AQA AS Biology B6 Exchange Kerboodle Answers

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6.1 Exchange between organisms and their environment AQA AS Biology B6  Exchange Kerboodle Answers

1Ans Four general things that need to be exchanged between organisms and environment are: respiratory gases (oxygen and carbon dioxide); nutrients (glucose, fatty acids, amino acids, vitamins, minerals); excretory products (urea and carbon dioxide) and heat. 2Ans Sides of the cube are 10 mm. Surface area of this cube will be: 6 X area of one side= 6 X (10X10) = 600 mm². Volume of cube will be: length X Height X width = 10 X 10 X 10 = 1000 mm³. Thus the ratio of surface area to volume will be = 600/ 1000 = 0.6. 3Ans Three factors that affect the rate of diffusion of substances across the cells are:

1. Surface area to volume ratio.
2. Distance between the exchange surface and the cells where diffusion need to take place.
3. Surrounding conditions such as temperature, concentration of the substances.

1Ans Single-celled organisms absorb and release gases by diffusion through their outer surface. They have a large surface area, a thin surface, and a short diffusion pathway so there’s no need for a gas exchange system. 2.Ans By increasing the ratio between surface area to volume whales can spend much of its life in cold water. There are two shapes a cuboid and a closed cylinder. Surface area of the cuboid will be: 2 (Length + Width + height) = 2 (6+5+12) = 2 X 23 = 46 cm². Volume of the cuboid will be: Length X Width X Height = 6 X 5 X 12 = 360 cm³. Surface area of closed cylinder = 2πrh+2πr² = (2 X 3.14 X 2 X 8) + (2 X 3.14 X 2²)= 100.48 + 25.12 = 125.6 cm². Volume of a closed cylinder: πr²h = 3.14 X 2² X 8 = 100.48 cm³. Surface area of the shape will be= (surface area of the cuboid + surface area of the cylinder) – πr² = (46 + 125.6) – 3.14 X 2 X 2= 171.6 -12.56 = 159.04 cm². Volume of the shape will be = volume of cuboid + volume of cylinder = 360 + 100.48 = 460.48 cm³. Ratio between surface area and volume will be: 159.04/460.48 = 0.34.

6.2 Gas exchange in single-celled organisms and insects AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans Diffusion. 2.Ans Insects evolve mechanisms to conserve water. Increase in surface area of exchange surface will affects the water conservation. Thus to overcome this issue insects evolved an internal network of tubes known as tracheae. 3.Ans For effective diffusion rate the diffusion pathway must be of small size. That is why insects are too small in size. Thus the length of the diffusion pathway restricted the size of the insects. Spiracle movements: 1.Ans When the spiracles are closed the oxygen in the tracheae are diffuses into the respiratory tissues. 2.Ans The oxygen is continuously used up by the respiratory tissues and cells. Thus when the spiracles are closed then the oxygen is diffused into them and result is changed concentration of oxygen in tracheae. 3.Ans Increased concentration of carbon dioxide and decreased concentration of oxygen in tracheae 4.Ans Spiracle movement is important to terrestrial insects as they prevent the water loss by closing them when there is no need to exchange gases. 5.Ans Millions of years ago, the air surrounding our planet was warmer, moister and contained more oxygen. Thus the insects are larger than today.

6.3 Gas exchange in fish AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans In case of fish, flow of water over the gill lamellae and the flow of blood within the gills are in opposite direction. This is known as a contercurrent flow. 2.Ans The essential feature of countercurrent is that the water and the blood flow in opposite directions. This means that:

• By having the blood flow in the opposite direction, the gradient is always such that the water has more available oxygen than the blood, and oxygen diffusion continues to the place after the blood has acquired more than 50% of the water’s oxygen content.
• Therefore, even when the blood is highly saturated, having flowed past most of the length of the lamellae, there is still a concentration gradient and it can continue to absorb oxygen from the water.

3.Ans Mackerel fish has more gill lamellae; gill filaments and larger surface area as compared to plaice fish, to absorb more oxygen for respiration and provide energy. 4.Ans One way flow is an advantage to fish because it requires less energy and need not to be reversed the flow.

6.4 Gas exchange in the leaf of a plant AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans Gas exchange in plants is similar to that of insects.

• All the cells are near to external air thus act as a source of oxygen and carbon dioxide.
• Diffusion take place in gas phase thus the rate of diffusion is high as compared to diffusion in water.

2.Ans Two differences between plant and insects gas exchange are:

1. In case of plants there is no specific exchange system is there. Only stomata are there which regulate the entry and exit of gases.
2. Throughout the mesophyll there are numnerous interconnecting spaces which cause rapid contact of gases with the mesophyll cells.

3.Ans Stomata are surrounded by a pair of guard cells. These cells regulate the opening and closing of stomata. Thus regulate the gaseous exchange through the cells. Beside this Terrestrial organisms lose water by evaporation. Plants in such environment need to maintain a balance between gas exchange and water evaporation. Stomata play important role here. Stomata become close when there is excessive water loss. Exchange of carbon dioxide: 1.Ans Respiration. 2.Ans Photosynthesis. 3.Ans At this point carbon dioxide produced by plants during photosynthesis is used up by plants for respiration. Extra carbon dioxide is not required at this light intensity. 4.Ans Percentage increase in carbon dioxide at 15000 lux will be (160-115/ 115)* 100 = 0.3913*100 = 39.13 %. 5.Ans With stomata closed little gas exchange with environment. Some interchange of gases as carbon dioxide produced in respiration can be used for photosynthesis and vice versa but in the absence of the supply of external air photosynthesis and respiration will cease and the plant will die. 7.Ans At zero light intensity only respiration will take place.

6.5 Limiting water loss AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans In case of terrestrial plants and insects there is a need to maintain a balance between the gas exchange and water evaporation. For an efficient gas exchange what features are required make it difficult to control water evaporation. Thus, it is a challenge for a terrestrial organism to limit water loss without compromising their efficiency of gas-exchange system. 2.Ans Waterproof covering 3.Ans Because plants synthesise their food by photosynthesis and in order to obtain good yield photosynthesis should be efficient. Photosynthesis requires large surface area to volume ratio to capture more light and to exchange the gases. So to limit water loss for terrestrial plant is a great challenge. 4.a)Ans Stomata are largely confined to the lower epidermis of the leaves. Rolling of leaves protect the water loss by creating an equal water potential zone by trapping the water in a region of still air. This region becomes saturated with water vapour and thus has high water potential. b)Ans Mostly stomata are present on the lower epidermis thus rolling up of leaf in the other direction not be effective in reducing water loss. Not only desert plants have problem obtaining water 1.)Ans Plants growing on sand dunes need to have xerophytic features even though there is plentiful rainfall. Reasons are as following:

• As rain water quickly drains away through the porous sand and thus become out of the reach for plants root.
• Rate of transpiration is high in sand dunes

2.Ans Plants found in salt marshes near the coast have water surrounding their roots but due to high salt concentration and low water potential than the roots. As it is clear that, water flow from high to low water potential regions. Thus at high concentration of salts in soil it is difficult for the plants roots to absorb water. 3.Ans In many places where the ground freezes and thaws each year, most water is frozen. Plants must survive the winter with little water. They push their roots deep down under the frozen topsoil to find liquid ground water. Plants that grow on seasonally frozen ground can grow quite tall. The deep layer of soil under the layer of frozen ground can support them. 4.Ans At lower temperature enzyme involved in photosynthesis will not work efficiently. Thus the process of photosynthesis is either shut down or become slow. Thus in cold regions temperature is the limiting factor not the light. This is why having a smaller leaf area does not reduce the rate of photosynthesis in the same way as it would for plants in warmer climates.

6.6 Structure of the human gas-exchange system AQA AS Biology B6  Exchange Kerboodle Answers

1 .Ans Volume of absorbed oxygen and produced carbon dioxide are large in mammals because of the following reasons:

• Mammals are larger organisms and have large living cell volume.
• To maintain the body temperature rate of metabolic processes and respiration is high.

2.Ans Air passage way from gas exchange surface of the lungs to the nose is: Alveoli àBronchioles àBronchi à tracheae à larynx à pharynx à nasal cavities à nostrils. 3.Ans Inner tracheal and bronchial walls are made up of muscles which are lined up with ciliated epithelium and goblet cells. The goblet cells secrete mucus, which forms a blanket over the tops of the ciliated cells. The regular, coordinated beating of the cilia sweeps the mucus up and out of the airways, carrying any debris that is stuck to it. Thus trachea and bronchus protect the alveoli from damage by sending the filtered air to them.

6.7 The mechanism of breathing AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans The time between two corresponding points is 3.5 second. This indicates one cycle and so is the time for one breath. The number of breaths per minute is therefore 60÷3.5 = 17.14

2. Ans 3000cm3 = 3 dm3.

The graph shows that the exhaled volume or tidal volume is 0.48dm³. less than the maximum inhaled volume (which is 3 dm3.) So the exhaled volume is 3-0.48 = 2.52 3.Ans The muscles in the diaphragm contract causing it to flatten and mover downwards. The external intercostal muscles contract moving the rib cage up and outwards. Both actions increase the volume of the lungs and so the pressure in the alveoli is reduced. Pulmonary ventilation: 1 .Ans Pulmonary ventilation rate (dm³ min^-1) = Tidal volume (dm³) X breathing volume (min^-1) 10.2 = 0.6 X x = x= 10.2/0.6 = 17 min^-1. Breathing rate = 17 min^-1. Banner 2

6.8 Exchange of gases in the lungs AQA AS Biology B6  Exchange Kerboodle Answers

1.a)Ans Exchange surfaces (alveolar and capillary) are very thin. Thus distance over which diffusion takes place is very short. bAns In each human lung there are about 300 million alveoli present. Total alveolar surface area is around 70 m² which is huge and about half the area of a tennis court. This provides large surface area for gas exchange. c.Ans Breathing movement and heart pupming, together these ensure that a steep concentration gradient of gases to be exchanged is maintained. Breathing movement constantly ventilate the lungs whereas the pumping of heart circulates blood around alveoli. Blood flow through the pulmonary capillaries maintains a concentration gradient. Diffusion rate is more rapid which will result in the greater concentration gradient. Pumping of blood through capillaries removes oxygen as it diffuses from the alveoli into the blood. The supply to news carbon dioxide as it diffuses out of the blood helps to maintain a concentration gradient that would otherwise disappear as the concentrations equalised. d.)because of the narrow pulmonary capillaries RBCs are slowed as they pass through these capillaries. This gives more time for diffusion. Flattening of the RBCs against the wall of capillaries reduces the distance between RBCs and alveolar air. 2.AnsThus when there is 300 million alveoli present then the diffusion will be difference in concentration = 3. Surface area = 300 and length of diffusion path = 10. Rate of diffusion = 90. For 600 million alveoli rate of diffusion= 180. Thus when there is 600 million alveoli then the rate of diffusion is increased by twice. Risk factors for lung disease 1 Ans Risk factors associated with lung disease are following:

• Smoking: 90% COPD cases are due to smoking.
• Air pollution: in areas of heavy industries air pollutant particles and gases increase the chance of COPD.
• Genetic make-up: In case of some people genetic make-up is responsible for COPD. This could be reason that why non-smoker get lung-diseases.
• Infection: People who frequently get lung infection are more likely to get COPD.

2.Ans 60% 3.Ans percentage alive for smoker at the age 70 is 30% and for non-smokers it is approximately 90%. Thus the likelihood will be 90/30 =3. That means 3 times greater is the likelihood of a non-smoker living to age 70 than someone who smokes over 25 cigarettes a day. 4.Ans life expectancy of the person after quitting smoking at the age 40 years will be upto 80-90 years. 5 .Ans In pulmonary fibrosis disease considerable volume of air space is occupied by fibrous tissue. This resulted in thickened epithelium which will further increase the diffusion pathway so the diffusion of oxygen into the blood is extremely slow. 6.Ans Pulomonary fibrosis decreases the value of FEV because of decreased alveolar spaces in lungs less air will be taken in and less will be blown out. Smoking and lung cancer Figure shows death from lung cancer in the UK correlated to the number of cigarettes smoked per year during a period in the last century. Study it carefully and then answer the following questions: 1.Ans For male 1945 and for female: 1980. 2 Ans there is increase in number of deaths as the cigarettes consumption is increased in both the sexes. 3.Ans 1. Cigarette smoking; 2. Air pollution; 3. Epidemic disease. 4.Ans Because it takes time for the development of lung cancer.

6.9 Enzymes an digestion AQA AS Biology B6  Exchange Kerboodle Answers

1.Ans Hydrolysis is a reaction in which molecules are broken down by adding water molecules. 2.Ans Salivary glands and pancreas. 3.Ans Villi and microvilli increase surface area to speed up the absorption of soluble molecules. As the food in the stomach has not yet been broken down into soluble molecules they cannot be absorbed and so villi and microvilli are unnecessary. 4Ans monosaccharide α-glucose 5.Ans Maltase, sucrase and lactase. Lactose intolerance 1.Ans (a) Respiration Ans (b) These microbes do the anaerobic respiration in which carbon dioxide is not produced. 2.Ans The ability to consume lactose is prevalent in modern humans because of many variables: genome mutation, significance to development, advantageous to survival, in cohesion with cultural practices, modern technology to extend tolerance ability/alternatives. 3.Ans Undigested lactose is digested by the microorganisms in the intestine. This microbial digestion of lactose produces small soluble molecules and a large volume of gas. Microbial digestion of lactose in intestine can cause diarrhoea because the soluble molecules lower the water potential of the material in the colon. Banner 3

6.10 Absorption of the products of digestion

1Ans Mitochondria, Endoplasmic reticulum and Golgi apparatus. 2.Ans Sodium ions 3.Ans Number of protein channels is high. Absorption of fatty acids   1Ans By the adding glycerol and phosphate to the fatty acids chylomicron are formed which are easily absorbed by the intestines. 2Ans It is clear that by adding phosphate and glycerol phosphate the rate of amount of absorbed oleic acid is increased. Thus it is evidence which supports the view that the absorption of fatty acids involves phosphorylation. Banner 4

Practice questions

1.i)Ans Diffusion ii)Ans In case of flatworm Cells become flattened in shape in order to decrease the distance between surface and each cell, by this way flatworm’s body allow efficient gas exchange. (b) Ans An organ is a collection of tissues joined together as a structural unit in order to perform a common function. The leaf is a collection of tissues which include:

1. The epidermis which covers the upper and lower surfaces.
2. The mesophyll inside the leaf which is rich in chloroplasts.
3. The veins contains the vascular tissue (where xylem and phloem are present).

ii)Ans Through diffusion carbon dioxide in the air outside a leaf reaches mesophyll cells inside the leaf. Q 2 (a) Ans P = Tracheae; Q= Bronchi (bAns 1. Increases the volume in the thorax

2. Lowers the pressure in the thorax
3. Pressure outside the lungs is greater so air is pushed inside.

Q 3 (a) Ans 1. Flattens

2. Contracts

(b) Ans 1. Diaphragm contracts and Flattens

2. Increases the volume in the thorax
3. Decreases the pressure inside the lungs.
4. Air moves from high to lower pressure.

(c) Ans Many alveoli and capillaries are there which provide a large surface area for gas exchange. So, in this way rate diffusion is increased. Flattened epithelium of alveoli and capillaries will decrease diffusion pathway in between thus gas exchange become faster and it is easy to maintain a concentration gradient. 4.a)Ans Mean number of times the air bubble was renewed per four at temperature are: For 10 °C it is 10+12+8 =30/3 =10 For 20°C it is 18+22+18 = 58/3 = 19.33 For 30°C it is 44+48+38 = 130/3 = 43.33 (b)Ans the name of the shape of the line obtained is exponential. (c) Ans With increasing temperature beetle’s need for oxygen to carry out aerobic respiration is increasing. This means that the efficiency of the gas exchange system and of respiration system is increased with increase in temperature. (a) Ans In case of group C FEV is 3.5 and for Group A people it is 5. Thus percentage decrease= 5-3.5/5*100 = 1.5/5*100 =0.3*100 = 30%. (b) Ans Group B people are suffering from fibrosis disease. Group B breathe out as quickly as the healthy group A. So bronchioles are not affected in fibrosis. However the total volume of air breathed out is reduced. 6(a) Ans

Model cell description	Surface area /µm2	Volume / µm3	Ratio of surface area to volume
Cube, side length 4 µm	96	64	1.5:1
Sphere, diameter 4 µm	50.3	33.5	1.5:1
Cube, side length 6 µm	216	216	1:1
Sphere, diameter 6 µm	113.04	113.04	1:1

(b) Ans As the size of a structure increases the surface area to volume ratio decreases. Thus size is a limiting factor for gas exchange. Q 7 (a) Ans Method:

1. Place 5g of the organism (maggots) into the tube and replace the bung.
2. Introduce a drop of dye into the glass tube.
3. Open the connection (three-way tap) to the syringe and move the fluid to a convenient place on the pipette (i.e. towards the end of the scale that is furthest from the test tube).
4. Mark the starting position of the fluid on the pipette tube with a permanent marker.
5. Isolate the respirometer by closing the connection to the syringe and the atmosphere and immediately start the stopwatch.
6. Mark the position of the fluid on the pipette at 1 minute intervals for 5 minutes.
7. At the end of 5 minutes open the connection to the outside air. Measure the distance travelled by the liquid during each minute (the distance from one mark to the next on your pipette).
8. If your tube does not have volumes marked onto it, you will need to convert the distance moved into the volume of oxygen used.
9. Record your results in a suitable table.

(b) Ans Calculate the mean (adjusted) distance moved by the manometer fluid per minute. If you know the diameter of the capillary tube, you can convert the distance moved to a volume: Volume of liquid in a tube – length x πr² This gives you a value for the volume of oxygen absorbed by the organisms per minute. 8 .Ans Forced expiration is the use of the diaphragm muscles and the intercostal muscles to achieve greater expiration. Normally the diaphragm relaxes to its dome shape, reducing the thorax volume and hence increasing the pressure, causing air to be forced out of the lungs down a pressure gradient. This is sufficient when someone is resting, however when you are exercising you are respiring at a far greater rate, thus more CO2 needs to be expelled from your lungs as the blood will be more CO2 rich as a product of respiration is CO2. Additionally, when exercising your muscles require more O2 thus more air is taken in and breathing rate increases. Thus, when exercising an additional mechanism is required as the diaphragm relaxing on its own is simply not enough to force out all the air. So, the inercostal muscles are involved during forced expiration. To expire, the internal intercostal muscles contract, pulling the rib cage down and inwards. This decreases the thorax volume and increases pressure, forcing air out, whilst simultaneously the diaphragm relaxes. (b) Ans Percentage decrease in the FEV for group B compared with group A for the first second: For A the value of FEV is 4.2 while for B it is 0.8. Thus percentage decrease = (4.2-0.8/4.2)* 4.2 = 3.4/ 4.2*100 =81%. (c) Ans Asthma caused the drop in the mean FEV. This happens due to the following reasons:

1. The muscular wall of the bronchioles contracts
2. The epithelium of the bronchioles secrete more mucus
3. Therefore the diameter of the airways reduces, becoming narrower.
4. So the flow of air is reduced Therefore the rate of air breathed out is decreased, and the curve is less steep The total volume of air breathed out in a certain time is decreased therefore the curve the curve is at a lower mean volume

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