Question 1
1
Cells in the alveolus wall produce a surfactant. Its function is to prevent alveoli collapse at the end of expiration. Surfactants are used in the treatment of respiratory system disease in premature babies. The table shows some of the components of different surfactant preparations.
structured0 marks
Question 1(c)
1(c)
State one feature of the alveoli, other than the presence of a film of moisture, that adapts them to gas exchange.
Easystructured1 marks
Answer
large total surface area; wall is single layer of cells/thin walls/short distance between alveoli and capillaries; alveoli surrounded by (many) capillaries; Do not accept answers that make reference to membrane.
Question 1
1
Hypoxia is a condition in which tissues of the body are deprived of an adequate oxygen supply. A study was carried out in rats to examine the effects of continuing hypoxia on the structure of the diaphragm, and to determine whether nitric oxide is implicated in adaptation of the diaphragm to hypoxia. The diaphragm helps to supply oxygen to tissues and organs in the body by ventilating the lungs. A group of 36 adult male rats were kept for 6 weeks in low oxygen while 36 adult male rats were kept in normal oxygen levels.
10 marks
Question 1(c)
1(c)
Using the data so far presented in this question, explain the effect of hypoxia on the body.
Mediumstructured2 marks
Answer
a. diaphragm more endurance/stronger/generates more force for more ventilation/inspiration b. right ventricle mass increases to pump more blood c. erythrocyte percentage increases to transport oxygen d. less growth/body mass which reduces oxygen demand Reject "loss of body mass" The physiological reason is required for each mark 2 max
Question 1(f)
1(f)
Using all relevant data in the question, evaluate the effectiveness of the rats' adaptation to hypoxia.
Hardstructured3 marks
Answer
a. not effective because body mass lost b. effective because body mass still increases/rats still grow c. not effective because contractions/force exerted by diaphragm decreases d. effective because more sodium-potassium pumps so more/faster rate of diaphragm/muscle contractions e. effective because endurance of diaphragm increases f. effective because mass of right ventricle increases g. effective because erythrocyte percentage increases For each marking point the candidate must make it clear whether they are arguing for adaptation being effective or not. This can be done by giving the physiological benefit of a change, for example greater mass of right ventricle so more blood pumped. 3 max
Question 1
1
A study was carried out to explore the impact of various sports on pulmonary functions in professional athletes. The forced vital capacity (FVC) is the maximum volume of air a person can expel from the lungs after full inspiration. Forced expiratory volume in one second (FEV) is the volume of air that can forcibly be blown out in one second, after full inspiration. The table below summarizes the mean results for the different sports including the standard deviation. Removed for copyright reasons
0 marks
Question 1(a)
1(a)
0 marks
Question 1(a)(i)
1(a)(i)
State one type of apparatus that could be used to measure lung capacity.
Easystructured1 marks
Answer
spirometer OR lung volume capacity bag/balloon OR chest belt OR pressure meter Do not accept respirometer
Question 1(a)(ii)
1(a)(ii)
Outline how the FVC could be measured using the apparatus in (a)(i).
Mediumstructured2 marks
Answer
a. set to zero mark / «re»calibrate b. sit up straight or stand up/same position c. inspire/inhale as deeply as possible «through mouthpiece» expire/exhale as completely as possible d. several times e. Detail specific to apparatus such as displacement of water when using a balloon 2 max
Question 1(b)
1(b)
Using the information in the table, suggest the likely impact of long distance running on the pulmonary functions of professional athletes.
Mediumstructured1 marks
Answer
a. «useful as» will increase FVC «over time» b. «not useful as» no effect on FEV «is similar to control / small increase» c. consequence not clear «maybe only runners with higher FVC succeed to professional level» 1 max
Question 1(c)
1(c)
147 professional athletes and 30 non-athletes were included in the study. State one variable that had to be controlled in this experiment.
Easystructured1 marks
Answer
age / sex / health / height / mass Do not accept BMI 1 max
Question 1
1
The Chinese soft-shelled turtle, Pelodiscus sinensis, lives in salt water marshes. The turtle can live under water and out of water. These turtles have fully developed lungs and kidneys, however, many microvilli have been discovered in the mouth of P. sinensis. A study was undertaken to test the hypothesis that oxygen uptake and urea excretion can simultaneously occur in the mouth. Initial experiments involved collecting nitrogen excretion data from P. sinensis. The turtle urinates both in water and out of water. When in water it allows waste products to be washed out of its mouth. When out of water it regularly dips its head into shallow water to wash its mouth. The table shows the mean rates of ammonia and urea excretion from the mouth and kidney over six days.
18 marks
Question 1(c)
1(c)
It was noted that during long periods out of water, turtles rhythmically moved their mouths to take in water from a shallow source and then discharge it. Changes in the dissolved oxygen and the quantity of accumulated urea in the rinse water discharged by the turtles were monitored over time as shown in this graph.
3 marks
Question 1(c)(i)
1(c)(i)
Describe the trends shown by the graph for dissolved oxygen in water discharged from the mouth.
Mediumstructured1 marks
Answer
decrease «when head is submerged» and increase when head is out of water
Question 1(c)(ii)
1(c)(ii)
Suggest reasons for these trends in dissolved oxygen. In order to test whether a urea transporter was present in the mouth tissues of the turtles, phloretin (a known inhibitor of membrane proteins that transport urea) was added to the water in which a further set of turtles submerged their heads. The results of that treatment are shown.
Mediumstructured2 marks
Answer
ii a. oxygen absorbed from water/exchanged for urea when head dipped in water «so oxygen concentration decreases» b. lungs cannot be used with head in water / can «only» be used with head out of water c. oxygen from water «in mouth» used in «aerobic cell» respiration d. oxygen from air dissolves in water when head out of water «so oxygen concentration increases» 2 max
Question 1
1
Measurements of the lung capacity of a student were recorded using a spirometer and displayed with a data logger. Initially the student was at rest, then changed to carrying out strenuous exercise. The results are displayed in the graph.
0 marks
Question 1(a)
1(a)
Calculate the ventilation rate at rest, giving the units.
Mediumstructured1 marks
Answer
12 breaths per minute/ 6 litres per minute Accept answers from 11 to 12 breaths per minute. Accept answers from 5.5 to 6 litres per minute. Answer must include breaths or litres and a standard unit of time. Correct: eg: 12 breaths / minute eg: \(0.1 L \sec ^{-1}\) or \(6 L \min ^{-1}\) Incorrect: eg: but 12 breaths = 0 marks
Question 1(b)
1(b)
Explain the changes in ventilation after 35 seconds.
Mediumstructured2 marks
Answer
a. the volume of air per breath increases OR the volume of each breath reaches a maximum/levels off OR frequency of ventilation/breaths per minute increases b. exercise increases «rate of cellular» respiration/energy use/blood \(\mathrm{CO}_{2}\) /acidity c. exercise causes increased demand for oxygen/removal of carbon dioxide d. maximum rate/depth of ventilation is determined by the capacity of the student 2 max
Question 1(c)
1(c)
Suggest how the total lung volume at rest would differ for a patient with emphysema.
Hardstructured1 marks
Answer
«total resting lung volume» would be greater
Question 1
1
The drawing shows part of a Thunbergia grandiflora plant. It has been widely cultivated as an ornamental garden plant.
7 marks
Question 1(b)
1(b)
The drawing shows a section through a T. grandiflora flower, which contains a honeybee (Apis mellifera).
3 marks
Question 1(b)(i)
1(b)(i)
Identify the structure labelled X .
Easystructured1 marks
Answer
ovule
Question 1
1
The stomatal density is the number of stomata per area of leaf epidermis. A study was done on the stomatal density of the lower epidermis of leaves from Protium decandrum, a tree found in the Amazon forest. The image shows the epidermis stripped from the lower surface of a leaf.
0 marks
Question 1(a)
1(a)
Calculate the magnification of the image, showing your working.
Mediumpractical2 marks
Answer
Calculation: size of bar \(\div 15 \mu \mathrm{~m}(1.5 \mathrm{~cm} \div 15 \mu \mathrm{~m}\) or \(15000 \mu \mathrm{~m} \div 15 \mu \mathrm{m}\) ); Answer: 1000 x ; First marking point is for division by 15 \(\mu m\); Second marking point is for the correct answer; accept 930 and 1070 x.
Question 1(b)
1(b)
The mean stomatal density for the lower epidermis of P. decandrum was around 600 per \(\mathrm{mm}^{2}\). Predict how the stomatal density for the upper epidermis would compare.
Mediumstructured1 marks
Answer
(upper surface/epidermis usually has) fewer stomata/lower stomatal density/no stomata/OWTTE Do not accept a numerical value only.
Question 1
1
Increases in the frequency and severity of drought are part of climate change in many areas of the world. Drought represents one of the major threats to food security as it can drastically decrease crop yield. Water stress occurs when the demand for water exceeds its availability. A water stress index of 0.0 indicates non-water-stressed plants with normal transpiration and 1.0 is maximum water stress with much less transpiration.
10 marks
Question 1(a)
1(a)
Define transpiration. A study was carried out on sorghum (Sorghum bicolor), an important cereal crop. The sorghum plants were grown for 15 weeks after the date of planting. Flowering occurred in week 9 . There were 3 treatment groups in the study: - Control: plants were watered throughout the study - Pre-flowering drought: no water until week 9, followed by normal watering - Post-flowering drought: normal amounts of water until week 9, but none after. Removed for copyright reasons
Easystructured1 marks
Answer
loss of water/evaporation through the stomata/leaves;
Question 1(b)
1(b)
4 marks
Answer
i a. in both groups drought/lack of water causes (significant) increase in water stress index; b. in both groups, with water, water stress index close to values for the control/not significantly different; c. both groups have no changes between weeks 14 and 15 / values remain constant in weeks 14 and 15; Only accept similarities;
Question 1(b)(i)
1(b)(i)
Compare the changes in water stress of the pre- and post-flowering drought plants over the period shown on the graph.
Mediumstructured2 marks
Question 1(b)(ii)
1(b)(ii)
Using the data, evaluate the hypothesis that sorghum plants are more vulnerable to drought after flowering. It was known that plant growth under certain drought conditions is intimately linked to microbial communities in the root and in the soil around the root. The scientists took samples from both the root and soil, identified the bacterial phyla present and classified them into two groups: Gram-positive and Gram-negative bacteria. The graph shows the abundance in the root of the three most common Gram-positive phyla, a, b and c , and the three most common Gram-negative phyla, d , e and f , found at week 8 (before flowering), under control conditions and pre-flowering drought conditions.
Mediumstructured2 marks
Answer
a. (the hypothesis is supported as) more immediate response to drought in post-flowering plants than pre-flowering; b. at week 5 of pre-flowering drought the stress index has only reached 0.15 whereas after two weeks of post-flowering drought it is 0.5 and/or after five weeks it is nearly 0.8 OR larger/higher/greater level response to drought in post-flowering plants than pre-flowering; c. stress index reaches a maximum of 0.56 pre-flowering but 0.78 post-flowering / much higher at week 15/end of study; d. stress index remains high for post flowering; OWTTE Accept OWTTE for valid contrasts 2 max
Question 1
1
Phytoremediation is the use of plants to extract and remove contaminants or lower their bioavailability (amount available to organisms) in soil. Tree species are used because they have deep root systems and fast growth rates that enable them to take up contaminants in larger amounts than plants such as grasses. The diagram shows the steps in phytoremediation in plants. A study in North Carolina, USA, used different tree species for phytoremediation of groundwater contaminated by leaked petrochemicals. Four types of hybrid poplar clones (Populus sp.), loblolly pine (Pinus taeda) and willow (Salix sp.) were planted from 2006 to 2008, and measurements were made of trees in 2010 and 2012. The graph shows percent survival of hybrid poplar clones, loblolly pine and willow trees on the phytoremediation site.
structured15 marks
Question 1(c)
1(c)
State the relationship between fuel concentration and transpiration rate in the hybrid poplar clone OP-367.
Easystructured1 marks
Answer
negative correlation / as fuel concentration increases rate of transpiration decreases;
Question 1(d)
1(d)
Compare and contrast the effect of fuel concentration on the hybrid poplar clone OP-367 and willow cuttings.
Mediumstructured2 marks
Answer
a. both have low/reduced transpiration at \(500 \mu \mathrm{l}\) or higher (at 48 / 72 hours) OR both have lowest transpiration/greatest decrease at \(1000 \mu \mathrm{l}\) (at 48 / 72 hours); b. with \(200 \mu \mathrm{l}\) willow transpiration is increased/maximal but poplar is decreased OR at 24 hours/initially willow transpiration is increased but poplar decreased (at all concentrations) OR transpiration always lower than control in poplar but sometimes higher in willow; mpa is for a similarity and mpb is for a difference. 2 max
Question 1(i)
1(i)
Explain the difference in results for immersed cuttings and leafless control.
Mediumstructured2 marks
Answer
a. roots of immersed cuttings absorbed dioxane; b. dioxane evaporated/volatilized/lost from leaves/stomata of immersed cuttings; c. no/less transpiration by cuttings without leaves; d. less water absorbed by cuttings without leaves; e. less xylem flow/water transport up the stem/transpiration stream/transpiration pull without leaves; f. without leaves there is no/less photosynthesis/less energy for dioxane uptake/life processes; 2 max
Question 1
1
The image shows a cast of the lower surface of a leaf from a busy Lizzie (Impatiens walleriana) plant as seen under a light microscope with a magnification of \(600 \times\).
structured4 marks
Question 1(a)
1(a)
On the image, label a guard cell.
EasyPractical1 marks
Answer
any guard cell correctly labelled; Line must touch part of guard cell. Do not allow a circle around the whole stoma.
Question 1(c)
1(c)
0 marks
Question 1(c)(i)
1(c)(i)
Outline how stomatal density in busy Lizzie leaves can be estimated within a known field of view.
Mediumstructured1 marks
Answer
(count) number of stomata (within field of view / image) AND divide by the area / field of view;
Question 1(c)(ii)
1(c)(ii)
Suggest how this estimate can be made more reliable.
Mediumstructured1 marks
Answer
repeat with other areas of cast/leaf AND calculate the mean/average OR repeat/replicate with more leaves AND calculate the mean/average; 1 max