GCE A LEVEL BIOLOGY SPECIMEN ASSESSMENT MATERIALS

A LEVEL BIOLOGY Specimen Assessment Materials 1

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For teaching from 2015 For award from 2017

GCE A LEVEL BIOLOGY

SPECIMEN ASSESSMENT MATERIALS

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Contents Page Question Papers COMPONENT 1: Energy for life 5 COMPONENT 2: Continuity of life 25 COMPONENT 3: Requirements for life 49 Mark Schemes COMPONENT 1: Energy for life 85 Summary of marks allocated to assessment objectives 97 COMPONENT 2: Continuity of life 98 Summary of marks allocated to assessment objectives 111 COMPONENT 3: Requirements for life 112 Summary of marks allocated to assessment objectives 129

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Candidate Name Centre Number Candidate Number

A LEVEL BIOLOGY COMPONENT 1 Energy for Life SPECIMEN PAPER 2 hours

For Examiner’s use only

Question Maximum Mark

Mark Awarded

1. 13

2. 7

3. 9

4. 9

5. 7

6. 10

7. 12

8. 13

9. 10

10. 10

Total 100

ADDITIONAL MATERIALS In addition to this examination paper, you will need a calculator. INSTRUCTIONS TO CANDIDATES Use black ink or black ball-point pen. Do not use gel pen. Do not use correction fluid. Write your name, centre number and candidate number in the spaces at the top of this page. Answer all questions. Write your answers in the spaces provided in this booklet. INFORMATION FOR CANDIDATES The number of marks is given in brackets at the end of each question or part-question. You are reminded of the need for good English and orderly presentation in your answers.

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Answer all questions.

1. The diagram below shows a set of apparatus used to investigate the rate of photosynthesis in Elodea (Canadian pondweed) at different light intensities. The apparatus was assembled as shown in a darkened room. The number of bubbles of oxygen, coming from the cut end of the pond weed, was counted for

1 minute. This was repeated for a second and third minute.

RESULTS

Distance of plant from light source (mm)

Number of bubbles counted in 1 minute

Mean number of bubbles counted

in 1 minute

600 32 29 33 31 400 45 46 44 45 300 54 53 58 200 71 68 73 71 100 114 87 90 97

(a) (i) Complete the table by calculating the mean number of bubbles

counted in 1 minute for 300 mm. [1] (ii) What problem was experienced with the 100 mm results and how

does this affect the experimental results for this distance? [2]

………………………………………………………………………………………… …………………………………………………………………………………………

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(iii) Give three ways in which you could improve this experiment so that it would produce more accurate results. [3]

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(b) Diuron is a weed-killer which is a very specific and sensitive inhibitor of photosynthesis.

It blocks the electron carrier binding site on Photosystem II. This stops the electron flow from where it is generated in Photosystem II, to the electron carrier. This reduces the ability of the plant to convert light energy into chemical energy.

Diuron only blocks electron flow from Photosystem II. It has no effect on Photosystem I or other reactions in photosynthesis, such as light absorption

or carbon fixation in the Calvin cycle. Explain the effects of Diuron on non-cyclic photophosphorylation and explain

why cyclic photophosphorylation is not affected. [4]

………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………………

(c) Suggest why a plant would die when the weed-killer Diuron is sprayed onto it.

[3]

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2. Muscle cells were broken up and separated into fractions. Samples of each fraction were incubated either with glucose or pyruvate (pyruvic

acid). Tests were carried out for the production of carbon dioxide and lactate (lactic acid) in

each sample. The results are given in the table below.

Cell fraction

Incubated with glucose Incubated with pyruvate Carbon dioxide Lactate Carbon dioxide Lactate

Ribosomes

absent

absent

absent

absent

Mitochondria

absent

absent

produced

absent

Cytoplasmic

residue

absent

produced

absent

absent

(a) Explain the results obtained with each cell fraction. (i) Ribosomes [1]

………………………………………………………………………………………… (ii) Mitochondria [2]

…………………………………………………………………………………............ …………………………………………………………………………………………

(iii) Cytoplasmic residue [2]

………………………………………………………………………………………… …………………………………………………………………………………………

(b) Explain why no carbon dioxide is evolved when cyanide is added to the

mitochondrial fraction prior to incubation. [2]

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3. An investigation into the effects of mineral deficiency was carried out using young maize seedlings.

The results of this work are shown below.

(a) Describe the appearance of a similar plant grown in the absence of

magnesium. [1]

…………………………………………………………………………………………………. (b) Give experimental details of how you would set up this experiment to obtain

reliable results over a four week period, including which variables should be controlled. [4]

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(c) Using the results, suggest a function that iron has in plant metabolism. [1]

…………………………………………………………………………………………………

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(d) The full nutrients plant grew to a height of 203 mm over the four weeks whilst the plant without nitrogen only grew to a height of 52 mm. Calculate the % decrease in growth of the plant deprived of nitrogen. [2]

% decrease = ……………………. (e) The seedling in distilled water died after 16 days. Suggest how the seedling

managed to survive this long. [1]

…………………………………………………………………………………………………

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4. The photograph below shows a species of Gram negative bacteria.

Dr Gary Gaugler / science photo library

(a) (i) What is the term used for bacteria of this shape? [1]

……………………………………………………………………

(ii) Draw a simple labelled diagram to show the structure of the cell. [4] (b) (i) Describe how you would recognise Gram negative bacteria under a

microscope. [1]

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(ii) Give two features of a Gram negative bacterial cell wall. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

(iii) Give one advantage to the bacteria of having this cell wall structure. [1]

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5. The diagram represents a molecule of ATP.

(a) Name the components Y and Z. [1]

Y: ……………………………… Z: ………………………………

(b) Describe how energy is released from ATP. [2]

………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………………

(c) Describe the production of ATP by chemiosmosis. [4]

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6. A student introduced a pure culture of bacteria into a nutrient medium and recorded the numbers of bacteria per cm3. The results are shown in the table.

Time hours) Number of bacteria /

millions per cm3 0 10 2 15 4 25 6 215 8 465

10 488 12 488

(a) (i) The rate of population growth in the first 4 hours was 2.5 million bacteria cm3 hr-1. Calculate the rate of population growth in the following four hours. [2] rate of population growth= ……………… (ii) Use the data to draw a graph to show the increase in number of

bacteria. [3]

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(iii) State the phase of population growth occurring from: [1]

0-6 hours ………………………………………. 6-12 hours ……………………………………….

(iv) The growth of the bacterial population didn’t completely follow the trend the student was expecting. The student was expecting the number of bacteria to fall towards the end of the experiment. Give two suggestions why the population did not decline and describe how the experiment could have been adapted to test your suggestions.

[4] ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

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7. Glucose oxidase is an enzyme, shown below, the normal substrate of which is glucose.

The molecule arabinose acts as an inhibitor for the glucose oxidase.

(a) Enzymes are types of protein. (i) On the diagram, label the position of the enzyme’s active site. [1] (ii) Explain why you chose that position. [2]

………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

(iii) State the level of protein structure responsible for maintaining the

shape of the active site. [1]

…………………………………………………………… (b) Explain how arabinose inhibits the activity of glucose oxidase. [3]

…………………………………………………………………………………………………

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(c) The molecular structures of glucose and arabinose are shown below. Using your knowledge of carbohydrate structure identify which molecule, A or

B is arabinose giving a reason for your answer. [2]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

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(d) If a person drinks methylated spirits (which contains methanol), blindness can

result because the body turns the methanol into formaldehyde, which is highly poisonous. One form of hospital treatment is for the patient to drink alcohol (ethanol).

What conclusion can be made regarding the relative structures of methanol

and ethanol and suggest how this treatment will reduce the toxic effects of the methanol? [3]

………………………………………………………………………………………………………

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A B

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8. The diagram shows the transfer of energy through an ecosystem.

All values given are in kJ m-2 yr-1. (a) Calculate the respiratory losses of the primary consumers. [1] Answer ……………… kJ m-2 yr-1

(b) The values given are for a marshland and are in kJ m-2 yr-1. The area of the

marshland is 17 000 m2. (i) Calculate the total amount of energy ‘expelled’ by this ecosystem in

one year. [2] Answer …………………….. (ii) Describe what happens to this expelled energy. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

KEY: R = respiratory loss C = consumed E = expelled

Primary Producer GPP 20 810 NPP

8863

Primary

Consumer

Secondary Consumer

C

C

E

E

E

R

R

R

3368

383

46

1095

5465

270

67

11 977

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(c) The diagram opposite shows that only a limited amount of energy is transferred between trophic levels. The efficiency of energy transfer varies between trophic levels. Analyse the information provided to explain whether herbivores or carnivores are more efficient. [4]

………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………………

(d) The principles of energy transfer through food chains can be employed to compare the efficiency of different farming methods.

(i) Intensively reared animals are kept indoors in high densities and

under temperature controlled conditions. Explain how this increases productivity. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

(ii) Using this model of energy flow, explain why most of the animals reared for food are herbivores rather than carnivores. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

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9. The table below shows the “planetary boundaries” for a number of global processes.

Global Process

Variable Units of

measurement Boundary

Pre-industrial

value

Recent data

Biodiversity

loss Extinction rate species becoming extinct per million species per year

10 1 100-1 000

Climate change

Atmospheric carbon dioxide

ppm (parts per million) 350 280 390

Nitrogen cycle

Nitrogen removed from atmosphere

million tonnes per year 35 0 120

Land use Land used for crops % 15 5 12

Fresh water Use of fresh water km3 per year 4 000 415 2 600

Adapted from Rockstrom et al (2009) and from Lynas (2011)

(a) The term “planetary boundary” has been applied to the global processes referred to in the table above. Explain what you understand by the term “planetary boundary”. [2]

………………………………………………………………………………………………… ………………………………………………………………………………………………….

(b) Use the data in the table above to determine which of the five processes

given have exceeded their boundaries. [1] …………………………………………………………………………………………………. ………………………………………………………………………………………………….

(c) The pre-industrial value for extinction rate was one species per million per

year. This is called the background rate. Apart from human impact, what led to species becoming extinct in those times? [2]

………………………………………………………………………………………………… …………………………………………………………………………………………………

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(d) (i) What is meant by the term biodiversity? [1]

………………………………………………………………………………………… …………………………………………………………………………………………

(ii) Describe one action that could delay biodiversity loss. [1]

………………………………………………………………………………………… …………………………………………………………………………………………

(e) Explain the increase in the recent data for atmospheric carbon dioxide concentration. [3]

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10. Answer one of the following questions. All diagrams included in your answer must be fully annotated. Either, (a) Explain what is meant by the viable count technique and describe how

it is used to monitor population growth of microorganisms. [10]

0r (b) The maintenance of soil fertility is mainly determined by bacteria. Describe the importance of bacteria in the nitrogen cycle. [10] Suggest how farming practices can encourage the activity of these

bacteria.

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A LEVEL BIOLOGY COMPONENT 2 Continuity of Life SPECIMEN PAPER 2 hours



Mark Awarded

1. 10

2. 10

3. 19

4. 10

5. 13

6. 14

7. 14

8. 10

Total 100


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Answer all questions. 1. The species is the basic unit by which biodiversity is measured.

(a) Define the term species. [1] ………………………………………………………………………………………………… …………………………………………………………………………………………………

(b) Some data on species richness are shown below.

Estimated number of species

Britain Borneo World Fish (freshwater) 38 394 >8500

Amphibians 6 100 >4000 Reptiles 6 105 6500 Birds (breeding residents) 210 600 9881

Mammals 48 288 4327

(i) Which vertebrate class shows the greatest species richness? [1]

…………………………………………………………………………………………

(ii) Calculate the percentage of the world’s species of reptiles found in the following: [2]

I Britain ………………………………………… II Borneo …………………………………………

(c) Species richness does not give as complete a measure of biodiversity as Simpson’s Diversity Index. Simpson’s Diversity Index also includes information about the number of individuals in each species.

Name the technique you would use to estimate the numbers of each species

of animal and briefly describe the procedure. [2]

……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………….………………………….. ………………………………………………………………….…………………………..

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(d) Of the 288 species of mammals in Borneo, 102 are bats. The following diagram is a phylogenic tree showing the evolutionary relationship between some of the bats.

(i) Mark clearly on the phylogenic tree, with the letter X, the position of an ancestor common to Penthetor lucasi and Cynopterus sphinx but not common to Mops mops. [1]

(ii) Amino acid sequences of cytochrome oxidase from blood samples from each species were prepared. The sequences for the other three bats were compared with the sequence for Cyanopterus brachyotis. The number of positions at which there was a different amino acid is recorded in the table below.

Bat A B C

Number of different amino acids 19 7 13

Use the information in the table and the phylogenic tree to identify the bats. [1]

A ……………………………………………………

B …………………………………………………..

C …………………………………………………..

(e) Complete the following classification of Penthetor lucasi: [2]

Kingdom ………………………………………

……………………………….. chordata

Class …………………………………………

……………………………. chiroptera

……………………………. pteropodidae

Genus …………………………………………….

Species ……………………………………………

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Mops mops

Penthetor lucasi

Cynopterus brachyotis

Cynopterus sphinx

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Q.2.

(a) Sickle cell anaemia is an inherited disease. People who have the disease inherit two genes for sickle haemoglobin—one from each parent.

Sickle haemoglobin causes red blood cells to develop a sickle shape. These block blood flow in the blood vessels causing pain and organ damage. It leads to much reduced life expectancy.

People who inherit a sickle haemoglobin allele from one parent and a normal allele from the other parent have sickle cell trait. They usually have few symptoms and lead normal lives. In the United States, the sickle cell disease occurs in 1 out of every 625 African Americans. The Hardy-Weinberg formula states that if alleles A and a are present in a population with the frequencies of p and q, the proportion of individuals homozygous for the dominant allele (AA) will be p2, the proportion of heterozygotes (Aa) will be 2pq, and the proportion or homozygous recessives (aa) will be q2, where p + q = 1.

Use the Hardy Weinberg formula to calculate the proportion of African Americans that have sickle cell trait. [3] (b) A survey was conducted on 143 people in a clinic in a village in sub-Saharan

Africa. The survey noted if the patients were suffering from malaria or another condition and whether they had sickle cell trait. The results are shown in the table below:

Suffering from

Malaria Suffering from

another condition Number of patients 67 76 Number with sickle cell trait 0 19

(i). Suggest a possible conclusion from the results. [1]

………………………………………………………………………………………… …………………………………………………………………………………………

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(ii) Give two improvements to the design of the survey to make the

conclusion more reliable. [2] ………………………………………………………………………………………… …………………………………………………………………………………………

(c) The Hardy-Weinberg Principle states that allele frequencies remain the same

from generation to generation in the absence of a list of conditions, including natural selection.

Explain the mechanism by which natural selection might be expected to change the frequency of the allele for sickle cell anaemia in the gene pool of Africans living in Nigeria. [4]

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

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3. The drawing below shows part of a cross-section through a seminiferous tubule.

(a) On the diagram, label the cell types A - D. [2] (b) (i) Name two structures visible in the drawing of cell E that indicate that

the cell is undergoing cell division. [2]

…………………………………………………………………………………… (ii) State the type of cell division occurring in E. [1]

…………………………………………………………………………………… (iii) State the number of daughter cells that cell E will eventually produce. [1]

…………………………………………………………………………………… (iv) Describe two differences between the daughter cells produced by cell

E and those produced by cell A. [2] …………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… ……………………………………………………………………………………

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(c) (i) Name the cell labelled F in the diagram. [1]

……………………………………………………………………………………

(ii) Cell F has a prominent nucleolus and many ribosomes. What does the presence of these structures indicate about function of cell F? [1]

……………………………………………………………………………………

(d) The diagrams below show how hormones regulate the menstrual cycle.

(i) Name the hormones [2] P ……………………………………….. Q …………………………………….. R ……………………………………… S ……………………………………..

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(ii) Describe the changes that happen to structure X over days 1 -12. [2]

…………………………………………………………………………………………. …………………………………………………………………………………………. ………………………………………………………………………………………….

(iii) State the evidence shown in the diagrams that hormone P triggers these changes. [1]

………………………………………………………………………………………….

(iv) Describe and explain the relationship between the development of the

corpus luteum and the concentration of hormone S in the blood. [2]

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

(v) Give two ways in which the diagrams would differ in the event of pregnancy. [2]

………………………………………………………………………………………….

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Q.4 The diagram below is a cross-section through a grain of barley, Hordeum vulgare. bran endosperm embryo (a) (i) Explain how the endosperm was formed in the embryo sac of a barley

flower to produce triploid cells. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

(ii) The endosperm cells contain 21 chromosomes. Calculate how many

chromosomes there are in the cells of the embryo. [1]

………………………………………………………………………………………… (b) The table below shows the results obtained from a study of the germination

and early growth of barley grains.

Time after sowing (days)

Dry mass of endosperm (mg)

Dry mass of embryo (mg)

0 42 2 2 41 2 4 32 8 6 20 16 8 9 24 10 6 34

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(i) What can be concluded from these results? [2]

…………………………………………………………………………………………

…………………………………………………………………………………………

(ii) The diagram below shows an experiment demonstrating the presence

of amylase in a germinating broad bean seed.

Using the apparatus shown above, design a method to investigate the effect of age of seed on the concentration of amylase produced. [5] …………………………………………………………………………………………

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At start of experiment

At end of experiment following the Petri dish being flooded with iodine

starch agar

Petri dish

germinating broad bean

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Q.5 (a) The diagram below is of a bacterial plasmid 3 450 base pairs in length with the restriction sites of four different enzymes, EcoRI, BamHI, HindIII and PSTII shown.

(i) If the plasmid was cut using EcoRI and PSTII only, what would be the size of the smallest fragment? Show your working. [2]

Answer …………………..

(ii) A gene was present between 550 and 1699 base pairs. Which two enzymes would be needed to cut out the gene leaving the smallest number of additional bases? [1]

…………………………………………………………………………………………

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(b) The polymerase chain reaction (PCR) is used to amplify small sections of DNA rapidly. Each stage of the reaction takes place at a different temperature.

State what events happen during each of the following stages:

(i) the DNA is first heated to 95 °C; [1]

………………………………………………………………………………………… ………………………………………………………………………………………… (ii) the temperature is then reduced to 50-60 °C; [1] ………………………………………………………………………………………… …………………………………………………………………………………………

(iii) the temperature is then increased to 70 °C. [1]

………………………………………………………………………………………… …………………………………………………………………………………………

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(c) “D7S280” is a short tandem repeat (STR) sequence of “gata” found on human chromsome 7. This STR was sequenced and found to be between 126 and 177 base pairs within the 341 base pair sequence shown below (gata shown in bold).

DNA Sequence

Starting base number

1 AATTTTTGTA TTTTTTTTAG AGACGGGGTT TCACCATGTT GGTCAGGCTG ACTATGGAGT

61 TATTTTAAGG TTAATATATA TAAAGGGTAT GATAGAACAC TTGTCATAGT TTAGAACGAA

121 CTAACGATAG ATAGATAGAT AGATAGATAG ATAGATAGAT AGATAGATAG ATAGATAGAT

181 TGATAGTTTT TTTTTATCTC ACTAAATAGT CTATAGTAAA CATTTAATTA CCAATATTTG

241 GTGCAATTCT GTCAATGAGG ATAAATGTGG AATCGTTATA ATTCTTAAGA ATATATATTC

301 CCTCTGAGTT TTTGATACCT CAGATTTTAA GGCC

Key

A adenine

T thymine

C cytosine

G guanine

In a separate experiment, PCR was used to amplify the section shown in bold and the resulting product was run on an agarose gel using gel electrophoresis. Gel result

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(i) Describe and explain how electrophoresis produced the results seen in the gel result. [4]

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

(ii) Evaluate how well the results of the electrophoresis match the sequencing results. [2]

………………………………………………………………………………………… …………………………………………………………………………………………

(iii) How could the gel result be made more accurate? [1]

………………………………………………………………………………………… …………………………………………………………………………………………

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6. In pea plant seeds the allele for wrinkled seed coat (R) is dominant to the allele for smooth seed coat (r) and the allele for yellow seed colour (Y) is dominant to the allele for green (y).

A plant producing yellow, smooth coated seeds (heterozygous for colour) was crossed with a plant producing green, wrinkled seeds (heterozygous for texture).

In the first generation the offspring had the following phenotypes: 27 yellow wrinkled coat; 22 green wrinkled coat; 28 yellow smooth coat; 23 albino smooth coat.

(a) Explain what is meant by heterozygous. [1]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

(b) Using a genetic diagram explain fully the results outlined above. [5]

parental genotype …………………………… ………………………

gamete genotype …………………………… ………………………

offspring genotype ………....… ………....… ………....… ………....…

offspring phenotype ………....… ………....… ………....… ………....…

………....… ………....… ………....… ………....…

ratio ………....… ………....… ………....… ………....…

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(c) Using χ2 test the null hypothesis that ‘there is a significant difference between the observed and expected numbers of offspring of the different phenotypes’. A Chi-squared table is provided below.

Phenotype O E O-E (O-E)2 (O-E)2 E

Σ Σ

(i) χ2 = …………………………. [3]

Chi-squared table

Degrees of freedom P = 0.10 P = 0.05 P = 0.02

1 2.71 3.84 5.41

2 4.61 5.99 7.82

3 6.25 7.82 9.84

4 7.78 9.49 11.67

5 9.24 11.07 13.39

(ii) Using your calculated χ2 results, what conclusions can be drawn about the inheritance of pea coat colour and texture? [5]

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

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7. During the cell cycle, DNA is replicated prior to cell division.

(a) What term describes the replication of DNA? [1]

………………………………………………………

(b) Identify the stage of the cell cycle where this replication occurs. [1]

………………………………………………………

(c) Name two enzymes responsible for replicating DNA. [2]

………………………………………………………………………………………………

(d) Name the three constituent molecules that make up a DNA nucleotide. [2] ………………………………………………………………………………………………

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(e) Bone marrow cells were taken from two different people. The number of cells in each sample were counted regularly and the results shown below.

Using the graph, answer the following questions.

(i) It was concluded that Person A was suffering from cancer. Justify this conclusion. [2]

………………………………………………………………………………………… ………………………………………………………………………………………… ………………………………………………………………………………………… …………………………………………………………………………………………

(ii) Suggest why the number of cells in healthy Person B would increase. [1] ………………………………………………………………………………………… …………………………………………………………………………………………

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(f) The following diagrams represent stages in one type of cell division.

(i) Name the type of cell division shown. [1]

………………………………………………………

(ii) Give two reasons for your choice [2]

…………………………..........……………………………………………………… ……………………………..........……………………………………………………

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(iii) In the space below, draw one of the cells that would be produced if cell division in the two cells shown was allowed to continue. [2]

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8. Answer one of the following questions.

Any diagrams included must be fully annotated.

Either (a) Describe how gene technology can be used to produce human insulin.

[10] Or (b) Describe the adaptations of flowering plants for wind and insect

pollination. Describe pollination and fertilisation in plants. [10]

……………………………………………………………………………………………………………

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A LEVEL BIOLOGY COMPONENT 3 Requirements for Life SPECIMEN PAPER 2 hours



Mark Awarded

Section A 1. 10

2. 16

3. 11

4. 17

5. 10

6. 6

7. 10

Section B 20

Total 100


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SECTION A

Answer all questions. 1. In insects, ventilation of the gas exchange system involves contractions of the

abdomen, called body breathing movements. In an experiment to investigate ventilation in insects, a locust was placed in a syringe as shown in the diagram below.

The locust was left in the syringe for five minutes. The number of body breathing movements during 30 seconds was then counted and was repeated to obtain three sets of readings. The air in the syringe was then replaced with gas mixtures of different percentage compositions of oxygen and carbon dioxide and the experiment repeated.

(i) Why did the student leave the locust in the syringe for five minutes

before they began the first count? [1]

………………………………………………………………………………………… …………………………………………………………………………………………

(ii) What step should the student have taken before changing the gas

mixture in the syringe? [1]

………………………………………………………………………………………… …………………………………………………………………………………………

The results of the experiment are shown below.

Approximate percentage composition of the gas

mixture

Number of body breathing movements in

30 s

Mean number of body breathing

movements in 30 s

oxygen carbon dioxide

Atmospheric air

21 0.04 26 29 26 27

Gas mixtures

83 17 54 49 46 50 88 12 41 46 51 46 94 6 48 42 46 45 97 3 39 43 41 43 100 0 7 7 10 8

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(b) (i) Describe how oxygen reaches the tissues in insects. [2]

…………………………………………………………………………………… ……………………………………………………………………………………

(ii) Explain why ventilation in the locust involved movements of the whole body. [1]

…………………………………………………………………………………… ……………………………………………………………………………………

(c) (i) Based on the results given in the table, which gas is having the greater effect on ventilation in the locust - oxygen or carbon dioxide? Use data from the table to explain how you reached your conclusion.

[2]

…………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… ……………………………………………………………………………………

(ii) Air exhaled from the human lungs contains approximately 16%

oxygen and 4% carbon dioxide. Estimate the mean number of body breathing movements in 30 s if the experiment was repeated using exhaled air. [1]

……………………………………………………………………………………

(iii) Suggest an explanation for your answer to (c) (ii). [2]

…………………………………………………………………………………… …………………………………………………………………………………… …………………………………………………………………………………… ……………………………………………………………………………………

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2. The diagram below shows a simple human reflex arc.

(a) Using letters from the diagram complete the following table. [3]

DESCRIPTION LETTER

is an effector organ

may have connections to neurones in the brain

its cell body lies in the dorsal root ganglion

(b) Structure B in the diagram is said to be myelinated. Explain what this means

and the effect that this has on the speed of transmission of an action potential by this neurone. [2]

……………………………………………………………………………………………… ……………………………………………………………………………………………….

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(c) An experiment was carried out to investigate the distance travelled by an action potential in a non-myelinated neurone following the application of a strong stimulus.

distance from the stimulus (mm)

time (ms)

0.5 1.1 1.0 2.3 1.5 3.2 2.0 4.5 2.5 5.7

Calculate the speed of transmission of the action potential in this axon in mm ms-1. Give your answer to three significant figures. Show your workings.

[3]

Answer ……………………… mm ms-1

(d) In this experiment the value of the action potential was the same at all

distances tested from the stimulus.

(i) Describe and explain how an action potential was generated in the neurone above. [4]

………………………………………………………………………………………… …………………………………………………………………………………………. ………………………………………………………………………………………… …………………………………………………………………………………………. ………………………………………………………………………………………… …………………………………………………………………………………………. ………………………………………………………………………………………… …………………………………………………………………………………………. ………………………………………………………………………………………… ………………………………………………………………………………………….

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(ii) Explain why the value for the action potential was the same at all distances from the stimulus. [2]

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

………………………………………………………………………………………

(e) The experiment to determine the speed of transmission of the action potential was carried out on an axon of a squid neurone. Suggest why the results of this experiment might not be comparable if the experiment was repeated using a mammalian neurone. [2]

……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………

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3. Mammals and Amoeba are both described as being heterotrophic organisms that digest their food through holozoic nutrition.

(a) Explain the difference between heterotrophic and autotrophic organisms. [2] ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… ……………………………………………………………………………………………… (b) To carry out holozoic nutrition mammals have a tube gut divided into different

regions.

(i) Describe how this is different to holozoic nutrition in Amoeba. [1] ………………………………………………………………………………………… ………………………………………………………………………………………….

(ii) Explain the advantage to mammals of having different regions in the

gut. [2]

………………………………………………………………………………………… …………………………………………………………………………………………. ………………………………………………………………………………………….

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(c) Passage of food through the gut is aided by the production of mucus. The electron micrograph below shows some cells that produce and secrete mucus in the small intestine.

Magnification = x 3700

(i) Name the cells shown in the electron micrograph. [1]

………………………………………………………………………………………… (ii) In which type of tissue would you find these cells? [1]

………………………………………………………………………………………… (iii) Identify the process by which mucus is being secreted at S in the

electron micrograph. [1]

………………………………………………………………………………………… (d) The height of one of these cells is shown by the line labelled XY. Calculate the actual size of the cell, showing how you carried out your

calculation. Give your answer using an appropriate unit. [3]

Answer = …………………………………………

11

S

X

Y

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4. The diagram below shows cells taken from the proximal convoluted tubule of a nephron.

(a) (i) Identify in which region of the kidney you would find these cells. [1] ………………………………………………………………

(ii) In the table below identify structures X, Y and Z and suggest how they are adapted to their function. [6]

structure name of structure adaptation for function

X

Y

Z

(b) (i) Name the process by which these cells exchange oxygen and carbon

dioxide with the blood in the capillaries. [1]

………………………………………………………………

(ii) Name the process by which water passes into these cells. [1]

………………………………………………………………

blood capillary

X

Y

Z

O2

CO2

H2O

Na+

P

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(iii) Explain why the addition of cyanide ions would inhibit the movement of sodium ions into the tissue fluid from these cells. [3]

……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ……………………………………………………………………………………… ………………………………………………………………………………………

(c) The structures labelled P in the diagram were formed by the infolding of the

plasma membrane.

(i) Name the process by which these structures were formed. [1] ………………………………………………………………

(ii) Describe the role of this process. [1]

………………………………………………………………………………………

………………………………………………………………………………………

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(d) The diagrams below show the relative sizes of nephrons from three mammals that live in different habitats.

The table shows urine concentrations from these three mammals.

name of mammal habitat urine concentration (mosmol dm-3)

kangaroo rat xeric* 5 500

domestic cat mesic 3 100

beaver freshwater aquatic 520

* xeric habitats have limited availability of water

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Using information in the diagram and the table, what conclusion can be made about the relationship between the length of part D of the nephron and the adaptation of the mammals to their habitats? Explain your answer. [3]

……………………………………………………………………………………………… ………………………………………………………………………………………………

……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………

………………………………………………………………………………………………

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5. A student used the apparatus shown in the diagram to compare the rate of absorption of water by two different plant shoots.

(a) (i) Name the apparatus used in this experiment. [1]

………………………………………………………………

(ii) Using the formula below calculate the rate of water absorption by the

plant in mm3 min-1. [2]

Volume of water = distance travelled by bubble x π x [radius]2 (π = 3.14)

Answer = ………………………..mm3 min-1

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(b) When the student carried out the same experiment on the other plant shoot she found that water was absorbed at a rate of 1.3 mm3 min-1.

(i) Identify four control variables that the student should have kept constant in her experiments. [2]

……………………………………………………………………………………........

………………………………………………………………………………………….

(ii) Suggest what further measurement and calculation the student should carry out before reaching a conclusion about the difference in the rate of water absorption by the two plant shoots. [2] ……………………………………………………………………………………........

………………………………………………………………………………………….

…………………………………………………………………………………………..

(c) The student realised that some of the water was absorbed to be used in

photosynthesis and that some was lost through transpiration. She modified her method to find the rate at which water was being lost through transpiration alone.

She repeated her experiment on a different plant by placing it in the dark to

prevent photosynthesis and obtained the following results. Light Dark

Rate of water absorption (mm3 min-1) 4.5 1.5

From these results the student concluded that the plant shoot was using

water at a rate of 3.0 mm3min-1 for photosynthesis alone.

Explain why the student was incorrect in reaching this conclusion. [3]

………………………………………………………………………………………….

………………………………………………………………………………………….

………………………………………………………………………………………….

…………………………………………………………………………………………. ………………………………………………………………………………………….

………………………………………………………………………………………….

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6. Haemoglobin is a protein found in red blood cells. It has a quaternary structure. There are three different types of haemoglobin found in adult humans, HbA, HbA2 and HbF as shown in the table.

haemoglobin type % found in adult

blood types of polypeptide

chain found in the haemoglobin

molecule

HbA 97 2 alpha chains 2 beta chains

HbA2 2 2 alpha chains 2 delta chains

HbF 1 2 alpha chains 2 gamma chains

(a) From this data suggest how many genes are involved in the production of the haemoglobins found in the adult human. [1]

………………………………………………………………

(b) Beta thalassaemia is a condition in humans caused by a change in the nucleotide sequence which codes for primary structure of the beta polypeptide chain. Scientists have now discovered a drug which can switch on the gene for the production of HbF (this is the type of haemoglobin found in the foetus). Explain, using your knowledge of protein synthesis, how this drug results in the production of HbF. [3]

……………………………………………………………………………………………… ………………………………………………………………………………………………

……………………………………………………………………………………………… ……………………………………………………………………………………………… ………………………………………………………………………………………………

………………………………………………………………………………………………

(c) Evaluate the advantages and disadvantages to an adult human of producing mainly fetal haemoglobin. [2]

………………………………………………………………………………………………

………………………………………………………………………………………………

………………………………………………………………………………………………

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7. Answer one of the following questions.

Any diagrams included must be fully annotated.

Either (a) Using a labelled diagram of a synapse, describe and explain how

impulses are transmitted across a synapse. [10] Or (b) Describe and explain how the nervous and endocrine systems are

both involved in the maintenance of constant water levels in the blood. [10] ……………………………………………………………………………………………………………

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SECTION B: OPTIONAL TOPICS

Option A: Immunology and Disease

Option B: Human Musculoskeletal Anatomy

Option C: Neurobiology and Behaviour

Option D: Food Science

Answer the question on one topic only. Place a tick () in one of the boxes above, to show which topic you are answering. You are advised to spend about 20 minutes on this section.

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OPTION A: IMMUNOLOGY AND DISEASE 8. Sore throats and chest infections can be caused by a number of pathogenic bacteria

and viruses. To treat the disease by using antibiotics, medical practitioners need answers to the following questions.

• Is the disease caused by a virus or bacterium? • If it is a bacterial infection, what organism is responsible for the disease? • What antibiotics are effective against the disease? • Is the infection caused by an antibiotic resistant strain of bacteria? (a) Why is it important to know if the disease is caused by a bacterium or virus? [1]

………………………………………………………………………………………………… …………………………………………………………………………………………………

(b) One method of identifying if an antibiotic is effective against a bacterium is shown

below.

(i) Explain what has caused the appearance of the clear zones around each paper disc. [1]

………………………………………………………………………………………………… …………………………………………………………………………………………………

(ii) Describe how the plate would look if the bacteria were resistant to one of the

antibiotics tested. [1]

………………………………………………………………………………………………… ………………………………………………………………………………………………… (iii) Describe how this method could be adapted to determine what concentration

of an antibiotic is needed to treat a bacterial infection. [2] ………………………………………………………………………………………………… …………………………………………………………………………………………………

…………………………………………………………………………………………………

clear zone paper discs soaked in different antibiotics

bacterial lawn

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(c) If it is found that a species or strain of a bacterium is resistant to a particular antibiotic it is also important to know the minimum inhibitory concentration. This is the lowest concentration of the antibiotic that is effective against the bacterium.

In the technique shown in diagram A below, each strip contains an accurately prepared gradient of antibiotic concentrations measured in μg cm-3

(i) What is the minimum inhibitory concentration for the antibiotic shown in diagram A against this bacterial strain? [1]

………………………………… μg cm-3

(ii) At this concentration the antibiotic is said to be bacteriostatic. What is meant

by this term? [1]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

(iii) Suggest why this method provides better quality data than the simple disc assay method. [1]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

(iv) Which of the antibiotics tested in figure B has the greatest potential for future

development? Justify your answer. [2]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

A B

Antibiotic 2

Antibiotic 1

Antibiotic 3 Antibiotic 4

Antibiotic 5

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(d) Tetracycline is a broad spectrum bacteriostatic antibiotic. It is effective against many species of bacteria but it does not affect human cells. Once inside the bacterial cell, tetracycline prevents protein synthesis by blocking a tRNA binding site in bacterial ribosomes. Scientists have recently discovered a worrying feature about tetracycline. It increases the transfer of antibiotic resistant genes from one species of bacterium to another.

(i) What is the meaning of the term broad spectrum? [1]

………………………………………………………………………………………………… …………………………………………………………………………………………………

(ii) Explain what is meant by the term antibiotic resistance. [1]

………………………………………………………………………………………………… …………………………………………………………………………………………………

(e) Penicillin is a narrow spectrum antibiotic that is bactericidal. It prevents the

formation of bonds in the peptidoglycan cell wall of bacteria.

Explain why tetracycline can be administered without knowing the exact identity of the bacterium causing an infection whilst penicillin should only be used if the infection is known to be caused by a Gram positive bacterium. [3]

…………………………………………………………………………………………………………… …………………………………………………………………………………………………………… …………………………………………………………………………………………………………… …………………………………………………………………………………………………………… …………………………………………………………………………………………………………… ……………………………………………………………………………………………………………

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(f) For a number of viral diseases immunisation is a way of avoiding infection.

The graph shows the change in antibody concentration in the plasma following a course of immunisation against a viral pathogen.

(i) Approximately how many times more antibody is produced in the second immune response than the first immune response? Explain how you arrived at your answer. [3]

………………………………………............................................................................ ………………………………………………………………………………………………… …………………………………………………………………………………………………

(ii) With reference to the graph and using your own knowledge, explain how successful immunisation protects you against infection. [2]

………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………………

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OPTION B: HUMAN MUSCULOSKELETAL ANATOMY 9. (a) The diagram represents a stage micrometer and an eye piece graticule.

(i) Use the diagram to calibrate the eye piece graticule for the objective lens used. Show all your working. Include units with your answer. [2]

1 eye piece unit = ……………………………..

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(ii) Using the calibrated eye piece graticule, a section of bone was viewed. The size of the Haversian system shown below was measured between A and B and calculated to be 450 μm.

Why would it be necessary to recalibrate the eye piece graticule if a higher power objective lens was then used ? [1]

………………………………………………………………………………………………… ………………………………………………………………………………………………..

(iii) In the space below draw the structure shown above, as it would appear in LS

(longitudinal section) between points A and B. [1] (b) (i) Many of the cells found in the lacunae have well developed rough

endoplasmic reticulum and Golgi bodies, they are also rich in RNA. What does this suggest about the function of these cells? [2]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

A

B

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(ii) Describe how oxygen and nutrients are supplied to these cells and carbon dioxide and waste products are removed from them. [4]

…………………………………………………………………………………………………

…………………………………………………………………………………………………

…………………………………………………………………………………………………

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(c) Since the 17th century there has been a decline in the incidence of rickets in the UK.

However, some health trusts in the UK have found that more than 20% of children tested in recent years show signs of rickets, results which are reminiscent of the 17th century. What can you conclude about the dramatic increase in the incidence of rickets in recent years? [3]

…………………………………………………………………………………………………

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(d) (i) Describe the role of calcium ions in the contraction of skeletal muscle. [3] …………………………………………………………………………………………………

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(ii) Explain why immediately after death muscles remain in a contracted state, a condition termed ‘rigor mortis’. [2]

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(e) Give an example of an antagonistic pair of muscles and explain why muscles are

arranged in antagonistic pairs. [2] …………………………………………………………………………………………………..............

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…………………………………………………………………………………………………..............

20

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OPTION C: NEUROBIOLOGY AND BEHAVIOUR

10. (a) If a head injury has occurred, bleeding in the brain is possible. Head scans can be performed to assess the amount of damage to the brain. Two types of scan can be performed, magnetic resonance imaging (MRI) and computerised axial tomography (CT).

(i) State one advantage of using a CT scan over an MRI scan for a

patient with a head injury. [1] ………………………………………………………………………………………… (ii) An MRI scan may detect excessive bleeding in the motor areas of the

cerebral cortex in the right hemisphere of the brain. Suggest the effect this may have on the individual. [2]

…………………………………………………………………………………………

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…………………………………………………………………………………………

………………………………………………………………………………………… The diagram below is a section through the brain.

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(b) Use the letters from the diagram to complete the table to indicate the part of the brain that is involved in the following activities. [4]

Activity Letter

regulating core body temperature

recognising a face in a picture

drinking from a glass

regulating the CO2 concentration of the blood (c) Heart rate increases during exercise. Describe the role of the chemoreceptors and

cardiovascular centre in increasing heart rate. [5] …………………………………………………………………………………………………

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(d) A student set up an experiment to investigate the behaviour of woodlice. Woodlice are crustaceans that have gills on their underside that they use for gas exchange. The student set up two petri dishes, one had moist filter paper in the bottom the other had had no filter paper. The student then observed the movements of the woodlice and counted the number of turns that occurred during a three minute period.

The data obtained is shown below.

Environment Number of turns in a three minute period

Trial 1 Trial 2 Trial 3 Mean

Moist conditions 9 7 6 7.3

Dry conditions 17 16 19

(i) What type of behaviour is being exhibited by the woodlice? [1]

…………………………………………………………………………………………………

(ii) Calculate the mean rate of turning per minute of woodlice in dry conditions. [2]

Answer ……………………………..

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(iii) What conclusion can be drawn about woodlice behaviour from the data opposite? [3]

………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… ………………………………………………………………………………………………… …………………………………………………………………………………………………

(iv) Explain two ways in which this experiment could have been improved. [2]

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…………………………………………………………………………………………………

20

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OPTION D: FOOD SCIENCE

11. The graph below shows changes in basal metabolic rate (BMR) in human males and females as their age increases.

(a) (i) Why is a person’s “basal metabolic rate” not necessarily the same as

their “metabolic rate”? [1]

…………………………………………………………………………………………

………………………………………………………………………………………… (ii) Suggest why there is a difference in the BMR of males and females. [2]

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

(b) Data from the UK Department of Health (2012) states that all age groups

have more sugar in their diets than the recommended maximum of 11% of food energy intake. This was especially true for the age group 11 – 18 years, with 15.3% of their energy coming from sugar.

(i) A 17 year old girl has a recommended maximum intake of 55 g sugar

each day, which is 11% of her recommended energy intake. She actually ingests 15.3% of her energy from sugar. How many grams of sugar is she eating each day? Show your working. [2]

Answer ……………………………

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(ii) Some people who have a consistently high sugar intake develop a high body mass index. Outline how you would measure BMI. [3]

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(c) Excess sugar may be consumed in processed foods such as soft drinks. The

information shown below was taken from a soft drink.

A high proportion of the high fructose corn syrup is obtained from maize.

(i) Name the plant family to which maize belongs. [1]

…………………………………………………………………………………………

(ii) Scientists state that maize should not be classed as a vegetable but a fruit. Explain their statement. [2]

…………………………………………………………………………………………

…………………………………………………………………………………………

…………………………………………………………………………………………

INGREDIENTS: high fructose corn syrup; sucrose; water; salt; caffeine; natural flavouring; flavour enhancers; various colourings (artificial and natural); sorbic acid; sodium benzoate

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(iii) Distinguish between a flavouring and a flavour enhancer. [2]

…………………………………………………………………………………………

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(iv) To comply with EU legislation, what additional information must be provided about these ingredients on the food label shown above? [2]

…………………………………………………………………………………………

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…………………………………………………………………………………………

(d) A study was carried out into the effects of soft drink consumption on mental

health in teenagers. A group of 16 year olds were asked to record the number of soft drinks they consumed and were then asked to give themselves a hyperactivity score using a standard scoring system. The results for boys and girls in the study are shown below. The number of teenagers in each sample is shown below each range bar.

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The study concluded that there was a direct correlation between the amount of sugar consumed and hyperactivity and that girls are more affected by high sugar levels than boys. Evaluate the validity of these conclusions. [5] …………………………………………………………………………………………

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COMPONENT 1 – ENERGY FOR LIFE

MARK SCHEME

GENERAL INSTRUCTIONS The mark scheme should be applied precisely and no departure made from it. Recording of marks Examiners must mark in red ink. One tick must equate to one mark. Question totals should be written in the box at the end of the question. Question totals should be entered onto the grid on the front cover and these should be added to give the script total for each candidate. Marking rules All work should be seen to have been marked. Marking schemes will indicate when explicit working is deemed to be a necessary part of a correct answer. Crossed out responses not replaced should be marked. Marking abbreviations The following may be used in marking schemes or in the marking of scripts to indicate reasons for the marks awarded. cao = correct answer only ecf = error carried forward bod = benefit of doubt

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Question Marking details Marks Available

AO1 AO2 AO3 Total Maths Prac

1 (a) (i) 55 (1) 1 1 1

(ii)

Anomalous result for first count (1) Unreliable data (1)

2 2 2

(iii) Any 3 (x1) from: Collect oxygen and measure volume (1) Place in thermostatically controlled water bath (1) Use known concentration of bicarbonate/dissolved CO2 (1) Use a growlamp with correct wavelength of light (1) Use known mass/weight/leaf area of Elodea (1)

3 3 3

(b) Any 4 (x1) from: It stops electrons from PS II being moved to PS I (1) So blocking the reduction of NADP+ to NADPH (1) No photolysis of water (1) CPP is not stopped as the electrons pass from PSI and return to PSI (1) And the carrier involved in this is not affected (1)

4 4

(c) Any 3 (x1) from: Plant cannot generate NADPH2 so Calvin cycle stops (1) Less ATP synthesised (1) No glucose will be formed (1) So growing parts of the plant/sinks will not have glucose for respiration (1)

3 3

Question 1 total 0 8 5 13 1 5

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2 (a) (i) (Ribosomes) are not involved in respiration (1) 1 1

(ii) Any 2 (x1) from (Mitochondria) cannot metabolise glucose/have no enzymes for glucose breakdown (1) Pyruvate is broken down in the Krebs cycle producing CO2 (1) No lactate with either substrate as mitochondria are not involved in anaerobic respiration (1)

2

2

(iii) (Cytoplasmic residue) breaks down glucose anaerobically to lactate (1) Enzymes for pyruvate breakdown are in mitochondria (1)

2

2

(b)

Non competitive inhibitor (1) Of the ETC/(last) proton pump/ Cyanide must stop Krebs cycle Since no CO2 is produced (1)

1

1

2

Question 2 total 1 1 5 7

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3 (a) Poor growth and yellow leaves (1) 1 1

(b) Any 4 (x1) from: Same species/ stage (1) Same volume of water (1) Same temperature (1) Same light intensity (1) Repeats (1) Air bubbled through (1) Covered in card/foil (1) Change solutions weekly (1)

4 4 4

(c) Involved in chlorophyll synthesis (1) 1 1

(d)

203-52=151 (1) 151 x 100 = 74.38% (1) 203

2 2 2

(e) Nutrients/ named nutrients stored in seed (1) 1 1


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4 (a) (i) Bacillus (1) 1 1

(ii) Rod shaped, accurate drawing (1) Any 3 (x1) Labels from: Cell wall (1) Mesosome (1) Ring of DNA/ nucleoid (1) Ribosomes (1) Membrane (1) Cytoplasm (1) Any one of flagellum/plasmid/pili/capsule (1)

4 4 1

(b) (i) Stains red/does not stain with crystal violet (1)

1 1 1

(ii) Has a thin murein/peptidoglycan cell wall (1) Has an additional lipopolysaccharide/lipoprotein layer outside the cell wall (1)

2 2

(iii) Increased resistance to antibiotics/penicillin/lysozyme (1) 1 1


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5 (a) Ribose and Adenine (1) 1 1

(b) Any 2 (x1) from: (Bond x) is hydrolysed (1) To form ADP and Pi (1) Catalysed by the enzyme ATPase (1) 30.6 kJ energy released (1)

2 2

(c)

Any 4 (x1) from: Reduced {NAD/FAD} provides a source of protons and high energy electrons (1) Electrons lose energy as they pass from carrier to carrier (1) (This energy is) used to pump protons into the {inter-membrane space/thylakoid space} (1) Protons diffuse down a electrochemical gradient (1) Through a {stalked particle/ATP synthase/ATP synthetase} (1) Which provides the energy to phosphorylate ADP (1)

4 4

Question 5 total 7 7

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6

(a)

(i)

440

6 (1)

73.3 million bacteria cm3 hr-1 (1)

2

2 2

(ii)

Axes correctly assigned with correct labels plus units (1)

Appropriate linear scales (1) All points correctly plotted and joined with a curve or ruled straight lines (tolerance ½ small square) (1)

3

3 3

3

(iii)

(0-6 hours) lag phase and (6-12 hours) {exponential/log} phase (1)

1

1

(iv)

The concentration of toxic by-products wasn’t high enough/ ORA (1) The experiment could be left for a longer period of time (1) The population may have contained both living and dead cells/total count(1) Viable count used so that only living cells counted (1)

4

4 4


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7 (a) (i) Correct part labelled/circled (1) 1 1 (ii) Any 2 (x1) from:

Complementary shape to glucose/ similar shape to substrate and active site (1) Substrate/glucose will {fit into/binds with} {active site)/enzyme} (1) (not: competes) Can form enzyme substrate complex (1) Glucose can't fit anywhere else (1)

2 2

(iii) tertiary 1 1

(b)

Any 3 (x1) from: Complementary/similar shape to active site (1) (not: ref. to substrate) Enters active site/ occupies active site (1) Blocks active site (1) (not: competes/ joins active site) No glucose can enter (1) Less/ no enzyme substrate complex formed/ fewer products formed (1) Glucose not metabolised/no products formed/ metabolised more slowly (1) Competitive inhibitor (1)

3

3

(c) A is arabinose (1) 5C / pentose sugar – glucose is a 6C / hexose sugar (1)

1

1

2

(d) Ethanol and methanol must have similar shapes (1) Any 2 (x1) from: Ethanol acts as a competitive inhibitor (1) Methanol can't be converted to formaldehyde/converted more slowly (1) No/reduced toxic effects (1)

2

1

3


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8 (a) 1 890 (kJ m-2 yr-1) (1) 1 1 9

(b) (i) 6 606 x 17 000 (1) 112 302 000 (kJ m-2 yr-1) (1)

2 2 9

(ii) Any 2 (x1) from: Passes to {decomposers / detritivores} (1) {Respired / released} by decomposers (1) lost as heat (1)

2 2

(c)

Any 4 (x1) from: Carnivores expel a smaller proportion of {biomass/energy} in their waste compared to herbivores (1) Therefore carnivores convert a greater proportion of (animal) biomass into new animal {tissue/biomass} than herbivores (1) This is due to protein being {easier to digest/more efficiently digested} than cellulose (1) (therefore) carnivores are more efficient (1) Any correct calculation (1) Carnivores 17.5% efficient compared to herbivores 11% / 12% energy expelled from carnivore compared to 33% expelled from herbivore (1)

4 4

(d) (i)

By controlling the temperature reduces energy loss via respiration to maintain body temperature (1) (High densities) restrict movement therefore reduces energy loss via respiration for muscle contraction (1)

2 2

(ii) Any 2 (x1) from: There is a loss of energy at each stage in the food chain, due to respiration and other energy losses (1) Carnivores are further up the food chain than herbivores, so there is more energy lost in respiration (1) Making carnivorous animals very expensive to produce for food (1)

2 2


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Question Marking details AO1 AO2 AO3 Total Maths Prac

9 (a)

Any 2 (x1) from: Value of factor under consideration (1) Above this value, the global process will not be stable / below this value the global process will be stable (1) Exceeding boundary leads to gross global environmental change (1)

2 2

(b)

Biodiversity loss, climate change and nitrogen cycle processes have exceeded boundary / land use and fresh water processes have not (1)

1 1

(c)

Any 2 (x1) from: This is the “background rate” of extinction (1) In a changing environment, some species’ adaptation became unsuitable (1) They were selected against (1) Unless they could move fast enough to another habitat where they were suitable adapted (1)

2 2

(d) (i) the number and variety of species in a particular area (1) 1 1 (ii) Any 1 from:

Seed / sperm banks (1) Captive breeding programmes (1) Fishing quota (1) Trade restrictions (1) Management of wild populations (1) Restrict habitat destruction / pollution / deforestation / other means of habitat destruction (1)

1 1

(e)

Burning fossil fuels puts carbon dioxide directly into the atmosphere (1) Deforestation reduces carbon dioxide uptake (1) Decay/ burning of felled trees puts carbon dioxide directly into the atmosphere (1)

3 3

Question 9 total 4 6 10

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10. (a) Any 10 (x1) from:

A. Viable count is counting living cells (1)

10

B. As opposed to a direct count which counts both living and dead cells (1)

C. One cell gives rise to one colony/so No of colonies = No of viable cells (1)

D. Sterilisation/use of sterile equipment and media (not just use of aseptic technique) (1)

E. One eg of sterilising technique – autoclave/oven/radiation (1)

F. Culture needs diluting by {ten-fold/100-fold} steps (1)

G. Serial dilution (1)

H. 1 cm3 of original sample added to 9 cm3 of (sterile) water (or 100-fold 0.1 + 9.9 cm3) (1)

I. Mixed and process repeated (1)

J. Known volume(or e.g. such as 1 cm3/0.5 cm3) of serial dilution is added to agar plates (1)

K. Incubated at 25 oC (up to 37 oC) (1)

L. for 24-36 hours (1)

M. Count No of colonies in appropriate plate/50-100 colonies (1)

N. Multiply by dilution factor to calculate No of cells per cm3 in original sample (1)

O. Some comment on unreliability with reference to clumping of cells (1)

Question 10a total 10 10

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Question Marking details

Marks Available


10. (b) Any 10 (x 1) from:

10

A. Breakdown of {proteins /amino acids /nucleic acids} into ammonium ions (1)

B. Decomposition / putrefaction (1)

C. Nitrifying bacteria (1)

D. Nitrosomonas converts ammonium compounds to nitrites (1)

E. Nitrobacter converts nitrites to nitrates (1)

F. Nitrogen fixing bacteria (1)

G. Atmospheric nitrogen converted into organic nitrogen / e.g. (1)

H. Free living azotobacter (1)

I. Rhizobium (1)

J. Root nodules of (legumes) (1)

K. Denitrification converts nitrates into atmospheric nitrogen (1)

L. Encourage aerobic conditions (to stop denitrifyers) / ploughing (1)

M. Drainage for aerobic conditions to stop denitrifyers (1)

N. Add {organic waste products / manure / urea} to encourage putrefaction and nitrification (1)

O. Grow leguminous crops to encourage nitrogen fixation (1)

Question 10b total 10 10

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COMPONENT 1: ENERGY FOR LIFE

SUMMARY OF MARKS ALLOCATED TO ASSESSMENT OBJECTIVES

Q AO1 AO2 AO3 TOTAL MARK MATHS PRAC

1 8 5 13 1 5

2 1 1 5 7

3 1 4 4 9 2 4

4 4 5 9 2

5 7 7

6 6 4 10 5 7

7 3 8 1 12

8 7 6 13

9 4 6 10

10 a/b 10 10

TOTAL 30 45 25 100 10 16

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COMPONENT 2 - CONTINUITY OF LIFE

MARK SCHEME


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1 (a) A group of individuals that can interbreed and produce fertile offspring (1)

1 1

(b) (i) Birds (1) 1 1

(ii) Britain - 0.09 (1) Borneo -1.62 (1)

2 2 2

(c) Capture/recapture (1) accept Lincoln Index {Trap/catch} a sample, mark and release, {trap/catch} second sample, calculate proportion of marked animals. (1)

1 1 2 2

(d) (i)

(1)

1 1

(ii) A- M mops and B- C sphinx and C- P lucasi (1) 1 1 (e) correct missing taxa –phylum,order,family (1)

correct missing names –animal(ia), mammal(ia), penthetor, lucasi (1)

2 2


X

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2 (a) q2 = 1/625 =0.0016, q = 0.04 (1)

p + q = 1, p = 1.0 – 0.04=0.96 (1) (1 for each of working out each of q and p if answer not correct) 2pq = 2 x 0.04 x 0.96 = 0.0768 (3 for correct answer alone)

3 3 3

(b) (i)

Sickle cell trait gives some protection from malaria (1)

1 1

(ii)

Larger sample size (1) Adjust groups for age/ sex (1)

2 2

(c) People with sickle cell trait are less likely to be killed by malaria (1) They are heterozygous/ carry the sickle allele (1) More will survive to pass on the sickle allele to the next generation (1) The sickle allele will become more frequent in the next generation (1)

4 4


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3 (a) 2 marks for all correct, 1 mark for 3 correct A- spermatogonia B- primary spermatocyte C- secondary spermatocyte D- spermatids

2

2

(b) (i) Chromosomes /chromatids (1) Spindle (fibres) (1)

2 2

(ii) Meiosis (correct spelling only) (1) 1 1

(iii) 4 (1) 1 1

(iv)

Half the number of chromosomes/ haploid (1) New/ different combinations of alleles on the chromosomes (1)

2 2

(c) (i) Sertoli (1) 1 1

(ii) It is synthesising protein /protein synthesis (1) 1 1

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(d) (i) P- FSH / follicle stimulating hormone, Q-LH luteinising hormone (1)

R- Oestrogen , S- progesterone (1)

2 2

(ii) Follicle grows/ primary oocyte splits to secondary (by meiosis) (1) Fluid filled cavity develops (1)

2 2

(iii) There is an initial increase in FSH level in the blood (1) 1 1

(iv) The larger the corpus luteum the higher the level of progesterone (1) because the corpus luteum produces it. (1)

2 2

(v) Any 2 (x1) from: The corpus luteum will not degenerate (1) The level of S/progesterone will stay high (1) Uterus lining maintained (1)

2 2


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4 (a) (i) By fertilisation (1) One male nucleus with 2 polar nuclei (1)

2 2

(ii) 14 (1) 1 1

(b) (i) The mass lost from the endosperm (1) Used to increase the mass of the embryo (1)

2 2

(ii) Any 5 (x1) from: Different age of seeds (1) Same mass of seeds (1) Same/ fixed time (1) Add iodine (1) Measure diameter/area of clear zone (1) Repeats (1) Reference to suitable control (1)

5 5 5

Question 4 total 2 1 7 10 5

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5 (a) (i) 3450-2675 = 775 (1)

+30 = 805 bp (1)

2 marks for correct answer with units

2 2 2

(ii) EcoRI and HindIII (1) 1 1

(b) (i) breaks hydrogen bonds/ causing DNA strands to separate (1)

1 1 1

(ii) allows {primers/short pieces of DNA} to {anneal/bind/join} (1) 1 1 1

(iii) Enzyme adds compelmentary nucleotides / extension/ forms the phosphodiester bonds in the sugar-phosphate backbone (1)

1 1 1

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(c) (i)

AO2 Any 2 (x1) from: Gel electrophoresis is a method of separating DNA fragments according to size (1) The gel is made from agarose which contains pores in its matrix OWTTE (1) DNA samples are loaded into wells at one end and a voltage is applied across the gel (1) DNA is attracted to the positive electrode due to its negative charge on the phosphate group (1) AO3 Smaller fragments find it easier to migrate through the pores in the gel and so travel further than large fragments in the same time (1) Fragment size can be estimated by using running a DNA ladder (which contains fragments of known size) alongside (1)

2

2

4 4

(ii) From sequencing the expected fragment size = 52bp and From gel result band produced is approximately 50bp/half way between 75 bp and 25 bp ladder band OWTTE (1) Therefore good agreement between the results of the two tests (1)

2 2

(iii) Use more DNA fragments of values between 25 and 75/ smaller size than 75/ OWTTE (1)

1


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6 (a) Contains two different alleles (1) 1 1

(b) Parental genotypes Yyww yyWw (1) Gametes Yw, yw yW, yw (1) Yw yw yW YyWw yyWw yellow,wrinkled green, wrinkled yw Yyww yyww yellow, smooth green, smooth Genotypes correct (1) Genotype linked to phenotype (1) ratio 1 : 1: 1: 1 (1) AO3

4 1 5

(c)

(i)

Category O E 0-E (O-E)2 (O-E)2

E

Yellow wrinkled

27 25 2 4 0.16

green wrinkled 22 25 3 9 0.36 Yellow smooth 28 25 3 9 0.36 Green smooth 23 25 2 4 0.16

Σ

100 Σ 1.04

E column (1) (O-E)2 column (1) χ

2 result = 1.04 (1)

3

3

3

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(c) (ii) Table value = 7.82 (1) Accept null hypothesis (1) The calculated value is less than the critical value and so there is no significant difference at p=0.05 (1) deviation due to chance (1.04 < 7.82) (1) genes are not linked (1)

5 5 1

Question 6 total 1 7 6 14 4

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7 (a) Semi-conservative (1) 1

(b) Interphase (1) 1

(c) Any 2 (x1) from: Helicase (1) DNA polymerase (1) Ligase (1)

2

(d) 3 correct = 2 marks, 2 correct = 1 mark Base, deoxyribose, phosphate

2

(e) (i) (After three days) number of cells is much higher in cancer patients/ ref to data (1) (Reason) unrestricted cell division/mitosis (leading to cancerous growth) (1)

2

(ii) Normal growth (1) 1

(f) (i) Meiosis (correct spelling) (1) 1

(ii) Any 2 (x1) from: Crossing over has occurred /bivalents align in {stage B/metaphase} (1) Homologous chromosomes pulled to opposite poles in {stage C/anaphase 1}/ centromeres don’t split in {stage C/anaphase 1} (1) {following stage D / telophase (9)} chromosome number is halved (1)

2

(iii) One cell with two chromosomes as single chromatids (1) presence of nuclear membrane (1)

2


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8 (a) Any 10 (x1) from:

A. Cells from Islets of Langerhans (1)

10

DNA method

B. Gene for insulin located (1)

C. Gene/DNA probe (1)

D. (Cut gene) restriction endonuclease /enzyme (1)

Or mRNA method

E. MRNA for insulin isolated (1)

F. Reverse transcriptase qual. (e.g. make DNA from RNA) (1)

G. DNA polymerase qual. (e.g. converts to a double strand for incorporation into plasmid) (1)

H. (Restriction enzyme) leaves {sticky ends/unpaired bases} on gene (1)

I. Plasmids from {a bacterium/E.coli} (1)

J. Are cut with same restriction endonuclease (1)

K. Sticky ends with complementary unpaired bases (1)

L. DNA ligase qual.; (e.g. anneals sticky ends) (1)

M. Ref marker/tracer genes/antibiotic resistance (1)

N. Recombinant DNA introduced back into bacterium/ E. coli (1)

O. Ref to reverse transcriptase method has an advantage as introns are removed (1)


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8 (b) Any 10 (x1) from: Pollination

10

A. Pollination transfer of pollen from anther to stigma (1) B. Cross pollination involves pollen transfer from anther to

stigma of different plants whereas self pollination involves pollen transfer from anther to stigma of same plant (1)

C. (Wind pollination) light, smooth pollen qual (e.g. to be carried by wind) (1)

D. (Wind pollination) large feathery stigma qual. (e.g. to catch pollen) (1)

E. (Insect pollination) large, sticky pollen qual. (e.g. to stick to insect) (1)

F. (Insect pollination) {nectarys/large brightly coloured flowers} to attract insects (1)

Fertilization G. (After landing on stigma) pollen grain absorbs water/ref.

sucrose (1)

H. Pollen tube grows controlled by tube nucleus (1) I. {Enzymes/pectinase} are secreted to digest path through

{style/carpel/ovary wall} (1)

J. Passes through micropyle (1) K. Male gamete fuses with egg cell (1) L. Forms zygote which develops into embryo plant (1) M. Ref double fertilization (1) N. Second male gamete fuses with two polar nuclei to form

triploid nucleus (1)

O. Which forms endosperm (1) Question 8b totals 10 10

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COMPONENT 2: CONTINUITY OF LIFE



1 2 6 2 10 2 2

2 7 3 10 3

3 5 14 19

4 2 1 7 10 5

5 3 5 5 13 2 7

6 1 7 6 14 4

7 7 5 2 14

8a/b 10 10

TOTAL 30 45 25 100 11 14 DRAFT


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COMPONENT 3 - REQUIREMENTS FOR LIFE

MARK SCHEME


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1 (a) (i) To become acclimatised/ adjusted to the surrounding (1) 1 1 1

(ii) Replace the air with normal air to restore normal breathing patterns (1)

1 1 1

(b) (i) Air through spiracles (1) {Diffuses/carried to} tissues through tracheal system / trachea (1)

2 2

(ii) Movement / contraction of body needed to force air in and out of tracheae (1)

1 1

(c) (i) Carbon dioxide (1) Increase of O2 from 83-97% decrease of 7 breathing movements per minute but increase of CO2 from 0 – 3% increase of 35 (1)

2 2 1

(ii) Accept a number between 44 and 46 (1) 1 1 1

(iii) Exhaled air contains approx 4% CO2 (1) Little difference between the effects of 3, 6 and 12% CO2 on breathing rate (1)

2 2

Question 1 total 2 2 6 10 1 3 DRAFT


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2 (a) Effector = D (1) Connections to neurones in brain = F (1) Cell body in dorsal root ganglion = B (1)

3 3

(b) Enclosed in a myelin sheath / layer of fatty insulation /correct ref to Schwann cell (1) Speeds up rate of transmission (1)

2 2

(c) 2.5 / 5.7 (1) 0.438….. (1) 0.44 (to 3 sig figures) (1)

3 3 3

(d) (i) Any 4 (x1) from: Na+ ion (voltage gated) channels open (1) Na+ {flood / diffuse rapidly} into cytoplasm (1) By facilitated diffusion (1) K+ ion channels remain closed (1) More +ions inside axon than outside (1) depolarisation of the axon membrane (1)

4 4

(ii) Depolarisation exceeded threshold potential (1) Correct ref to all or nothing (1)

2 2

(e) Any 2 (x1) from:

Could have been myelinated (1) Different diameter of axons (1) Different threshold potentials (1) Different distribution / concentration of ion channels (1)

2 2


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3 (a) Autotrophs use energy to combine simple inorganic materials and manufacture their own complex organic compounds (1) Heterotrophs have to consume food material that contains complex organic molecules both as a source of raw materials / energy (1)

2 2

(b) (i) Amoeba carries out digestion intracellularly (1) 1 1

(ii) Different regions specialised to digest specific foods / carry out different functions (1) Provide different optimum conditions for different enzymes (1)

2 2

(c) (i) Goblet cells (1) 1 1

(ii) Epithelium (1) 1 1

(iii) Exocytosis (1) 1 1

(d) 77 / 3 700 (1) x 1 000 (1) 20.8 μm (1)

3 3 2 3


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4 (a) (i) Cortex (1)

1 1

(ii) X Mitochondrion (1) ATP for active transport (1) Y Basement

membrane (1) Thin pores to aid diffusion (1)

Z Microvilli (1) Increase surface area for absorption (1)

6 6

(b) (i) Diffusion (1) 1 1

(ii) Osmosis (1) 1 1

(iii) Inhibit aerobic respiration/ oxidative phosphorylation (1) ATP not produced (1) Active transport of Na+ stops (1)

3 3

(c) (i) Pinocytosis (1) 1 1

(ii) Uptake of fluid/ large soluble molecules (1) 1 1

(d) Identify D as the loop of Henle (1) Any 2 (x1) from: Longer the loop of Henle increased reabsorption of water/ ORA (1) (therefore) urine more concentrated / ORA (1) Enables kangaroo rat to live in xeric / dry habitat or beaver does not need to control water loss as lives in aquatic habitat (1)

3 3


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5 (a) (i) Potometer (1) 1 1 1

(ii) 0.52 x 3.14 x 50 (1) = 39.25/10 3.925 mm3 min- 1(1)

2 2 2

(b) (i) Any 4 for 2 marks/ OR any 2/3 for 1 mark from: Temperature Humidity Wind speed Light intensity

2 2 2

(ii) Measure surface area (1) Calculate result as water loss per mm2 of leaf / express result in mm3 min-1mm-2 (1)

2 2 2

(c) In the dark, pressure potential/ turgor pressure in guard cells falls (1) Stomata close (1) Reduces water lost by transpiration (1)

3 3 3


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6 (a) 4 (1) 1 1

(b) Transcription of gene for HbF/ gamma chain (1) Produces mRNA (1) Translation by ribosomes (to produce HbF) (1)

3 3

(c) More oxygen absorbed (1) Less oxygen released at low partial pressures of oxygen (1)

2 2

Question 6 total 4 2 6

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7 (a) Any 10 (x1) from: A. Accurate drawing of a synapse (1)

10

B.

C. D.

6 labels (2 correct labels per mark)

E. When action potential arrives at axon terminal Ca2+ moves into synaptic knob (1)

F. Synaptic vesicles fuse with presynaptic membrane (1) G. Neurotransmitter/ACh/eq released by exocytosis (1) H. Diffuses across synaptic cleft (1) I. ACh binds to receptors on post synaptic membrane (1) J. Causing the protein channels to open/Na gates open (1) K. Na+ flows in through channels (1) L. So depolarising the post synaptic membrane (1) M. With sufficient depolarisation, an impulse/action potential is

generated in the post synaptic neurone (1)

N. role of ACh-esterase in the breakdown of Ach (1) O. Diffusion back into axon terminal through presynaptic

membrane (1)

P. Use of ATP for re-synthesis and packaging / active transport of Ca2+ ions out of presynaptic knob (1)


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7 (b) Any 10 (x1) from: A. Osmoreceptors (1)

10

B. Hypothalamus in brain (1) C. Detect changes in water potential of blood (1) D. Water potential down (1) E. Loss of water or increase in salt (1) F. Nerve impulse (1) G. Posterior pituitary (1) H. Cells release ADH into blood (1) I. (ADH) produced by neurosecretory cells and stored in

pituitary (1)

J. Negative feedback (1) K. Glycoprotein receptors collecting duct walls (1) L. Membrane more permeable to water (1) M Water reabsorbed by osmosis (1) N. Into tissue fluid/ blood (1) O. Small quantity of concentrated urine (1) P. Water potential increases, ADH not released and less water

reabsorbed/ large volumes of dilute urine (1)

Question 7b total 10 10 DRAFT


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Option A

8 (a) Bacteria can be treated with antibiotics + viruses not affected (1) Accept: only bacteria affected by antibiotics

1 1

(b) (i) Antibiotics have killed bacteria in region around discs (1) 1 1

(ii) No clear zone observed (1) 1 1

(iii) Use different concentrations of same antibiotic (1) measure {diameter / area} of bacteria killed (1)

2 2 2

(c) (i) 1.5 μg cm-3 (1) 1 1

(ii) Prevent the growth of bacteria (1) Reject kill bacteria

1 1

(iii) Gives quantitative data (1) Actual concentration at different points known (1)

1 1 1

(iv) Antibiotic 1 (1) Causes inhibition /death of bacteria at lowest concentration (1)

2 2

(d) (i) Affect metabolic processes common to most bacteria (1) 1 1

(ii) A microorganism which should be affected by an antibiotic is no longer susceptible to it OWTTE (1)

1 1

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8 (e) A01 Tetracycline affects protein synthesis / process common to all bacteria (1) A02 Gram -ve bacteria have an {outer / protective} layer of {lipopolysaccharide / lipoprotein} (1) Prevents penicillin reaching murein cell wall (1)

1 2 3

(f) (i) 1 000 000 times more (1) Antibody concentration is a log scale (1) 108 = 100 000 000 102 = 100 (1)

3 3 2

(ii) Stimulates production of memory cells (1) Antibodies produced faster if exposed to the infection (1)

2 2

Option A Question total 6 9 5 20 2 3

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Option B

9 (a) (i) 80 eye piece units = 98 stage micrometer units (1) 1 epu = 98 x 10 = 12.25μm (1) 80

2 2 2 2

(ii) Eye piece unit has different values for each calibration (1) 1 1 1

(iii) Haversian canal in LS at centre, lacunae in parallel strips (1) 1 1

(b) (i) Protein synthesis (1) For release / secretion from cell (1)

2 2

(ii) Any 4 (x1) from: Arteries and veins in Haversian canals (1) Haversian canals connected by Volkmanns canals (1) Capillaries in canaliculi (1) Diffusion from capillaries into cells (1) Ref. lymphatic vessels in Haversian canals (1)

2 2 4

(c) Any 3 (x1) from: Lack of Vitamin D (1) Low fat diets (1) Protection from UV / Sun, not enough time exposed to sun (1) Vitamin D needed for absorption of calcium from gut (1)

3 3 DRAFT


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9 (d) (i) Any 3 (x1) from: Released from sarcoplasmic reticulum binds to troponin (1) Troponin changes shape (1) Moves tropomyosin away from binding sites on actin (1) Cross bridges can then form (1)

3 3

(ii) Energy from ATP needed to break cross bridge (1) No respiration in muscle after death therefore ATP not synthesised (1)

2 2

(e) Biceps and Triceps, or any correct alternative (1) Muscles can only contract and relax, cannot act as pistons and return themselves to original position (1)

1 1 2

Option B Question total 6 9 5 20 2 3

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Option C 10 (a) (i) Produces an image/diagnosis faster (1) 1 1

(ii) Any 2 (x1) from: May cause a stroke (1) Resulting in paralysis (1) In the left side of the body (1)

2 2

(b)

Activity Letter

Regulating core body temperature. B (1) Recognising a face in a picture. A (1) Drinking from a glass. D (1) Regulating the CO2 concentration of the blood. C (1)

4

4

(c)

Any 5 (x1) from: Chemoreceptors in the carotid/aortic sinus (1) Detect a decrease in blood pH (1) Due to an increase (rate of) respiration (1) Send nerve impulses to the medulla/cardiovascular centre (1) More nerve impulses (1) Are sent to the SAN (1) Along the sympathetic neurone (1) Results in the release of more noradrenalin (1) SAN discharges at a higher frequency (1)

5 5

(d) (i) Kinesis (1) 1 1

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10 (d) (ii) 52/3 = 17.3 (1) 5.8 turns per minute (1)

2 2 2

(iii) Woodlice turn more frequently in drier conditions than in moist conditions (1) In dry conditions the gills of the woodlice would dry out (1) In dry conditions the woodlice turns more frequently to try to leave the area of dry conditions and find moist conditions (1)

3 3 1

(iv) Use dry filter paper in the bottom of the petri dish in dry conditions/ use a range of humidity (1) Record the speed of movement of the woodlice (1)

2 2 2

Option C Question total 6 9 5 20 2 3

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Option D

11 (a) (i) BMR = rate of metabolism when the body is at rest and metabolic rate refers to the rate of metabolism at any given moment/varies with activity/age/health/ORA (1)

1 1

(ii) Males have {a lower proportion of body fat / higher proportion of muscle tissue than females} (1) Adipose tissue is less metabolically active than muscle tissue (1)

2 2

(b) (i) 11% energy intake = 55 g / 1% = = 5 g (1) 15.3% = 15.3 x 5 = 77 / 76.5 g (1)

2 2 2

(ii) Find mass in kg and height in metres (1) correct units for mass (kg) and height (m) (1) Calculate BMI = (1) BMI = (3)

3 3 3

(c) (i) Gramineae / poaceae (1)

1 1

(ii) Maize grains are fruits (1) Produced by fertilisation (1)

2 2

(iii) A flavouring gives a food a specific flavour (1) A flavour enhancer {has no flavour of its own / only enhances the taste or smell of a food (1)

2 2 DRAFT


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Option D

11 (c) (iv) Any 2 (x1) of: European Registration Numbers / E numbers (1) Category of additive (1) Presence / possible presence of allergens (1)

2 2

(d) Any 5 (x1) from: Large differences trend in graph agrees with the conclusion in numbers teenagers in each group (1) Don’t know volume of drinks / size of glass / mass of sugar consumed / brand of soft drink (1) Different brands contain different masses of sugar (1) Study does not differentiate between types of sugar (1) Other additives / ingredients linked to hyperactivity (1) No reference to other medical / dietary causes of hyperactivity (1) Hyperactivity only assessed by self-assessment (1)

5 5

Option D Question total 6 9 5 20 2 3

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COMPONENT 3: REQUIREMENTS FOR LIFE



1 2 2 6 10 1 3

2 2 12 2 16 3 0

3 5 6 0 11 2 3

4 4 10 3 17 0 0

5 1 2 7 10 2 6

6 0 4 2 6 0 0

7 10 0 0 10 0 0

SECTION A TOTAL 24 36 20 80 8 12

SECTION B TOTAL 6 9 5 20

2 3

OVERALL TOTAL

30

45

25

100

10

15

A Level Biology Specimen Assessment Materials/HT ED 24/06/14

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GCE A LEVEL BIOLOGY SPECIMEN ASSESSMENT MATERIALS