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AQA AS BIOLOGY TOPIC 3 3 EXCHANGE AND TRANSPORT www.examqa.com

A Q A AS B I O L O G Y TOPIC 3

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A Q A A S B I O L O G Y

TOPIC 3

3

EXCHANGE AND TRANSPORT

www.examqa.com

Large insects contract muscles associated with the abdomen to force air in and out of thespiracles. This is known as ‘abdominal pumping’. The table shows the mean rate of abdominalpumping of an insect before and during flight.

 

 Stage of flight

Mean rate ofabdominal pumping

/ dm 3 of air kg −1 hour −1

  Before 42

  During 186

(a)     Calculate the percentage increase in the rate of abdominal pumping before and duringflight. Show your working.

 

 

 

 

 

Answer ..................................... %(2)

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(b)     Abdominal pumping increases the efficiency of gas exchange between the tracheoles andmuscle tissue of the insect. Explain why.

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(c)     Abdominal pumping is an adaptation not found in many small insects. These small insectsobtain sufficient oxygen by diffusion.

Explain how their small size enables gas exchange to be efficient without the need forabdominal pumping.

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The graph shows the concentration of oxygen inside the tracheoles of an insect when atrest. It also shows when the spiracles are fully open.

(d)     Use the graph to calculate the frequency of spiracle opening. Show your working.

 

 

 

 

 

Frequency ..................................... times per minute(2)

(e)     The insect opens its spiracles at a lower frequency in very dry conditions. Suggest oneadvantage of this.

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(f)     The ends of tracheoles connect directly with the insect’s muscle tissue and are filled withwater. When flying, water is absorbed into the muscle tissue. Removal of water from thetracheoles increases the rate of diffusion of oxygen between the tracheoles and muscletissue. Suggest one reason why.

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(Total 9 marks)

Doctors investigated the effect of the smoking habits of men on their non-smoking wives.

The doctors recruited 540 non-smoking women aged 40 or older. They divided these women intogroups according to the smoking habits of their husbands.After 14 years, the doctors recorded how many of the wives had died and their cause of death.

They used these data to determine the relative risk of a wife dying from a particular diseaseaccording to her husband’s smoking habit.

In this comparison, they gave the relative risk to the wife of a non-smoker as 1.00. A valuegreater than 1.00 shows an increased risk compared to the wife of a non-smoker.

The results are shown in the table below.

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Cause ofdeath

Relative risk of wife dying

 

Husbandnon-smoker

Husbandsmokes 1 to 19cigarettes /day

Husbandsmokes more

than 19cigarettes / day

  Lung cancer 1.00 1.61 2.08

  Emphysema 1.00 1.29 1.49

  Cervical cancer 1.00 1.15 1.14

  Stomach cancer 1.00 1.02 0.99

  Heart disease 1.00 0.97 1.03

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A journalist concluded from these data that if a husband smoked, it greatly increased the risk ofhis wife dying of certain diseases. Evaluate this statement.

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Some people have a medical condition called pancreatitis. This can lead to their pancreatic ductbecoming blocked. As a result, a high concentration of amylase is found in their blood.

At 12-hour intervals, a doctor measured the concentration of amylase in the blood of a personsuffering from a blocked pancreatic duct. He also measured the concentration of amylase in theblood of a healthy person.

The figure below shows his results. 

    Concentration of amylase in the blood / arbitrary units

  Time / hours Person with blocked pancreatic duct Healthy person

  0 1800 800

  12 2200 750

  24 2500 700

  36 2000 750

  48 1400 800

(a)     (i)      The changes in concentration of amylase in the blood of a person with a blockedpancreatic duct are different from those of a healthy person during the period shownin the figure above.

Describe two of these differences.

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(ii)     In a person with a blocked pancreatic duct, starch digestion is affected.Explain how.

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(b)     Healthy people have amylase in their blood. This does not cause any harmful effects in thebody.Explain why.

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(c)     Pancreatitis can lead to the release of protein-digesting enzymes into the blood. This isharmful to the body.Suggest one reason why.

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(Total 8 marks)

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Biological washing powders contain enzymes which hydrolyse substances that cause stains onclothes.

A manufacturer tested the ability of two types of the same brand of washing powder to removedifferent food substances that stain clothes.

•        Type A contained an enzyme.•        Type B was identical to A except it did not contain the enzyme.

Figure 1 shows the results.

Figure 1

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A scientist worked for a company that wanted to develop a biological washing powder that waseffective over a range of temperatures. He investigated the effect of temperature on the rates ofthe reaction catalysed by two enzymes, P and S used in biological washing powders.

Figure 2 shows his results.

Figure 2

(a)     Many of the substances causing the food stains are large, insoluble proteins.Suggest how a biological washing powder removes this type of stain.

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(b)     The manufacturer of type A and type B washing powder claimed that these results showedthat biological washing powders are better at removing stains from clothes.

Use the information in Figure 1 to evaluate this claim.

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(c)     Most customers want a washing powder which removes stains from clothes over a range oftemperatures. After obtaining the results shown in Figure 2, which enzyme should thescientist recommend for use in a biological powder?

Give reasons for your answer.

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(d)     Biological washing powders often contain a number of different enzymes. This enablesthem to remove a wider range of stains from clothes.Explain why a number of enzymes are required to remove a wider range of stains.

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(Total 12 marks)

Environmental factors can affect the density of stomata in the lower epidermis of leaves of plantsof the same species.

Scientists investigated how growing plants at different temperatures affected the density ofstomata in the lower epidermis of leaves. They grew plants of the same species from seeds.Their method is outlined below.

•        They took 8 trays containing soil and planted 50 seeds in each tray.•        They put each tray in a controlled environment at a different temperature.•        When the plants had grown from the seeds, they selected 20 fully grown leaves from the

plants in each tray.

•        They determined the mean number of stomata per mm 2 in the lower epidermis for eachgroup of leaves.

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Their results are shown in the graph.

 

(a)     Give three environmental variables, other than temperature, that the scientists would havecontrolled when growing the plants.

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(b)     The scientists used a range of temperatures from 6 to 20 °C.Using their data, explain why they did not use temperatures above 20 °C.

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(c)     The scientists only selected fully grown leaves from the plants.

Suggest why.

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(d)     The plants grown at higher temperatures had a lower number of stomata per mm2.This would be an advantage to the plant because the transpiration rate increases as thetemperature increases.

Explain why the transpiration rate increases when the temperature increases.

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(Total 7 marks)

Scientists used fossil leaves from one species of pine tree to investigate whether changes in theconcentration of carbon dioxide in the air over long periods of time had led to changes in thenumber of stomata in the leaves.

Their method is outlined below.

•        They selected sites of different ages.•        They collected between 11 and 24 fossil leaves from each site.

•        They found the mean number of stomata per mm 2 on the leaves from each site.•        They estimated the age of each sample by dating organic remains around the leaves at

each site.

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They compared results from the fossil leaves with leaves from the same species of pine treegrowing today.

They knew the concentration of carbon dioxide in the air at different times in the past.

Their results are shown in the table. 

 Age of sample

/ years

Concentration ofcarbon dioxide in the

air / %

Mean number of stomata

per mm2

(± standard deviation)

  present day 0.0350 92 (±2)

     5000 0.0270 87 (±4)

  10 000 0.0250 95 (±2)

  15 000 0.0205 108 (±6)

  20 000 0.0195 115 (±4)

  25 000 0.0188 118 (±6)

  30 000 0.0190 130 (±6)

(a)     The concentration of carbon dioxide in the air has changed with time. Use the data todescribe how.

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(b)     The scientists calculated the mean number of stomata per mm2 and the standard deviation.

What does the standard deviation show?

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(c)     The scientists found the age of the fossil leaves by dating the organic remains aroundthem.Would this have affected the accuracy of their data? Explain your answer.

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(d)     30 000 years ago the mean number of stomata per mm2 on the lower epidermis of pine treeleaves was much higher than it is today. This would have enabled the plant to grow fasterwhen the carbon dioxide concentration of the air was low.

Explain why.

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(e)     A student who saw these results concluded that as the carbon dioxide concentration of theair had increased the number of stomata per mm2 in leaves had decreased.Do the results support this conclusion?

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(f)      The leaves of plants that grow in dry areas usually have a low number of stomata per mm2.Use your knowledge of leaf structure to suggest three other adaptations that the leavesmight have that enable the plants to grow well in dry conditions.

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(Total 12 marks)

(a)    The diagram shows the structure of the human gas exchange system.

Name organs

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Q ...................................................(1)

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(b)     Explain how downward movement of the diaphragm leads to air entering the lungs.

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(Total 3 marks)

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Some people have a form of heart failure where their heart is not pumping blood as well as itused to. Some people with heart failure are given an artificial heart to improve circulation of bloodfrom the left ventricle.Figure 1 shows where this type of artificial heart is connected.

 

Figure 1

 

(a)     Name the blood vessel to which the artificial heart is connected.

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(b)     In these patients, the right ventricle still produces sufficient blood flow to keep the patientalive.

Suggest why the left ventricle requires the help of the artificial heart but the right ventricledoes not.

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(c)     Figure 2 shows the internal structure of this type of artificial heart.

 

Figure 2

Valves A and B have the same functions as heart valves involved in the cardiac cycle.Name the heart valve that has the same function as:

valve A............................................................................................................

valve B............................................................................................................(2)

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(d)     There are different designs of artificial heart. Doctors compared results for patients whoreceived two different types of artificial heart, X and Y.

They recorded information 2 years after the artificial hearts were implanted. Their resultsare shown in Figure 3.

Figure 3 

   Information recorded 2 years after artificial heart

implanted

 Type of artificial

heart

Number ofpatients surviving

withoutreplacement ofartificial heart

Number ofpatients survivingbut who required

repair orreplacement ofartificial heart

Number ofpatients who

died

 X

(119 patients)62 13 44

 Y

(58 patients)7 24 27

Which type of artificial heart was the more successful? Use calculations to support youranswer.

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(Total 8 marks)

Some substances can cross the cell-surface membrane of a cell by simple diffusion through thephospholipid bilayer. Describe other ways by which substances cross this membrane.

(Total 5 marks)

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(a)     (i)      An arteriole is described as an organ. Explain why.

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(ii)     An arteriole contains muscle fibres. Explain how these muscle fibres reduce bloodflow to capillaries.

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(b)     (i)      A capillary has a thin wall. This leads to rapid exchange of substances between theblood and tissue fluid. Explain why.

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(ii)     Blood flow in capillaries is slow. Give the advantage of this.

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(c)     Kwashiorkor is a disease caused by a lack of protein in the blood. This leads to a swollenabdomen due to a build up of tissue fluid.

Explain why a lack of protein in the blood causes a build up of tissue fluid.

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(Total 8 marks)

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Students investigated the effect of removing leaves from a plant shoot on the rate of wateruptake. Each student set up a potometer with a shoot that had eight leaves. All the shoots camefrom the same plant. The potometer they used is shown in the diagram.

 

(a)     Describe how the students would have returned the air bubble to the start of the capillarytube in this investigation.

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(b)     Give two precautions the students should have taken when setting up the potometer toobtain reliable measurements of water uptake by the plant shoot.

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(c)     A potometer measures the rate of water uptake rather than the rate of transpiration. Givetwo reasons why the potometer does not truly measure the rate of transpiration.

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(d)     The students’ results are shown in the table. 

 Number of leaves removed

from the plant shootMean rate of water uptake /

cm3 per minute

  0 0.10

  2 0.08

  4 0.04

  6 0.02

  8 0.01

Explain the relationship between the number of leaves removed from the plant shoot andthe mean rate of water uptake.

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(Total 8 marks)

(a)    There are ethical and economic arguments for maintaining biodiversity.

(i)      Suggest one ethical argument for maintaining biodiversity.

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(ii)     Suggest one economic argument for maintaining biodiversity.

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Ecologists calculated the percentage of bird species that have become extinct on sixislands in the last one hundred years. They also calculated the percentage of original forestarea remaining on each island after the same time period. The graph shows their results.

 

Percentage of original forest arearemaining on each island

(b)     Explain the relationship between the percentage of original forest area remaining and thepercentage of bird species that have become extinct.

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(c)     What two measurements would the ecologists have needed to obtain to calculate the indexof diversity of birds on each island?

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(d)     The ecologists noted that the species of birds surviving on the coldest islands had a largerbody size than those surviving on warmer islands.

Explain how a larger body size is an adaptation to a colder climate

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(Total 8 marks)

The ‘placebo effect’ describes the improvement in patients’ symptoms due to psychologicaleffects. Scientists investigated the placebo effect in patients with asthma. They divided a largenumber of asthma patients into three groups, 1, 2 and 3.

•        Group 1 inhaled a spray containing albuterol every day. Albuterol is a drug used to treatasthma.

•        Group 2 inhaled a placebo spray every day. This was identical to the spray given togroup 1 but it did not contain albuterol.

•        Group 3 did not receive any spray treatment.

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(a)     Describe one way the scientists could have allocated the patients to each group.

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The scientists measured the forced expiratory volume (FEV1 ) of each patient at regularintervals. The forced expiratory volume (FEV1 ) is the volume of air forced out of the lungsin the first second when breathing out. The scientists recorded each patient’s FEV 1 beforetreatment started and after 60 days of treatment. They then calculated the mean increasein FEV1 for each group. Their results are shown in the graph. The bars show the standarddeviation.

 

Patient group

 

(b)     What do the standard deviation bars suggest about the difference in the mean increase inFEV1 between Group 1 and the other groups? Explain your answer.

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(c)     What do the data suggest about the ‘placebo effect’ in this investigation? Explain youranswer.

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(d)     On each occasion that a patient’s FEV 1 was measured, a doctor repeated themeasurement several times. Explain why.

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(e)     All the patients continued with their normal treatment for asthma. The normal treatmentwas the same for all patients and its effects were short-lived. The patients were told to stopthis treatment 24 hours before FEV1 measurements were taken.

(i)      Suggest why all the patients were allowed to continue with their normal asthmatreatment in this investigation.

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(ii)     Suggest why the patients were told to stop their normal asthma treatment 24 hoursbefore their FEV1 measurements were taken.

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(f)     After 60 days, the patients in each group were asked to give themselves an ImprovementScore from 0-10 to show how much they felt their symptoms had improved. This was donebefore their FEV1 was measured. The scientists calculated the mean Improvement Scorefor each group.

(i)      The scientists concluded that the data obtained for the Improvement Scores wereless reliable than the data obtained measuring FEV1 . Suggest why they concludedthis.

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(ii)     Group 3 reported the lowest mean Improvement Score. Suggest one explanation forthis.

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(Total 15 marks)

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          Scientists compared the results of three investigations, A, B and C. These investigations wereinto the effect of drinking different amounts of alcohol on the risk of developing heart disease.

The graph shows the results of these investigations.

 

(a)     Describe the relationship between increasing the number of alcoholic drinks per day andthe risk of heart disease in investigation A.

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(b)     All the volunteers who took part in investigation C were aged between 40 and 50 years old.Explain how choosing volunteers of a similar age improved this investigation.

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(c)     A newspaper headline used the information in the graph to claim ‘Alcohol is good for you.’Evaluate this claim.

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(Total 6 marks)

 

          The diagram shows the position of the diaphragm at times P and Q.

 

(a)     Describe what happens to the diaphragm between times P and Q to bring about thechange in its shape.

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(b)     Air moves into the lungs between times P and Q. Explain how the diaphragm causes this.

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(c)     Describe how oxygen in air in the alveoli enters the blood in capillaries.

 

 

 

 

 (2)

(Total 7 marks)

 

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