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Few, if any, questions left unanswered.

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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.
autoshape 7311

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).

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.

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.

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.

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.

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 Normal rats (normal control group)

Group II Diabetic rats (diabetic control group).

Group III Diabetic rats fed with Murraya koenigii (curry leaf tree)

Group IV Diabetic rats fed with Aegle marmelos (Bengal quince)

Group V Diabetic rats fed with Ocimum sanctum (Holy Basil)

Group VI 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



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).




autoshape 5
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Which plant extract is the most effective at reducing fasting BGC? (see key). (1 mark)

* Murraya koenigii (curry leaf tree).




Table 1: 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

Mean Blood Glucose Concentration (BGC) (mg/dL)

Fasting

30 min

60 min

90 min

120 min

(I) Normal control

79.5

149.6

168.0

128.6

87.4

(II) Diabetic control

252.4

319.4

359.4

336.5

311.4

(VI) Diabetic + peppermint treatment

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. (2 marks)
* 252.4 – 79.5 X 100 = 217.5 % higher in diabetic control.

79.5

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 = 46.2 % decrease.

252.4
(c) At what time does mean BGC peak in all three rat groups? (1 mark) * 60 minutes.
(d) Explain why mean BCG takes time to peak, following glucose ingestion. (1 mark)

* Takes time to absorb glucose from gut/small intestine into the blood capillaries.


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:
group 7875

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.


Table 2: Mean glycogen, enzyme and insulin levels in normal, untreated diabetic

and peppermint-treated diabetic rats after 30 days of treatment




Concentration measured


Group I

(normal control)



Group II

(diabetic control: untreated)



Group VI

(diabetic: peppermint treated)



Insulin in blood plasma (µmol/mL)

16.6

4.3

7.4

Glycogen synthase in liver (µmol/mg)

816.5

561.8

784.4

Glycogen phosphorylase in liver (µmol/mg)

637.2

876.5

716.5

Glycogen in liver (mg/g)

52.8

23.7

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% (2 marks)

4.3


% increase in glycogen synthase 784.4 – 561.8 X 100 = 39.6% (2 marks)

561.8


% decrease in glycogen phosphorylase 876.5 – 716.5 X 100 = 18.3% (2 marks)

876.5


% increase in glycogen 37.7 – 23.7 X 100 = 59.1% (2 marks)

23.7
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. (1 mark)

* Same age.

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 L-menthol 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
W
Water film
ith reference to the structure of alveoli, suggest three reasons why L-menthol from cigarette smoke is absorbed into the bloodstream quickly. (3 marks)




  1. *
    Alveolus
    autoshape 7842 Short diffusion distance (< 1µm)

- Alveoli/capillary walls are only one cell thick

autoshape 7841 - Made of squamous epithelial cells (flattened & very thin)

- Capillaries are in very close contact with walls of alveoli




  1. * A thin layer of moisture lines the alveoli-

L-menthol dissolves in it, which helps diffusion out of alveoli.

autoshape 7843


  1. *
    Blood capillary
    High diffusion gradient maintained –

by ventilation of alveoli and blood carrying away

L-menthol as soon as it is absorbed.


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.

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.
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).

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).


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 Table 4: Approximate standard energy values

for peppermint creams for different respiratory substrates



Standard Energy Values

Carbohydrates 16 kJg-1

Fats 39 kJg-1

Proteins 17 kJg-1




Nutritional Information

Content per 100g



Energy 1886 kJ

Carbohydrates 76.9 g

Fats 14.4 g

Proteins 2.7 g




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)



autoshape 7836

* Carbohydrates: 76.9 X 16 = 1230.4



* Fats 14.4 X 39 = 561.6 Total = 1837.9 kJ using standard

* Proteins 2.7 X 17 = 45.9 energy values
% difference = 1886 – 1837.9 X 100 OR 1886 – 1837.9 X 100 = 2.6%

1886 1837.9
Suggest two reasons for any difference between the two values. (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 Men: 10250 = 33 pmc Women 7030 = 22 pmc

12 314.3 314.3


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.

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).
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 CP = allele for purple flowers

alleles using superscripts. Cw = allele for white flowers
STEP 2

Write down the phenotypes and genotypes CPCw X CPCw

of the parents lilac flowers lilac flowers

autoshape 7697autoshape 7698autoshape 7699autoshape 7700

meiosis meiosis



STEP 3

Woval 7690oval 7690oval 7690oval 7690rite down the gametes produced by each parent:

Circle them, and indicate that meiosis has occurred.
oval 7690oval 7690


Sautoshape 7696TEP 4

Use a Punnet Square to show how the

gautoshape 7695oval 7690ametes might combine

CPCP CPCw


oval 7690

CPCw CwCw


STEP 5

State phenotypes & genotypes of

offspring (including ratios & percentages). 2 CPCw : 1 CPCP : 1 CwCw
2 Lilac : 1 Purple : 1 White

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.

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.


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.

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)


* Domain

Eukaryotae

Kingdom

Plantae


Table 5

Taxonomic classification of Mentha spicata
* Phylum

Magnoliophyta

* Class

Magnoliopsida

* Order

Lamiales

* Family

Lamiaceae

Genus

* Mentha

Species

* spicata

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.

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.
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)? (1 mark)

*Mentha longfolia.
Which two species of mint are the “oldest” species in evolutionary terms? (2 marks)

* Mentha suaveolens & Mentha aquatica.


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.

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.

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.
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.

Gene Technology (A2-5.2.3)

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