

Additional Comments/Guidelines

Mark

Repeating pattern/trends (of physical or chemical properties/reactions)

Allow named property Penalise groups

Bromine/Br

Not Br2 Accept Kr or Krypton

1

Potassium /K

If Na or Rb lose M1 but allow access to M2 and M3 If other incorrect elements 0/3

1

Allow same shielding

1 1

Smallest number of protons/smallest nuclear charge Similar shielding / same number of shells (as other elements in period 4)

03.4

Amphoteric

03.5

As2O3 + 6 Zn + 12 HNO3 → 2 AsH3 + 6 Zn(NO3)2 + 3 H2O

1

1 Accept multiples

1

14

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019 Question

Answers

Additional Comments/Guidelines

[Fe(OH)3(H2O)3]

Mark 1

M2: Allow red-brown 04.1

Brown

1

2 [Fe(H2O)6]3+ + 3 CO32–

 2 [Fe(OH)3(H2O)3] + 3 CO2 + 3 H2O

M3: Allow correct equations with Na2CO3 M3: Ignore state symbols

[FeCl4]− 04.2

1 1

–

−

[Fe(H2O)6]3+ + 4 Cl  [FeCl4] + 6 H2O

M2: Allow correct equations with HCl

1

04.3

(XS) Zn (in acid or HCl or H2SO4)

Allow KI/potassium iodide

1

04.4

[Fe(OH)2(H2O)4] green

1 1

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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

This question is marked using levels of response. Refer to the Mark Scheme Instructions for Examiners for guidance on how to mark this question.

Indicative Chemistry content

Level 3 5–6 marks All stages are covered and the description of each stage is generally correct and virtually complete. Answer is communicated coherently and shows a logical progression from stage 1 to stage 2 and stage 3

1a octahedral or 6 co-ordinate diagram

Stage 1: shapes of complexes

1b tetrahedral or square planar or 4 co-ordinate diagram

6

Stage 2: cis/ trans isomerism (or E-Z or geometric) Answer is illustrated using diagrams of at least 2 specific examples of pairs of cobalt or platinum complex isomers.

2a cis/trans isomerism in either square planar and/or octahedral complexes

04.5

Level 2 3–4 marks All stages are covered but the description of each stage may be incomplete or may contain inaccuracies OR two stages are covered and the explanations are generally correct and virtually complete. Answer is mainly coherent and shows progression from stage 1 to stage 2 and/or stage 3.

2b Diagrams showing cis and trans isomerism in a square planar complex 2c Diagrams showing cis and trans isomerism in both isomers of octahdedral complexes eg draw cis and trans M(H2O)4(OH)2 or [M(NH3)4(H2O)2]2+

Answer is illustrated using diagrams of at least 1 specific example of a pair of cobalt or platinum complex isomers. Level 1 1–2 marks Two stages are covered but the description of each stage may be incomplete or may contain inaccuracies, OR only one stage is covered but the explanation is generally correct and virtually complete. Answer includes isolated statements and these are presented in a logical order. Answer is illustrated using at least 1 appropriate diagram or formula.

Stage 3: optical isomerism 3a optical isomerism / non superimposable mirror images in octahedral complexes -

3b occurs with a specific bidentate ligands eg.C2O42 or NH2CH2CH2NH2 3c draw both optical isomers of eg [M(NH2CH2CH2NH2)3 ]2+

Level 0 0 marks Insufficient correct chemistry to gain a mark. 16

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

17

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

Answers

Additional Comments/Guidelines

Mark

NaCl + H2SO4  NaHSO4 + HCl

Allow 2 NaCl + H2SO4  Na2SO4 + 2 HCl

1

Proton donor

Allow (Bronsted-Lowry) acid

1

2 NaBr + 2 H2SO4  Na2SO4 + SO2 + Br2 + 2 H2O Or 2 NaBr + 3 H2SO4  2 NaHSO4 + SO2 + Br2 + 2 H2O Or 2 H+ + 2 Br - + H2SO4  SO2 + Br2 + 2 H2O Or 4 H+ + 2 Br - + SO42  SO2 + Br2 + 2 H2O

Ignore 2 NaBr + H2SO4  Na2SO4 + 2 HBr Ignore NaBr + H2SO4  NaHSO4 + HBr

1

brown gas or brown fumes or orange gas or orange fumes

Do not accept yellow solid Ignore fizzing and misty fumes

1

Oxidising agent

Allow electron acceptor Ignore acid / proton donor

1

05.1

05.2

05.3 05.4

05.5

1

(+)5 and -1 Is oxidised and reduced

Allow undergoes disproportionation Allows gains and loses electrons

1

D AgBr E Ag2CO3 F CO2 2 Ag+ + CO32-  Ag2CO3 AgBr + 2 NH3  Ag(NH3)2+ + Br –

Ignore state symbols

1 1 1 1 2

Or  Ag(NH3)2Br One mark for Ag(NH3)2+ and 1 mark for equation If D = AgCl, then allow 2 marks for – AgCl + 2 NH3  Ag(NH3)2+ + Cl

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MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

19

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

Answers M1 Amount of S2O32- = 9.00 x 0.0800 1000

06.1

Additional Comments/Guidelines

= 7.20 x 10– 4 mol

Mark 1

(From equations mol S2O32- = mol Cu2+ ) M2 Amount of Cu2+ in 25 cm3 = 7.20 x 10– 4 mol

M2 = answer to M1 (1:1 ratio)

1

M3 Amount of Cu2+ in 250 cm3 = 7.20 x 10– 4 x10 = 7.20 x 10– 3 mol

M3 = M2 x 10

1

M4 Mass of copper = 7.20 x 10– 3 mol x 63.5 = 0.457 g

M4 = M3 x 63.5

1

M5 mass = 0.985 g

M5 converting 985mg to g

1

M6 % Cu = 0.457 x 100 = 46.4 % 0.985

M6 is for the answer to 3 sf

1

Allow % Cu = 457 x 100 = 46.4 % for M5 and M6 985 Allow (M4 x1000)/985 x 100 for M5 and M6

06.2

Use more of the alloy Use a lower concentration of the thiosulfate solution/lower mass of Na2S2O3 to make solution

1 1

06.3

Oxidizing agent

Allow electron acceptor

1

06.4

1s2 2s2 2p6 3s2 3p6 3d9

Do not allow [Ar]3d9

1

06.5

Full (3)d (sub)shell or (3)d10 No (d-d) transitions possible/ cannot absorb visible/white light

M2 is dependent on M1 Ignore reflects visible/white light

1 1 20

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019 M1: n = (5.00/253.8) = 0.0197 mol

Allow 254 If 126.9 or 127 used lose M1 only

M2: T = 458 K and P = 100 000 Pa

06.6

M3: V = nRT or 0.0197 x 8.31 x 458 P 100 000

1 1

or 7.50 x 10-4 (m3)

M3 If rearrangement incorrect can only score M1 and M2

M4: V =750 (cm3)

1 1

6

M4: Allow M3 x 10 M4: Allow 749

21

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

07.1

Answers

Additional Comments/Guidelines

Mark

Moles SO2 eqbm (=6.08/64.1 = 0.0949) so moles O2 eqbm = 0.0474 Mass of oxygen (= 0.0474 x 32(.0)) = 1.52 g

Allow 0.0475 Allow M1 x 32

1 1

M1: Mole fraction SO3 = 0.15 Mole fraction SO2 = 0.57 Mole fraction O2 = 0.28

Accept fractions for M1

1

Do not accept [ ] λ = mole fraction

1

M3 is for rearrangement with or without numbers If incorrect rearrangement allow correct M1 and M2 only

1

( = (λSO2)2 P2 x (λO2) P ) (λSO3)2 P2

M2: Kp = (pSO2)2 x (pO2) (pSO3)2 07.2 M3: P = Kp x (λSO3)2 (λSO2)2 x (λO2) M4 P = 1.91 x 105 (Pa)

or

Kp x (0.15)2 (0.57)2 x (0.28)

Allow range 1.88 x 105 to 1.94 x 105

1

M1 Kp is higher at higher temperature or converse 07.3

07.4

1

M2 At higher temperature more dissociation occurs / more products are formed / equilibrium shifts to the right/forward direction

M2: Allow converse arguments M2 dependent on M1.

1

(√3.94 x 104 Pa) = 198.5

Allow 198 – 198.5 (answer is 198.49)

1

Pa1/2 or Pa0.5

If √7.62 x 105 = 873 then lose M1 but allow M2

1

22

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

Answers

Additional Comments/Guidelines

O C2H5

ᵟ–

Hᵟ+

ᵟ–

08.1

O

ᵟ

Mark

M1 two lone pairs on each O atom and δ+ and δ− on each H-O bond

1

M2 dotted/broken line shown between lone pair on one molecule and the correct H on another

1

M3 O…….H–O in straight line, dependent on M2

1

Ignore any partial charges on C–H or C–O bonds

+

C2H5

H

For straight line in M3, allow a deviation of up to 15° If a different molecule containing hydrogen bonding due to O–H bond drawn (e.g. methanol, water) or an incorrect attempt at the structure of ethanol, then maximum of 2 marks (i.e. only penalise if would score all three marks otherwise) 1

Hydrogen bonds (between ethanol molecules)

08.2

(permanent) dipole-dipole OR van der Waals force (between methoxymethane molecules)

Allow vdW

1

Hydrogen bonds are stronger/est intermolecular force

Allow more energy to break/overcome hydrogen bonding Allow converse arguments

1

23

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

O F

F

P Cl

Cl

Cl

Cl

F

F

POCl3: allow any shape showing 1 double bond between P and O and 3 P-Cl bonds

1

ClF4-: allow any shape showing 4 Cl-F bonds and 2 lone pairs

1

08.3 (distorted) Tetrahedral Square planar 90o

1 1 1

24

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

Question

09.1

Answers

Additional Comments/Guidelines

Mark -

M1: [H+] = [OH-]

M1: accept equal number/amounts of H+ and OH

M2: [H+] (= 10-pH) = 2.138 x 10-7

M2: allow 2.14 x 10

1

M3: Kw = [H+]2 or (2.138 x 10-7)2

M3: allow (M2)2

1

M4: Kw = 4.57 x 10-14

M4: allow 4.58 x 10 M4 is dependent on (an answer)2 in M3

View with Figure X (ie graph) as they may show working there.

Ignore calculations of mols of salt or acid

M1: Determines volume at half equivalence (=

M2: pH = 4.80 to 4.95

-7

1

-14

19.5 cm3 ) 2

= 9.75 (cm3)

M1: Allow reading on graph to be from 19.4 to 19.7 giving M1 = 9.7 to 9.85

1

1

1 M2: Reads off pH at half equivalence

09.2 M3: Ka (= 10-pH) = 10-4.9 = 1.26 x 10–5

Alternative method M1: pH of pure acid = 3 M2: Ka = (10–3)2 / 0.080 M3: = 1.25 x 10–5 09.3

cresolphthalein

M3: Allow 1.12 x 10–5 to 1.58 x 10-5 M3: Allow 2sf or more

1

Alternative M1 if calculation incorrect: Allow pH = pKa or [H+] = Ka at half equivalence

1

25

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/1 – JUNE 2019

M1:

Ka = [H+][X-] or [HX]

[H+] = Ka x [HX] [X-]

M1: allow [H+] = Ka x [acid] [salt]

1

M2: amount of HX = 0.0500 mol M3: amount of HX after addn of KOH = 0.05 - 3 x 10-4 = 0.0497 mol

M3: = M2 – 3 x 10-4

M5:

[H+] =( 1.41 x 10-5 x 0.0497 ) = 5.04(15) x 10-5 0.0139

M6: pH = -log10 5.04(15) x 10-5

=

4.30

1 1

M4: amount of KX after addn of KOH = 0.0136 + 3 x 10-4 = 0.0139 mol 09.4

1

1 Answer to 2 decimal places

1

If no attempt at M3 and M4 max 2 marks If M3 or M4 attempted using 3 x 10-4 max 4 (M1, M2, M3 or M4 and M6)

Allow inverse expression 09.5

1 ratio

remains (almost) constant

26

A-level BIOLOGY 7402/2 Paper 2 Mark scheme June 2019 Version: 1.0 Final

*196A7402/2/MS*

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

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MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Mark scheme instructions to examiners 1. General The mark scheme for each question shows: • • • •

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent.

2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for the same mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of errors / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the ‘Comments’ column of the mark scheme) are not penalised.

3

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

3.2

Marking procedure for calculations Full marks can be given for a correct numerical answer, without any working shown. However, if the answer is incorrect, mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.4

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the mark scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the mark scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.5

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term.

3.6

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.7

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

4

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark

Comments

1. (Colonisation by) pioneer species;

4 max

2. Accept example of change e.g. forms soil/humus/organic matter/nutrients.

2. Pioneers/species/organisms change the environment/habitat/conditions/factors; 3. (Environment becomes) less hostile for other/new species

2. Must convey idea of change being caused by pioneers/species/organisms

OR 01.1

(Environment becomes) more suitable for other/new species OR

3. Accept previous species out-competed.

(Environment becomes) less suitable for previous species;

4.Ignore increase in genetic diversity.

4. Change/increase in diversity/biodiversity; 5. (To) climax community;

01.2

0.155;

1

Accept standard form e.g. -2 15.5 x 10

1. Answer of 180/178/177.5 = 2 marks;; 01.3

2. Incorrect answer but shows use of numbers 57 and 127 (with decimal points in any position) within the calculation = 1 mark;

2

1. Ignore any numbers following 177.5

5

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark 3

1. Change in (sequence of) amino acid(s)/primary structure; 02.1

Comments 1. Reject amino acids are formed. 1. Reject amino acids code.

2. Change in hydrogen/ionic/disulfide bonds;

3. Reject active site.

0

3. Alters tertiary/3 structure;

3. Ignore quaternary. 3. Ignore 3D.

1. Produce healthy (red blood) cells

3

OR Produce (normal) polypeptide/haemoglobin; 2. No sickle/faulty/SCD (red blood) cells (produced)

1. Accept produce ‘normal’/non-SCD cells.

OR 02.2

No defective polypeptide/haemoglobin;

1. and 3. Ignore type of stem cell e.g. pluripotent.

3. Stem/marrow cells (continuously) divide/replicate OR

3. Differentiate is not equivalent to divide/replicate.

Less chance of rejection (from brother/sister);

(For gene therapy) 1. No destruction of bone marrow OR No destruction of stem cells; 2. Donors are not required; 3. Less/no chance of rejection (own stem cells); 02.3

3 max

Max 2 marks for marking points 1, 2 and 3 1. Accept no destruction of faulty bone marrow unless context indicates this is against gene therapy.

4. Sickle/faulty (red blood) cells still produced

2. Stating ‘only own cells used’ is not equivalent.

5. Immune response against genetically modified cells/virus

5. Accept ‘virus could cause problems’ or ‘risk(s) with virus’.

(Against gene therapy)

OR Long-term effect not known (as is new treatment) OR Virus could cause side effects;

6

1. Produce only healthy (red blood) cells is only equivalent to mark point 1.

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark

Comments

2 max

Mark in pairs 1 and 2 or 3 and 4.

1. Tip produces IAA;

1 and 2. Accept auxin for IAA.

2. Affects concentration of IAA

1. Ignore contains/stores IAA.

OR 03.1

Affects (shoot) length/growth/elongation;

1. Accept affects amount of IAA.

3. Mitosis/division occurs in shoot tips;

2. Accept affects independent variable.

4. Affects (shoot) length/growth/elongation;

2 and 4. Ignore affects results. 2

1.Ignore photosynthesis. 1. Ignore aerobic/anaerobic (respiration).

1. For respiration; 03.2

1. Reject glucose used in photosynthesis.

2. Provide ATP/energy (for growth);

2. Reject produce energy. 2. Do not credit photosynthesis provides ATP. 2 1. To prevent/reduce evaporation; 2. (Which) alters concentration of (IAA) solution 03.3

OR

2. Accept auxin for IAA.

(Which) alters water potential;

1. Ignore contamination. 3

1. Increase in IAA concentration the higher/greater the mean (change in) length; 03.4

1. Accept evaporation of (IAA/glucose) ‘solution’.

2. (High) IAA stimulates cell elongation; 3. In roots, growth/elongation less/inhibited;

1, 2 and 3. Accept auxin for IAA. 3. Accept decrease in (mean) change in length but reject ‘decreases length’ on its own.

7

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

3. Accept ‘opposite results or ‘negative correlation’. 1 03.5

8

0.4 and 39.6;

Both numbers required and must be in order shown.

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark

Comments

1. Lower (force of contraction) in mouse/B (than control/100%) below 29 °C

4 max

1. Accept any temperature below 29 °C for mouse/B or any specified temperature below 26.5 °C for rabbit/D.

OR Lower (force of contraction) in rabbit/D (than control/100%) below 26.5 °C;

2. Accept any temperature above 29 °C for mouse/B or any temperature above 26.5 °C for rabbit/D.

2. Higher (force of contraction) in mouse/B (than control/100%) above 29 °C OR Higher (force of contraction) in rabbit/D (than control/100%) above 26.5 °C; 04.1

OR

1. and 2. Accept 27 °C for 26.5 °C and accept 28.5 °C for 29 °C.

No other organism/species used;

3. Accept only two animals/species used.

3. Only (used) mouse and rabbit

4. Body temperature of mouse/rabbit higher (than temperatures investigated);

4. Accept body temperature of mouse/rabbit not known

5. Only used one/0.5 pH (below typical pH) OR

7. Ignore SD.

(Should) use more pH values; 6. (Used) isolated muscle tissue; 7. No stats test to see if (difference is) significant; 1. (Less/No) tropomyosin moved from binding site OR Shape of tropomyosin not changed so binding site not exposed/available; 2. (Fewer/No) actinomyosin bridges formed; 3. Myosin head does not move OR 04.2

Myosin does not pull actin (filaments) OR (Less/No) ATP (hydrol)ase (activation);

3

1 and 2. Reject active site only once. 1. Ignore troponin. 2. Accept actin and myosin do not bind. 3. Reject ATP synthase. Do not penalise reference to calcium rather than calcium ions. Credit all mark points even if context relates to what happens when calcium ions are present.

9

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

1. Regenerates/produces NAD OR oxidises reduced NAD; 2. (So) glycolysis continues; 04.3

2

1. Reject NADP and any reference to FAD. 1. Accept descriptions of oxidation e.g. loss of hydrogen. 2. Accept description of glycolysis e.g. glucose to pyruvate. 2. Accept ‘for oxidising/converting triose phosphate to pyruvate’.

10

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance 1. (Attaches to receptors on target cells and) activates/stimulates enzymes;

Mark 2

2. Glycerol/amino acids/fatty acids into glucose;

Comments 1. Reject ‘produces enzymes’. 2. Reject ‘glucagon converts’ as context suggests enzyme action.

05.1

2. Ignore lipids/fats/proteins but reject glycogen. 2. Reject occurs in pancreas. 1. Correct answer of 3.24 = 2 marks;;

2

2. Incorrect but multiplies by 34 (with decimal point in any position) = 1 mark 05.2

OR Incorrect but shows sequence 324 = 1 mark OR 3.2 = 1 mark; 1. (More) insulin binds to receptors;

2

2. (Stimulates) uptake of glucose by channel/transport proteins

2. Reject active transport.

OR

05.3

2. Accept activates enzymes for glycogenesis.

Activates enzymes which convert glucose to glycogen;

2. Accept carrier proteins or GLUT 4 for channel proteins. 2. Accept insulin stimulates addition of channel proteins in membranes.

1. Less/no ATP is converted to cyclic AMP/cAMP; 2. Less/no kinase is activated; 3. Less/no glycogen is converted to glucose 05.4

OR Less/no glycogenolysis;

3

If no indication of less/no for any of the mark points award max 2 marks. Accept all marks in context of adrenaline. Ignore gluconeogenesis.

11

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark

06.1 GgXRXr ;

06.2

1

Comments 1. Accept alleles in any order. 1. Accept GgRr with alleles in any order.

If it were recessive all flies of 3 and 4 would be grey OR 3 and 4 produce 9/black (fly)

1

OR Grey parents produce black (fly); 06.3

Mark in pairs 1 and 2 or 3 and 4.

1. (Fly) 3 (and 4) produce 9/black (fly) OR

2. Accept allele for grey colour would be passed on by 3.

(Fly) 9 would not be black OR (Fly) 9 would be grey OR Grey parents/male produce black female (fly); 2. (Fly) 3 would pass dominant allele to 9; 3. (Fly) 2 (and 1) produce 5/grey (fly)

2 max

OR Black female produces grey male

4. Accept allele for black colour would be passed on by 2.

OR (Fly) 5 could not be grey OR (Fly) 5 would be black; 4.(Fly) 5 would receive recessive allele from 2; 06.4

1. GgXrXr and ggXRY; R

r

R

r

r

r

2. GgX X , ggX X , GgX Y and ggX Y; 3. Grey-bodied red-eyed female, black-bodied red-eyed female, grey-bodied white-eyed

3

1 and 2. Accept the following alternative notations for sexlinked crosses e.g. for mp 1

male, black-bodied white-eyed male and ratio

Ggrr and ggRY or

1 : 1 : 1 : 1;

Ggrr and gg R- or Ggrr and ggR i.e. space for Y

12

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

If 1, 2 and 3 incorrect allow one mark for correct gametes from incorrect dihybrid parental genotypes.

chromosome. 2 and 3. Accept any order of genotypes and phenotypes. 3. Accept sequence of phenotypes does not need to mirror genotypes but must be correct. 3. Accept alternative ratios in correct proportions e.g. 4:4:4:4

06.5

1. Correct answer of 48% = 2 marks;;

1. Accept 0.48 for 1 mark.

2. Incorrect answer but shows understanding that 2pq = heterozygous/carriers = 1 mark 2

OR Incorrect answer but shows understanding that 2

1 – (p + q

2

)

2. Accept understanding of 2pq by using a calculation involving 2 x two different numbers.

= heterozygous/carriers = 1 mark;

13

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question

Marking Guidance

Mark

07.1

 NADP, ADP, Pi and water;

1

07.2

1. Chlorophyll absorbs light

2

OR

2. Electron/s are lost

2. Accept electrons go to electron transport/carrier chain for ‘electrons lost’.

OR (Chlorophyll) becomes positively charged;

Ink and (leaf) pigments would mix

1. Ignore photosystems. 2. Ignore site/molecule from where electrons are lost.

Light excites/moves electrons in chlorophyll;

07.3

Comments

1

OR (With ink) origin/line in different position OR (With pencil) origin/line in same position OR (With pencil) origin/line still visible; 07.4

1. Level of solvent below origin/line;

2

2. Remove/stop before (solvent) reaches top/end;

1. Reject water or any named aqueous solution. 1. Accept named organic solvent.

07.5

Accept any answer in range of 0.58 to 0.62;

1

Accept 0.58 or 0.62. Ignore any numbers which follow numbers in range.

07.6

(Absorb) different/more wavelengths (of light) for photosynthesis;

1

Accept wider/larger range of wavelengths. Accept frequency for wavelength. Accept lightdependent reaction /photophosphorylation /photoionisation for photosynthesis.

14

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question 08.1

Marking Guidance 1. (Short) single strand of DNA;

Mark

Comments

2

2. Bases complementary (with DNA/allele/gene); 08.2

1. Restriction endonuclease/enzyme;

2

2. (Cuts DNA at specific) base sequence

2. Accept palindromic sequence.

OR (Breaks) phosphodiester bonds OR (Cuts DNA) at recognition/restriction site; 08.3

(So DNA) probe binds/attaches/anneals;

1

08.4

1. (Lane 1 has DNA fragments) of known sizes/lengths;

2

2. Compare (position of viral fragment/s); 08.5

3, 4, 5 with these numbers in any sequence;

1

All three numbers required. Reject if more than three numbers given.

15

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

Question 09.1

Marking Guidance 1. Use a grid

Mark 5

OR Divide area into squares/sections;

Comments 1. Accept use of tape measures/map/area with coordinates. 1. Accept Belt transect.

2. Method of obtaining random coordinates/numbers e.g. calculator/computer/random numbers table/generator;

2. If transect method used accept quadrats at regular intervals or current mark point 2.

3. Count number/frequency in a quadrat/section;

3. Accept % cover in quadrat/section.

4. Large sample and calculate mean/average number (per quadrat/section);

3. Ignore amount/abundance.

5. Valid method of calculating total number of sundews, e.g. mean number of plants per quadrat/section/m2 multiplied by number of quadrats/sections/m2 in marsh;

4. Accept large sample and calculate mean %. 4. Accept large sample and method of calculating mean. 4. Accept many/multiple for large sample but ignore several. 4.If a specific number is given it must be 10 or more. 5. Do not allow ‘scale up’ without further qualification. 5. Do not award if % cover determined.

09.2

1. Digestion/breakdown of proteins; 2. Provides amino acids OR (Sundew can) produce a named (organic) nitrogen-containing compound e.g. proteins, amino acids, DNA, ATP;

16

2 max

Mark in pairs 1 and 2, or 3 and 4. Ignore carbohydrates, lipids or named carbohydrate/ lipid.

3. Digestion/breakdown of named (organic) phosphate-containing compound e.g. DNA, RNA;

2. Ignore if nitrate or ammonium ions given as products.

4. Provides named (organic) phosphate-containing product e.g. nucleotides

4. Accept phosphate as a named product.

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

OR (Sundew can) produce a named phosphatecontaining compound e.g. ATP, DNA;

Question 10.1

Marking Guidance 1. (Refers to) saltatory conduction

Mark 3

OR (Nerve) impulses/depolarisation/ions pass to other neurones

Comments 1. Accept suitable description that refers to (transmission) from node to node (of Ranvier). 1 and 2. Accept action potentials for impulses.

OR

1. Accept action potential for depolarisation.

Depolarisation occurs along whole length (of axon);

1, 2 and 3. Reject first mark awarded if answer refers to messages/signals for impulses. Reject even if impulse/s also referred to.

2. (Nerve) impulses slowed/stopped; 3. (Refers to) neuromuscular junction OR (Refers to) sarcolemma;

10.2

1. Slower/fewer impulse(s) along sympathetic/parasympathetic (pathway/neurones);

3

1. Accept action potentials for impulses. 1.Reject no impulses. 1, 2 and 3. Ignore ‘information’ but reject first mark awarded if answer refers to messages/signals for impulses. Reject even if impulse/s also referred to.

2. (Impulses) from cardiac centre OR (Impulses) from medulla; 3. To SAN;

10.3

1. It/DNA is complementary to (m)RNA; 2. Binds to mRNA (for huntingtin); 3. Prevents translation;

3

1. Accept (transcription) results in complementary (m)RNA. Ignore miRNA/siRNA/transcriptional factors. 3. Ignore transcription.

17

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

10.4

1. Small sample size

2

max

OR

3. Accept huntington for huntingtin. Ignore miRNA/siRNA/transcriptional factors.

Only 46; 2. Only four-months OR short period (of trial); 3. Huntingtin/protein reduced OR Huntingtin/protein still produced OR Huntingtin/protein not removed; 4. Allele/gene/mutation/mRNA (for Huntington’s) still present OR (May be) temporary OR Drug treatment repeated; 5. Brain already damaged OR Brain damage may continue; 10.5

1. (Drug/DNA) will directly/quickly reach brain

2

2. Reject protein is digested.

OR

2. Ignore location of enzymes.

(Cerebrospinal) fluid bathes the brain;

2. Accept Drug/DNA denatured.

2. (Drug/DNA) not destroyed by acid OR (Drug/DNA) not digested (by enzymes); 10.6

1. (Increased) methylation of DNA/gene/allele; 2. Inhibits/prevents transcription; OR

18

2 max

Mark in pairs but if no mark credited allow one mark for any reference to transcription or gene expression being affected.

MARK SCHEME – A LEVEL BIOLOGY – 7402/2 – JUNE 2019

1. Reject acetylation of DNA. 3. Decreased methylation of DNA/gene/allele; 4. Stimulates/allows transcription;

Accept gene expression for transcription but ignore gene switched on/off.

OR

Ignore methylation of histones.

5. Decreased acetylation of histone(s);

Accept DNA-histone complex as equivalent to histone(s).

6. Inhibits transcription;

.

OR 7. Increased acetylation of histone(s); 8. Stimulates/allows transcription;

19

A-level CHEMISTRY 7405/2 Paper 2 Organic and Physical Chemistry Mark scheme June 2019 Version: 1.0 Final

*196A74052/MS*

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

AS and A-Level Chemistry Mark Scheme Instructions for Examiners 1. General The mark scheme for each question shows:    

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. You should mark according to the contents of the mark scheme. If you are in any doubt about applying the mark scheme to a particular response, consult your Team Leader. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which might confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent. The use of M1, M2, M3 etc in the right-hand column refers to the marking points in the order in which they appear in the mark scheme. So, M1 refers to the first marking point, M2 the second marking point etc. 2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for a mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general ‘List’ principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of error / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the mark scheme) are not penalised.

3

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

For example, in a question requiring 2 answers for 2 marks:

3.2

Correct answers

Incorrect answers (i.e. incorrect rather than neutral)

Mark (2)

1

0

1

1

1

1

They have not exceeded the maximum number of responses so there is no penalty.

1

2

0

They have exceeded the maximum number of responses so the extra incorrect response cancels the correct one.

2

0

2

2

1

1

2

2

0

3

0

2

The maximum mark is 2

3

1

1

The incorrect response cancels out one of the two correct responses that gained credit.

3

2

0

Two incorrect responses cancel out the two marks gained.

3

3

0

Comment

Marking procedure for calculations Full marks should be awarded for a correct numerical answer, without any working shown, unless the question states ‘Show your working’ or ‘justify your answer’. In this case, the mark scheme will clearly indicate what is required to gain full credit. If an answer to a calculation is incorrect and working is shown, process mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the marking scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the marking scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.4

Equations In questions requiring students to write equations, state symbols are generally ignored unless otherwise stated in the ‘Comments’ column. Examiners should also credit correct equations using multiples and fractions unless otherwise stated in the ‘Comments’ column.

4

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

3.5

Oxidation states In general, the sign for an oxidation state will be assumed to be positive unless specifically shown to be negative.

3.6

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.7

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term or if the question requires correct IUPAC nomenclature.

3.8

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.9

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

3.10

Marking crossed out work Crossed out work that has not been replaced should be marked as if it were not crossed out, if possible. Where crossed out work has been replaced, the replacement work and not the crossed out work should be marked.

3.11

Reagents and Observations The command word “Identify”, allows the student to choose to use either the name or the formula of a reagent in their answer. In some circumstances, the list principle may apply when both the name and the formula are used. Specific details will be given in mark schemes. The guiding principle is that a reagent is a chemical which can be taken out of a bottle or container. Failure to identify complete reagents will be penalised, but follow-on marks (e.g. for a subsequent equation or observation) can be scored from an incorrect attempt (possibly an incomplete reagent) at the correct reagent. Specific details will be given in mark schemes. For example, no credit would be given for  the cyanide ion or CN– when the reagent should be potassium cyanide or KCN;  the hydroxide ion or OH– when the reagent should be sodium hydroxide or NaOH; 5

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

 the Ag(NH3)2+ ion when the reagent should be Tollens’ reagent (or ammoniacal silver nitrate). In this example, no credit is given for the ion, but credit could be given for a correct observation following on from the use of the ion. Specific details will be given in mark schemes. In the event that a student provides, for example, both KCN and cyanide ion, it would be usual to ignore the reference to the cyanide ion (because this is not contradictory) and credit the KCN. Specific details will be given in mark schemes. Where an observation is required, the answer must state clearly what is seen, heard or detected by smell. Statements such as ‘carbon dioxide is given off’ or ‘barium sulfate is formed’ would not gain marks as observations. Credit would be given for descriptions such as ‘effervescence’ or ‘fizzing’ or for ‘white precipitate or white ppt’. Where relevant, ‘no visible change’ is an acceptable answer, but the statement ‘no observation’ would not gain a mark.

3.12

Organic structures Where students are asked to draw organic structures, unless a specific type is required in the question and stated in the mark scheme, these may be given as displayed, structural or skeletal formulas or a combination of all three as long as the result is unambiguous. In general  Displayed formulae must show all of the bonds and all of the atoms in the molecule, but need not show correct bond angles.  Skeletal formulae must show carbon atoms by an angle or suitable intersection in the skeleton chain. Functional groups must be shown and it is essential that all atoms other than C atoms are shown in these (except H atoms in the functional groups of aldehydes, secondary amines and N-substituted amides which do not need to be shown).  Structures must not be ambiguous, e.g. 1-bromopropane should be shown as CH3CH2CH2Br and not as the molecular formula C3H7Br which could also represent the isomeric 2-bromopropane.  Bonds should be drawn correctly between the relevant atoms. This principle applies in all cases where the attached functional group contains a carbon atom, e.g nitrile, carboxylic acid, aldehyde and acid chloride. The carbon-carbon bond should be clearly shown. Wrongly bonded atoms will be penalised on every occasion. (see the examples below)  The same principle should also be applied to the structure of alcohols. For example, if students show the alcohol functional group as C ─ HO, they should be penalised on every occasion.  Latitude should be given to the representation of C ─ C bonds in alkyl groups, given that CH3─ is considered to be interchangeable with H3C─ even though the latter would be preferred.  Similar latitude should be given to the representation of amines where NH2─ C will be allowed, although H2N─ C would be preferred.  Poor presentation of vertical C ─ CH3 bonds or vertical C ─ NH2 bonds should not be penalised. For other functional groups, such as ─ OH and ─ CN, the limit of tolerance is the half-way position between the vertical bond and the relevant atoms in the attached group.

6

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

By way of illustration, the following would apply.

C

CH3 C

C

C

C

CH3CH2

CH3

allowed

OH

allowed

not allowed

OH

not allowed

NH2

not allowed NO2

C

NH2 C

NH2

NH2

allowed

CN

allowed

COOH

C

C

allowed

CHO

not allowed

C

C

C COOH

not allowed

COCl

C

CHO

CHO

not allowed

C COOH

not allowed

C

not allowed

C

CN

not allowed

allowed

not allowed

not allowed

not allowed

C COCl

not allowed

not allowed

 Representation of CH2 by CH2 will be penalised  Some examples are given here of structures for specific compounds that should not gain credit (but, exceptions may be made in the context of balancing equations)

CH3COH

for

ethanal

CH3CH2HO OHCH2CH3 C2H6O

for for for

ethanol ethanol ethanol

CH2CH2 CH2.CH2 CH2:CH2

for for for

ethene ethene ethene

 Each of the following should gain credit as alternatives to correct representations of the structures. CH2 = CH2 CH3CHOHCH3

for for

ethene, H2C=CH2 propan-2-ol, CH3CH(OH)CH3 7

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

 In most cases, the use of “sticks” to represent C ─ H bonds in a structure should not be penalised. The exceptions to this when “sticks” will be penalised include  when a displayed formula is required  when a skeletal structure is required or has been drawn by the candidate. 3.13

Organic names As a general principle, non-IUPAC names or incorrect spelling or incomplete names should not gain credit. Some illustrations are given here. Unnecessary but not wrong numbers will not be penalised such as the number ‘2’ in 2-methylpropane or the number ‘1’ in 2-chlorobutan-1-oic acid.

8

but-2-ol

should be butan-2-ol

2-hydroxybutane

should be butan-2-ol

butane-2-ol

should be butan-2-ol

2-butanol

should be butan-2-ol

ethan-1,2-diol

should be ethane-1,2-diol

2-methpropan-2-ol

should be 2-methylpropan-2-ol

2-methylbutan-3-ol

should be 3-methylbutan-2-ol

3-methylpentan

should be 3-methylpentane

3-mythylpentane

should be 3-methylpentane

3-methypentane

should be 3-methylpentane

propanitrile

should be propanenitrile

aminethane

should be ethylamine (although aminoethane can gain credit)

2-methyl-3-bromobutane

should be 2-bromo-3-methylbutane

3-bromo-2-methylbutane

should be 2-bromo-3-methylbutane

3-methyl-2-bromobutane

should be 2-bromo-3-methylbutane

2-methylbut-3-ene

should be 3-methylbut-1-ene

difluorodichloromethane

should be dichlorodifluoromethane

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

3.14

Organic reaction mechanisms Curly arrows should originate either from a lone pair of electrons or from a bond. The following representations should not gain credit and will be penalised each time within a clip.

H3C

Br

H3C

..

H3C

Br

_ : OH

. . Br

_ OH

..

For example, the following would score zero marks H H3 C

C

HO

H

Br

When the curly arrow is showing the formation of a bond to an atom, the arrow can go directly to the relevant atom, alongside the relevant atom or more than half-way towards the relevant atom. In free-radical substitution  the absence of a radical dot should be penalised once only within a clip.  the use of half-headed arrows is not required, but the use of double-headed arrows or the incorrect use of half-headed arrows in free-radical mechanisms should be penalised once only within a clip Mechanisms may be drawn using structural, displayed or skeletal formulae. However, if skeletal formulae are used in mechanisms such as elimination reactions (from halogenoalkanes or alcohols) or in electrophilic substitutions, any hydrogen atoms that are essential to a step in the mechanism must be shown. 3.15 Extended responses For questions marked using a ‘Levels of Response’ mark scheme: Level of response mark schemes are broken down into three levels, each of which has a descriptor. Each descriptor contains two statements. The first statement is the Chemistry content statement and the second statement is the communication statement. Determining a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the Chemistry content descriptor for that level. The descriptor for the level indicates the qualities that might be seen in the student’s answer for that level. If it meets the lowest level, then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer.

9

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level. Once the level has been decided, the mark within the level is determined by the communication statement: • If the answer completely matches the communication descriptor, award the higher mark within the level. • If the answer does not completely match the communication descriptor, award the lower mark within the level. The exemplar materials used during standardisation will help you to determine the appropriate level. There will be an exemplar in the standardising materials which will correspond with each level of the mark scheme and for each mark within each level. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the exemplar to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the exemplar. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other chemically valid points. Students may not have to cover all of the points mentioned in the indicative content to reach the highest level of the mark scheme. The mark scheme will state how much chemical content is required for the highest level. An answer which contains nothing of relevance to the question must be awarded no marks.

For other extended response answers: Where a mark scheme includes linkage words (such as ‘therefore’, ‘so’, ‘because’ etc), these are optional. However, a student’s marks for the question may be limited if they do not demonstrate the ability to construct and develop a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. In particular answers in the form of bullet pointed lists may not be awarded full marks if there is no indication of logical flow between each point or if points are in an illogical order. The mark schemes for some questions state that the maximum mark available for an extended response answer is limited if the answer is not coherent, relevant, substantiated and logically structured. During the standardisation process, the Lead Examiner will provide marked exemplar material to demonstrate answers which have not met these criteria. You should use these exemplars as a comparison when marking student answers.

10

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

Answers

Additional Comments/Guidelines

Br-(CH2)6-Br + 4 NH3 → H2N-(CH2)6-NH2

M1 both organic compounds correct (not molecular formulae)

+ 2 NH4Br

Br

01.1

OR

+ 4 NH3

Br



Allow one correct structural formula and the other correct molecular formula of type XC6H12X

NH2

+ 2 NH4Br

H2N

:NH3

M1 arrow & lone pair

Mark

M2 balanced Or with structural formulae, Br(CH2)6NH2 etc

NH 2 Br

2

3

Allow SN1 Penalise incorrect partial charges in M1

H

NH2 N

H

H

:NH 3

01.2

M2 structure

M3 arrow NH3 removal need not be shown but penalise Br- removal

allow

Impurity

allow

N H

11

(or as structural formula)

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

01.3

M1

Stage 1 reagent

KCN or NaCN

Not HCN this loses M1 and M2 Any mention of acid loses M1 & M2

1

M2

Stage 1 condition

aqueous alcohol

M2 dependent on correct M1 (allow condition if only CN- ions)

1

M3

Stage 2 reagent & condition H2 and Ni or Pt or Pd

M3 only accessible if a cyanide is used in stage 1

1

Allow LiAlH4 (in dry ether) – acidic/aqueous = CE, but allow followed by acid. NOT NaBH4 NOT Sn/HCl or Fe/HCl Ignore heat and reflux and pressure Apply list principle to incorrect reagents/conditions In 3-aminopentane 01.4

01.5

12

Allow converse for ammonia +

Lone pair on N more available or Lone pair on N accepts H better

Or greater stability of protonated N

1

because of alkyl electron pushing /inductive effect

Mark independently

1

No carbon (atom is) attached to 4 different groups

Allow central carbon has two alkyl groups Allow symmetrical molecule

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

Answers Thermometer and bung in flask with bulb level with side arm.

02.1

Must be cross section diagram with no gaps at joints

Condenser jacket with water in at bottom and out at top.

Liquids are immiscible 02.2

02.3

Additional Comments/Guidelines

Liquid goes clear / not cloudy

Mark 1 1

Allow don’t mix, forms two layers (stated or implied) Allow it is insoluble Ignore density or reference to solutions

1

Ignore colourless

1

13

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Via moles Amount cyclohexanol (= 14.4/100) = 0.144 mol

Via mass Amount cyclohexanol (= 14.4/100) = 0.144 mol

Via volume Amount cyclohexanol (= 14.4/100) = 0.144 mol

Mass cyclohexene formed = 4.15 x 0.81 = 3.36 g

Mass cyclohexene formed = 4.15 x 0.81 = 3.36 g

Mass of cyclohexene expected (= 0.144 × 82.0 = 11.808 g ) OR M1 × 82

02.4

amount cyclohexene obtained

mass of cyclohexene expected

volume of cyclohexene expected

(= 3.36/82.0 = 0.0410 mol )

(= 0.144 × 82.0

(= 11.808/0.810 = 14.577cm3 )

OR M2/82.0

OR =

%Yield = 0.0410 x 100 0.144 OR M3 x 100 M1

%Yield = 3.36 x 100 11.808 OR M2 x 100 M3

%Yield = 4.15 x 100 14.577

= 28.5% (must be 3 sf)

= 28.5% (must be 3 sf)

= 28.5% (must be 3 sf)

= 11.808 g )

M1 × 82.0

M1

M2

M3

OR M2/0.810

OR

M4

4.15 x 100 M3 M5

Only award M5 if answer is to 3sf and follows some attempt at % yield calculation in M4

Br

M1 arrow

02.5

Br

M2 structure

+ Br

: Br M3 arrow & lone pair on bromide

14

Lose M1 if Full charges on BrBr OR Wrong partial charges on BrBr OR Arrow is to Br+ ion (formed in a preliminary step)

Any C shown in the ring must have the correct number of hydrogens attached to score M2

3

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

Answers

%mass mol

03.1

÷ smaller

x5

C 88.2 88.2 12

H 11.8 11.8 1

=7.35

=11.8

7.35 7.35

11.8 7.35

=1 =5

1.61 =8

Empirical formula = molecular formula C5H8

H C

OR

Mark

M1 for amounts 7.35 and 11.8

1

M2 for process dividing M1 by smaller

1

M3 for answer C5H8 only

1

Allow alternatives

M4 (must be branched) H 2C

Additional Comments/Guidelines

CH 3

CH3

C CH 2

CH2

C

1

C CH3

HCCCH(CH3)2

Must be skeletal

1

M2 can only be this and is independent of M1

1

OR

03.2 Buta-1,3-diene

15

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

CH2

H C

H

Must show trailing bonds Ignore brackets and n

1

C H2 C

03.3

Allow skeletal – with brackets

Must be E ‘trans’ Mark independently Restricted rotation about the C=C or double bond

Allow lack of rotation/no rotation/limited rotation about the C=C or double bond

1

Ignore different groups on each carbon of the C=C double bond

03.4

16

Carbon Carbon bonds are non polar or (too) strong or not attacked by nucleophiles Or Carbon Carbon bonds cannot be hydrolysed

Allow carbon chains ….. OR Bonds between repeating units ……… Ignore CH bonds

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

04.1

04.2

Answers

Additional Comments/Guidelines

Expt 2 3.2 × 104

Mark 1

Both needed

4

Expt 3 3.2 × 10 P order = 1

These answers only, not consequential on 4.1 Allow if 4.1 blank.

Q order = 2

1 1

( Rate = k[R]2[S] 2 ) k = Rate/[R]2[S] 2 04.3

OR 1.20 × 103/(1.00 × 102)2( 2.45 × 102) 2

M1 for rearrangement

M1

k = 19992 = 2.00 × 104

M2 for answer (Allow 1.99 × 104)

M2

Units mol-3 dm9 s1

Allow conseq units for their expression in M1

M3

17

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

Answers Regrettably, this question contained a typographical error

05

18

which affected some students’ ability to answer it. All students were awarded full marks for this question.

Additional Comments/Guidelines

Mark

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

Answers

Additional Comments/Guidelines

If incorrect reagent then no marks

Must be a single test-tube reaction M1 Reagent: acidified potassium dichromate OR K2Cr2O7/H2SO4 OR K2Cr2O7/H+ OR acidified K2Cr2O7

06.1

M2 ….-1-ol

(orange to) green solution OR goes green

M3 ….-2-ol

no (visible/observed) reaction/change or NVR or stays orange

OR M1 Reagent: acidified potassium manganate(VII) or KMnO4/H2SO4 OR KMnO4/H+ OR acidified KMnO4 M2….-1-ol (purple to) colourless solution OR goes colourless M3….-2-ol no (visible/observed) reaction/change or stays purple

A C

OR

1 1 1

For acidified potassium manganate(VII): If “manganate” or “(potassium manganate(IV)” or incorrect formula or no acid, penalise M1 but mark on Credit alkaline / neutral KMnO4 for possible full marks but M2 gives brown precipitate or solution goes green

CH3 CH 2 CH 3

H3C

Br

06.2

For acidified potassium dichromate: if “dichromate” or “(potassium) dichromate(IV)” or incorrect formula or no acid, penalise M1 but mark on - ignore dichromate described as “yellow” or “red”.

B

Br H 3C

Mark

C

CH2 Br

Br

OR

2

Br

Br

Br

Br

19

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Allow Kekulé structures C

D

Br

Br

06.3

Br

F A 06.4

G E

20

Penalise missing aromatic ring each time Br

Br

2

Br

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

G

07.1

Answers Cyclopentanone

Allow cyclopentan -1-one but no other numbers Ignore spaces, commas and hyphens

This question is marked using Levels of Response. Refer to the Mark Scheme Instructions for Examiners for guidance.

Indicative Chemistry content

Level 3

1a) Y has a higher bp 1b) Y has H-bonds between molecules and X has dip-dip imf 1c) More energy required to overcome H-bonds Mention of covalent bond breaking loses 1c

5-6 marks

All stages are covered and each stage is generally correct and virtually complete. Answer is well structured with no repetition or irrelevant points. Accurate and clear expression of ideas with no errors in use of technical terms.

07.2

Level 2 3-4 marks

Level 1 1-2 marks

1

Stage 1: boiling points

All stages are covered but stage(s) may be incomplete or 13 may contain inaccuracies OR two stages are covered and Stage 2: C NMR are generally correct and virtually complete. 2a) Both have 3 peaks/absorptions in their 13C NMR 2b) X has peaks at 20-50 OR 190-220ppm Answer shows some attempt at structure 2c) Y has peaks at 50-90 OR 90-150ppm Ideas are expressed with reasonable clarity with, perhaps, (Ignore peaks at 5-40ppm - present in both) some repetition or some irrelevant points. Some minor errors in use of technical terms

Stage 3: ir

Two stages are covered but stage(s) may be incomplete or may contain inaccuracies OR only one stage is covered but is generally correct and virtually complete.

3a) X has a peak (for C=O) at 1680-1750 cm-1 3b) Y has peak (for OH) at 3230-3550 cm-1 OR peak (for C=C) at 1620-1680 cm-1 3c) They would have different fingerprint regions (below 1500 cm-1)

Answer includes isolated statements and these are presented in a logical order. Answer may contain valid points which are not clearly linked. Errors in the use of technical terms. 0 mark

Ma rk

Additional Comments/Guidelines

6

Insufficient correct chemistry to gain a mark.

.

21

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

Answers

Additional Comments/Guidelines Allow minor spelling errors e.g. electrophillic or subsitution Ignore nitration

Electrophilic substitution both words needed G 08.1

08.2

+ 3 H2 ……………… + 2 H2O

Allow 6 [H]

O

08.3

CH 3

C

Cl

R

N

H

H

O CH 3

]

C

1

1

M1 for structure of ion including 2 charges (+ on N must be correct in both cases if drawn twice)

M2 for 3 arrows and lp

[

Mark

NHR

M2 for 3 arrows and lp on O - may be scored in two steps

2

RNH 2 M1 for structure

Corrosive OR forms strong acid/HCl (fumes) OR vulnerable to hydrolysis OR dangerous (to use)

Ignore use of RNH2 to remove H+ in M2, but penalise use of Cl

Allow anhydride is less corrosive OR does not form strong acid fumes OR less vulnerable to hydrolysis OR ethanoyl chloride is more expensive

08.4

Allow reacts violently / extremely exothermic / extremely vigorous Ignore toxic / harmful / hazardous

22

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

O

O

C O

C O

1 CH 3

CH3

08.5 OR HN

NH2

08.6

+ CH3COONH4

CH 3 C O

………………………..

+ 2H2O

Via moles Amount paracetamol = 250 × 103 / 151.0 = 1655.6 mol OR (250 × 103) / M1 08.7

-

+

1

OR via mass

M1 Mr paracetamol = 151(.0) M2

-

Allow CH3COO / CH3CO2 and NH4 Allow NH4CH3COO

(= amount hydroquinone used)

M1 Mr paracetamol = 151(.0) So 110 g hydroquinone forms 151 g paracetamol

M1

M2 Mass hydroquinone needed 250 × 110 / 151.0 OR 250 × 110 / M1

M2

M3 Mass hydroquinone = 1655.6 × 110.0 = 182119 g = 182 kg OR correct answer to M2 × 110.0 / 1000

= 182 kg

M3

Min 2sf If Mr values used wrong way round can score M2

23

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

09.1 09.2 G 09.3

Answers

Allow concentrations of 5M or higher Allow conc sulfuric or conc strong alkalis

1

Using ninhydrin or ultraviolet light

Allow I2 (vapour)

1 1

7 or seven

OR Some amino acids have the same Rf value or have the same affinity with the first/either solvent

24

Mark

Conc HCl

Some of the amino acids did not separate/dissolve with the first/either solvent 09.4

Additional Comments/Guidelines

Not amino acids have different Rf values in different solvents

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

10.1

Answers

Additional Comments/Guidelines

Mark

(COOH)2 = C2H2O4= 90

M1

118 - 90 = 28 OR C2H4

M2

C4H6O4

Must be molecular formula Structural formula can score M1 & M2

M3

M1 for answer (to 3sfs min)

M1

Amount H2A in 25cm3 = 1.177 × 103 mol

M2 = 0.5 × M1

M2

Amount H2A in 250 cm3 = 1.177 × 102 mol

M3 = M2 × 10

M3

Mass =

M4 =

M4

Amount NaOH = (21.60 × 103 ) × 0.109 = 2.3544 × 103 mol 10.2

G 10.3

10.4

1.39 g (Must be 3sf)

1

4 or four

CH3

CH 3

CH 3

C

C

OH

OH

1

CH 3

OR

OH

OH

The precise (relative molecular) masses are different or wtte 10.5

answer to (M3 × 118) and must be 3sf

Allow Mr are different to 2 or more or several dp Ignore different molecular formula Ignore accuracy Penalise fragments

1

25

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

Answers

Additional Comments/Guidelines

Mark

O CH3CH2

C

11.1

CH3CH2

+ CH3CH2OH

C

M1 Structure of ester (allow C2H5CO2C2H5)

1

M2 propanoic acid formula (allow C2H5CO2H) and correctly balanced equation

1

O

O CH3

CH2

C

+ CH3CH2COOH OCH2CH3

O

1

Ethyl propanoate only M3 Ethyl propanoate only O H 2C

O

C

C 17H 31

O

11.2

HC

O

C

C 17H 31

26

O

C

C 17H 33

Allow –O2C–, –OOC–, –OCO– 1 Not –CO2– , –COO– 1

M2 for two C17H31 and one C17H33 in any order top to bottom

O H 2C

M1 for all except C17H3x (i.e. all to the left of the dotted line)

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

9

11.3

M1 for skeleton

12

C9 – C14 shown with double bonds in the correct place

M2 for both Z correct Independent marks

Other representations include

Ignore structure beyond carbon 14

1

1

If hydrogens shown or not skeletal can only score M2 12

12 12 9

9

9

11.4

C19H34O2 + 26½ O2

19 CO2 + 17 H2O

11.5

Absorption in spectrum at 2350 cm-1 does not correspond to data booklet value of 1680 – 1750 cm-1 or for C=O bonds in organic compounds)

Allow 53/2 or all doubled Allow would expect a peak at 1680 – 1750 cm-1

1

C=O Bonds in CO2 absorb infrared radiation (of 2350 cm-1)

11.6

IR radiation emitted by the earth does not escape (from the atmosphere) OR This energy is transferred to other molecules in the atmosphere by collisions (so all atmosphere is warmed)

1

1 Ignore IR reflected

1

27

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019 Question

Answers

Additional Comments/Guidelines

M1 scored for the 2 H ‘bonds’ between A and T

1

M2 scores for the 3 H ‘bonds’ between C and G

1

Lose 1 for each extra ‘bond’ H bonds must be linear 12.1

Penalise the use of full bonds instead of dashed lines once only Ignore lone pairs and partial charges even if wrong

28

Mark

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

12.2

M1 scored for correct selection of cytosine and associated sugar

1

M2 scored for selection of correct (upper) phosphate

1

M1 & M2 can be scored with one ‘ring’ Allow ring either side of the top O of either phosphate

If wrong base circled, can score M2 for correct phosphate conseq to their base, i.e. top left, Thymine it’s the upper phosphate top right, Adenine it’s the lower phosphate bottom right, Guanine it’s the lower phosphate

12.3

(Complementary means the two strands must have base sequences) that match (all) A to T and C to G

Ignore reference to (hydrogen) bonding

1

29

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/2 – JUNE 2019

Question

Answers M1 for structure of 2-methylbutanal

Additional Comments/Guidelines

Mark

Allow C2H5 for CH3CH2 1

M2 for 2 curly arrows and lp on hydride, i.e. CH 3 CH3 CH 2

C

O

O C

H

H

H H

13.1

1 O

OR

Penalise M2 for wrong partial charges on C=O

Explanation: M3 H ion / nucleophile is attracted to + C

Ignore product 1

M4 electron rich C=C

1

M5 H ion / nucleophile is repelled by C=C

1

OR C=C only attacked by/reacts with electrophiles

NOT dichromate

13.2

Tollens’ (reagent) OR ammoniacal silver nitrate OR description of making Tollens’ Silver mirror/ppt OR black solid / precipitate / deposit

30

Fehling’s/ Benedict’s (solutions)

red solid / precipitate (allow orange or brown)

For Tollens’ reagent: for M1 ignore either AgNO3 or [Ag(NH3)2+] or “the silver mirror test” on their own, or “Tolling’s reagent”, but mark on

1

For Fehling’s/Benedict’s solution:

1

for M1 Ignore Cu2+(aq) or CuSO4 or “Fellings” on their own, but mark on

A-level BIOLOGY 7402/3 Paper 3 Mark scheme June 2019 Version: 1.0 Final

*196A7402/3/MS*

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

2

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Level of response marking instructions Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme.

Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content.

Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks.

3

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Mark scheme instructions to examiners 1. General The mark scheme for each question shows: • • • •

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent.

2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for the same mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of errors / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the ‘Comments’ column of the mark scheme) are not penalised.

4

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

3.2

Marking procedure for calculations Full marks can be given for a correct numerical answer, without any working shown. However, if the answer is incorrect, mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.4

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the mark scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the mark scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.5

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term.

3.6

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.7

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

5

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

Marking Guidance

Mark

Comments

1. High blood/hydrostatic pressure;

3 max

Ignore references to podocytes

2. Two named small substances pass out eg water, glucose, ions, urea;

1. Ignore references to afferent and efferent arterioles 1. Ignore ‘increasing/higher blood pressure’ as does not necessarily mean high

3. (Through small) gaps/pores/fenestrations in (capillary) endothelium; 4. (And) through (capillary) basement membrane;

01.1

2. Accept correct named ions 2. Accept mineral ions/minerals 2. Accept amino acids/small proteins 2. Ignore references to molecules not filtered 3. Accept epithelium for endothelium

01.2

01.3

6

 Glucose by facilitated diffusion and active transport and water down a water potential gradient 17.4;

1

1

Accept any number of fours after the decimal point.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

3

1. Thicker medulla means a longer loop (of Henle); 2. (The longer the loop of Henle means) increase in sodium ion concentration (in medulla)

2. Must have idea of increase/longer/m ore

OR (The longer the loop of Henle means) sodium ion gradient maintained for longer (in medulla) OR (The longer the loop of Henle means) more sodium ions are moved out (into medulla);

01.4

3. (Therefore) water potential gradient maintained (for longer), so more water (re)absorbed (from loop and collecting duct); OR More water is (re)absorbed from the loop (of Henle) / collecting duct by osmosis;

Question

Marking Guidance

Mark

3. Reject water being reabsorbed into the loop of Henle 3. Direction is important 3. Accept Ψ for water potential

Comments

7

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

1. (They use enzymes to) decompose proteins/DNA/RNA/urea;

2

2. Producing/releasing ammonia/ammonium compounds/ammonium ions;

1. Accept any named molecule containing nitrogen eg enzymes, NAD, ATP, amino acids 1. Accept digest/breakdown/hydrolyse for decompose

02.1

1. Ignore ‘nitrogen -containing compounds’ unqualified 2. Accept (they) perform ammonification 2. Accept named ammonium compound 3

Principle is 1. Named apparatus

Accept any valid method, for example 1. Use of colorimeter; 02.2

2. Measure the absorbance/transmission (of light); 3. Example of how method can be standardised eg same volume of water, zeroing colorimeter, same wavelength of light, shaking the sample;

2. What is measured 3. Standardisation of method 1. Reject calorimeter 2. Reject if samples are filtered unless filtering to remove debris 2. Accept descriptions 3. Ignore references to calibration curves

8

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

Marking Guidance

Valve A

Mark 1

(Left) atrioventricular 03.1

Comments Reject right side in either context Accept mitral/bicuspid for Valve A.

Chamber B

Reject tricuspid for Valve A

Left ventricle;

Ignore AV for Valve A Accept any two suitable safety precautions for 1 mark, eg;

1 max

Disinfect bench/equipment

Ignore take care with scalpel/scissors or keep away from fingers

Cover any cuts

Ignore goggles

Use a sharp scalpel/scissors Wash hands/wear gloves 03.2

Cut away from self/others/on a hard surface Safe disposal 1. Pressure in (left) atrium is higher than in ventricle/B causing valve to open;

2 1. and 2. Ignore pressure in front of/behind valve

OR (When) pressure above valve is higher than below valve it opens; 03.3

1. and 2. As long as direction of opening/closing of valve is correct, ignore ‘semi lunar’

2. Pressure in (left) ventricle/B is higher than in atrium causing valve to close;

2. Accept cords/tendons prevent valve turning inside out

OR (When) pressure in below valve is higher than above valve it closes;

1. More impulses/action potentials along sympathetic (nervous system pathway/branch); 03.4

03.5

2

1. Ignore signals/informatio n/ messages 1. Idea of more impulses/action potentials is required

1

Accept any correct rounding except 73

2. To SAN increasing the heart rate (seen in Figure 3);

73.4; (73.4375)

9

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Group to be given 1. Sugar solution (only) OR

03.6

1. Accept ‘glucose’ for sugar

A drink with sugar (and no caffeine);

1. Ignore named drinks unless qualified

Reason

1. Ignore ‘sugar’ by itself

2. To show/prove that sugar (alone) is not causing the increases (in HR) OR To show that sugar does not have an effect;

10

2

1. Ignore references to use of a placebo tablet 2. Accept ‘to see the effect of sugar’

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

04.1

Marking Guidance Substitution;

Mark 1

Comments Accept inversion or translocation Ignore ‘point mutation’

1. (VO2 max and CS activity) increased for both groups;

3 max

2. No statistical test, so do not know if differences are significant

2. Ignore standard deviation 2. Accept correct named statistical test eg t-test

OR No statistical test, so differences could be due to chance;

04.2

Max 2 marks for mark points 2, 3 and 4

3. Only 8 weeks training OR Training did not last long; 4. Might not be true for all types of training/exercise/females; In Group C: 1. Less mitochondrial replication/production; 04.3

2. Less transcription (of genes) for mitochondrial proteins/CS OR

2 1. and 2. Accept converse for Group T 2. Accept less CS/enzyme is produced

Less translation of (mRNA into) mitochondrial proteins;

11

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

For (no mark) 1. (From Figure 5 Group T have) increased CS activity for Krebs cycle; 2. (from Figure 4 Group T have) increased VO2 max so more oxygen (available) OR (from Figure 4 Group T have) increased VO2 max so more aerobic respiration OR (from Figure 4 Group T have) increased VO2max so delayed anaerobic respiration; 04.4

Against (no mark)

3 max

Max 2 marks for mark points 3, 4 and 5 Ignore any answers relating to sample size or duration of investigation Ignore ‘correlation does not mean causation’ unless qualified 2. Accept ‘less lactate’ for delayed anaerobic respiration

3. No correlation between (percentage change in) VO2 max and CS activity OR No correlation on Figure 6; 4. It might not be thymine causing the increase OR There may be other differences in the control region (of Group T) that cause the increase; 5. VO2 max/CS activity not the only measures of ability to exercise for longer;

12

4. Ignore ‘could be due to lifestyle/diet changes’ 5. Accept ideas that they did not measure ability to exercise for longer

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

Marking Guidance Accept suitable null hypothesis that includes type of light and behaviour, eg

05.1

Mark 1

The type of light has no effect on the behaviour/movement of COTS OR

Comments Ignore general null hypotheses, or example ‘there is no difference between observed and expected’

There is no difference in behaviour/movement with constant/flashing light; Accept any two factors for one mark from the list below;

1 max

Ignore humidity

Salinity / salt concentration of the water

Ignore type of coral

Temperature (of the water) 05.2

List rule applies

Ignore depth of water

Amount / distribution of food pH (of the water) Oxygen/carbon dioxide concentration Intensity/wavelength of (constant and flashing) light Yes (no mark)

3

1. Movement is away from either type/both types of light OR Negative (photo) taxis to both types/either types of light; 2. Significant movement away from constant light as p=0.02/<0.05/=2%/<5% 05.3

OR

2. and 3. Ignore ‘results’ in the context of significance or chance

Movement away from constant light is not due to chance as p=0.02/<0.05/=2%/<5%; No (no mark) 3. Movement away from flashing light is not significant as p=0.69/>0.05/=69%/>5% OR Movement away from flashing light is due to chance as p=0.69/>0.05/=69%/>5%;

13

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Correct answer of 3 hours = 2 marks;; Allow 1 mark for distance of 48 000 mm in working 05.4

1 max for answer of 185 minutes/3 hours and 5 minutes/3.09 hours 1 max for answer of 1 hour (ie answers that use 564 in their calculation);

14

2 max

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question Marking Guidance

Mark

06.1

0.1;

1

06.2

Accept answer in the range of 4.7 to 4.9;

1

1. (Trexall acts as a) competitive inhibitor

3

OR (Trexall) competes (with folic acid/substrate) for/is able to fit into/binds at active site (on dihydrofolate reductase / enzyme);

Comments

Accept 1. Reject Trexall and folic acid have the same shape

2. Accept folic acid/substrate is prevented from binding

2. Less folic acid/substrate attaches OR Fewer enzyme-substrate complexes;

06.3

3. Accept fewer/not enough nucleotides available during interphase/for semi-conservative replication/to add to (all) template strands/for transcription

3. Fewer/not enough nucleotides available for DNA replication;

Percentage change

2

1. To allow comparison as tumours may differ in volume/size (at the start of the investigation); Tumour volume 2. (As) tumours may differ in length/width/shape 06.4

OR (As) volume is (best) indication of the number of cells in tumour;

2. Accept ‘as tumours are three dimensional’ 2. Ignore answers relating to density/thickness

15

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Answer in the range 32 015.93682 to 32 045 = 2 marks

2 max

OR

Accept any suitable rounding

3.20 x 104 = 2 marks;; 06.5

Allow 1 mark for correct calculation of volume after treatment in range of 24 011.95261 to 24 034 /2.40 x 104 Allow 1 mark if student uses diameter throughout instead of radius, in range of 256 127 to 256 361/2.56 x 105 For (the use of 30 mg)

2 max

1. There is a significantly greater reduction (in tumour size with 30 mg), as SD (bars) do not overlap;

1. Accept ‘significant difference’ for ‘significantly greater reduction’

2. In some cases (with 30 mg) there was a 100% reduction in size/tumours would have been eradicated; 06.6

Accept converse arguments for all mark points.

Against (the use of 30 mg)

3. Ignore 30 mg has a lot of deviation/large standard deviation’ unqualified

3. There is too much/a lot of variation in effectiveness with 30 mg (in contrast with 20 mg); 4. (No idea of) extra cost of providing 30 mg per week; 5. (Increased risk of) side effects with higher doses;

Accept any two suitable suggestions for one mark, eg; Severity/duration of arthritis 06.7

Current/other medication Type of arthritis Weight/body mass Ethnicity

16

1 max

Reject age/health as they are directly in the stem Ignore gender/sex Ignore general answers such as diet/activity/lifestyle

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

3 max For 1. Pain decreases more with Trexall/Group R compared with the control group/Group S OR Pain decreases by 4.6 with Trexall/Group R and by 2 with the control group/Group S;

2 max for answer only giving reasons against 1. Ignore numbers stated from Table 3, eg 9.7 to 5.1 and 9.8 to 7.8

Against 06.8

2. Small sample size/only 12 people/only studied females / effects in males could be different; 3. (Mean score for severity of) pain in control group/Group S is (also) lower; 4. No statistical testing, so do not know if decrease/difference is significant; 5. Pain is (a) subjective (measurement);

3. Could be subsumed within MP1 4. Ignore ‘do not know if results are significant’ 5. Accept ‘patients might lie about pain’

17

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question 7 Level of response marking guidance Level of response marking instructions Level of response mark schemes are broken down into levels, each of which has a descriptor. The descriptor for the level shows the average performance for the level. There are marks in each level. Before you apply the mark scheme to a student’s answer read through the answer and annotate it (as instructed) to show the qualities that are being looked for. You can then apply the mark scheme.

Step 1 Determine a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the descriptor for that level. The descriptor for the level indicates the different qualities that might be seen in the student’s answer for that level. If it meets the lowest level then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer. With practice and familiarity you will find that for better answers you will be able to quickly skip through the lower levels of the mark scheme. When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level and then use the variability of the response to help decide the mark within the level, ie if the response is predominantly level 3 with a small amount of level 4 material it would be placed in level 3 but be awarded a mark near the top of the level because of the level 4 content.

Step 2 Determine a mark Once you have assigned a level you need to decide on the mark. The descriptors on how to allocate marks can help with this. The exemplar materials used during standardisation will help. There will be an answer in the standardising materials which will correspond with each level of the mark scheme. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the example to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the example. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other valid points. Students do not have to cover all of the points mentioned in the Indicative content to reach the highest level of the mark scheme. An answer which contains nothing of relevance to the question must be awarded no marks.

18

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Extended Abstract

21–25

Response shows holistic approach to the question with a fully integrated answer which makes clear links between several different topics and the theme of the question.

beyond specific

Biology is detailed and comprehensive A-level content, uses appropriate terminology, and is very well written and always clearly explained.

context

No significant errors or irrelevant material.

Generalised

For top marks in the band, the answer shows evidence of reading beyond specification requirements. Relational Integrated into a 16–20

whole

Response links several topics to the main theme of the question, to form a series of interrelated points which are clearly explained. Biology is fundamentally correct A-level content and contains some points which are detailed, though there may be some which are less well developed, with appropriate use of terminology. Perhaps one significant error and, or, one irrelevant topic which detracts from the overall quality of the answer.

Multistructural Several aspects 11–15

covered but they are unrelated

Response mostly deals with suitable topics but they are not interrelated and links are not made to the theme of the question. Biology is usually correct A-level content, though it lacks detail. It is usually clearly explained and generally uses appropriate terminology. Some significant errors and, or, more than one irrelevant topic.

Unistructural

6–10

Only one or few aspects covered

Response predominantly deals with only one or two topics that relate to the question. Biology presented shows some superficial A-level content that may be poorly explained, lacking in detail, or show limited use of appropriate terminology. May contain a number of significant errors and, or, irrelevant topics.

Unfocused

1–5

Response only indirectly addresses the theme of the question and merely presents a series of biological facts which are usually descriptive in nature or poorly explained and at times may be factually incorrect. Content and terminology is generally below A-level. May contain a large number of errors and, or, irrelevant topics.

0

Nothing of relevance or no response.

19

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Commentary on terms and statements in the levels mark scheme The levels mark scheme for the essay contains a number of words and statements that are open to different interpretations. This commentary defines the meanings of these words and statements in the context of marking the essay. Many words and statements are used in the descriptions of more than one level of response. The definitions of these remain the same throughout. Levels mark scheme word/statement Holistic A fully integrated answer which makes clear links between several different topics and the theme of the question

Definition Synoptic, drawing from different topics (usually sections of the specification) All topics relate to the title and theme of the essay; for example, explaining the biological importance of a process. When considering, for example, the importance of a process, the explanation must be at A-level standard.

Biology is detailed and comprehensive A-level content, uses appropriate terminology, and is very well written and always clearly explained.

‘Several’ here is defined as at least four topic areas from the specification covered. This means some sentences, not just a word or two. It does not mean using many examples from one topic area. Detailed and comprehensive A-level content is the specification content. Terminology is that used in the specification.

No significant errors or irrelevant material.

For top marks in the band, the answer shows evidence of reading beyond specification requirements. Response mostly deals with suitable topics but they are not interrelated and links are not made to the theme of the question.

20

Well written and clearly explained refers mainly to biological content and use of terminology. Prose, handwriting and spelling are secondary considerations. Phonetic spelling is accepted, unless examiners are instructed not to do so for particular words; for example, glucagon, glucose and glycogen. A significant error is one which significantly detracts from the biological accuracy or correctness of a described example. This will usually involve more than one word. Irrelevant material is several lines (or more) that clearly fails to address the title, or the theme of the title. An example that is relevant to the title and is not required in the specification content. The example must be used at A-level standard. Not addressing the biological theme of the essay (eg importance) at A-level standard.

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

Marking Guidance The importance of DNA as an information carrying molecule and its use in gene technologies

07.1

            

3.1.5.1 3.1.5.2 3.2.1.1 3.2.1.2 3.2.2 3.4.1 3.4.2 3.4.3 3.4.4 3.4.7 3.7.1 3.7.3 3.8.1

     

3.8.2.1 3.8.2.2 3.8.2.3 3.8.3 3.8.4.1 3.8.4.2

 3.8.4.3

Mark [25 marks]

Structure of DNA DNA replication DNA in mitochondria (and chloroplasts) Prokaryotic DNA DNA replication in interphase and binary fission DNA, genes and chromosomes DNA and protein synthesis Genetic diversity and meiosis Genetic diversity and adaptation Investigating diversity Inheritance Evolution may lead to speciation Alteration of the sequence of bases in DNA can alter the structure of proteins Most of a cell’s DNA is not translated Regulation of transcription and translation Gene expression and cancer Using genome projects Recombinant DNA technology Differences in DNA between individuals of the same species can be exploited for identification and diagnosis of heritable conditions Genetic fingerprinting

In order to fully address the question and reach the highest mark bands students must also include at least four topics in their answer, to demonstrate a synoptic approach to the essay. Students may be able to show the relevance of other topics from the specification. Note, other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard.

21

MARK SCHEME – A-LEVEL BIOLOGY – 7402/3 – JUNE 2019

Question

Marking Guidance The importance of bonds and bonding in organisms

07.2

                           

3.1.1 3.1.2 3.1.3 3.1.4.1 3.1.4.2 3.1.5.1 3.1.5.2 3.1.6 3.1.7 3.2.2 3.2.3 3.2.4 3.3.3 3.3.4.1 3.3.4.2 3.4.2 3.4.3 3.5.1 3.5.2 3.5.4 3.6.2.2 3.6.3 3.6.4.2 3.6.4.3 3.8.1 3.8.2.2 3.8.2.3 3.8.4.1

Mark [25 marks]

Monomers and polymers Carbohydrates Lipids General properties of proteins Many proteins are enzymes Structure of DNA and RNA DNA replication ATP Water – cohesion Mitosis Transport across cell membranes Cell recognition and the immune system Digestion and absorption Mass transport in animals – haemoglobin Mass transport in plants DNA and protein synthesis Mutation and meiosis Photosynthesis Respiration Nutrient cycles Synaptic transmission Skeletal muscles Control of blood glucose concentration Control of blood water potential Mutations Regulation of transcription and translation Gene expression and cancer Recombinant DNA technology

In order to fully address the question and reach the highest mark bands students must also include at least four topics in their answer, to demonstrate a synoptic approach to the essay. Students may be able to show the relevance of other topics from the specification. Note, other topics from beyond the specification can be used, providing they relate to the title and contain factually correct material of at least an A-level standard. Credit should not be given for topics beyond the specification which are below A-level standard.

22

A-level CHEMISTRY 7405/3 Paper 3 Mark scheme June 2019 Version: 1.0 Final

*196A74053/MS*

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

Mark schemes are prepared by the Lead Assessment Writer and considered, together with the relevant questions, by a panel of subject teachers. This mark scheme includes any amendments made at the standardisation events which all associates participate in and is the scheme which was used by them in this examination. The standardisation process ensures that the mark scheme covers the students’ responses to questions and that every associate understands and applies it in the same correct way. As preparation for standardisation each associate analyses a number of students’ scripts. Alternative answers not already covered by the mark scheme are discussed and legislated for. If, after the standardisation process, associates encounter unusual answers which have not been raised they are required to refer these to the Lead Assessment Writer. It must be stressed that a mark scheme is a working document, in many cases further developed and expanded on the basis of students’ reactions to a particular paper. Assumptions about future mark schemes on the basis of one year’s document should be avoided; whilst the guiding principles of assessment remain constant, details will change, depending on the content of a particular examination paper.

Further copies of this mark scheme are available from aqa.org.uk

Copyright © 2019 AQA and its licensors. All rights reserved. AQA retains the copyright on all its publications. However, registered schools/colleges for AQA are permitted to copy material from this booklet for their own internal use, with the following important exception: AQA cannot give permission to schools/colleges to photocopy any material that is acknowledged to a third party even for internal use within the centre.

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

AS and A-Level Chemistry Mark Scheme Instructions for Examiners 1. General The mark scheme for each question shows:    

the marks available for each part of the question the total marks available for the question the typical answer or answers which are expected extra information to help the examiner make his or her judgement and help to delineate what is acceptable or not worthy of credit or, in discursive answers, to give an overview of the area in which a mark or marks may be awarded.

The extra information in the ‘Comments’ column is aligned to the appropriate answer in the lefthand part of the mark scheme and should only be applied to that item in the mark scheme. You should mark according to the contents of the mark scheme. If you are in any doubt about applying the mark scheme to a particular response, consult your Team Leader. At the beginning of a part of a question a reminder may be given, for example: where consequential marking needs to be considered in a calculation; or the answer may be on the diagram or at a different place on the script. In general the right-hand side of the mark scheme is there to provide those extra details which might confuse the main part of the mark scheme yet may be helpful in ensuring that marking is straightforward and consistent. The use of M1, M2, M3 etc in the right-hand column refers to the marking points in the order in which they appear in the mark scheme. So, M1 refers to the first marking point, M2 the second marking point etc. 2. Emboldening 2.1

In a list of acceptable answers where more than one mark is available ‘any two from’ is used, with the number of marks emboldened. Each of the following bullet points is a potential mark.

2.2

A bold and is used to indicate that both parts of the answer are required to award the mark.

2.3

Alternative answers acceptable for a mark are indicated by the use of OR. Different terms in the mark scheme are shown by a / ; eg allow smooth / free movement.

3. Marking points 3.1

Marking of lists This applies to questions requiring a set number of responses, but for which students have provided extra responses. The general ‘List’ principle to be followed in such a situation is that ‘right + wrong = wrong’. Each error / contradiction negates each correct response. So, if the number of error / contradictions equals or exceeds the number of marks available for the question, no marks can be awarded. However, responses considered to be neutral (often prefaced by ‘Ignore’ in the mark scheme) are not penalised.

3

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

For example, in a question requiring 2 answers for 2 marks:

3.2

Correct answers

Incorrect answers (i.e. incorrect rather than neutral)

Mark (2)

1

0

1

1

1

1

They have not exceeded the maximum number of responses so there is no penalty.

1

2

0

They have exceeded the maximum number of responses so the extra incorrect response cancels the correct one.

2

0

2

2

1

1

2

2

0

3

0

2

The maximum mark is 2

3

1

1

The incorrect response cancels out one of the two correct responses that gained credit.

3

2

0

Two incorrect responses cancel out the two marks gained.

3

3

0

Comment

Marking procedure for calculations Full marks should be awarded for a correct numerical answer, without any working shown, unless the question states ‘Show your working’ or ‘justify your answer’. In this case, the mark scheme will clearly indicate what is required to gain full credit. If an answer to a calculation is incorrect and working is shown, process mark(s) can usually be gained by correct substitution / working and this is shown in the ‘Comments’ column or by each stage of a longer calculation.

3.3

Errors carried forward, consequential marking and arithmetic errors Allowances for errors carried forward are most likely to be restricted to calculation questions and should be shown by the abbreviation ECF or consequential in the marking scheme. An arithmetic error should be penalised for one mark only unless otherwise amplified in the marking scheme. Arithmetic errors may arise from a slip in a calculation or from an incorrect transfer of a numerical value from data given in a question.

3.4

Equations In questions requiring students to write equations, state symbols are generally ignored unless otherwise stated in the ‘Comments’ column. Examiners should also credit correct equations using multiples and fractions unless otherwise stated in the ‘Comments’ column.

4

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

3.5

Oxidation states In general, the sign for an oxidation state will be assumed to be positive unless specifically shown to be negative.

3.6

Interpretation of ‘it’ Answers using the word ‘it’ should be given credit only if it is clear that the ‘it’ refers to the correct subject.

3.7

Phonetic spelling The phonetic spelling of correct scientific terminology should be credited unless there is a possible confusion with another technical term or if the question requires correct IUPAC nomenclature.

3.8

Brackets (…..) are used to indicate information which is not essential for the mark to be awarded but is included to help the examiner identify the sense of the answer required.

3.9

Ignore / Insufficient / Do not allow Ignore or insufficient is used when the information given is irrelevant to the question or not enough to gain the marking point. Any further correct amplification could gain the marking point. Do not allow means that this is a wrong answer which, even if the correct answer is given, will still mean that the mark is not awarded.

3.10

Marking crossed out work Crossed out work that has not been replaced should be marked as if it were not crossed out, if possible. Where crossed out work has been replaced, the replacement work and not the crossed out work should be marked.

3.11

Reagents The command word “Identify”, allows the student to choose to use either the name or the formula of a reagent in their answer. In some circumstances, the list principle may apply when both the name and the formula are used. Specific details will be given in mark schemes. The guiding principle is that a reagent is a chemical which can be taken out of a bottle or container. Failure to identify complete reagents will be penalised, but follow-on marks (e.g. for a subsequent equation or observation) can be scored from an incorrect attempt (possibly an incomplete reagent) at the correct reagent. Specific details will be given in mark schemes. For example, no credit would be given for  the cyanide ion or CN– when the reagent should be potassium cyanide or KCN;  the hydroxide ion or OH– when the reagent should be sodium hydroxide or NaOH; 5

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

 the Ag(NH3)2+ ion when the reagent should be Tollens’ reagent (or ammoniacal silver nitrate). In this example, no credit is given for the ion, but credit could be given for a correct observation following on from the use of the ion. Specific details will be given in mark schemes. In the event that a student provides, for example, both KCN and cyanide ion, it would be usual to ignore the reference to the cyanide ion (because this is not contradictory) and credit the KCN. Specific details will be given in mark schemes. 3.12

Organic structures Where students are asked to draw organic structures, unless a specific type is required in the question and stated in the mark scheme, these may be given as displayed, structural or skeletal formulas or a combination of all three as long as the result is unambiguous. In general  Displayed formulae must show all of the bonds and all of the atoms in the molecule, but need not show correct bond angles.  Skeletal formulae must show carbon atoms by an angle or suitable intersection in the skeleton chain. Functional groups must be shown and it is essential that all atoms other than C atoms are shown in these (except H atoms in the functional groups of aldehydes, secondary amines and N-substituted amides which do not need to be shown).  Structures must not be ambiguous, e.g. 1-bromopropane should be shown as CH3CH2CH2Br and not as the molecular formula C3H7Br which could also represent the isomeric 2-bromopropane.  Bonds should be drawn correctly between the relevant atoms. This principle applies in all cases where the attached functional group contains a carbon atom, e.g nitrile, carboxylic acid, aldehyde and acid chloride. The carbon-carbon bond should be clearly shown. Wrongly bonded atoms will be penalised on every occasion. (see the examples below)  The same principle should also be applied to the structure of alcohols. For example, if students show the alcohol functional group as C ─ HO, they should be penalised on every occasion.  Latitude should be given to the representation of C ─ C bonds in alkyl groups, given that CH3─ is considered to be interchangeable with H3C─ even though the latter would be preferred.  Similar latitude should be given to the representation of amines where NH2─ C will be allowed, although H2N─ C would be preferred.  Poor presentation of vertical C ─ CH3 bonds or vertical C ─ NH2 bonds should not be penalised. For other functional groups, such as ─ OH and ─ CN, the limit of tolerance is the half-way position between the vertical bond and the relevant atoms in the attached group.

6

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

By way of illustration, the following would apply.

C

CH3 C

C

OH

CH3CH2

CH3

allowed

C

OH C

allowed

not allowed

not allowed

NH2

not allowed NO2

C

NH2 C

NH2

NH2

allowed

CN

allowed

COOH

C

C

allowed

CHO

not allowed

C

C

C COOH

not allowed

COCl

C

not allowed

not allowed

not allowed

C COCl

CHO

CHO

not allowed

C COOH

not allowed

C

not allowed

C

CN

not allowed

allowed

not allowed

not allowed

 Representation of CH2 by CH2 will be penalised  Some examples are given here of structures for specific compounds that should not gain credit (but, exceptions may be made in the context of balancing equations)

CH3COH

for

ethanal

CH3CH2HO OHCH2CH3 C2H6O

for for for

ethanol ethanol ethanol

CH2CH2 CH2.CH2 CH2:CH2

for for for

ethene ethene ethene

 Each of the following should gain credit as alternatives to correct representations of the structures. CH2 = CH2 CH3CHOHCH3

for for

ethene, H2C=CH2 propan-2-ol, CH3CH(OH)CH3

7

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

 In most cases, the use of “sticks” to represent C ─ H bonds in a structure should not be penalised. The exceptions to this when “sticks” will be penalised include  structures in mechanisms where the C ─ H bond is essential (e.g. elimination reactions in halogenoalkanes and alcohols)  when a displayed formula is required  when a skeletal structure is required or has been drawn by the candidate 3.13

Organic names As a general principle, non-IUPAC names or incorrect spelling or incomplete names should not gain credit. Some illustrations are given here. Unnecessary but not wrong numbers will not be penalised such as the number ‘2’ in 2methylpropane or the number ‘1’ in 2-chlorobutan-1-oic acid.

8

but-2-ol

should be butan-2-ol

2-hydroxybutane

should be butan-2-ol

butane-2-ol

should be butan-2-ol

2-butanol

should be butan-2-ol

ethan-1,2-diol

should be ethane-1,2-diol

2-methpropan-2-ol

should be 2-methylpropan-2-ol

2-methylbutan-3-ol

should be 3-methylbutan-2-ol

3-methylpentan

should be 3-methylpentane

3-mythylpentane

should be 3-methylpentane

3-methypentane

should be 3-methylpentane

propanitrile

should be propanenitrile

aminethane

should be ethylamine (although aminoethane can gain credit)

2-methyl-3-bromobutane

should be 2-bromo-3-methylbutane

3-bromo-2-methylbutane

should be 2-bromo-3-methylbutane

3-methyl-2-bromobutane

should be 2-bromo-3-methylbutane

2-methylbut-3-ene

should be 3-methylbut-1-ene

difluorodichloromethane

should be dichlorodifluoromethane

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

3.14

Organic reaction mechanisms Curly arrows should originate either from a lone pair of electrons or from a bond. The following representations should not gain credit and will be penalised each time within a clip.

H3C

Br

H3C

..

H3C

Br

_ : OH

. . Br

_ OH

..

For example, the following would score zero marks H H3 C

C

HO

H

Br

When the curly arrow is showing the formation of a bond to an atom, the arrow can go directly to the relevant atom, alongside the relevant atom or more than half-way towards the relevant atom. In free-radical substitution  the absence of a radical dot should be penalised once only within a clip.  the use of half-headed arrows is not required, but the use of double-headed arrows or the incorrect use of half-headed arrows in free-radical mechanisms should be penalised once only within a clip The correct use of skeletal formulae in mechanisms is acceptable, but where a C-H bond breaks, both the bond and the H must be drawn to gain credit. 3.15 Extended responses For questions marked using a ‘Levels of Response’ mark scheme: Level of response mark schemes are broken down into three levels, each of which has a descriptor. Each descriptor contains two statements. The first statement is the Chemistry content statement and the second statement is the communication statement. Determining a level Start at the lowest level of the mark scheme and use it as a ladder to see whether the answer meets the Chemistry content descriptor for that level. The descriptor for the level indicates the qualities that might be seen in the student’s answer for that level. If it meets the lowest level, then go to the next one and decide if it meets this level, and so on, until you have a match between the level descriptor and the answer.

9

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

When assigning a level you should look at the overall quality of the answer and not look to pick holes in small and specific parts of the answer where the student has not performed quite as well as the rest. If the answer covers different aspects of different levels of the mark scheme you should use a best fit approach for defining the level. Once the level has been decided, the mark within the level is determined by the communication statement: • If the answer completely matches the communication descriptor, award the higher mark within the level. • If the answer does not completely match the communication descriptor, award the lower mark within the level. The exemplar materials used during standardisation will help you to determine the appropriate level. There will be an exemplar in the standardising materials which will correspond with each level of the mark scheme and for each mark within each level. This answer will have been awarded a mark by the Lead Examiner. You can compare the student’s answer with the exemplar to determine if it is the same standard, better or worse than the example. You can then use this to allocate a mark for the answer based on the Lead Examiner’s mark on the exemplar. You may well need to read back through the answer as you apply the mark scheme to clarify points and assure yourself that the level and the mark are appropriate. Indicative content in the mark scheme is provided as a guide for examiners. It is not intended to be exhaustive and you must credit other chemically valid points. Students may not have to cover all of the points mentioned in the indicative content to reach the highest level of the mark scheme. The mark scheme will state how much chemical content is required for the highest level. An answer which contains nothing of relevance to the question must be awarded no marks.

For other extended response answers: Where a mark scheme includes linkage words (such as ‘therefore’, ‘so’, ‘because’ etc), these are optional. However, a student’s marks for the question may be limited if they do not demonstrate the ability to construct and develop a sustained line of reasoning which is coherent, relevant, substantiated and logically structured. In particular answers in the form of bullet pointed lists may not be awarded full marks if there is no indication of logical flow between each point or if points are in an illogical order. The mark schemes for some questions state that the maximum mark available for an extended response answer is limited if the answer is not coherent, relevant, substantiated and logically structured. During the standardisation process, the Lead Examiner will provide marked exemplar material to demonstrate answers which have not met these criteria. You should use these exemplars as a comparison when marking student answers.

10

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

Question

Answers S2O32– + 2 H+  SO2 + S + H2O

Additional Comments/Guidelines

Mark

ALLOW S2O32– + 2H3O+  SO2 + S + 3H2O ALLOW ½S2O32– + H+  ½SO2 + ½S + ½H2O ALLOW ½S2O32– + H3O+ ½SO2 + ½S + 1½H2O

01.1

1

IGNORE state symbols NOT multiples NOT if any spectator ions included (unless crossed out)

M1

acid(ic) / acidity / produces H+

M1 ALLOW low(ers) pH

1

IGNORE toxic / soluble IGNORE sulfurous / sulfuric / H2SO4 IGNORE rain IGNORE proton donor (unless qualified, e.g. reacts with water to form a proton donor) 01.2

NOT any other named acid M2

SO2 + H2O  H2SO3 / H+ + HSO3–

M2 ALLOW

SO2 + H2O  2 H+ + SO32– SO2 + 2 H2O  H3O +

ALLOW

SO2 + 3 H2O  2 H3O+ + SO32–

ALLOW multiples IGNORE state symbols

11

HSO3–

ALLOW

+

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

M1 bent shape and 2 lone pairs on O

H H

ALLOW any suitable representation of lone pairs (e.g. dots, crosses, lobes with/without dots/crosses)

M1

O

M2

104½°

M2 ALLOW 104-105°

M3

lone pairs repel more (strongly) than bond(ing) pairs

M3 ALLOW non-bonding pair for lone pair

1 1 1

ALLOW covalent bond for bond(ing) pair ALLOW shared pair for bond(ing) pair 01.3

ALLOW OH bond for bond(ing) pair ALLOW bond for bond(ing) pair NOT OH or O-H without the word bond for bond(ing) pair M4

so bond angle reduced from/less than 109½° / tetrahedral

M4 ALLOW bond angle reduced from 120° if bent with one lone pair in M1 ALLOW reduced from 109° ALLOW reduced by 2.5° per lone pair or 5° if M2 correct

12

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

This question is marked using levels of response. Refer to the Mark Scheme Instructions for examiners for guidance on how to mark this question

Indicative Chemistry content

Level 3

(1b) Puts acid or thiosulfate into container on/with cross or in colorimeter (1c) Add second reactant and start timing

5-6 marks

Level 2 3-4 marks

All stages are covered and the explanation of each stage is correct and virtually complete. (6 v 5) Answer is well structured, with no repetition or irrelevant points. Accurate and clear expression of ideas with no errors in use of technical terms. All stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR two stages covered and the explanations are generally correct and virtually complete (4 v 3) Answer has some structure. Ideas are expressed with reasonable clarity with, perhaps, some repetition or some irrelevant points. If any, only minor errors in use of technical terms.

01.4

Level 1 1-2 marks

Two stages are covered but the explanation of each stage may be incomplete or may contain inaccuracies OR only one stage is covered but the explanation is generally correct and virtually complete (2 v 1) Answer includes statements which are presented in a logical order and/or linked.

Level 0

Stage 1 Method (1a) Idea of using disappearing cross or colorimetry

Stage 2 Measurements (2a) Repeat at different temperatures (if number of temperatures stated, must be more than two) (2b) Record time, t, for cross to disappear / defined reading on colorimeter (2c) Idea of ensuring other variables (cross, volumes, concentrations) kept constant (apart from T) 6

Stage 3 Use of Results (3a) 1/t (or 1000/time, etc) is a measure of rate (3b) plot of rate (or 1/t etc) (y-axis) against T (xaxis) (can come from labelled axes on sketch) (IGNORE T against rate) (3c) sketch of plot as shown (ALLOW 3c if axes not labelled but NOT if incorrectly labelled) rate (or 1/t etc)

Insufficient correct Chemistry to warrant a mark

0 marks T 13

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

Question

Answers

O

02.1

H

O S

O

Additional Comments/Guidelines

IGNORE shape / bond angles

Mark

1

IGNORE lone pair(s) on O atoms O

NOT lone pair(s) on S atom

H

IGNORE state symbols in both equations ALLOW multiples in both equations 02.2

Equation 1: H2SO4  HSO4– + H+ / H2SO4 + H2O  HSO4– + H3O+

Equation 1:

NOT ⇌

1

Equation 2: HSO4– ⇌ SO42– + H+ / HSO4– + H2O ⇌ SO42– + H3O+

Equation 2:

NOT  𝑜𝑟 ⟷

1

ALLOW ⇋ 𝑜𝑟 ⇄ 𝑜𝑟 ⇆

14

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 M1

02.3

weigh solid and transfer using a method that allows exact mass to be known (there should be two weighings, one of which could be zeroing, and method could be by difference or with washings or directly weighed into container) 3

M2

dissolve in water in suitable container (NOT in 250 cm of water)

M3

transfer with washings into 250 cm3 volumetric/graduated flask

M4

make up to mark / 250 cm3 AND THEN shake / invert / mix

M1 IGNORE any mass quoted

1

NOT if any other solid added M2 NOT if any other solution added

1

3

M3 Reference to 250 cm can appear anywhere M4 ALLOW if conical flask used NOT if beaker used

1 1

Alternative method (M2-4) M2 in 250 cm3 volumetric/graduated flask M3 dissolve (NOT in 250 cm3 of water) M4 make up to mark / 250 cm3 AND THEN shake/invert/mix

15

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 M1 [H+] = 10–1.72

(= 0.0191 (mol dm–3))

Correct answer scores M1-5 (must be 3sf)

M2 amount NaHSO4 = 0.605/120.1 (= 5.04 x 10–3 (mol)) M3 initial [NaHSO4] = M4 𝐾a =

+

[HSO4–]

2−

[H ][SO4 ] − [HSO4 ]

or

𝐾a

–2

= M2 x 10 (= 5.04 x 10 (mol dm )) 2

=

[H+ ] − [HSO4 ]

0.01912 0.0504 − 0.0191 M5 Ka = 1.17 x 10–2 (1.15 – 1.18 x 10–2) must be 3sf 𝐾a

02.4

1 –3

Alternative method that does not subtract 0.0191: –3

–3

7.21 x 10 (7.15 – 7.26 x 10 ) scores M1-5 (where M4

𝐾a

=

1 1

0.01912 ) 0.0504

=

M6 mol dm–3

1

If not correct answer: For M1-3, if answer is shown, it must be correct (ignore sf)

1 1

ALLOW ECF from M1/2/3 to M4/5 (but not from M3 to M5 if omission of M3 gives negative M5) NOT ECF from incorrect Ka expression in M4 to M5 M6 If not mol dm–3, ALLOW ECF for units from incorrect Ka expression in M4 7.21 x 10–2 (7.15 – 7.26 x 10–2) gives M1,2,4,5 (by alternative method omitting M3)

02.5

M1

(HSO4– ⇌ SO42– + H+) equilibrium moves/shifts left (to counteract / remove increased [SO42–])

M2

so [H+] decreases

M1 ALLOW H+ reacts with SO42–/sulfate IGNORE favours the reverse / left / backwards reaction NOT base / A– / sodium sulfate in place of SO42–/sulfate +

+

M2 ALLOW fewer H (ions) or amount of H lower or removes H+ M2 independent of M1

16

1 1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

Question

Answers

salt bridge

Additional Comments/Guidelines

ALLOW description of salt bridge, e.g. filter paper / string / wick soaked in suitable solution

03.1

Mark

U tube (NOT YouTube) filled with suitable solution / gel

1

NOT U tube alone

complete the circuit

ALLOW ions to flow / move / transfer ALLOW to balance charge / to maintain electrical neutrality

03.2

1

IGNORE current / charge to flow NOT electrons to flow

03.3

B = platinum

ALLOW Pt / platinum black

1

17

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 Identity

Conditions

M1

C

HCl

1 mol dm–3

M2

D

H2 / hydrogen

100 kPa

M3

E

FeCl2 and FeCl3

1 mol dm

NOT incorrect state symbols ALLOW M or molar or mol/dm3 for mol dm–3 –3

+

M1 ALLOW 1 mol dm H

1

–3

1

ALLOW 0.5 mol dm H2SO4 –3

1

ALLOW 1 mol dm–3 HNO3 IGNORE 100 kPa

M4

298 K (any mention)

M2 ALLOW 1 bar

1

NOT 1 atm / 101 kPa

03.4

NOT H for hydrogen NOT 1 mol dm–3 M3 ALLOW 1 mol dm–3 Fe2+ and Fe3+ ALLOW other identified Fe(II) and Fe(III) compounds with appropriate concentrations, e.g. 1 mol dm–3 FeSO4 and 0.5 mol dm–3 Fe2(SO4)3 IGNORE 100 kPa

M1 H2 + 2 Fe3+  2 H+ + 2 Fe2+

M1 IGNORE state symbols ALLOW multiples / fractions ALLOW equation with equilibrium sign if forward reaction shown is in this direction

03.5 M2 replace voltmeter with lamp/wire/ammeter owtte

18

M2 ALLOW remove voltmeter

1 1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

M1

missing value (+) 2.3(0)

1 1.16 1.15 1.14 1.13 1.12 1.11

Ecell / V

1.1 1.09

03.6

1.08

1.07 1.06 1.05 -5

-4

-3

-2

1.04 -1 0 1 2+ 2+ ln ([Zn ]/[Cu ])

2

3

4

5

M2

suitable scales (plotted points use at least half of grid)

M3

points plotted correctly (± ½ small square per point) and best fit line drawn (within one small square of each point)

M2 ALLOW scales which use half the grid for plotted points

1 1

M3 If M1 incorrect, should be plotted accordingly and best fit line ignore if anomalous

19

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 M1

gradient = 0.013 (must be negative)

M1 ALLOW –0.0125 to –0.0136

1

ALLOW ECF from graph if outside this range

03.7

M2

M1 = () 4.3 x 10–5 T or

M3

T = 302 or 303 (K)

𝐌𝟏

M3 temperature must match gradient unless 0.016 used (ALLOW positive temperature if positive gradient used)

T = (−)4.3 𝑥 10−5

1 1

at least 2sf Correct M3 also scores M2 NOT negative temperature M3 (Alternate gradient = 0.016 gives) T = 372 (K)

03.8

M1

E = 0.8(0) V

M2

non standard conditions or

1 M2 ALLOW temperature is not 298K

concentration (of Zn2+) not 1 (mol dm–3) or 2+

2+

–3

concentration (of Zn ) less than 1 (mol dm )

20

NOT concentration (of Zn ) greater than 1 (mol dm–3) NOT concentration (of Zn2+) is different

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 Question

04.1

04.2

Answers

nucleophilic addition

Additional Comments/Guidelines

both words needed NOT any additional names

M1

racemic (mixture) / racemate

M2

planar (around) carbonyl / C=O

M3

(equal chance of) attack from each side (by CN–)

M4

a correct structure of 2-hydroxypropanenitrile

M5

correct 3D representations of both isomers, e.g.

Mark

1

1 M2 NOT molecule is planar ALLOW flat for planar

1 1

M4 any correct 2D or 3D structure

1

M5 must show at least one wedge bond and one dash bond in each structure and any bonds in the plane cannot be at 180º to each other

1

second structure could be drawn as mirror image of first or with same orientation with two groups swapped round, e.g.

ALLOW ECF for second structure from incorrect first structure, providing molecule is chiral

21

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

M1

conc H2SO4 or conc H3PO4

M1 ALLOW conc to come from conditions line

1

M2

heat / 170°C

M2 depends on attempt at correct reagent in M1

1

ALLOW high temperature / hot / 100-300°C / 373 – 573 K / reflux IGNORE references to pressure 04.3

IGNORE warm NOT ethanolic / alcoholic Alternative answer M1 Al2O3 M2 pass vapour over hot Al2O3

MUST show trailing bonds IGNORE any brackets or n 04.4

NOT C–N or C=N if CN group displayed ALLOW structures with CN on either C in each of the three units ALLOW –CH2–CH(CN)–CH2–CH(CN)–CH2–CH(CN)–

22

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

Question

Answers Fe + H2SO4  FeSO4 + H2

Additional Comments/Guidelines

Mark

ALLOW Fe + 2H+  Fe2+ + H2 ALLOW Fe + 2H+ + SO42–  Fe2+ + SO42– + H2 ALLOW Fe + H2SO4  Fe2+ + SO42– + H2

05.1

ALLOW Fe + 2H+ + SO42–  FeSO4 + H2

1

ALLOW multiples IGNORE state symbols

05.2

22.65 (cm3)

5 Fe2+ + MnO4– + 8 H+  5 Fe3+ + Mn2+ + 4 H2O

1

ALLOW multiples IGNORE state symbols

05.3

1

NOT if electrons shown

05.4

05.5

colourless / (pale) green to (hint of) pink

NOT …. to purple ALLOW …. to pale / hint of purple

pipette

both needed

burette

ALLOW (graduated/volumetric) pipette ALLOW (graduated/volumetric) burette

1

1

NOT dropping pipette

23

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

05.6

24

1.47(%)

ALLOW 1.5(%)

1

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019 Question

Marking Guidance

Mark

6

C

1

7

A

1

8

D

1

9

A

1

10

C

1

11

B

1

12

B

1

13

A

1

14

B

1

15

B

1

16

A

1

17

A

1

18

D

1

19

B

1

20

B

1

Comments

25

MARK SCHEME – A-LEVEL CHEMISTRY – 7405/3 – JUNE 2019

26

21

A

1

22

C

1

23

B

1

24

A

1

25

B

1

26

C

1

27

D

1

28

A

1

29

D

1

30

C

1

31

D

1

32

A

1

33

D

1

34

D

1

35

D

1