1
Answer pack
“Peppermint” by Jessie Etsy
Reproduced courtesy of the artist and Southern
Exposure Seed Exchange, Virginia USA
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Grading
Excellent
Thorough piece of work.
Excellent knowledge, understanding and
application of AS & A2 knowledge.
Good
Most parts completed to a high standard, but there
are still some areas you need to work on.
Few, if any, questions left unanswered.
Satisfactory
Good in parts, but a significant number of answers
require improvement, and/or several questions
were not attempted
Unsatisfactory
Not enough effort put into the booklet.
Too many gaps and/or not enough detail in most
questions. Gives the impression of being rushed,
or not thought out properly.
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Revision pack 1
Why mint?
All mint plants belong to the genus Mentha, There are 18 recognised species of Mentha, with
numerous varieties and a further 11 hybrids (created by crossing two different species). They grow
in the wild on every continent except Antarctica, and are noted for their production on essential
oils. They have been used traditionally for more than 2000 years as a source of medical remedies,
flavourings, fragrances and even as a source of insecticides.
Mint plants are currently one of the most important commercial herbs grown, as the essential oils
they produce are of high economic value. The most important commercial mint species are
Mentha x piperita (peppermint), Mentha spicata (Native spearmint), Mentha x gracilis (Scotch
spearmint) and Mentha canadensis (cornmint). Both Native and Scotch Spearmint are mainly
produced in North America and China; peppermint oil is produced in the USA and India, while
cornmint oil is mainly produced in India. Globally, over 40,000 tonnes of mint oils are produced per
annum, with a value of over $800 million.
The essential oils produced by different mint plants are a complex mix of alcohols, monoterpenes
and other hydrocarbons, oxides, esters, aldehydes and ketones and to a lesser extent carboxylic
acids. Peppermint oil is mainly composed of L-L-menthol and L-menthone; cornmint oil is mainly
composed of L-menthol and spearmint oil is mainly composed of L-carvone.
Other important mint oils include pennyroyal oil (an oil rich in the ketone D-pulegone), produced by
Mentha pulegium, and bergamot oil, produced by Mentha aquatic var. citrata oil, which is rich in
linalool and linalyl acetate.
The essential oils produced by mint plants are secreted by specialised cells and stored in circular
structures called trichomes. These are small glandular outgrowths of the leaf epidermis. The
natural role of these oils is to inhibit the growth of other competitor plant species (allelopathy), and
to protect the mint plants from attack by insects and pathogens such as bacteria and fungi.
Wrigley’s peppermint flavoured chewing gum was first
marketed in 1893, in the same year that Colgate
introduced mint-flavoured toothpaste.
Information kindly supplied in 2013 by Brian Lawrence, editor of Mint -The Genus Mentha. CRC Press (2007).
Lawrence (ed).
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Find out more about the historical and commercial uses of mint plants and the essential oils they
produce. Place your summary here – in any form that suits you (e.g. written; pictorial).
Consider food and drink, fragrances, medicines and any other uses you can find. (10 marks)
Mint hint: do an internet search using key words on page 1 to help you.
There are plenty of mint sources out there!
* Food & Drink:
- mint tea.
- mint sauce.
- mint jelly.
- mint chocolate.
- mint sweets.
- crème de menthe.
- mint lemonade.
- chewing gum.
- as a preservative.
* Fragrances:
- aromatherapy oils.
- soaps, bath oils, shampoo, shower gel.
- air freshener.
- cosmetics.
* Medicines:
- cough remedies.
- decongestants.
- treatment of insect bites.
- indigestion remedies.
- treatment of IBS.
- antiseptics.
* Other:
- insecticides.
- menthol cigarettes.
- mouth wash.
- breath freshener.
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Cell Membranes (AS-1.1.2)
Biological Molecules (AS-2.1.1)
Have you ever wondered why mints make your mouth feel cold? In common with many other
parts of your body, your mouth contains sensory receptor cells called thermoreceptors. The cell
surface membrane of these cells contains a protein ion channel called theTransient receptor
potential cation channel subfamily M member 8, or TRPM8. This channel opens in the presence of
cold temperature, possibly because changes in temperature cause the receptor to change shape,
allowing Na+ ions into the cell. The resulting depolarisation leads to the transmission of a nerve
impulse to the thermoregulatory centre in brain, which interprets the impulse as a cold sensation.
Researchers have found that L-L-menthol, an organic compound found in a few mint oils, activates
the same receptor by causing it to change shape, leading to the “cold” sensation we feel when
tasting mint. What’s more, if you suck on a mint and drink cold water at the same time, the “double
dose” of sensory information causes an extra cold sensation. Try it!
Source: http://boundlessthicket.blogspot.co.uk/2012/04/that-cool-mint-feeling-when-i-was-in.html
(which includes research citations).
Figure 1. Structure of a cell membrane
http://en.wikipedia.org/wiki
Figure 2. Structure of L-menthol
http://en.wikipedia.org/wiki/Menthol
Suggest why Na+ ions enter cells through protein channels in the cell membrane rather than
directly through the phospholipid bilayer. (1 mark)
* Charged ions cannot penetrate phospholipid bilayer easily (as it is non-polar).
Suggest how L-menthol changes TRPM8 channel’s shape, with reference to changes inTRPM8’s
tertiary structure. (2 marks)
* Binds to the protein.
*Altering bonds holding together its 3D shape.
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Nerves (A2-4.1.2)
Describe how influx of Na+ ions into a sensory receptor sets up an action potential, with reference
to generator potentials and the “all-or-nothing” response.
(5 marks)
* Stimulus causes non-voltage gated sodium channels to open in receptor cells.
* Na+ ions diffuse into the receptor cell and cause it to depolarise.
* Generator potential is set up (small positive charge across the membrane).
* If enough non-voltage gated sodium channels open, threshold potential (-50mV) is reached.
* Once this threshold potential is reached, nearby voltage-gated sodium channels open.
* This sets off an action potential in the receptor cells, transmitted as a nerve impulse.
* This is a all-or-nothing response: the stimulus either sets off an
action potential (+40 mV) or it doesn’t.
Outline how an action potential is transmitted along a neurone.
(4 marks)
* Increase in Na+ ion conc. in the sensory receptor sets up a local current of Na+ ions.
* Na+ ions diffuse sideways along the inside of neurone away from the region of higher conc.
* This causes voltage-gated sodium channels to open further down the neurone, setting off
another action potential.
* The local current moves down the neurone as a wave of depolarisation (followed by a wave of
repolarisation) – this is the nerve impulse.
Why do myelinated neurones transmit action potentials faster than unmyelinated? (3 marks)
* Myelin is impermeable to Na+ and K+ ions.
* Myelin provides electrical insulation.
* In myelinated neurones, depolarisation occurs only at the gaps in the myelin sheath.
* Called the nodes of Ranvier.
* So action potentials jumps from node to node (saltatory conduction).
Name the type of neurone which transmits action potentials from the sensory receptor to the brain.
(1 mark)
* Sensory neurone.
Describe two distinguishing features of this type of neurone. (2 marks)
* Short axon.
* Long dendron.
* Cell body along one side / outside the CNS / in the dorsal root ganglion.
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Communication (A2-4.1.1)
Mice which lack the gene that codes for the TRMP8 receptor are called TRMP8-null mutants.
Three independent research groups have reported that these mutants are severely impaired in
their ability to detect cold temperatures.
Suggest how and why the inability of TRT8-null mutant mice to detect cold could lead to death of
these mice in a cold environment. Include reference to the following in your answer:
*homeostasis
(1 mark)
*endotherms
(2 marks)
*thermoreceptors (2 marks)
*how mice normally thermoregulate in the cold (3 marks)
*whyTRT8-null mutant mice die in the cold
(2 marks)
* Homeostasis is maintenance of a constant internal environment.
* Endotherms are organisms that can use internal sources of heat to regulate body temp.
* Endotherms keep a constant core temp.(temp. of blood flowing through the brain & vital organs)
at an optimum for enzyme activity.
* Thermorecepors are normally responsible for detecting changes in the core temp. of blood
running through the hypothalamus.
* Thermoreceptors also monitor changes in skin temp. to detect changes in external temp.
* The thermoregulatory centre in the hypothalamus processes the input from thermoreceptors to
bring about physiological ad behavioural responses which help maintain core temperature.
* Normal thermoregulation: if core temperature falls too low, responses include mechanisms to
increase heat production and to reduce heat loss.
* These include:
-contraction of erector pili muscles (hairs on skin raise to trap a layer of insulating air).
-vasoconstriction of arterioles in the skin, to reduce heat loss by radiation.
-increased rate of metabolism to generate more heat, especially in the liver.
-behavioural responses to reduce heat loss, e.g. huddling, seeking sun or seeking shelter.
* ForTRT8-null mutant mice, failure to detect falling environmental temp. and falling core body
temp. will lead to failure of mechanisms to increase heat production/decrease heat loss.
* Leads to hypothermia /death.
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Hormones (A2-4.1.3)
Diabetes mellitus is a serious metabolic disorder associated with hyperglycaemia (high blood
glucose concentration). To investigate claims that herbal (plant) remedies can reduce blood
glucose concentration in diabetics, researchers from the University of Madras in India carried out
experiments to study the efficacy (effectiveness) of extracts of four types of plant on the
concentration of blood glucose, blood insulin, liver enzyme and liver glycogen levels in diabetic
rats.
All four plants are used on Indian cooking, have religious significance and have a history of
medicinal use in India. One of the plants tested was Mentha x piperita (peppermint).
For the investigation, male albino rats of the Wistar strain were used. Some rats were treated with
the drug streptozotocin to induce diabetes, and some rats were left untreated (normal). Rats were
divided into six groups, each group containing at least six rats:
Group I
Group II
Group III
Group IV
Group V
Group VI
Normal rats (normal control group)
Diabetic rats (diabetic control group).
Diabetic rats fed with Murraya koenigii (curry leaf tree)
Diabetic rats fed with Aegle marmelos (Bengal quince)
Diabetic rats fed with Ocimum sanctum (Holy Basil)
Diabetic rats fed with Mentha x piperita
Treated rats were fed daily with plant extracts for 29 days.
On the 29th day, they were fasted, then fed a fixed amount of glucose. The changes in blood
glucose concentration (BGC) over the next 2 hours were measure every 30 minutes (glucose
tolerance test – Table 1).
On the 30th day, the rats were fasted again, and then killed.
Various assays were done to measure the concentration of glycogen and enzymes in the rats’
livers and glucose and insulin in the rats’ blood, amongst other things. The results for groups I, II
and VI will be considered here to ascertain the effect of oral peppermint extract.
Source: R. T. Narendhirakannan et al., “Biochemical evaluation of antidiabetogenic properties of some commonly used Indian plants on
streptozotocin-induced diabetes in experimental rats,” Clinical & Experimental Pharmacology & Physiology, vol 33, no 12, pp 1150–1157, 2006
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Graph 1
Average fasting blood glucose concentration (BGC) in normal rats (Group I),
untreated diabetic rats (Group II) and diabetic rats treated with plant extracts for 30
days (Groups III to VI).
Which plant extract is the most effective at reducing fasting BGC? (see key).
* Murraya koenigii (curry leaf tree).
Table 1:
(1 mark)
Glucose tolerance test: changes in blood glucose concentration (BGC) in normal,
untreated diabetic and peppermint-treated diabetic rats after 30 days of treatment, at
30 minute intervals following ingestion of 2g/kg glucose.
Rat Group
(I) Normal control
(II) Diabetic control
(VI) Diabetic + peppermint treatment
Mean Blood Glucose Concentration (BGC) (mg/dL)
Fasting
30 min
60 min
90 min
120 min
79.5
149.6
168.0
128.6
87.4
252.4
319.4
359.4
336.5
311.4
135.7
178.8
215.5
201.8
143.8
(a) Calculate % difference in fasting BGC in group (II) diabetic control rats versus
group (I) normal control rats.
marks)
* 252.4 – 79.5 X 100 =
79.5
(2
217.5 % higher in diabetic control.
OR
252.4 – 79.5 X 100 = 68.5 % lower in normal control
252.4
(b) Calculate % decrease in fasting BGC in group (VI) diabetic + peppermint rats versus
group (II) diabetic control rats.
(2 marks)
* 252.4 – 135.7 X 100 =
252.4
46.2 % decrease.
(c) At what time does mean BGC peak in all three rat groups?
* 60 minutes.
(d) Explain why mean BCG takes time to peak, following glucose ingestion.
* Takes time to absorb glucose from gut/small intestine into the blood capillaries.
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(1 mark)
(1 mark)
10
Hormones (A2-4.1.3)
Draw a flow diagram to outline how an increase in BGC leads to increased insulin secretion in
normal rats.
(5 marks)
* Increased BGC  more glucose diffuses into β-cells through channel proteins.
* Glucose is used to make ATP in respiration.
* ATP closes the K+ channels.
* K+ ions accumulate in β-cells, so they depolarise.
* Voltage-gated Ca2+ gates open, so Ca2+ ions diffuse into the β-cells.
* Vesicles containing insulin migrate to plasma membrane of β-cells and fuse.
* Insulin released by exocytosis.
Figure 3 shows some of the reactions involved in the processing of glucose by hepatocytes:
glycogen
glycogen
synthetase
* glycogenesis
glycogen
phosphorylase
glucose-1-phosphate
* glycogenolysis
glucose
* gluconeogenesis
amino acids & glycerol
Fill in the names of the missing processes in the correct boxes: (3 marks)
gluconeogenesis
glycogenolysis
glycogenesis
Name the process stimulated by insulin: (1 mark)
* glycogenesis.
Name two hormones which increase blood glucose: (2 mark)
* glucagon.
* adrenaline.
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Table 2:
Mean glycogen, enzyme and insulin levels in normal, untreated diabetic
and peppermint-treated diabetic rats after 30 days of treatment
Concentration measured
Insulin in blood plasma (µmol/mL)
Glycogen synthase in liver (µmol/mg)
Glycogen phosphorylase in liver (µmol/mg)
Glycogen in liver (mg/g)
Group I
Group II
Group VI
(normal control)
(diabetic control:
untreated)
(diabetic: peppermint
treated)
16.6
816.5
637.2
52.8
4.3
561.8
876.5
23.7
7.4
784.4
716.5
37.7
(a) Explain whether the diabetic rats have Type 1 or Type 2 diabetes.
(2 marks)
* Type 1.
* Low insulin in blood plasma suggests they are unable to make or secrete insulin
(rather than Type II, which is an inability to respond to insulin).
(b) When insulin binds to insulin receptors on the plasma membrane of a liver cell, it causes
production of more glucose carriers in the plasma membrane so that more glucose enters the
cell. Use the information in Figure 3 & Table 2 to suggest two ways in which insulin helps
remove glucose from liver cells.
(2 marks)
* Stimulates production/activity of glycogen synthase.
* Stimulates conversion of glucose  glycogen (glycogenesis).
* Inhibits production/activity of glycogen phosphorylase.
* Inhibits conversion of glycogen  glucose (glycogenolysis).
(c) To quantify the effects of peppermint extract, compare the data for group (VI) diabetic
+ peppermint rats versus group (II) diabetic control rats and calculate the following:
% increase in blood plasma insulin 7.4 – 4.3 X 100 = 72.1%
4.3
% increase in glycogen synthase 784.4 – 561.8 X 100 = 39.6%
561.8
% decrease in glycogen phosphorylase 876.5 – 716.5 X 100 = 18.3%
876.5
% increase in glycogen 37.7 – 23.7 X 100 = 59.1%
23.7
(2 marks)
(2 marks)
(2 marks)
(2 marks)
Suggest why rats of the same strain and gender (male Wistrar) were used for all groups in this
investigation.
(1 mark)
* To control these factors / eliminate them as possible causes of variation.
Suggest one other way in which the rats should be matched.
* Same age.
(1 mark)
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Revision Pack 2
Exchange Surfaces & Breathing (AS-1.2.1)
Health & Disease (AS-2.2.2)
L-menthol is a major component of peppermint oil and cornmint oil. It is widely used to produce
medicines such as decongestants and cough sweets, and to make “menthol cigarettes”. According
to the Food and Drugs Administration (FDA) in the USA, the cooling and anaesthetic effects of Lmenthol reduce the harshness of cigarette smoke, which could facilitate deeper and more
prolonged inhalation of tobacco smoke, resulting in greater smoke intake per cigarette.
Source:http://www.fda.gov/downloads/AdvisoryCommittees/CommitteesMeetingMaterials/TobaccoProductsScientificAdvisoryCommittee/UCM244975.pdf
With reference to the structure of alveoli, suggest three reasons why L-menthol from cigarette
smoke is absorbed into the bloodstream quickly.
(3 marks)
(1)
* Short diffusion distance (< 1µm)
- Alveoli/capillary walls are only one cell thick
- Made of squamous epithelial cells (flattened & very thin)
- Capillaries are in very close contact with walls of alveoli
Water film
Alveolus
(2)
* A thin layer of moisture lines the alveoliL-menthol dissolves in it, which helps diffusion out of alveoli.
(3)
* High diffusion gradient maintained –
by ventilation of alveoli and blood carrying away
L-menthol as soon as it is absorbed.
Blood capillary
Image Source: http://wikieducator.org/File:
Gas_exchange_in_the_alveolus_unlabeled_diagram.JPG
Outline three health risks of smoking which could be increased by smoking “menthol cigarettes”.
(3 marks)
* Lung cancer – carcinogens cause malignant carcinoma.
* Emphysema –alveoli burst; ventilation reduced.
* Bronchitis – chest infections; caused by mucus accumulation (nicotine paralyses cilia).
* Atherosclerosis – fatty deposits in artery walls.
* Thrombosis – blood clots.
* Increased carbon monoxide inhalation – reducing ability of blood to carry O2.
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Excretion (A2-4.2.1)
Inhaled L-menthol is absorbed into the blood via the lungs, ingested L-menthol is
absorbed into the blood via the stomach and small intestine.
The liver is responsible for converting L-menthol in the blood into an excretory product. Most of the
L-menthol which arrives at the liver is converted to menthol glucuronide. This molecule (RMM =
332) is transported to the kidneys, where it is filtered out of the blood by nephrons and is
eventually excreted in urine.
Name the following blood vessels associated with the liver:
(a) The artery entering the liver which carries L-menthol absorbed from smoking (inhaled).
(1 mark)
* Hepatic artery.
(b) The vein entering the liver which carries L-menthol absorbed from the gut (ingested).
(1 mark)
* Hepatic portal vein.
(c) The vein that carries L-menthol glucoronide out of the liver.
(1 mark)
* Hepatic vein.
With reference to the structure of the glomerulus capillaries and Bowman’s capsule, describe how
nephrons filter menthol glucuronide out of blood by nephrons.
(3 marks)
* Afferent arteriole (entering glomerulus) is wider than efferent arteriole (exiting).
* so blood in glomerulus capillaries has high hydrostatic pressure.
* so small molecules with RMM <69,000 (including menthol glucuronide RMM = 332) are
forced out of blood in glomerulus into the lumen of Bowman’s capsule.
* by ultrafiltration.
* via pores in capillary (endothelium) wall / basement membrane / gaps between
podocytes (epithelium) of Bownan’s capsule wall.
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Photosynthesis (A2-4.3.1)
Transport in Plants (AS-1.2.3)
In 1771 Joseph Priestley, an English Chemist and clergyman, put a sprig of mint into a transparent
closed space with a candle that burned out the air until it soon went out. After 27 days, he relit the
extinguished candle again by focusing sun light beams with a mirror onto the candle wick and it
burned perfectly well in the air that previously would not support it. In carrying out several other
experiments, scientists such as Priestley and his contemporaries (Karl Scheele from Sweden and
Antoine Lavoisier from France) discovered oxygen.
Source: http://www.juliantrubin.com/bigten/oxygenexperiments.html
Explain in detail why oxygen is a bi-product of non-cyclic photophosphorylation.
(4 marks)
* Photons of light are absorbed by PSI and PSII.
* Electrons of primary pigments (P700 and P680) become excited to higher energy levels.
* Electrons from P700 in PSI are used to reduce NADP.
* Electrons lost from PSI are replaced by electrons from P680 in PSII.
* Electrons lost from PSII are replaced by electrons released by photolysis of water.
* Photolysis also releases H+ ions (used to make ATP in chemiosmosis)
and oxygen (as a bi-product).
Although most mint plants are green, they contain a range of photosynthetic pigments.
Name three photosynthetic pigments found in mint plants.
(3 marks)
* Chlorophyll a.
* Chlorophyll b.
* Carotene.
* Xanthophyll.
Explain why plants with several photosynthetic pigments can make more starch than plants with
just one photosynthetic pigment.
(3 marks)
* More wavelengths of visible spectrum absorbed.
* More energy to make ATP and reduced NADP (in the LDS).
* Faster conversion of glycerate-3-phosphate to triose phosphate
(in the LIS).
*More triose phosphate to convert to starch (in the LIS).
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In relation to photosynthesis, what is a limiting factor?
(2 marks)
* Limiting factor is that factor which limits the rate of photosynthesis.
* If that factor is increased, then the rate of photosynthesis increases.
Explain fully how each of the following factors can limit the rate of photosynthesis:
(a) Low temperature (3 marks)
* Reduced kinetic energy of enzymes and substrates involved in LIS (Calvin cycle).
* Fewer successful collisions between enzymes and substrates.
* E.g. rubisco, RuBP and CO2.
* Less starch made.
(b) Low light intensity (3 marks)
* Fewer photons absorbed by photosynthetic pigments.
* Fewer excited electrons from primary pigments.
* Less ATP and reduced NADP made in the LDS (photophosphorylation).
* Less conversion of GP to TP and TP to RuBP during LIS (Calvin cycle).
* Less TP made, so less starch made.
(c) Low CO2 concentration (3 marks)
* Lower rate of carboxylation of RuBP to GP (using CO2).
* Less conversion of GP to TP and TP to RuBP during LIS (Calvin cycle).
* Less TP made, so less starch made.
Suggest why, on a hot day, the rate of photosynthesis may be less at 12 noon (when light intensity
peaks) compared to at 10 a.m. (3 marks)
* Increased temperature.
* Increased rate of transpiration / water evaporation from soil.
* Reduced water availability.
* Stomata close (so less uptake of CO2).
* Increased temperature.
* Increased rate of enzyme-controlled reactions.
* Increased respiration rate.
* More CO2 produced by the leaf (so less uptake needed for pts.).
* Temperature above optimum for enzymes (involved in pts.).
* Increased light intensity.
* Damages chlorophyll/chloroplasts.
* Chloroplasts moved away (from damaging light).
* So rate of pts. is lower (so less uptake of CO2).
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Respiration (4.4.1)
A box of twelve chocolate peppermint creams contains 200g net of creams. The creams contain
glucose syrup, cocoa butter, emulsifier, soya lecithin, invertase and peppermint oil.
Table 3: Nutritional information
for peppermint creams
Nutritional Information
Content per 100g
Energy
1886 kJ
Carbohydrates
76.9 g
Fats
14.4 g
Proteins
2.7 g
Table 4: Approximate standard energy values
for different respiratory substrates
Standard Energy Values
Carbohydrates
16 kJg-1
Fats
39 kJg-1
Proteins
17 kJg-1
Use the information in tables 3 and 4 to work out the % difference between the energy content per
100g as stated on the box, and the energy value calculated using standard energy values and
nutritional information. Show your working.
(5 marks)
* Carbohydrates:
* Fats
* Proteins
76.9 X 16 = 1230.4
14.4 X 39 = 561.6
2.7 X 17 = 45.9
Total = 1837.9 kJ using standard
energy values
% difference = 1886 – 1837.9 X 100 OR 1886 – 1837.9 X 100
1886
1837.9
Suggest two reasons for any difference between the two values.
= 2.6%
(3 marks)
* Energy values will vary for different types of
carbohydrate, protein and fat (standard energy values are averages).
* Standard energy values are approximate (to nearest kJg-1).
According to the Rowett Research Institute, the average daily energy intake in the UK is 10250kJ
for men and 7030kJ for women. Source:http://www.rowett.ac.uk/edu_web/sec_pup/energy_expenditure.pdf
Given that a box of 12 peppermint creams has a net weight of 200g and an energy content of
1886 kJ per 100g, how many peppermint creams could the average man and woman eat per day
before exceeding their average daily intake?
(3 marks)
* 1886 X 2 = 314.3 kJ per pmc
12
Men: 10250 = 33 pmc
314.3
Women 7030 = 22 pmc
314.3
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Explain why different respiratory substrates have different energy values, with reference to ATP
production by mitochondria. (4 marks)
* the more H atoms a respiratory substrate contains per g;
* the more reduced NAD/ reduced FAD you get;
* so more H+ ions are released for chemiosmosis;
* so more ATP is generated in oxidative phosphorylation;
* so we say it has “more energy it per g” (units: kJg-1).
Finally, back to Joseph Priestley and his mint.
“I took a quantity of air, made thoroughly noxious, by mice breathing and dying in it, and divided it into two parts; one
of which I put into a phial immersed in water; and to the other (which was contained in a glass jar, standing in water) I
put a sprig of mint. This was about the beginning of August 1771, and after eight or nine days, I found that a mouse
lived perfectly well in that part of the air, in which the sprig of mint had grown, but died the moment it was put into the
other part of the same original quantity of air; and which I had kept in the very same exposure, but without any plant
growing in it”.
Source: Priestley ‘Observations on Different Kinds of Air', Philosophical Transactions (1772), 62, 193-4.
Explain Priestley’s findings:
(a) With reference to the role of oxygen (supplied by the mint plant) in aerobic respiration.
(3 marks)
* Aerobic respiration requires O2.
* O2 is the final electron acceptor at the end of the ETC (oxidative phosphorylation).
* O2 combines with electrons and protons to make water: 2H+ + 2e- + 1/2O2 - H2O
* Most of the ATP produced in aerobic respiration is produced by oxidative phosphorylation.
(a) With reference to anaerobic respiration (in the absence of the mint plant).
(3 marks)
* Oxidative phosphorylation cannot occur.
* Therefore, reduced NAD and reduced FAD from Kreb’s cycle cannot be re-oxidised.
* So NAD and FAD are not available to accept more H atoms.
* Kreb’s cycle stops (so no ATP made by this process either)
* Glycolysis can still occur, but only produces 2 ATP per glucose.
* Anaerobic respiration does not produce enough ATP to keep the mouse alive.
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Revision pack 3
Transport in Plants (AS-1.2.3)
Cellular Control (A2-5.1.1)
In the early 20th Century, the sole source of peppermint oil in the USA was a cultivar of Mentha x
piperita called Black Mitcham. However, the variety fell victim to a wilt disease caused by the
fungusVerticillium. This fungus infects and blocks the xylem of plants, so causing mint leaves to
turn brown, dry, curl and drop. Soil fumigation and crop burning were only partially successful in
reducing the incidence of this disease.
In the late 1950s, a number of disease-resistant mutants which maintained the original flavour of
peppermint oil were created by irradiating peppermint plants with γ-radiation. These mutants along
with some selections from the original gene pool are now in general use. The savings achieved
through its introduction amount to millions of dollars yearly.
With reference to water transport in plants, suggest how Verticillium infection leads to the withering
of mint leaves. (2 marks)
* The fungus blocks xylem vessels, so stops the transpiration stream (which helps draw water into
leaves from the xylem).
* Cells in the leaf are short of water, and start to plasmolyse, so the leaf wilts.
Other:
* Leaf is also deprived of mineral ions (carried dissolved in water) so cannot make chlorophyll. The
leaf starts to turn brown.
* Cytokinin/auxin production stops; ethene production increases; cellulase enzymes
digest cell walls in the abscission zone; petioles snap and leaves fall.
Explain whether you would categorise disease-resistance in peppermint as:
a spontaneous (random) mutation or an induced mutation? (1 mark)
* Induced mutation – caused by irradiation with γ-rays.
a harmful, neutral or beneficial mutation? (1 mark)
* Beneficial – increases survival chances (resistance to wilt disease).
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Meiosis & Variation (A2-5.1.2)
Supposing, in particular species of mint, a single gene, C, controls flower colour, and that this
gene has 2 alleles: purple (CP) and white (CW). The two alleles are codominant, so that a
heterozygous plant (CRCW) has pale lilac flowers.
What is meant by the term “codominance”?
(1 mark)
* Where two alleles both contribute to the phenotype if they occur together.
Set out a full genetic diagram to work out the expected ratio of genotypes and phenotypes
amongst the offspring produced by crossing two plants with pale lilac flowers . (5 marks)
STEP 1
Write down the symbols representing the
alleles using superscripts.
STEP 2
Write down the phenotypes and genotypes
of the parents
CP = allele for purple flowers
Cw = allele for white flowers
CPCw
lilac flowers
X
CPCw
lilac flowers
meiosis
STEP 3
Write down the gametes produced by each parent:
Circle them, and indicate that meiosis has occurred.
STEP 4
Use a Punnet Square to show how the
gametes might combine
CP
CP
Ww
CP
CP
Ww
w
C
STEP 5
State phenotypes & genotypes of
offspring (including ratios & percentages).
meiosis
Cw
Cw
CP
Ww
Ww
Ww
Cw
CP
CP
PCP
CWw
CPCw
CPCw
CwCw
Ww
Ww
2 CPCw :
1 CPCP
:
1 CwCw
2 Lilac : 1 Purple : 1 White
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Cellular Control (A2-5.1.1)
Meiosis & Variation (A2-5.1.2)
Figure 8 shows the metabolic pathway for L-menthol synthesis. The key gives the names of the
enzymes involved in converting precursor molecules into L-menthol, starting with an organic
compound called IPP.
Figure 8
L-menthol pathway
http://en.wikipedia.org/wiki/Menthol
Pennyroyal oil, produced by Mentha pulegium, is mainly made up of the ketone pulegone.
Suggest which enzyme in the L-menthol synthesis pathway that this mint species is unable to
produce. (1 mark)
* PR (pulegone reductase) –cannot convert pulegone to menthone without this enzyme.
Pregnant women are advised to avoid using products containing pennyroyal oil because pulegone
is known to have teratogenic properties (it causes deformities in fetuses). With reference to
homeobox genes, suggest how pulegone may cause deformity. (2 marks)
*Pulegone may interferes with normal expression of homeobox genes.
*e.g. activates them too early/ in the wrong order/ in the wrong cells.
*Causing birth defects.
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A mutant species of Mentha canadensis has the genotype AaBb.
Gene A/a codes for the synthesis of the enzyme GPPS
Allele A codes for ability to synthesise enzyme GPPS
Allele a codes for inability to synthesise enzyme GPPS
Gene B/b codes for the synthesis of the enzyme MR
Allele B codes for ability to synthesise enzyme MR
Allele b codes for inability to synthesise enzyme MR
Can a plant of genotype AaBb make L-menthol? Explain your answer. (2 marks)
* Yes; can synthesise GPPS & MR (both required to complete L-menthol synthesis pathway).
* As it has one dominant allele for each of the genes which code for these enzymes.
Two plants of genotype AaBb are crossed (AaBb X AaBb). Add numbers to the boxes to complete
the following:
Expected ratio of genotypes of offspring: (2 marks)
9
A-B-
:
3
A-bb
:
3
aaB- :
1
aabb
Expected ratio of phenotypes of offspring: (1 mark)
9
can make L-menthol :
7
can’t make L-menthol
Explain what type of gene interaction these results demonstrate. (3 marks)
* Complementary epistasis.
* At least one dominant allele of both the A/a and B/b gene loci is needed to make
L-menthol (phenotype).
A third gene, C/c, codes for the synthesis of enzyme iPR. Allele C codes for ability to synthesise
enzyme iPR whereas allele c codes for the inability to synthesis enzyme iPR.
Plants of genotype AaBbCc can make L-menthol, as can all other plants which carry at least one
dominant allele of each of the three genes.
Write down at least 6 genotypes of a plant which cannot make L-menthol (there are 19!)
(3 marks)
aabbcc; aabbCc; aabbCC; aaBbcc; aaBbCc; aaBbCC; aaBBcc;
aaBBCc; aaBBCC; Aabbcc; AabbCc; AabbCC; AaBbcc; AaBBcc;
AAbbcc; AAbbCc; AAbbCC; AABbcc; AABBcc.
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Evolution (AS-2.3.3)
Cellular Control (A2-5.1.1)
Two different populations of the same species of mint may show considerable variation in
phenotypes (continuous variation), even when the mint plants making up the two different
populations are genetically identical.
State three abiotic factors which affect the height of two genetically identical mint plant
populations growing in different places. (3 marks)
* Water availability.
* CO2 concentration.
* Light intensity / hours of daylight.
* Temperature.
* Mineral ion concentration of soil.
* pH of soil.
Mentha pulegium (pennyroyal mint) contains a high percentage of D-pulegone, a ketone with
insecticidal properties. This chemical protects the plant from damage by many different insect
pests, including weevils, aphids, bore beetles and crane flies.
Supposing a single pennyroyal mint plant spontaneously undergoes a point mutation in the gene
which codes for the enzyme iPl. This results in the creation of a new allele of the gene, which
codes for a slightly altered iPl enzyme with a single amino acid change. The altered enzyme
converts the normal precursor of pulegone (cis-isopulegone) into a new compound which is 10X
more toxic to insects than pulegone itself.
What type of point mutation is most likely to lead to a single amino acid change in a protein?
(1 mark)
* Substitution.
Explain why change in even just one amino acid can drastically alter the function of a protein.
(3 marks)
* Changes primary structure of a protein.
* Which may change its tertiary structure.
* Which may change its 3D shape, and therefore its function.
With reference to the theory of evolution by natural selection, explain how, in time, a new allele
arising in a single plant can spread through the whole pennyroyal population.
(3 marks)
* The best-adapted individuals in mint population can out-compete those individuals that are less
well-adapted [in this case, more likely to survive damage by grazing insects].
* The best adapted are more likely to survive to reproduce.
* They pass on their favourable allele to the next generation.
* Over many generations the favourable allele becomes more frequent in the population.
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Revision pack 4
Classification (AS-2.3.2)
In the binomial system of naming organisms, spearmint is called Mentha spicata.
What is the “binomial system of classification”?
(2 marks)
* In the Binomial System, organisms are given two Latin names;
* a genus name and a species name.
*Mentha spicata (genus name = Mentha; species name = spicata).
*The system was devised by the Swedish Botanist Carl Linnaeus (1707 – 1778) at a
time when Latin was the universal language.
What are the benefits of this system?
(2 marks)
*Its use avoids confusion: in different countries.
*Different common names used for the same organism.
*Same common name is used for different organisms.
Organisms can be classified into 8 major taxon levels. Mentha spicata is in the kingdom Plantae.
State three key features of all the organisms found in this kingdom. (3 marks)
* Eukaryotic.
* Multicellular.
* Cell walls made of cellulose.
* Autotrophic nutrition.
Table 5 shows the full taxonomic classification of Mentha spicata. Fill in all the gaps.
(7 marks)
Table 5
Taxonomic classification of
Mentha spicata
* Domain
Eukaryotae
Kingdom
Plantae
* Phylum
Magnoliophyta
* Class
Magnoliopsida
* Order
Lamiales
* Family
Lamiaceae
Genus
* Mentha
Species
* spicata
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Meiosis & Variation (A2-5.1.2)
There are 18 different species of mint, and 11 hybrids. Peppermint (Mentha x piperita) is the best
known hybrid. In the plant world, peppermint is the equivalent of a mule, since it is a cross
between two distinct species (Mentha aquatica X Mentha spicata) and thus is infertile. It can only
be reproduced by taking cuttings.
Chromosome analysis of peppermint has shown that it is polyploid. The plant is triploid (3n), being
the product of fertilisation of a diploid gamete (2n) from one parent species and a haploid gamete
(n) from the other parent species.
With reference to the “Biological species concept” explain why Mentha x piperita cannot be
classified as a true species. (1 mark)
* Members of the same species CAN interbreed freely to produce fertile offspring.
With reference to the events which normally take place in meiosis, explain why Mentha x piperita
plants are sterile. (3 marks)
* All the chromosomes do not exist in homologous pairs.
[chromosomes from the haploid gamete do not have homologous pairs although
those from the diploid gamete do]
* Therefore cannot form bivalents during prophase I of meiosis.
* Therefore cannot make gametes.
Mentha x piperita plants could be made fertile by chromosome doubling to create a hexaploid
plant. Explain how chromosome doubling would make peppermint plants fertile. (2 marks)
* All the chromosomes would now exist in homologous pairs so could form bivalents.
* So could complete meiosis successfully to form gametes.
Suggest why hexaploid Mentha x piperita plants would produce a higher yield of peppermint oil
than triploid plants. (3 marks)
* Triploid plants could have 3 copies of a gene to make oil.
* Hexaploid plants would have 6 copies.
* In theory hexaploids could produce twice as much oil as triploids.
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Meiosis & Variation (A2-5.1.2)
Diet & Food Production (AS-2.2.1)
Amongst the 18 mint species and 11 mint hybrids, there are many cultivars of the various mint
species and hybrids. Here’s just one example: Mentha sauveolens (apple mint) has a cultivar
called pineapple mint. The leaves of apple mint are green; whereas the leaves of pineapple mint
are variegated (i.e. they are green with bands of white). As the common names suggest, apple
mint leaves smell like apples whereas pineapple mint leaves smell like pineapples.
Suggest how plant breeders used selective breeding/artificial selection to establish a variety of
apple mint which smelt like pineapple.
(3 marks)
*Choose plants which smell of pineapple for breeding.
*Selective cross-breeding of these plants will increase production of the chemical
which smells like pineapple.
*Over many generations, this will produce plants making high concentrations of the
chemical which smells like pineapple.
Suggest why, in selecting for a pineapple aroma, plant breeders may have inadvertently selected
for variegated leaves.
(2 marks)
*Alleles of the genes responsible for variegated leaves could be (closely) linked to
alleles of the genes for pineapple aroma/ carried on the same chromosome.
*So in selecting for alleles for pineapple aroma, alleles for variegation are
selected for at the same time.
Some of the novelty mint varieties, such as Mentha avensis (cultivar banana mint) suffer from
“lack of vigour”. Suggest why, in selecting for a banana aroma through inbreeding, plant breeders
have produced a weakened plant variety.
(2 marks)
*Smaller gene pool.
*Increased frequency of unfavourable recessive alleles.
*Increased frequency of unfavourable homozygous recessive genotypes.
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Meiosis & Variation (A2-5.1.2)
Classification (AS-2.2.1)
Figure 9 shows a cladistic tree showing the phylogenetic relationships of the 18 different
Mentha (mint) species. The tree is based on morphology, phytochemicals (such as ingredients of
the oils the mint plants produce) and DNA analysis.
Figure 9
Cladistic tree for
Mentha species
Image courtesy of Dr Brian Lawrence
Mint -The Genus Mentha. CRC Press (2007). Brian Lawrence (ed).
Which species of mint is most closely related to spearmint (Mentha spicata)?
*Mentha longfolia.
(1 mark)
Which two species of mint are the “oldest” species in evolutionary terms?
* Mentha suaveolens & Mentha aquatica.
(2 marks)
Is Mentha pulegium more closely related to Mentha cervina or Mentha diemenica? (2 marks)
How do you know?
* Mentha diemenica.
* Mentha pulegium shares a more recent common ancestor with this species.
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Cloning (A2-5.2.1)
Mentha species can reproduce sexually (by producing seeds) or asexually (by vegetative
propagation). They produce flowers, but also spread themselves by producing horizontal stems
called rhizomes or stolons, which form roots to help the plant spread.
State two advantages and two disadvantages of reproducing asexually.
(4 marks)
* Asexual reproduction – pros:
- rapid colonisation.
- no need for a second plant (for cross-pollination).
- offspring are clones of parent, so if parent plant is well-adapted, offspring are too.
* Asexual reproduction – cons:
- genetic uniformity means all offspring are equally susceptible to same disease.
- lack of genetic variety mean the gene pool is small, so population is less able to adapt
to changing environment.
State two advantages and two disadvantages of reproducing sexually.
(4 marks)
* Sexual reproduction – pros:
- genetically varied population with larger gene pool.
- decreases susceptibility of whole population to the same disease.
- increases the ability of population to evolve to changing environment.
* Sexual reproduction – cons:
- slower colonisation.
- there is a need for a second plant (for cross-pollination).
Write a concluding statement to explain why the survival of any mint species is enhanced by its
ability to reproduce both sexually and asexually.
(2 marks)
* Maximises chances of survival of the species because the advantages of one method of
reproduction counteract the disadvantages of the other.
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Cloning (A2-5.2.1)
To propagate your mint plants at the beginning of this project, you took cuttings and rooted them.
Outline how you could micropropagate your mint plants using tissue culture. (7 marks)
* Remove explants from meristems in shoot tips.
*Sterilise by dipping in alcohol.
*Place explants on sterile nutrient agar containing cytokinins to stimulate mitosis.
*Explants grow into a ball of cells called a callus.
*Subdivide the callus and place onto sterile nutrient agar containing auxins to
stimulate differentiation.
*Callus cells develop roots and shoots and grow into tiny cloned plants called plantlets.
*Transfer plantlets to soil to grow plants (“pricking out”).
State three advantages micropropagation for large scale plant production.
(3 marks)
*crops grown independent of season.
*disease-free crops.
*uniform crops all with desirable characteristics.
*conservation of rare species.
State three disadvantages micropropagation for large scale plant production. (3 marks)
*sterile conditions required (contamination destroys the crop) – expensive.
*labour-intensive – expensive.
*requires laboratory-trained staff – expensive.
*genetic uniformity; so entire crop can be lost to the same disease or pest.
*some calluses do not form roots very readily.
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Biotechnology (A2-5.2.2)
Think back to your summary of the commercial uses of mint plants (page 4).
With reference to the definition of “biotechnology”, explain how mint oil production is an example of
biotechnology. (3 marks)
* Biotechnology is the industrial use of living organisms, or parts of living organisms;
* to produce food, drugs or other products.
* Mint plants are grown on an industrial scale to isolate mint oils.
* Mint oils have a very wide range of uses and are commercially important.
The ingredients of the essential oils of mint plants are known to have antibacterial properties.
Outline how you would to compare the effectiveness of peppermint oil versus spearmint oil in
inhibiting bacterial growth. (5 marks)
*Use a micropipette to place 100 µL of bacterial culture onto a sterile agar plate.
*Spread plate bacteria over the surface using a sterilised glass spreader.
*Cut two wells out of the agar, using a cork borer (sterilised by flaming).
*Place 100µLof peppermint oil in one well and 100 µL of spearmint oil in the second.
*Incubate 30oC overnight.
*Clear halos around the wells show that the oils have inhibited bacterial growth.
*The bigger the halo, the greater the inhibition.
Describe three aseptic techniques used. (3 marks)
*Swab the work surface with disinfectant.
*Work near a Bunsen flame to reduce contamination.
*Dip a glass spreader in alcohol and flame to sterilise.
*Hold petri dish lid ajar to minimise aerial contamination
during spread plating.
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Gene Technology (A2-5.2.3)
Recently, scientists from Rothamsted Research Station in Hertfordshire have genetically engineered a new
variety of wheat which contains a gene isolated from peppermint. The gene codes for a pheromone which
both repels aphids and attracts natural predators of aphids such as ladybirds and parasitic wasps. The
pheromone produced naturally by peppermint plants mimics the action of a pheromone which aphids
produce themselves when they are under threat of predation, and which warns other aphids to stay away to
avoid danger.
Aphids are a major pest of cereal crops, causing an estimated £120m worth of damage to cereal crops per
year in the UK alone. The Rothamsted scientists say that the new wheat variety could lead to increased
crop yield and reduced use of insecticides, which are costly to apply and environmentally damaging.
Anti-GM campaigners disagree. They say that manipulation of plant genomes, such as introducing genes
from other plants, could have unpredictable consequences, and there are safety concerns about eating
wheat which produces new phytochemicals. They say that, in the long term, aphids could become
habituated to the pheromone, and could learn not to respond to it. They also say the peppermint gene
might spread into wild grasses by hybridisation.
Scientists from Rothamstead, however, argue that benefits far outweigh the risks. Further information on
the project is at http://www.rothamsted.ac.uk/Content-Section=AphidWheat.html
Pick out two suggested benefits and two suggested risks of introducing the pheromone gene from
peppermint into wheat by genetic engineering.
(4 marks)
* Suggested benefits:
- Increased crop yield (reduced crop damage).
- Reduced use of insecticides (saving money & reducing environmental damage).
* Suggested risks:
- Unpredictable consequences (safety concerns/habituation of aphids).
- Hybridisation with wild grasses.
Outline the role of Ti plasmid, Agrobacterium tumefaciens and micropropagation in the creation of
GM crops such as wheat containing a gene from peppermint. (4 marks)
* Gene coding for pheromone isolated from peppermint.
* Gene introduced into Ti plasmid.
*Ti plasmid placed into Agrobacterium tumefaciens (bacterium).
*This bacterium used to infect wheat plant explant cells.
*Callus cells subdivided and used to grow GM wheat plantlets.
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Revision pack 5
Maintaining Biodiversity (AS-2.3.4)
Populations (A2-5.3.2)
Conservation of biological resources can be of economic benefit, both directly and
indirectly…there is potential to make a mint out of mint!
Suggest the potential economic benefits of conserving an area where:
(a) A new variety of an existing mint species is discovered.
(2 marks)
* There may be varieties of existing species carrying disease resistance alleles
or which produce higher oil yields than plants currently grown.
* These could be interbred with cultivated stock to improve disease resistance / yield.
(b) A new mint species is discovered. (1 mark)
* Undiscovered species could be a valuable source of new mint oils,
food, medicines, insecticides.
Outline two ways in which in-situ conservation could maintain mint biodiversity.
(2 marks)
In-situ Conservation
* Legislation – pass laws to stop clearing land for development.
* Conservation Parks – set up national parks, nature reserves or Sites of Special
Scientific Interest (SSSIs).
* Repopulation – Re-create wildlife habitats that have been destroyed.
Outline two ways in which ex-situ conservation could maintain mint biodiversity.
(2 marks)
Ex-situ Conservation
* Conserving endangered mint species or varieties by activities that take place outside
its normal environment using botanical gardens and seed banks.
* Grow endangered mint plant species in botanical gardens.
* Collect seeds from the wild and store long-term in seed banks.
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Cell Structure (AS-1.1.1)
Cell Membranes (AS-1.1.2)
Figure 4
Diagram of trichome on a mint leaf surface
Figure 5
Electron micrograph of a mint trichome
www.plantphysiol.org/content/136/4/4215/F1.full
www.sciencedirect.com/science/article/pii/S1369526602002
649
The essential oils produced by mint plants are secreted by specialised cells and stored in circular
structures called trichomes The natural role of mint oils is to inhibit the growth of other competitor
plant species (allelopathy), and to protect the mint plants from attack by insects and pathogens
such as bacteria and fungi. Many or the constituent ingredients of these oils are known to disrupt
cell membranes, this making them more leaky.
Explain which type of electron microscope was used to take the picture in Figure 5. (2 marks)
* Scanning electron microscope.
* 3D image (surface view).
Calculate the actual width of the storage cavity in Figure 5, given that the magnification of the
electron micrograph is X2000. Give your answer in µm. (2 marks)
A=I
M
Actual size in µm
=
Image size
Magnification
=
30mm
2000
=
15 µm
X 1000
X1000
State two advantages and two disadvantages of using electron microscopes compared to using
light microscopes. (4 marks)
* Advantages:
- Higher resolution.
- Higher magnification.
- More detail visible.
* Disadvantages
- Black & white images.
- Kills specimens.
- Very expensive.
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- Difficult to use.
Suggest which components of cell membranes could be altered or damaged by mint oils.
(1 mark)
* Proteins / phospholipids / cholesterol.
[probably phospholipids – components of mint oil may dissolve them]
Suggest why the cell surface membranes of plants, insects, bacteria and fungi are all equally
susceptible to damage by mint oils. (1 mark)
* All have the same structure (phospholipid bilayer).
Suggest how mint oils could interfere with biochemical reactions such as respiration and
photosynthesis. (3 marks)
* Respiration is controlled by mitochondria.
* Photosynthesis is controlled by chloroplasts.
* Both are membrane-bound organelles.
* Mint oils could damage the membranes of these organelles (including the internal
photosynthetic membranes of chloroplasts).
* Organelles could burst / no longer carry our chemiosmosis.
Suggest why mint plants store the oils in specialised trichomes, rather than throughout the mint
plant. (2 marks)
* Mint oils may damage organelles and/or the cell plasma membrane.
* Mint oils are stored in trichomes so they can't damage the cells of
the mint plant which produces them.
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Diet & Food Production (AS-2.2.1)
Ecosystems (A2-5.3.1) & Populations (A2-5.3.2)
Peppermint, spearmint and cornmint are grown on a
vast agricultural scale in North America, China and
India. In recent years, farmers such as Sir Michael
Colman have re-established peppermint as a
commercial crop in the UK. Figure 6 shows one of
his peppermint fields.
Peppermint and spearmint thrive in sunny but moist
soil preferably in a temperate climate, whereas
cornmint prefers a subtropical climate. Mint plants
are susceptible to attack by a wide range of pests
(e.g. beetles, nematodes & weevils) and pathogens
(e.g. mildew), as well as competition from weeds
such as thistles and buttercups.
Figure 6
Peppermint field at Summerdown Farm
in Hampshire, UK
Image courtesy of Sir Michael Colman
Mint plants occupy the first trophic level in a food chain.
(a) What term is used to describe organisms which occupy this tropic level? (1 mark)
* Producers.
(b) Name the type of nutrition is used by organisms of this trophic level. (1 mark)
* Autotrophic nutrition.
Only a small percentage of the light energy in sunlight which hits the surface of mint plants is used
in photosynthesis. State three ways in which light energy is “lost” at this stage.
(3 marks)
* Light hits parts of the plant which can’t photosynthesise (e.g. bark).
* Some light energy is used to heat up leaf surface / evaporate water.
* Some light doesn’t hit chloroplasts.
* Some light is the wrong wavelength to get absorbed by photosynthetic pigments /
chlorophyll reflects green light.
Distinguish between gross primary productivity and net primary productivity. (2 marks)
* GPP = rate at which plants convert light energy into chemical energy (fixed biomass)
during photosynthesis.
* NPP = Energy available to plants after some fixed biomass is used in respiration (R).
* NPP = GPP – R.
When a farmer harvests the mint plants, which of these is he or she harvesting? (1 mark)
* NPP.
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Explain how each of the following farming practices helps to increase primary productivity:
Herbicides
(2 marks)
* Kill weeds.
* Weeds compete for light, water & mineral ions.
Pesticides
(2 marks)
* Kill pests.
* Pests graze on mint leaves/roots.
Irrigation
(2 marks)
* Provides water.
* Plants can continue to grow if water in short supply (e.g. drought).
Fungicides (2 marks)
* Kill fungal pathogens.
* Fungi infect damage mint plants by causing disease
Crop rotation with legumes (2 marks)
* Improves nitrate ion content of soil
* Nitrogen source to make amino acids/nucleic acids/ ATP/chlorophyll
Many mint species grow in damp, marshy places alongside rivers, streams and ponds. A gardener
planted Mentha suaveolens and Mentha aquatica in a wet area of his garden. Initially, both
species thrived, but after five years, only one of the two species remained.
What type of competition has occurred here? Explain.
* Interspecific competition – between two species.
(2 marks)
What important population concept is illustrated by eventual loss of one species?
Why has it happened?
(3 marks)
* Principle of competitive exclusion.
* Two species competing for same resources in same habitat.
* One species out-competes the other, which becomes extinct.
Describe a method that could be used to study the abundance of Mentha suaveolens in the wild,
in an area where the plant is uniformly distributed.
(4 marks)
* Random sampling.
* Place two long tape measures at right angles.
* Use random numbers as coordinates to place quadrat.
* Use a key to identify Mentha suaveolens plants.
* Count number of /estimate % cover of Mentha suaveolens in the quadrat
* Repeat for reliability (e.g. 10 times).
Science & Plants for Schools: www.saps.org.uk (2013) – Developed by Dr Richard Spencer
Revising A-level biology – answer pack
36
Plant Responses (A2-5.4.1)
When you took your original mint cuttings from the parent plant, you were instructed to remove the
apical bud and to dip the base of the shoot in a rooting powder containing auxins.
With reference to the effects of auxins, suggest why:
(a) the apical bud was removed. (3 marks)
* Remove apical bud to make mint plant bushier.
* Apical bud produces auxins which inhibit growth of lateral buds.
* This is called apical dominance.
* Removal of bud stops auxin production so removes the inhibition
* So allowing lateral buds to grow into lateral shoots.
(b) the shoots were dipped in rooting powder. (1 mark)
* Auxins promote the growth of lateral roots, (so cuttings develop roots).
Over the coming months, you will have to make sure your mint plants get an adequate supply of
light.
With reference to auxins and tropisms, explain what would happen to your mint plants if they were
illuminated only from one side and why.
(3 marks)
* Mint shoots illuminated from one side will bend towards the light.
* Because auxins from the apical bud accumulate on the shaded side.
* This causes cells on the shaded side to elongation.
* So shaded side grows longer than illuminated side, causing the bending.
How does this tropic response help the plant to avoid abiotic stress?
(2 marks)
* This helps plants grow towards a light source.
* Important as mint plants are autotrophs and need to absorb light for photosynthesis
Science & Plants for Schools: www.saps.org.uk (2013) – Developed by Dr Richard Spencer
Revising A-level biology – answer pack
37
Animal Behaviour (A2-5.4.3)
Many of the phytochemicals produced by mint plants have a direct toxic effect on insects and
other primary consumers which eat mint plants, and so are classed as natural pesticides.
However, peppermint plants (Mentha x piperita) have evolved an additional and more indirect
means of protection. They secrete a pheromone which is given off by the peppermint plant leaves,
which both repels aphids and attracts natural predators of aphids such as ladybirds and parasitic
wasps.
The pheromone mimics the action of a pheromone which aphids produce themselves when they
are under threat of predation, and which warns other aphids to stay away to avoid danger. Note
that natural predators have evolved to respond to this pheromone, so that although it acts as a
warning to other aphids to keep away from predators, it actually attracts predators too!
The pheromone produced by peppermint is released in minute quantities. It deters aphids when
there is no real danger, and attracts aphid predators so that any aphids which are present are
more likely to get eaten. In short, it protects the plant.
Production of the aphid pheromone by peppermint plants is finely tuned. With reference to
habituation, suggest whether or not increased production of the pheromone would lead to
increased or decreased damage by aphids. Remember, the pheromone is not actually toxic to
aphids. (5 marks)
* Aphids could learn not to respond to the pheromone (habituation) because there is no real
danger, so more aphids are present to damage the crop.
* However, the pheromone could attract more predators, which now have more aphids to
feed on. In this case, those aphids which have become habituated are more likely to get
eaten, so the habituation will only be short term.
* Aphids will be deterred from eating the crop, decreasing damage.
* This could lead to habituation of the predators, which learn not
to respond to the pheromone as they find that there are fewer aphids for them to eat.
* Then, the aphids could learn not to respond to the pheromone (habituation) because there
is no real danger, so more aphids are present to damage the crop. Then more predators are
attracted, and so on……
* A cycle of increased / decreased damage could be established!
Science & Plants for Schools: www.saps.org.uk (2013) – Developed by Dr Richard Spencer
Revising A-level biology – answer pack
38
Animal Responses (A2-5.4.2)
Peppermint oil is said to relieve a wide range of human ailments. The oil is extracted from Mentha
x piperita by steam distillation. It is comprised mainly of L-menthol, a volatile organic compound
which has a cooling effect the skin. Peppermint oil helps clear the nasal passages in people
suffering from respiratory tract infections; it has antibacterial properties which help to freshen
breath and it even helps remove head lice.
There is some evidence that peppermint oil eases stomach and bowel spasms by relaxing
muscles in the alimentary (digestive) tract and so may be effective in treating of IBS (irritable
bowel syndrome).
If the claimed effects of peppermint oil on the human digestive system are true, suggest which
type of muscle peppermint helps relax. (1 mark)
* Smooth (involuntary) muscle – found in digestive system.
Some researchers hypothesise that peppermint helps relax muscles by inhibiting the release of
calcium ions in muscle cells. With reference to the sliding filament theory, explain why inhibition of
calcium ion release would prevent muscle contraction.
(3 marks)
* Release of calcium ions from the sarcoplasmic reticulum in muscle fibres is necessary for
muscle contraction.
* Calcium ions bind to troponin, which changes shape and moves tropomyosin away from actin
– myosin binding sites on the actin filaments.
* This allows myosin heads to attach to the binding sites on the actin filaments, leading to a
power stroke and muscle contraction.
* If peppermint inhibits release of calcium ions from the sarcoplasmic reticulum, then the
actin – myosin binding sites will remain blocked by tropomyosin, and the muscle will stay
relaxed.
Science & Plants for Schools: www.saps.org.uk (2013) – Developed by Dr Richard Spencer
Revising A-level biology – answer pack
39
Animal Responses (A2-5.4.2)
Carvone is an organic compound which exists in
two forms called R-carvone and S-carvone.
As shown in figure 7, the two forms of carvone
are mirror images of each other (they are called
enantiomers).
R-carvone is produced by spearmint plants
(Mentha spicata) whereas S-carvone is produced
by caraway plants (Carum carvi).
Figure 7
Structures of R-carvone and S-carvone
R-carvone and S-carvone are thought to bind to different olfactory receptors in the nose. The brain
interprets R-carvone as a minty smell whereas S-carvone is interpreted as a more spicy smell.
Some people cannot differentiate between these two smells!
Suggest why R-carvone cannot bind to an S-carvone olfactory receptor, and why S-carvone
cannot bind to an R-carvone olfactory receptor (think about hands fitting into gloves!).
(2 marks)
* R-carvone and S-carvone molecules have different shapes and can only fit into receptors
with the correct complementary shape.
* Therefore, they do not fit into each other’s receptors.
20% of people cannot differentiate between the smells of R-carvone and S-carvone (these people
perceive both as smelling like spearmint). Suggest whether these people:
(a) Lack R-carvone receptors
(b) Lack S-carvone receptors
(c) Possess S-carvone receptors with an altered shape which bind to both R-carvone & S-carvone.
(d) Possess R-carvone receptors with an altered shape which bind to both R-carvone & S-carvone.
(e) A combination of (a) and (c)
(f) A combination (b) and (d)
Briefly, explain your choice.(2 marks)
* (f) They must have R-carvone receptors to detect spearmint smell but lack
S-carvone receptors so can’t detect caraway.
* However, as they perceive the smell of caraway to be the same as spearmint,
their R-carvone receptors must be able to bind to S-carvone too.
Science & Plants for Schools: www.saps.org.uk (2013) – Developed by Dr Richard Spencer
Revising A-level biology – answer pack