Solved Examples — Class 11 Biology Chapter 11 Notes

How to Use This Section

This chapter is examined in a very particular way. Almost every question is "name the molecule, the enzyme or the scientist", "give the number", "which cell, or which side of which membrane", or a $\mathrm{C_3}$ against $\mathrm{C_4}$ comparison. There is very little to derive. The marks sit in exact names, exact numbers and exact locations, and they are lost by a single swapped word.

That makes this one of the easiest chapters in the book to score full marks in, and one of the easiest to lose them in. $\mathrm{P_{700}}$ belongs to PS I and $\mathrm{P_{680}}$ to PS II. PS II works first even though PS I is numbered first. Water splitting is associated with PS II, never with PS I. Protons gather in the lumen, not in the stroma. $\mathrm{CF_0}$ is the channel in the membrane and $\mathrm{CF_1}$ is the knob facing the stroma. The acceptor in a $\mathrm{C_3}$ plant has 5 carbons and the first product has 3; the acceptor in a $\mathrm{C_4}$ plant has 3 carbons and the first product has 4. Every $\mathrm{CO_2}$ costs 3 ATP and 2 NADPH, not 2 and 2. Each of those pairs has cost more marks than any reasoning question in the chapter.

Work through the three tiers in order.

Tier 1 - Concept Checks. One fact per question, straight from the chapter. If you cannot answer one of these without stopping, go back to the section it came from before moving on.
Tier 2 - Application and Identification. You are told something about a plant, a leaf section or an isolated chloroplast, and you have to say what follows. This tier also carries the Calvin cycle arithmetic, which is the one place in the chapter where you actually calculate. This is exactly how the chapter is applied, so spend the most time here.
Tier 3 - Comparisons and Long Answers. The two-column differences and the joined-up descriptions - light reaction against dark reaction, absorption spectrum against action spectrum, PS I against PS II, cyclic against non-cyclic flow, the Z scheme end to end, chemiosmosis in full, the Calvin cycle with its numbers, $\mathrm{C_3}$ against $\mathrm{C_4}$ twice over, photorespiration, and Blackman's law.

A working habit that pays in almost every question of this chapter: before you answer, ask where the reaction is happening and which molecule is being handed to which. Grana or stroma. Lumen side or stroma side. Mesophyll or bundle sheath. Those three questions alone settle most of the paper.

The nine chapter-end exercises, which come to eleven questions once the three parts of exercise 9 are counted separately, are all already answered inside the twelve teaching sections of this chapter. The last block lists exactly where each one is answered, so there is nothing left over to work out here.

The Facts These Questions Draw On

What the simple experiments show. A variegated leaf, or a leaf partially covered with black paper, tests positive for starch only in the green parts that were in the light - so photosynthesis occurred only in the green parts of the leaves in the presence of light. When half a leaf is enclosed in a test tube with $\mathrm{KOH}$ -soaked cotton, which absorbs $\mathrm{CO_2}$ , the exposed half tests positive for starch and the enclosed half tests negative - this showed that $\mathrm{CO_2}$ was required for photosynthesis. The three requirements together are chlorophyll, light and $\mathrm{CO_2}$ .

The five names. Joseph Priestley (1733-1804), in 1770, showed the essential role of air in the growth of green plants: a candle burning in a bell jar goes out and a mouse in a closed space suffocates, but a mint plant placed in the same bell jar keeps the mouse alive and the candle burning. His hypothesis was that plants restore to the air whatever breathing animals and burning candles remove. He discovered oxygen in 1774. Jan Ingenhousz (1730-1799) used a similar setup placed once in the dark and once in the sunlight and showed that sunlight is essential for the plant process that purifies the air; in an aquatic plant, bubbles formed around the green parts in bright sunlight but not in the dark, and he later identified these bubbles to be of oxygen, showing that only the green part of the plants could release oxygen. Julius von Sachs, in about 1854, provided evidence for the production of glucose when plants grow, and that glucose is usually stored as starch; he found that the green substance in plants is located in special bodies, later called chloroplasts, and that the green parts are where glucose is made. T. W. Engelmann (1843-1909) used a prism to split light into its spectral components and illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria - the bacteria were used to detect the sites of $\mathrm{O_2}$ evolution, they accumulated mainly in the region of blue and red light, and a first action spectrum of photosynthesis was thus described, one that resembles roughly the absorption spectra of chlorophyll a and b. Cornelius van Niel (1897-1985), a microbiologist, working on purple and green bacteria, demonstrated that photosynthesis is essentially a light-dependent reaction in which hydrogen from a suitable oxidisable compound reduces carbon dioxide to carbohydrates.

The three equations. The empirical equation for oxygen-evolving organisms, with $\mathrm{[CH_2O]}$ standing for a carbohydrate such as glucose, a six-carbon sugar:

$\mathrm{CO_2 + H_2O} \xrightarrow{\text{light}} \mathrm{[CH_2O] + O_2}$

van Niel's general equation, in which $\mathrm{H_2A}$ is the hydrogen donor:

$\mathrm{2H_2A + CO_2} \xrightarrow{\text{light}} \mathrm{2A + CH_2O + H_2O}$

In green plants $\mathrm{H_2O}$ is the hydrogen donor and is oxidised to $\mathrm{O_2}$ , while in purple and green sulphur bacteria $\mathrm{H_2S}$ is the donor and the oxidation product is sulphur or sulphate, depending on the organism, and not $\mathrm{O_2}$ - so some organisms do not release $\mathrm{O_2}$ during photosynthesis. Hence he inferred that the $\mathrm{O_2}$ evolved by the green plant comes from $\mathrm{H_2O}$ , not from carbon dioxide, and this was later proved by using radioisotopic techniques. That correction gives the correct overall equation:

$\mathrm{6CO_2 + 12H_2O} \xrightarrow{\text{light}} \mathrm{C_6H_{12}O_6 + 6H_2O + 6O_2}$

Twelve waters go in because the six $\mathrm{O_2}$ released come from water, and six new waters appear on the product side. This is not a single reaction but the description of a multistep process.

The chloroplast and the division of labour. Photosynthesis takes place in the green leaves of plants, but it does so also in other green parts. The mesophyll cells in the leaves have a large number of chloroplasts, and usually the chloroplasts align themselves along the walls of the mesophyll cells, such that they get the optimum quantity of the incident light. Within the chloroplast there is a membranous system consisting of the grana, the stroma lamellae and the matrix stroma, and there is a clear division of labour: the membrane system is responsible for trapping the light energy and for the synthesis of ATP and NADPH, while in the stroma, enzymatic reactions synthesise sugar, which in turn forms starch. The membrane reactions are directly light driven and are called light reactions or photochemical reactions; the stroma reactions are not directly light driven but are dependent on the products of the light reactions - ATP and NADPH - and are called, by convention, dark reactions or carbon reactions. This should not be construed to mean that they occur in darkness or that they are not light-dependent, so calling the biosynthetic phase the "dark reaction" is arguably a misnomer.

The four pigments. A chromatographic separation of the leaf pigments shows that the colour we see in leaves is not due to a single pigment but due to four pigments - chlorophyll a, which is bright or blue green, chlorophyll b, which is yellow green, xanthophylls, which are yellow, and carotenoids, which are yellow to yellow-orange. Pigments are substances that have an ability to absorb light, at specific wavelengths.

Absorption spectrum and action spectrum. An absorption spectrum plots how much light a pigment absorbs at each wavelength; an action spectrum plots the rate of photosynthesis at each wavelength. Chlorophyll a shows maximum absorption in the blue and the red regions, and the wavelengths at which there is maximum absorption by chlorophyll a also show a higher rate of photosynthesis - hence chlorophyll a is the chief pigment associated with photosynthesis. But there is no complete one-to-one overlap between the absorption spectrum of chlorophyll a and the action spectrum of photosynthesis. Together the graphs show that most of the photosynthesis takes place in the blue and red regions of the spectrum, and that some photosynthesis does take place at the other wavelengths of the visible spectrum. The gap exists because other thylakoid pigments - chlorophyll b, xanthophylls and carotenoids - which are called accessory pigments, also absorb light and transfer the energy to chlorophyll a. They enable a wider range of wavelengths of incoming light to be utilised for photosynthesis, and they protect chlorophyll a from photo-oxidation.

The photochemical phase. Light reactions, or the photochemical phase, include light absorption, water splitting, oxygen release, and the formation of high-energy chemical intermediates, ATP and NADPH. Several protein complexes are involved in the process. No $\mathrm{CO_2}$ is fixed and no sugar is made in this phase.

The two photosystems. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction - so PS II acts first and PS I second. The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments, except one molecule of chlorophyll a, forming a light harvesting system also called antennae, and these pigments help to make photosynthesis more efficient by absorbing different wavelengths of light. The single chlorophyll a molecule forms the reaction centre, and the reaction centre is different in both the photosystems: in PS I it has an absorption peak at $700\ \mathrm{nm}$ and is called $\mathrm{P_{700}}$ , in PS II it has absorption maxima at $680\ \mathrm{nm}$ and is called $\mathrm{P_{680}}$ .

The Z scheme. In photosystem II the reaction centre chlorophyll a absorbs $680\ \mathrm{nm}$ wavelength of red light, causing electrons to become excited and jump into an orbit farther from the atomic nucleus. These electrons are picked up by an electron acceptor which passes them to an electron transport system consisting of cytochromes. This movement of electrons is downhill, in terms of an oxidation-reduction or redox potential scale. The electrons are not used up as they pass through the electron transport chain, but are passed on to the pigments of photosystem PS I. Simultaneously, electrons in the reaction centre of PS I are also excited when they receive red light of wavelength $700\ \mathrm{nm}$ , and are transferred to another acceptor molecule that has a greater redox potential. These electrons then are moved downhill again, this time to a molecule of energy-rich $\mathrm{NADP^+}$ , and the addition of these electrons reduces $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ . This whole scheme of transfer of electrons is called the Z scheme, due to its characteristic shape, and this shape is formed when all the carriers are placed in a sequence on a redox potential scale.

Splitting of water. The electrons that were moved from photosystem II must be replaced, and this is achieved by electrons available due to splitting of water. The splitting of water is associated with the PS II; water is split into $\mathrm{2H^+}$ , $\mathrm{[O]}$ and electrons, and this creates oxygen, one of the net products of photosynthesis. The electrons needed to replace those removed from photosystem I are provided by photosystem II.

$\mathrm{2H_2O \rightarrow 4H^+ + O_2 + 4e^-}$

The water splitting complex is associated with the PS II, which itself is physically located on the inner side of the membrane of the thylakoid, so the protons and the oxygen formed are released into the lumen.

Cyclic and non-cyclic photophosphorylation. Phosphorylation is the process through which ATP is synthesised by cells, in mitochondria and chloroplasts, and photo-phosphorylation is the synthesis of ATP from ADP and inorganic phosphate in the presence of light. When the two photosystems work in a series, first PS II and then the PS I, a process called non-cyclic photo-phosphorylation occurs, and the two photosystems are connected through an electron transport chain, as in the Z scheme; both ATP and $\mathrm{NADPH + H^+}$ are synthesised by this kind of electron flow. When only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons. A possible location where this could be happening is in the stroma lamellae, because the membrane or lamellae of the grana have both PS I and PS II while the stroma lamellae membranes lack PS II as well as the NADP reductase enzyme. The excited electron does not pass on to $\mathrm{NADP^+}$ but is cycled back to the PS I complex through the electron transport chain, and the cyclic flow hence results only in the synthesis of ATP, but not of $\mathrm{NADPH + H^+}$ . Cyclic photophosphorylation also occurs when only light of wavelengths beyond $680\ \mathrm{nm}$ are available for excitation.

The chemiosmotic hypothesis. The chemiosmotic hypothesis has been put forward to explain the mechanism of ATP synthesis. Like in respiration, in photosynthesis too, ATP synthesis is linked to development of a proton gradient across a membrane - this time the membranes of the thylakoid. There is one difference though: here the proton accumulation is towards the inside of the membrane, that is, in the lumen, whereas in respiration, protons accumulate in the intermembrane space of the mitochondria. Three things build the gradient. (a) Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions that are produced by the splitting of water accumulate within the lumen of the thylakoids. (b) As electrons move through the photosystems, protons are transported across the membrane; this happens because the primary acceptor of electrons, which is located towards the outer side of the membrane, transfers its electron not to an electron carrier but to an H carrier. Hence this molecule removes a proton from the stroma while transporting an electron, and when this molecule passes on its electron to the electron carrier on the inner side of the membrane, the proton is released into the lumen side of the membrane. (c) The NADP reductase enzyme is located on the stroma side of the membrane; along with electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ , and these protons are also removed from the stroma. The result is that within the chloroplast, protons in the stroma decrease in number, while in the lumen there is accumulation of protons, and this creates a proton gradient across the thylakoid membrane as well as a measurable decrease in pH in the lumen.

ATP synthase. It is the breakdown of this gradient that leads to the synthesis of ATP. The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the $\mathrm{CF_0}$ of the ATP synthase. $\mathrm{CF_0}$ has a transmembrane channel that carries out facilitated diffusion of protons across the membrane, while $\mathrm{CF_1}$ protrudes on the outer surface of the thylakoid membrane on the side that faces the stroma. The breakdown of the gradient provides enough energy to cause a conformational change in the $\mathrm{CF_1}$ particle of the ATP synthase, which makes the enzyme synthesise several molecules of ATP. Chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase. Along with the NADPH produced by the movement of electrons, the ATP will be used immediately in the biosynthetic reaction taking place in the stroma, responsible for fixing $\mathrm{CO_2}$ and synthesis of sugars.

The biosynthetic phase. The products of the light reaction are ATP, NADPH and $\mathrm{O_2}$ ; the $\mathrm{O_2}$ diffuses out of the chloroplast while ATP and NADPH are used to drive the processes leading to the synthesis of food, more accurately sugars. This process does not directly depend on the presence of light but is dependent on the products of the light reaction, that is ATP and NADPH, besides $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . Immediately after light becomes unavailable, the biosynthetic process continues for some time, and then stops, and if light is then made available, the synthesis starts again.

Calvin, PGA and OAA. Just after world war II, among the several efforts to put radioisotopes to beneficial use, the work of Melvin Calvin is exemplary. His use of radioactive $\mathrm{{}^{14}C}$ in algal photosynthesis studies led to the discovery that the first $\mathrm{CO_2}$ fixation product was a 3-carbon organic acid, and he also contributed to working out the complete biosynthetic pathway, and hence it was called the Calvin cycle after him. The first product identified was 3-phosphoglyceric acid, or in short PGA. Experiments conducted over a wide range of plants led to the discovery of another group of plants, where the first stable product of $\mathrm{CO_2}$ fixation was again an organic acid, but one which had 4 carbon atoms in it - this acid was identified to be oxaloacetic acid, or OAA. Since then, $\mathrm{CO_2}$ assimilation during photosynthesis was said to be of two main types - the $\mathrm{C_3}$ pathway and the $\mathrm{C_4}$ pathway, named after the number of carbon atoms in the first fixation product.

The acceptor. The studies very unexpectedly showed that the acceptor molecule was a 5-carbon ketose sugar - ribulose bisphosphate, $\mathrm{RuBP}$ . Since the first product was a $\mathrm{C_3}$ acid, scientists believed that the primary acceptor would be a 2-carbon compound, and they spent many years trying to identify a 2-carbon compound before they discovered the 5-carbon $\mathrm{RuBP}$ . The arithmetic works because 5 carbons plus 1 carbon give 6 carbons, which appear as two molecules of the 3-carbon PGA.

The Calvin cycle. Calvin and his co-workers worked out the whole pathway and showed that the pathway operated in a cyclic manner - the $\mathrm{RuBP}$ was regenerated. The Calvin pathway occurs in all photosynthetic plants; it does not matter whether they have $\mathrm{C_3}$ or $\mathrm{C_4}$ or any other pathway. For ease of understanding, the Calvin cycle can be described under three stages: carboxylation, reduction and regeneration. Carboxylation is the fixation of $\mathrm{CO_2}$ into a stable organic intermediate, and it is the most crucial step of the Calvin cycle, in which $\mathrm{CO_2}$ is utilised for the carboxylation of $\mathrm{RuBP}$ , catalysed by the enzyme $\mathrm{RuBP}$ carboxylase, resulting in the formation of two molecules of 3-PGA.

$\mathrm{RuBP + CO_2} \xrightarrow{\text{RuBisCO}} 2 \times \mathrm{3PGA}$

Since this enzyme also has an oxygenation activity, it would be more correct to call it $\mathrm{RuBP}$ carboxylase-oxygenase, or RuBisCO. Reduction is a series of reactions that lead to the formation of glucose, and the steps involve the utilisation of 2 molecules of ATP for phosphorylation and two of NADPH for reduction, per $\mathrm{CO_2}$ molecule fixed. Regeneration of the $\mathrm{CO_2}$ acceptor molecule $\mathrm{RuBP}$ is crucial if the cycle is to continue uninterrupted, and the regeneration steps require one ATP for phosphorylation to form $\mathrm{RuBP}$ .

The arithmetic. For every $\mathrm{CO_2}$ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH are required - 2 ATP and 2 NADPH in reduction, and 1 ATP in regeneration. The fixation of six molecules of $\mathrm{CO_2}$ and 6 turns of the cycle are required for the formation of one molecule of glucose from the pathway, so one glucose costs 18 ATP and 12 NADPH, and the cycle gives back 18 ADP and 12 NADP. It is probably to meet this difference in the number of ATP and NADPH used in the dark reaction that the cyclic phosphorylation takes place.

The $\mathrm{C_4}$ pathway. Plants that are adapted to dry tropical regions have the $\mathrm{C_4}$ pathway. Though these plants have the $\mathrm{C_4}$ oxaloacetic acid as the first $\mathrm{CO_2}$ fixation product, they use the $\mathrm{C_3}$ pathway or the Calvin cycle as the main biosynthetic pathway. They are special in that they have a special type of leaf anatomy, they tolerate higher temperatures, they show a response to high light intensities, they lack a process called photorespiration and they have greater productivity of biomass. The particularly large cells around the vascular bundles of the $\mathrm{C_4}$ plants are called bundle sheath cells, and the leaves which have such anatomy are said to have Kranz anatomy - Kranz means wreath, and is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles, and they are characterised by having a large number of chloroplasts, thick walls impervious to gaseous exchange, and no intercellular spaces. Maize and sorghum are the named examples, and the presence of the bundle sheath would help you identify the $\mathrm{C_4}$ plants.

The Hatch and Slack pathway. This pathway has been named the Hatch and Slack pathway, and it is again a cyclic process. The primary $\mathrm{CO_2}$ acceptor is a 3-carbon molecule, phosphoenol pyruvate (PEP), and it is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase, or PEPcase, and it is important to register that the mesophyll cells lack the RuBisCO enzyme. The $\mathrm{C_4}$ acid OAA is formed in the mesophyll cells, and it then forms other 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells itself, and these are transported to the bundle sheath cells. In the bundle sheath cells these $\mathrm{C_4}$ acids are broken down to release $\mathrm{CO_2}$ and a 3-carbon molecule, and the 3-carbon molecule is transported back to the mesophyll, where it is converted to PEP again, thus completing the cycle. The $\mathrm{CO_2}$ released in the bundle sheath cells enters the $\mathrm{C_3}$ or the Calvin pathway, a pathway common to all plants. The bundle sheath cells are rich in the enzyme ribulose bisphosphate carboxylase-oxygenase (RuBisCO), but lack PEPcase. The Calvin pathway occurs in all the mesophyll cells of the $\mathrm{C_3}$ plants; in the $\mathrm{C_4}$ plants it does not take place in the mesophyll cells, but does so only in the bundle sheath cells.

Photorespiration. RuBisCO is the most abundant enzyme in the world. RuBisCO is characterised by the fact that its active site can bind to both $\mathrm{CO_2}$ and $\mathrm{O_2}$ - hence the name. RuBisCO has a much greater affinity for $\mathrm{CO_2}$ when the $\mathrm{CO_2}$ to $\mathrm{O_2}$ ratio is nearly equal, this binding is competitive, and it is the relative concentration of $\mathrm{O_2}$ and $\mathrm{CO_2}$ that determines which of the two will bind to the enzyme. In $\mathrm{C_3}$ plants some $\mathrm{O_2}$ does bind to RuBisCO, and hence $\mathrm{CO_2}$ fixation is decreased. Here the RuBP, instead of being converted to 2 molecules of PGA, binds with $\mathrm{O_2}$ to form one molecule of phosphoglycerate and one molecule of phosphoglycolate, which is a 2-carbon compound - this pathway is called photorespiration. In the photorespiratory pathway there is neither synthesis of sugars, nor of ATP; rather it results in the release of $\mathrm{CO_2}$ with the utilisation of ATP. In the photorespiratory pathway there is no synthesis of ATP or NADPH. The biological function of photorespiration is not known yet. In $\mathrm{C_4}$ plants photorespiration does not occur, because they have a mechanism that increases the concentration of $\mathrm{CO_2}$ at the enzyme site: the $\mathrm{C_4}$ acid from the mesophyll is broken down in the bundle sheath cells to release $\mathrm{CO_2}$ , and this results in increasing the intracellular concentration of $\mathrm{CO_2}$ , which in turn ensures that the RuBisCO functions as a carboxylase, minimising the oxygenase activity. Since $\mathrm{C_4}$ plants lack photorespiration, productivity and yields are better in these plants, and in addition these plants show tolerance to higher temperatures.

The completed comparison.

Characteristic	$\mathrm{C_3}$ Plants	$\mathrm{C_4}$ Plants
Cell type in which the Calvin cycle takes place	Mesophyll	Bundle sheath
Cell type in which the initial carboxylation reaction occurs	Mesophyll	Mesophyll
Cell types in the leaf that fix $\mathrm{CO_2}$	One: mesophyll	Two: bundle sheath and mesophyll
Primary $\mathrm{CO_2}$ acceptor	RuBP	PEP
Carbons in the primary acceptor	5	3
Primary $\mathrm{CO_2}$ fixation product	PGA	OAA
Carbons in the primary fixation product	3	4
Does the plant have RuBisCO	Yes	Yes
Does the plant have PEPcase	No	Yes
Which cells have RuBisCO	Mesophyll	Bundle sheath
$\mathrm{CO_2}$ fixation rate under high light	Low	High
Photorespiration at low light intensities	Negligible	Negligible
Photorespiration at high light intensities	High	Negligible
Photorespiration at low $\mathrm{CO_2}$ concentrations	High	Negligible
Photorespiration at high $\mathrm{CO_2}$ concentrations	Negligible	Negligible
Temperature optimum	$20^\circ$ to $25^\circ\mathrm{C}$	$30^\circ$ to $40^\circ\mathrm{C}$
Examples	Most temperate plants - wheat and rice are typical	Maize and sorghum

Factors affecting photosynthesis. The rate of photosynthesis is very important in determining the yield of plants, including crop plants, and photosynthesis is under the influence of several factors, both internal (plant) and external. The plant or internal factors are the number, size, age and orientation of leaves, mesophyll cells and chloroplasts, internal $\mathrm{CO_2}$ concentration and the amount of chlorophyll, and they are dependent on the genetic predisposition and the growth of the plant. The external factors are the availability of sunlight, temperature, $\mathrm{CO_2}$ concentration and water. Though several factors interact and simultaneously affect photosynthesis or $\mathrm{CO_2}$ fixation, usually one factor is the major cause or is the one that limits the rate, and hence, at any point the rate will be determined by the factor available at sub-optimal levels. Blackman's (1905) Law of Limiting Factors states that if a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value: it is the factor which directly affects the process if its quantity is changed. The worked example is that despite the presence of a green leaf and optimal light and $\mathrm{CO_2}$ conditions, the plant may not photosynthesise if the temperature is very low, and this leaf, if given the optimal temperature, will start photosynthesising.

Light. Light quality, light intensity and the duration of exposure to light are the three separate things the word covers. There is a linear relationship between incident light and $\mathrm{CO_2}$ fixation rates at low light intensities, and at higher light intensities, gradually the rate does not show further increase, as other factors become limiting. Light saturation occurs at 10 per cent of the full sunlight, and hence, except for plants in shade or in dense forests, light is rarely a limiting factor in nature. Increase in incident light beyond a point causes the breakdown of chlorophyll and a decrease in photosynthesis.

Carbon dioxide. Carbon dioxide is the major limiting factor for photosynthesis. The concentration of $\mathrm{CO_2}$ is very low in the atmosphere, between 0.03 and 0.04 per cent, and an increase in concentration up to 0.05 per cent can cause an increase in $\mathrm{CO_2}$ fixation rates; beyond this the levels can become damaging over longer periods. At low light conditions neither $\mathrm{C_3}$ nor $\mathrm{C_4}$ plants respond to high $\mathrm{CO_2}$ conditions, but at high light intensities, both $\mathrm{C_3}$ and $\mathrm{C_4}$ plants show an increase in the rates of photosynthesis. $\mathrm{C_4}$ plants show saturation at about $360\ \mu\mathrm{L\,L^{-1}}$ , while $\mathrm{C_3}$ plants respond to increased $\mathrm{CO_2}$ concentration and saturation is seen only beyond $450\ \mu\mathrm{L\,L^{-1}}$ . Thus, current availability of $\mathrm{CO_2}$ levels is limiting to the $\mathrm{C_3}$ plants. The fact that $\mathrm{C_3}$ plants respond to higher $\mathrm{CO_2}$ concentration by showing increased rates of photosynthesis, leading to higher productivity, has been used for some greenhouse crops such as tomatoes and bell pepper, which are allowed to grow in a carbon dioxide enriched atmosphere, which leads to higher yields.

Temperature and water. The dark reactions, being enzymatic, are temperature controlled, and though the light reactions are also temperature sensitive, they are affected to a much lesser extent. The $\mathrm{C_4}$ plants respond to higher temperatures and show a higher rate of photosynthesis, while $\mathrm{C_3}$ plants have a much lower temperature optimum, and the temperature optimum for photosynthesis of different plants also depends on the habitat that they are adapted to - tropical plants have a higher temperature optimum than the plants adapted to temperate climates. Even though water is one of the reactants in the light reaction, the effect of water as a factor is more through its effect on the plant, rather than directly on photosynthesis: water stress causes the stomata to close, hence reducing the $\mathrm{CO_2}$ availability, and water stress also makes leaves wilt, thus reducing the surface area of the leaves and their metabolic activity as well.

Tier 1 - Concept Checks

Question 1

Q. What do the variegated-leaf experiment and the $\mathrm{KOH}$ experiment together prove?

Answer. They prove that chlorophyll, light and $\mathrm{CO_2}$ are required for photosynthesis. In the first, a variegated leaf, or a leaf that was partially covered with black paper, is exposed to light, and on testing these leaves for the presence of starch it is clear that photosynthesis occurred only in the green parts of the leaves in the presence of light. In the second, part of a leaf is enclosed in a test tube containing $\mathrm{KOH}$ -soaked cotton, which absorbs $\mathrm{CO_2}$ , while the other half is exposed to air; afterwards the exposed part tests positive for starch while the portion inside the tube tests negative, which showed that $\mathrm{CO_2}$ was required for photosynthesis. Starch is the indicator in both.

Question 2

Q. What did Joseph Priestley show, and in which years?

Answer. Joseph Priestley (1733-1804), in 1770, performed a series of experiments that revealed the essential role of air in the growth of green plants. He saw that a candle burning in a closed space - a bell jar - soon gets extinguished, and that a mouse would soon suffocate in a closed space, so a burning candle or an animal that breathes the air both somehow damage the air. But when he placed a mint plant in the same bell jar, he found that the mouse stayed alive and the candle continued to burn. His hypothesis was that plants restore to the air whatever breathing animals and burning candles remove. He discovered oxygen in 1774.

Question 3

Q. What did Jan Ingenhousz add to Priestley's work?

Answer. He added light. Jan Ingenhousz (1730-1799) used a similar setup to Priestley's, but placed it once in the dark and once in the sunlight, and so showed that sunlight is essential for the plant process that purifies the air. In an elegant experiment with an aquatic plant he showed that in bright sunlight small bubbles were formed around the green parts, while in the dark they did not. He later identified these bubbles to be of oxygen, and so showed that it is only the green part of the plants that could release oxygen.

Question 4

Q. What was Julius von Sachs' contribution?

Answer. Julius von Sachs, in about 1854, provided evidence for the production of glucose when plants grow, and that glucose is usually stored as starch. His later studies showed that the green substance in plants - chlorophyll, as we now call it - is located in special bodies, later called chloroplasts, within plant cells. He found that the green parts in plants are where glucose is made.

Question 5

Q. Describe Engelmann's experiment and name what it produced.

Answer. T. W. Engelmann (1843-1909) used a prism to split light into its spectral components and then illuminated a green alga, Cladophora, placed in a suspension of aerobic bacteria. The bacteria were used to detect the sites of $\mathrm{O_2}$ evolution, and he observed that they accumulated mainly in the region of blue and red light of the split spectrum. A first action spectrum of photosynthesis was thus described, and it resembles roughly the absorption spectra of chlorophyll a and b. Note the word carefully - Engelmann gave the action spectrum, which belongs to the process; the absorption spectrum belongs to the pigment.

Question 6

Q. What did van Niel demonstrate, and what is the hydrogen donor in green plants and in sulphur bacteria?

Answer. Cornelius van Niel (1897-1985), a microbiologist, based on his studies of purple and green bacteria, demonstrated that photosynthesis is essentially a light-dependent reaction in which hydrogen from a suitable oxidisable compound reduces carbon dioxide to carbohydrates. His general equation is

$\mathrm{2H_2A + CO_2} \xrightarrow{\text{light}} \mathrm{2A + CH_2O + H_2O}$

In green plants $\mathrm{H_2O}$ is the hydrogen donor and is oxidised to $\mathrm{O_2}$ . In purple and green sulphur bacteria $\mathrm{H_2S}$ is the hydrogen donor, and the oxidation product is sulphur or sulphate, depending on the organism, and not $\mathrm{O_2}$ - so some organisms do not release $\mathrm{O_2}$ during photosynthesis. Hence he inferred that the $\mathrm{O_2}$ evolved by the green plant comes from $\mathrm{H_2O}$ , not from carbon dioxide, and this was later proved by using radioisotopic techniques.

Question 7

Q. Write the correct overall equation of photosynthesis, and say why twelve molecules of water appear on the left.

Answer. The correct equation is

$\mathrm{6CO_2 + 12H_2O} \xrightarrow{\text{light}} \mathrm{C_6H_{12}O_6 + 6H_2O + 6O_2}$

where $\mathrm{C_6H_{12}O_6}$ represents glucose. Twelve waters appear because the oxygen released comes from water, not from $\mathrm{CO_2}$ . Six molecules of $\mathrm{O_2}$ need twelve molecules of water to supply them, and six new water molecules appear on the product side. The older empirical equation for oxygen-evolving organisms was written as $\mathrm{CO_2 + H_2O \rightarrow [CH_2O] + O_2}$ , where $\mathrm{[CH_2O]}$ represented a carbohydrate. Remember also that this is not a single reaction but the description of a multistep process.

Question 8

Q. State the division of labour inside a chloroplast.

Answer. Within the chloroplast there is a membranous system consisting of the grana, the stroma lamellae and the matrix stroma, and there is a clear division of labour within the chloroplast. The membrane system is responsible for trapping the light energy and also for the synthesis of ATP and NADPH. In the stroma, enzymatic reactions synthesise sugar, which in turn forms starch. Add the two supporting facts: the mesophyll cells in the leaves have a large number of chloroplasts, and usually the chloroplasts align themselves along the walls of the mesophyll cells, such that they get the optimum quantity of the incident light.

Question 9

Q. Why is the name "dark reaction" a misnomer?

Answer. Because the dark reactions neither require darkness nor are they independent of light. The stroma reactions are not directly light driven but are dependent on the products of the light reactions - ATP and NADPH. To distinguish them they are called, by convention, dark reactions, or carbon reactions, but this should not be construed to mean that they occur in darkness or that they are not light-dependent. The proof is simple: immediately after light becomes unavailable, the biosynthetic process continues for some time, and then stops, and if light is then made available, the synthesis starts again. Hence calling the biosynthetic phase the "dark reaction" is arguably a misnomer.

Question 10

Q. Name the four leaf pigments and give the colour of each.

Answer. A chromatographic separation of the leaf pigments shows that the colour we see in leaves is not due to a single pigment but due to four pigments. They are chlorophyll a, which is bright or blue green; chlorophyll b, which is yellow green; xanthophylls, which are yellow; and carotenoids, which are yellow to yellow-orange. Pigments are substances that have an ability to absorb light, at specific wavelengths.

Question 11

Q. Why is chlorophyll a called the chief pigment of photosynthesis?

Answer. Because the wavelengths at which there is maximum absorption by chlorophyll a also show a higher rate of photosynthesis. Chlorophyll a shows maximum absorption in the blue and the red regions, and the action spectrum, which is the rate of photosynthesis at each wavelength, peaks in the same blue and red regions. Hence we can conclude that chlorophyll a is the chief pigment associated with photosynthesis. Note the qualification the chapter adds at once - there is no complete one-to-one overlap between the absorption spectrum of chlorophyll a and the action spectrum of photosynthesis.

Question 12

Q. Name the accessory pigments and give their two functions.

Answer. The accessory pigments are chlorophyll b, xanthophylls and carotenoids. Though chlorophyll is the major pigment responsible for trapping light, these other thylakoid pigments also absorb light and transfer the energy to chlorophyll a. Their two functions are:

They enable a wider range of wavelengths of incoming light to be utilised for photosynthesis.
They protect chlorophyll a from photo-oxidation.

Give both functions - a question asking why plants have chlorophyll b and other accessory pigments wants the pair, not just the first.

Question 13

Q. What does the light reaction consist of?

Answer. Light reactions, or the photochemical phase, include light absorption, water splitting, oxygen release, and the formation of high-energy chemical intermediates, ATP and NADPH. Several protein complexes are involved in the process. Note what is not in the list: no $\mathrm{CO_2}$ is fixed and no sugar is made in the light reaction. Both of those belong to the biosynthetic phase in the stroma.

Question 14

Q. On what basis are the two photosystems numbered, and which of them functions first?

Answer. The pigments are organised into two discrete photochemical light harvesting complexes (LHC) within the Photosystem I (PS I) and Photosystem II (PS II). These are named in the sequence of their discovery, and not in the sequence in which they function during the light reaction. So PS I was discovered first and took the number one, but in the light reaction PS II functions first and PS I second - the electrons begin at PS II, travel down an electron transport chain, and only then reach PS I. The numbers are a historical label, not a running order.

Question 15

Q. What is the antenna, what is the reaction centre, and what are $\mathrm{P_{700}}$ and $\mathrm{P_{680}}$ ?

Answer. The LHC are made up of hundreds of pigment molecules bound to proteins. Each photosystem has all the pigments, except one molecule of chlorophyll a, forming a light harvesting system also called antennae, and these pigments help to make photosynthesis more efficient by absorbing different wavelengths of light. The single chlorophyll a molecule forms the reaction centre, and the reaction centre is different in both the photosystems. In PS I the reaction centre chlorophyll a has an absorption peak at $700\ \mathrm{nm}$ , hence is called $\mathrm{P_{700}}$ . In PS II the reaction centre chlorophyll a has absorption maxima at $680\ \mathrm{nm}$ , and is called $\mathrm{P_{680}}$ . P stands for the pigment and the number is the wavelength in nanometres at which that reaction centre absorbs most strongly.

Question 16

Q. Write the water-splitting reaction, say which photosystem it is associated with, and say where the products go.

Answer. The splitting of water is associated with the PS II; water is split into $\mathrm{2H^+}$ , $\mathrm{[O]}$ and electrons.

$\mathrm{2H_2O \rightarrow 4H^+ + O_2 + 4e^-}$

The electrons that were moved from photosystem II must be replaced, and this is achieved by electrons available due to splitting of water. This creates oxygen, one of the net products of photosynthesis. The water splitting complex is associated with the PS II, which itself is physically located on the inner side of the membrane of the thylakoid, so the protons and the $\mathrm{O_2}$ formed are released into the lumen. Never attach water splitting to PS I.

Question 17

Q. What is the Z scheme, and why does it have that name?

Answer. This whole scheme of transfer of electrons is called the Z scheme, due to its characteristic shape. This shape is formed when all the carriers are placed in a sequence on a redox potential scale. Trace the path and the letter appears - up at PS II where $\mathrm{P_{680}}$ absorbs $680\ \mathrm{nm}$ light, a long slope down through the cytochromes, up again at PS I where $\mathrm{P_{700}}$ absorbs $700\ \mathrm{nm}$ light, then down to $\mathrm{NADP^+}$ . Turned on its side that is a Z. It is not the physical arrangement of the complexes in the membrane, and it is not a graph of time or of wavelength.

Question 18

Q. Define phosphorylation and photo-phosphorylation.

Answer. Phosphorylation is the process through which ATP is synthesised by cells, in mitochondria and chloroplasts. Photo-phosphorylation is the synthesis of ATP from ADP and inorganic phosphate in the presence of light.

$\mathrm{ADP + iP} \xrightarrow{\text{light}} \mathrm{ATP}$

The prefix photo is the only thing that separates the two. Keep the three parts of the definition - from ADP, and inorganic phosphate, in the presence of light.

Question 19

Q. Where are protons accumulated during photosynthesis, and what are $\mathrm{CF_0}$ and $\mathrm{CF_1}$ ?

Answer. In photosynthesis the proton accumulation is towards the inside of the membrane, that is, in the lumen of the thylakoid; in respiration, protons accumulate in the intermembrane space of the mitochondria. As for the enzyme, the ATP synthase enzyme consists of two parts. $\mathrm{CF_0}$ has a transmembrane channel that carries out facilitated diffusion of protons across the membrane. $\mathrm{CF_1}$ protrudes on the outer surface of the thylakoid membrane on the side that faces the stroma, and the breakdown of the gradient provides enough energy to cause a conformational change in the $\mathrm{CF_1}$ particle of the ATP synthase, which makes the enzyme synthesise several molecules of ATP. Chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase.

Question 20

Q. What did Melvin Calvin use, what did he find, and what is the primary acceptor of $\mathrm{CO_2}$ ?

Answer. Just after world war II, among the several efforts to put radioisotopes to beneficial use, the work of Melvin Calvin is exemplary. His use of radioactive $\mathrm{{}^{14}C}$ in algal photosynthesis studies led to the discovery that the first $\mathrm{CO_2}$ fixation product was a 3-carbon organic acid, and he also contributed to working out the complete biosynthetic pathway, and hence it was called the Calvin cycle after him. The first product identified was 3-phosphoglyceric acid, or in short PGA. The acceptor was the surprise: the studies very unexpectedly showed that the acceptor molecule was a 5-carbon ketose sugar - ribulose bisphosphate, $\mathrm{RuBP}$ . Since the first product was a $\mathrm{C_3}$ acid, scientists believed that the primary acceptor would be a 2-carbon compound, and they spent many years trying to identify a 2-carbon compound before they discovered the 5-carbon $\mathrm{RuBP}$ .

Question 21

Q. What is Kranz anatomy, and what are the three characters of a bundle sheath cell?

Answer. The particularly large cells around the vascular bundles of the $\mathrm{C_4}$ plants are called bundle sheath cells, and the leaves which have such anatomy are said to have Kranz anatomy. Kranz means wreath, and the name is a reflection of the arrangement of cells. The bundle sheath cells may form several layers around the vascular bundles, and they are characterised by a large number of chloroplasts, thick walls impervious to gaseous exchange, and no intercellular spaces. Each character has a reason worth carrying: many chloroplasts because this is the only place the Calvin cycle runs in a $\mathrm{C_4}$ plant, and thick impervious walls with no intercellular spaces so that the $\mathrm{CO_2}$ released inside cannot leak back out.

Question 22

Q. In the Hatch and Slack pathway, name the primary acceptor, the enzyme, the first product and the cell in which each step happens.

Answer. This pathway has been named the Hatch and Slack pathway, and it is again a cyclic process. The primary $\mathrm{CO_2}$ acceptor is a 3-carbon molecule, phosphoenol pyruvate (PEP), and it is present in the mesophyll cells. The enzyme responsible for this fixation is PEP carboxylase, or PEPcase, and the mesophyll cells lack the RuBisCO enzyme. The $\mathrm{C_4}$ acid OAA is formed in the mesophyll cells, then other 4-carbon compounds like malic acid or aspartic acid are formed in the mesophyll cells itself and transported to the bundle sheath cells. In the bundle sheath cells these $\mathrm{C_4}$ acids are broken down to release $\mathrm{CO_2}$ and a 3-carbon molecule, the 3-carbon molecule is transported back to the mesophyll, where it is converted to PEP again, thus completing the cycle, and the $\mathrm{CO_2}$ released enters the $\mathrm{C_3}$ or the Calvin pathway. The bundle sheath cells are rich in RuBisCO but lack PEPcase.

Tier 2 - Application and Identification

Question 23

Q. A destarched variegated leaf is partly covered with black paper and left in sunlight for a few hours, then tested for starch. Mark on a diagram of that leaf the one region that will turn blue-black, and say what each of the other three regions proves.

Answer. Only the part that is both green and uncovered will test positive. The leaf has four kinds of region, and each one removes a different variable.

Region of the leaf	Starch test	What it proves
Green and exposed to light	Positive	This is the control - all three requirements present
Green but covered by the black paper	Negative	Light is required
Non-green (white) and exposed to light	Negative	Chlorophyll is required
Non-green and covered	Negative	Both are missing, so it adds nothing new

The conclusion the chapter draws is exactly this: on testing these leaves for the presence of starch it is clear that photosynthesis occurred only in the green parts of the leaves in the presence of light. To add the third requirement you need the second experiment - $\mathrm{KOH}$ -soaked cotton absorbs $\mathrm{CO_2}$ , and the enclosed part tests negative for starch while the exposed part tests positive, which showed that $\mathrm{CO_2}$ was required for photosynthesis.

Question 24

Q. A chloroplast is broken open and separated into a membrane fraction and a soluble stroma fraction. Each fraction is supplied with light, $\mathrm{CO_2}$ and water. Which fraction releases oxygen, which one makes sugar, and why does neither fraction alone make sugar for long?

Answer. The membrane fraction - the grana and stroma lamellae - releases the oxygen, because the membrane system is responsible for trapping the light energy and also for the synthesis of ATP and NADPH, and the splitting of water, which creates oxygen, is associated with PS II in that membrane. The stroma fraction is the one that can make sugar, because in the stroma, enzymatic reactions synthesise sugar, which in turn forms starch.

Neither works alone for long. The stroma has the enzymes but no supply of ATP and NADPH, and the biosynthetic phase is dependent on the products of the light reaction, that is ATP and NADPH, besides $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . The membrane fraction makes ATP and NADPH but has no Calvin cycle enzymes to spend them on. This separation is the whole meaning of the clear division of labour within the chloroplast.

Question 25

Q. A chloroplast is illuminated with light of wavelengths beyond $680\ \mathrm{nm}$ only. Which photosystem works, which type of photophosphorylation runs, and which products appear?

Answer. Only PS I works. Look at the two reaction centres: $\mathrm{P_{700}}$ of PS I absorbs at $700\ \mathrm{nm}$ and can still be excited by such light, while $\mathrm{P_{680}}$ of PS II absorbs at $680\ \mathrm{nm}$ and cannot. So PS II falls silent. This is exactly the stated condition - cyclic photophosphorylation also occurs when only light of wavelengths beyond $680\ \mathrm{nm}$ are available for excitation.

What follows is a chain of consequences. When only PS I is functional, the electron is circulated within the photosystem and the phosphorylation occurs due to cyclic flow of electrons. The excited electron does not pass on to $\mathrm{NADP^+}$ but is cycled back to the PS I complex through the electron transport chain. The cyclic flow hence results only in the synthesis of ATP, but not of $\mathrm{NADPH + H^+}$ . And since water splitting is associated with PS II, and PS II is not running, no water is split and no oxygen is released. So the answer is: ATP only; no NADPH, no oxygen.

Question 26

Q. A membrane fraction isolated from a chloroplast is found to lack PS II and also to lack the NADP reductase enzyme. Name the membrane, and say what kind of electron flow it can support.

Answer. It is the stroma lamellae. The membrane or lamellae of the grana have both PS I and PS II, whereas the stroma lamellae membranes lack PS II as well as the NADP reductase enzyme. That is why a possible location where cyclic photophosphorylation could be happening is in the stroma lamellae.

Both missing pieces point the same way. With no PS II there is no second photosystem to work in series with, so non-cyclic flow, which needs the two photosystems working in a series, first PS II and then PS I, is impossible. With no NADP reductase there is nothing to hand the electron on to $\mathrm{NADP^+}$ , so the excited electron is cycled back to the PS I complex through the electron transport chain. The fraction can therefore support cyclic electron flow only, giving ATP but not $\mathrm{NADPH + H^+}$ , and no oxygen.

Question 27

Q. A plant photosynthesising steadily in bright light is suddenly put into complete darkness. Does sugar synthesis stop at that instant? Explain what happens over the next few minutes and what it proves.

Answer. No, it does not stop at that instant. Immediately after light becomes unavailable, the biosynthetic process continues for some time, and then stops. If light is then made available, the synthesis starts again.

The reason is that the biosynthetic phase does not directly depend on the presence of light, but is dependent on the products of the light reaction, that is ATP and NADPH, besides $\mathrm{CO_2}$ and $\mathrm{H_2O}$ . At the moment the light goes off, the stroma still holds the ATP and NADPH already made, so the Calvin cycle keeps turning until that stock is spent, and then it stops because no fresh ATP and NADPH are being delivered.

That short lag is the whole proof of two things: the biosynthetic phase runs on the products of the light reaction and not on light itself, and therefore calling it the "dark reaction" is arguably a misnomer - it neither requires darkness nor is it independent of light.

Question 28

Q. Isolated thylakoids are illuminated, and the pH inside the thylakoid lumen is measured before and during illumination. What change is seen, and name the three processes that cause it.

Answer. The pH of the lumen falls during illumination - the chapter records a measurable decrease in pH in the lumen - because within the chloroplast, protons in the stroma decrease in number, while in the lumen there is accumulation of protons. Three separate processes cause it, and all three widen the same gradient.

(a) Water splitting. Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions that are produced by the splitting of water accumulate within the lumen of the thylakoids.

(b) Proton transport during electron movement. As electrons move through the photosystems, protons are transported across the membrane. The primary acceptor of electrons, which is located towards the outer side of the membrane, transfers its electron not to an electron carrier but to an H carrier, so this molecule removes a proton from the stroma while transporting an electron, and when this molecule passes on its electron to the electron carrier on the inner side of the membrane, the proton is released into the lumen side of the membrane.

(c) NADP reductase. The NADP reductase enzyme is located on the stroma side of the membrane, and along with electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ ; these protons are also removed from the stroma.

Note the pattern - (a) puts protons into the lumen, (b) moves them from the stroma into the lumen, and (c) only takes them out of the stroma.

Question 29

Q. A chemical is added that makes the thylakoid membrane freely permeable to protons, so that the proton gradient cannot be maintained. The light reaction still absorbs light and still splits water. What stops, and why?

Answer. ATP synthesis stops. It is the breakdown of this gradient that leads to the synthesis of ATP - not the existence of the gradient by itself, but protons moving across the membrane to the stroma through the transmembrane channel of the $\mathrm{CF_0}$ of the ATP synthase. The breakdown of the gradient provides enough energy to cause a conformational change in the $\mathrm{CF_1}$ particle of the ATP synthase, which makes the enzyme synthesise several molecules of ATP. If protons leak back everywhere across the membrane instead of being forced through $\mathrm{CF_0}$ , then no gradient builds up, nothing is driven through the channel and $\mathrm{CF_1}$ never changes conformation.

Two things do not stop. Light absorption and water splitting continue, so oxygen is still released; and NADPH can still be formed, because NADP reductase sits on the stroma side and reduces $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ from the electrons arriving from PS I. But the biosynthetic phase soon halts anyway, because the Calvin cycle needs 3 ATP as well as 2 NADPH for every $\mathrm{CO_2}$ fixed. Remember the checklist - chemiosmosis requires a membrane, a proton pump, a proton gradient and ATP synthase, and this chemical destroys the third.

Question 30

Q. A chloroplast fixes 12 molecules of $\mathrm{CO_2}$ through the Calvin cycle. How many turns of the cycle is that, how much ATP and NADPH is spent, and how much glucose is made?

Answer. Work from the per- $\mathrm{CO_2}$ ladder. For every $\mathrm{CO_2}$ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH are required - 2 ATP and 2 NADPH in reduction, plus 1 ATP in regeneration - and the fixation of six molecules of $\mathrm{CO_2}$ and 6 turns of the cycle are required for the formation of one molecule of glucose.

Quantity	Working	Answer
Turns of the cycle	One turn fixes one $\mathrm{CO_2}$	12 turns
ATP	$12 \times 3$	36 ATP
NADPH	$12 \times 2$	24 NADPH
Glucose	$12 \div 6$	2 molecules of glucose

The products released back are 36 ADP and 24 NADP. Set it out this way every time - turns first, then multiply by 3 and by 2, then divide by 6.

Question 31

Q. A leaf synthesises 5 molecules of glucose by the Calvin cycle. How many molecules of $\mathrm{CO_2}$ were fixed, how many turns were needed, and how much ATP and NADPH was used?

Answer. Six turns and six $\mathrm{CO_2}$ per glucose, so multiply everything for one glucose by five.

$\mathrm{CO_2}$ fixed: $5 \times 6 =$ 30 molecules
Turns of the cycle: $5 \times 6 =$ 30 turns
ATP: $5 \times 18 =$ 90 ATP
NADPH: $5 \times 12 =$ 60 NADPH

Check the figures against the per- $\mathrm{CO_2}$ rate as a safeguard: 30 $\mathrm{CO_2}$ times 3 ATP is 90, and 30 $\mathrm{CO_2}$ times 2 NADPH is 60. Both routes agree, so the arithmetic is sound. The single glucose figures to memorise are 18 ATP and 12 NADPH.

Question 32

Q. The stroma of a chloroplast holds 48 molecules of ATP and 48 molecules of NADPH. Assuming nothing else is limiting, how many molecules of $\mathrm{CO_2}$ can be fixed, and how many complete glucose molecules can be made? Which of the two currencies runs out first?

Answer. Test each currency separately against the per- $\mathrm{CO_2}$ rate of 3 ATP and 2 NADPH.

On ATP: $48 \div 3 =$ 16 $\mathrm{CO_2}$ molecules.
On NADPH: $48 \div 2 =$ 24 $\mathrm{CO_2}$ molecules.

The smaller figure decides, so ATP runs out first and only 16 $\mathrm{CO_2}$ molecules can be fixed. Fixing those 16 uses 48 ATP and 32 NADPH, leaving 16 NADPH unused.

For complete glucose, 6 turns are required for the formation of one molecule of glucose, and $16 \div 6$ is 2 whole molecules of glucose with 4 turns left over. Those 2 glucose molecules account for 36 ATP and 24 NADPH, and the remaining 12 ATP and 8 NADPH carry the cycle through 4 more turns that do not complete a third glucose.

This is exactly why cyclic flow exists. The two numbers are not equal - 3 ATP against 2 NADPH - and it is probably to meet this difference in the number of ATP and NADPH used in the dark reaction that the cyclic phosphorylation takes place, since cyclic flow makes ATP without making NADPH.

Question 33

Q. A student writes that one molecule of glucose costs 12 ATP and 12 NADPH. Find the mistake, and state the correct figures with the reason.

Answer. The NADPH figure is right and the ATP figure is wrong - the correct answer is 18 ATP and 12 NADPH.

The student has counted only the ATP used in reduction. Reduction uses 2 molecules of ATP for phosphorylation and two of NADPH for reduction, per $\mathrm{CO_2}$ molecule fixed, which over 6 turns gives 12 ATP and 12 NADPH - and there the student stopped. What has been dropped is the third stage: regeneration of the $\mathrm{CO_2}$ acceptor molecule $\mathrm{RuBP}$ is crucial if the cycle is to continue uninterrupted, and the regeneration steps require one ATP for phosphorylation to form $\mathrm{RuBP}$ . That extra 1 ATP per $\mathrm{CO_2}$ , over 6 turns, is the missing 6 ATP.

So the correct ladder is 3 ATP and 2 NADPH per $\mathrm{CO_2}$ , then 6 turns, giving 18 ATP and 12 NADPH per glucose, with 18 ADP and 12 NADP coming back out. Remember also that NADPH is used only in reduction - regeneration uses no NADPH at all, which is why the NADPH figure was the one the student got right.

Question 34

Q. A vertical section of a leaf shows particularly large cells arranged in several layers around each vascular bundle. These cells are packed with chloroplasts, have thick walls and show no intercellular spaces. Identify the plant type, name the anatomy, and say where the Calvin cycle runs in this leaf.

Answer. It is a $\mathrm{C_4}$ plant, and the anatomy is Kranz anatomy. The particularly large cells around the vascular bundles of the $\mathrm{C_4}$ plants are called bundle sheath cells, and the leaves which have such anatomy are said to have Kranz anatomy; Kranz means wreath, and the name is a reflection of the arrangement of cells. Every detail in the description matches: the bundle sheath cells may form several layers around the vascular bundles, and they are characterised by a large number of chloroplasts, thick walls impervious to gaseous exchange, and no intercellular spaces.

The Calvin cycle runs only in the bundle sheath cells. The Calvin pathway occurs in all the mesophyll cells of the $\mathrm{C_3}$ plants; in the $\mathrm{C_4}$ plants it does not take place in the mesophyll cells, but does so only in the bundle sheath cells. That is what the many chloroplasts and the impervious walls are for - the $\mathrm{CO_2}$ released inside must not leak back out. Maize and sorghum are the examples the chapter names, and the presence of the bundle sheath would help you identify the $\mathrm{C_4}$ plants.

Question 35

Q. The mesophyll cells of a plant are found to contain PEPcase but no RuBisCO. What kind of plant is it, what is the first product of fixation in those cells, where does the fixed carbon go next, and what does this arrangement buy the plant?

Answer. It is a $\mathrm{C_4}$ plant. The primary $\mathrm{CO_2}$ acceptor is a 3-carbon molecule, phosphoenol pyruvate (PEP), and it is present in the mesophyll cells, where the enzyme responsible for this fixation is PEP carboxylase, or PEPcase, and the mesophyll cells lack the RuBisCO enzyme. The $\mathrm{C_4}$ acid OAA is formed in the mesophyll cells - so the first product of fixation is oxaloacetic acid, a 4-carbon organic acid.

Where it goes next: OAA forms other 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells itself, and these are transported to the bundle sheath cells, where these $\mathrm{C_4}$ acids are broken down to release $\mathrm{CO_2}$ and a 3-carbon molecule. The 3-carbon molecule is transported back to the mesophyll, where it is converted to PEP again, thus completing the cycle, and the $\mathrm{CO_2}$ released in the bundle sheath cells enters the $\mathrm{C_3}$ or the Calvin pathway. The bundle sheath cells are rich in RuBisCO but lack PEPcase.

What it buys: releasing $\mathrm{CO_2}$ inside the bundle sheath increases the intracellular concentration of $\mathrm{CO_2}$ , which ensures that the RuBisCO functions as a carboxylase, minimising the oxygenase activity, so photorespiration does not occur. Hence productivity and yields are better in these plants, and they show tolerance to higher temperatures, with an optimum of $30^\circ$ to $40^\circ\mathrm{C}$ against $20^\circ$ to $25^\circ\mathrm{C}$ for a $\mathrm{C_3}$ plant.

Question 36

Q. A greenhouse tomato crop is given optimal light and is watered well, but the air temperature is held at $10^\circ\mathrm{C}$ and the $\mathrm{CO_2}$ is at the ordinary atmospheric level. The grower wants a higher rate of photosynthesis. Which factor should be corrected first, and what should be done after that?

Answer. Correct the temperature first. Blackman's (1905) Law of Limiting Factors states that if a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value: it is the factor which directly affects the process if its quantity is changed. Here light is already optimal and water is not short, so raising either of those changes nothing; the temperature is the factor at a sub-optimal level, and at any point the rate will be determined by the factor available at sub-optimal levels. This is the chapter's own illustration - despite the presence of a green leaf and optimal light and $\mathrm{CO_2}$ conditions, the plant may not photosynthesise if the temperature is very low, and this leaf, if given the optimal temperature, will start photosynthesising. Temperature matters here because the dark reactions, being enzymatic, are temperature controlled, while the light reactions are temperature sensitive but affected to a much lesser extent.

Then raise the $\mathrm{CO_2}$ . Once temperature is no longer the constraint, carbon dioxide is the major limiting factor for photosynthesis, because the concentration of $\mathrm{CO_2}$ is very low in the atmosphere, between 0.03 and 0.04 per cent. Tomato is a $\mathrm{C_3}$ plant, and $\mathrm{C_3}$ plants respond to increased $\mathrm{CO_2}$ concentration, saturation being seen only beyond $450\ \mu\mathrm{L\,L^{-1}}$ , whereas $\mathrm{C_4}$ plants saturate at about $360\ \mu\mathrm{L\,L^{-1}}$ ; thus, current availability of $\mathrm{CO_2}$ levels is limiting to the $\mathrm{C_3}$ plants. Exactly this is done commercially - greenhouse crops such as tomatoes and bell pepper are allowed to grow in a carbon dioxide enriched atmosphere, which leads to higher yields - though an increase in concentration up to 0.05 per cent can cause an increase in $\mathrm{CO_2}$ fixation rates; beyond this the levels can become damaging over longer periods.

Tier 3 - Comparisons and Long Answers

Question 37

Q. Distinguish between the light reaction and the dark reaction of photosynthesis.

Answer. Give the anchor difference first: the light reaction is directly light driven and makes the energy currencies, while the dark reaction is not directly light driven and spends them to make sugar.

Point of comparison	Light reaction	Dark reaction
Other names	Photochemical reactions, the photochemical phase	Carbon reactions, the biosynthetic phase
Where it occurs	The membrane system - the grana and stroma lamellae	The stroma
Dependence on light	Directly light driven	Not directly light driven, but dependent on the products of the light reactions - ATP and NADPH
What happens in it	Light absorption, water splitting, oxygen release, and the formation of high-energy chemical intermediates, ATP and NADPH	Enzymatic reactions synthesise sugar, which in turn forms starch
Raw materials	Light and water	$\mathrm{CO_2}$ and $\mathrm{H_2O}$ , plus the ATP and NADPH from the light reaction
Products	ATP, NADPH and $\mathrm{O_2}$	Sugars, and hence starch; ADP and NADP are returned
Is $\mathrm{CO_2}$ fixed here	No	Yes - this is where carboxylation happens
Effect of switching light off	Stops at once	Continues for some time, and then stops

The trap in the naming. Calling the biosynthetic phase the "dark reaction" is arguably a misnomer, because it neither requires darkness nor is it independent of light. The chapter itself warns that the convention should not be construed to mean that they occur in darkness or that they are not light-dependent. If a question asks whether dark reactions occur at night, the answer is that they run only as long as the ATP and NADPH already in the stroma last, and then stop.

Question 38

Q. Distinguish between an absorption spectrum and an action spectrum, and explain what the imperfect overlap between them tells us.

Answer. An absorption spectrum plots how much light a pigment absorbs at each wavelength. An action spectrum plots the rate of photosynthesis at each wavelength. One is a property of a pigment, the other of a process.

Point of comparison	Absorption spectrum	Action spectrum
What is plotted	The light absorbed by a pigment at each wavelength	The rate of photosynthesis at each wavelength
What it belongs to	A pigment, such as chlorophyll a, chlorophyll b or the carotenoids	The whole process of photosynthesis in a living tissue
How it is obtained	By measuring the light absorbed by an extracted pigment	By measuring photosynthesis wavelength by wavelength - Engelmann used a prism, Cladophora and aerobic bacteria to detect the sites of $\mathrm{O_2}$ evolution
Peaks	Chlorophyll a shows maximum absorption in the blue and the red regions	A higher rate of photosynthesis at the same blue and red wavelengths
Who described the first one for photosynthesis	Belongs to the pigment chemists	T. W. Engelmann - a first action spectrum of photosynthesis was thus described

What the comparison establishes. The wavelengths at which there is maximum absorption by chlorophyll a also show a higher rate of photosynthesis, and hence we can conclude that chlorophyll a is the chief pigment associated with photosynthesis. Engelmann's own result fits - the bacteria accumulated mainly in the region of blue and red light of the split spectrum, and the action spectrum resembles roughly the absorption spectra of chlorophyll a and b.

What the mismatch establishes. There is no complete one-to-one overlap between the absorption spectrum of chlorophyll a and the action spectrum of photosynthesis. Together the graphs show that most of the photosynthesis takes place in the blue and red regions of the spectrum, and that some photosynthesis does take place at the other wavelengths of the visible spectrum. Photosynthesis is happening at wavelengths chlorophyll a hardly absorbs, so something else must be absorbing there. That something is the accessory pigments - chlorophyll b, xanthophylls and carotenoids - which also absorb light and transfer the energy to chlorophyll a, and which enable a wider range of wavelengths of incoming light to be utilised for photosynthesis while also protecting chlorophyll a from photo-oxidation. The imperfect overlap is the evidence for the accessory pigments, not a flaw in the measurement.

Question 39

Q. Compare Photosystem I and Photosystem II.

Answer.

Point of comparison	PS I	PS II
Why it carries that number	Discovered first	Discovered second - the photosystems are named in the sequence of their discovery, and not in the sequence in which they function
Order of working in the light reaction	Second - it receives the electrons sent down the chain	First - it starts the electron flow
Reaction centre	Chlorophyll a with an absorption peak at $700\ \mathrm{nm}$ , called $\mathrm{P_{700}}$	Chlorophyll a with absorption maxima at $680\ \mathrm{nm}$ , called $\mathrm{P_{680}}$
Splitting of water	Not associated with PS I	The splitting of water is associated with the PS II
Release of $\mathrm{O_2}$	No	Yes - this creates oxygen, one of the net products of photosynthesis
Source of replacement electrons	From PS II - the electrons needed to replace those removed from photosystem I are provided by photosystem II	From the splitting of water
Where the electron goes next	To another acceptor molecule that has a greater redox potential, then downhill to $\mathrm{NADP^+}$ , reducing it to $\mathrm{NADPH + H^+}$	To an electron acceptor which passes them to an electron transport system consisting of cytochromes, and on to the pigments of PS I
Location in the membranes	Present in both the grana lamellae and the stroma lamellae	Present in the grana lamellae only - the stroma lamellae membranes lack PS II
Can it work alone	Yes - when only PS I is functional, cyclic photophosphorylation occurs	No - it must pass its electrons on

The one line to write first in an exam. PS I is numbered first because it was discovered first, but PS II acts first in the light reaction. $\mathrm{P_{700}}$ goes with PS I and $\mathrm{P_{680}}$ with PS II, and the easy way to keep that straight is that the photosystem found first carries the larger number, $700\ \mathrm{nm}$ .

Question 40

Q. Give a comparison between cyclic and non-cyclic photophosphorylation.

Answer. Start with the definition both share. Photo-phosphorylation is the synthesis of ATP from ADP and inorganic phosphate in the presence of light. The two kinds differ in how many photosystems take part and where the electron ends up.

Point of comparison	Cyclic photo-phosphorylation	Non-cyclic photo-phosphorylation
Photosystems involved	Only PS I is functional	Both PS II and PS I
Do they work in series	No - a single photosystem acts alone	Yes - first PS II and then the PS I, connected through an electron transport chain, as in the Z scheme
Path of the electron	The electron is circulated within the photosystem and is cycled back to the PS I complex through the electron transport chain	The electron travels from PS II, down the electron transport chain, to PS I, and finally to $\mathrm{NADP^+}$ - it does not return
External electron donor	None needed - the same electron comes back	Water, which replaces the electrons removed from PS II
Splitting of water	Does not occur	Occurs - the splitting of water is associated with the PS II
Release of $\mathrm{O_2}$	No	Yes
Is $\mathrm{NADP^+}$ reduced	No - the excited electron does not pass on to $\mathrm{NADP^+}$	Yes - the addition of these electrons reduces $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$
Products	ATP only - the cyclic flow hence results only in the synthesis of ATP, but not of $\mathrm{NADPH + H^+}$	Both ATP and $\mathrm{NADPH + H^+}$
Likely site	The stroma lamellae	The membrane or lamellae of the grana
What that membrane contains	The stroma lamellae membranes lack PS II as well as the NADP reductase enzyme	The membrane or lamellae of the grana have both PS I and PS II
Wavelength condition	Also occurs when only light of wavelengths beyond $680\ \mathrm{nm}$ are available for excitation	Needs both $\mathrm{P_{680}}$ and $\mathrm{P_{700}}$ to be excited

Why the plant keeps both. The Calvin cycle needs 3 molecules of ATP and 2 of NADPH for every $\mathrm{CO_2}$ fixed - an unequal demand. It is probably to meet this difference in the number of ATP and NADPH used in the dark reaction that the cyclic phosphorylation takes place, since cyclic flow makes ATP without making any NADPH.

This comparison answers chapter-end exercise 9 (b).

Question 41

Q. Trace the Z scheme from the absorption of a photon to the reduction of $\mathrm{NADP^+}$ , and explain the name.

Answer. Follow one electron all the way.

Step 1 - excitation at PS II. In photosystem II the reaction centre chlorophyll a absorbs $680\ \mathrm{nm}$ wavelength of red light, causing electrons to become excited and jump into an orbit farther from the atomic nucleus. That is the whole of what light does chemically - it lifts an electron to a higher energy level.

Step 2 - the first downhill run. These electrons are picked up by an electron acceptor which passes them to an electron transport system consisting of cytochromes. This movement of electrons is downhill, in terms of an oxidation-reduction or redox potential scale. The electrons are not used up as they pass through the electron transport chain, but are passed on to the pigments of photosystem PS I - the chain is a relay, not a consumer. It is during this passage that protons are transported across the membrane, which is how the chain contributes to ATP synthesis.

Step 3 - excitation at PS I. Simultaneously, electrons in the reaction centre of PS I are also excited when they receive red light of wavelength $700\ \mathrm{nm}$ , and are transferred to another acceptor molecule that has a greater redox potential. Note simultaneously - the two photosystems are being hit by light at the same moment; it is the electrons that travel from one to the other.

Step 4 - the second downhill run. These electrons then are moved downhill again, this time to a molecule of energy-rich $\mathrm{NADP^+}$ , and the addition of these electrons reduces $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ . The NADP reductase enzyme is located on the stroma side of the membrane, and the protons needed for that reduction are removed from the stroma.

Step 5 - closing the gap at PS II. The electrons that were moved from photosystem II must be replaced, and this is achieved by electrons available due to splitting of water. The splitting of water is associated with the PS II; water is split into $\mathrm{2H^+}$ , $\mathrm{[O]}$ and electrons, and this creates oxygen, one of the net products of photosynthesis.

$\mathrm{2H_2O \rightarrow 4H^+ + O_2 + 4e^-}$

Since the water splitting complex is associated with PS II, which is physically located on the inner side of the membrane of the thylakoid, the protons and the oxygen formed are released into the lumen.

The name. This whole scheme of transfer of electrons is called the Z scheme, due to its characteristic shape, and this shape is formed when all the carriers are placed in a sequence on a redox potential scale. Plotted that way the path goes up at PS II, a long slope down through the cytochromes, up again at PS I, then down to $\mathrm{NADP^+}$ - turned on its side, a Z. It is not the physical arrangement of the complexes in the membrane.

Question 42

Q. Explain the chemiosmotic hypothesis as it applies to the chloroplast.

Answer. The chemiosmotic hypothesis has been put forward to explain the mechanism of ATP synthesis. The idea in one line: build a proton gradient across a membrane, then let it collapse through ATP synthase.

The membrane and the direction. Like in respiration, in photosynthesis too, ATP synthesis is linked to development of a proton gradient across a membrane. This time these are the membranes of the thylakoid. There is one difference though. Here the proton accumulation is towards the inside of the membrane, that is, in the lumen. In respiration, protons accumulate in the intermembrane space of the mitochondria.

How the gradient is created - three causes.

(a) Since splitting of the water molecule takes place on the inner side of the membrane, the protons or hydrogen ions that are produced by the splitting of water accumulate within the lumen of the thylakoids.

(b) As electrons move through the photosystems, protons are transported across the membrane. This happens because the primary acceptor of electrons, which is located towards the outer side of the membrane, transfers its electron not to an electron carrier but to an H carrier. Hence this molecule removes a proton from the stroma while transporting an electron. When this molecule passes on its electron to the electron carrier on the inner side of the membrane, the proton is released into the lumen side of the membrane.

(c) The NADP reductase enzyme is located on the stroma side of the membrane. Along with electrons that come from the acceptor of electrons of PS I, protons are necessary for the reduction of $\mathrm{NADP^+}$ to $\mathrm{NADPH + H^+}$ . These protons are also removed from the stroma.

The result. Within the chloroplast, protons in the stroma decrease in number, while in the lumen there is accumulation of protons. This creates a proton gradient across the thylakoid membrane as well as a measurable decrease in pH in the lumen.

How the gradient makes ATP. It is the breakdown of this gradient that leads to the synthesis of ATP. The gradient is broken down due to the movement of protons across the membrane to the stroma through the transmembrane channel of the $\mathrm{CF_0}$ of the ATP synthase.

Part of ATP synthase	Where it sits	What it does
$\mathrm{CF_0}$	Embedded in the thylakoid membrane	A transmembrane channel that carries out facilitated diffusion of protons across the membrane
$\mathrm{CF_1}$	Protrudes on the outer surface of the thylakoid membrane, on the side that faces the stroma	Undergoes the conformational change that makes the enzyme synthesise ATP

The break down of the gradient provides enough energy to cause a conformational change in the $\mathrm{CF_1}$ particle of the ATP synthase, which makes the enzyme synthesise several molecules of ATP.

What chemiosmosis requires. A membrane, a proton pump, a proton gradient and ATP synthase.

Where the ATP goes. Along with the NADPH produced by the movement of electrons, the ATP will be used immediately in the biosynthetic reaction taking place in the stroma, responsible for fixing $\mathrm{CO_2}$ and synthesis of sugars.

Question 43

Q. Describe the three stages of the Calvin cycle and work out the ATP and NADPH needed for one molecule of glucose.

Answer. Calvin and his co-workers worked out the whole pathway and showed that the pathway operated in a cyclic manner - the $\mathrm{RuBP}$ was regenerated. Note before anything else that the Calvin pathway occurs in all photosynthetic plants; it does not matter whether they have $\mathrm{C_3}$ or $\mathrm{C_4}$ or any other pathway. For ease of understanding, the Calvin cycle can be described under three stages: carboxylation, reduction and regeneration.

Stage 1 - Carboxylation. Carboxylation is the fixation of $\mathrm{CO_2}$ into a stable organic intermediate, and it is the most crucial step of the Calvin cycle. $\mathrm{CO_2}$ is utilised for the carboxylation of $\mathrm{RuBP}$ , this reaction is catalysed by the enzyme $\mathrm{RuBP}$ carboxylase, and it results in the formation of two molecules of 3-PGA.

$\mathrm{RuBP + CO_2} \xrightarrow{\text{RuBisCO}} 2 \times \mathrm{3PGA}$

Since this enzyme also has an oxygenation activity, it would be more correct to call it $\mathrm{RuBP}$ carboxylase-oxygenase, or RuBisCO. The carbon count works out because the 5 carbons of $\mathrm{RuBP}$ plus the 1 carbon of $\mathrm{CO_2}$ give 6 carbons, which appear as two molecules of the 3-carbon PGA.

Stage 2 - Reduction. These are a series of reactions that lead to the formation of glucose. The steps involve the utilisation of 2 molecules of ATP for phosphorylation and two of NADPH for reduction, per $\mathrm{CO_2}$ molecule fixed.

Stage 3 - Regeneration. Regeneration of the $\mathrm{CO_2}$ acceptor molecule $\mathrm{RuBP}$ is crucial if the cycle is to continue uninterrupted. The regeneration steps require one ATP for phosphorylation to form $\mathrm{RuBP}$ . No NADPH is used in regeneration at all.

The arithmetic.

Stage	ATP per $\mathrm{CO_2}$	NADPH per $\mathrm{CO_2}$
Carboxylation	None	None
Reduction	2	2
Regeneration	1	None
Total	3	2

Hence for every $\mathrm{CO_2}$ molecule entering the Calvin cycle, 3 molecules of ATP and 2 of NADPH are required. The fixation of six molecules of $\mathrm{CO_2}$ and 6 turns of the cycle are required for the formation of one molecule of glucose from the pathway, so multiply by six:

$6 \times 3 = 18\ \text{ATP} \qquad 6 \times 2 = 12\ \text{NADPH}$

In	Out
Six $\mathrm{CO_2}$	One glucose
18 ATP	18 ADP
12 NADPH	12 NADP

Notice that the two numbers are not equal - 3 ATP against 2 NADPH, and it is probably to meet this difference in the number of ATP and NADPH used in the dark reaction that the cyclic phosphorylation takes place. The favourite wrong answer is 12 ATP and 12 NADPH, which comes from forgetting the one extra ATP spent in regeneration.

Question 44

Q. Give a comparison between the $\mathrm{C_3}$ and $\mathrm{C_4}$ pathways.

Answer. Say the anchor sentence first, because it prevents the commonest error: both types run the Calvin cycle. Though $\mathrm{C_4}$ plants have the $\mathrm{C_4}$ oxaloacetic acid as the first $\mathrm{CO_2}$ fixation product, they use the $\mathrm{C_3}$ pathway, or the Calvin cycle, as the main biosynthetic pathway, and the basic pathway that results in the formation of the sugars, the Calvin pathway, is common to the $\mathrm{C_3}$ and $\mathrm{C_4}$ plants. The $\mathrm{C_4}$ route is an extra pump placed in front of the Calvin cycle, not a replacement for it.

Characteristic	$\mathrm{C_3}$ pathway	$\mathrm{C_4}$ pathway
Other name	The Calvin cycle	The Hatch and Slack pathway
Primary $\mathrm{CO_2}$ acceptor	RuBP - ribulose bisphosphate	PEP - phosphoenol pyruvate
Carbons in the acceptor	5	3
Enzyme of the first fixation	RuBisCO	PEP carboxylase, or PEPcase
First stable product	PGA - 3-phosphoglyceric acid	OAA - oxaloacetic acid
Carbons in the first product	3	4
Cell of the initial carboxylation	Mesophyll	Mesophyll
Cell in which the Calvin cycle runs	All the mesophyll cells	Only the bundle sheath cells
Number of cell types that fix $\mathrm{CO_2}$	One: mesophyll	Two: mesophyll and bundle sheath
Does it have RuBisCO	Yes, in the mesophyll	Yes, in the bundle sheath; the mesophyll cells lack RuBisCO
Does it have PEPcase	No	Yes, in the mesophyll; the bundle sheath cells lack PEPcase
Photorespiration	Present - high at high light intensities and at low $\mathrm{CO_2}$ concentrations	Absent - negligible under all conditions
$\mathrm{CO_2}$ fixation rate under high light	Low	High
Temperature optimum	$20^\circ$ to $25^\circ\mathrm{C}$	$30^\circ$ to $40^\circ\mathrm{C}$
Productivity of biomass	Lower	Greater
Habitat	Most temperate plants	Plants adapted to dry tropical regions
Examples	Wheat and rice are typical	Maize and sorghum

How the $\mathrm{C_4}$ route runs, in order. PEP in the mesophyll fixes $\mathrm{CO_2}$ using PEPcase to give OAA; OAA forms other 4-carbon compounds like malic acid or aspartic acid in the mesophyll cells itself; these are transported to the bundle sheath cells; there the $\mathrm{C_4}$ acids are broken down to release $\mathrm{CO_2}$ and a 3-carbon molecule; the 3-carbon molecule is transported back to the mesophyll, where it is converted to PEP again, thus completing the cycle; and the $\mathrm{CO_2}$ released in the bundle sheath cells enters the $\mathrm{C_3}$ or the Calvin pathway, a pathway common to all plants.

Read the carbon numbers slowly, because they run backwards from what students expect. The $\mathrm{C_3}$ plant has the 5-carbon acceptor and the 3-carbon product; the $\mathrm{C_4}$ plant has the 3-carbon acceptor and the 4-carbon product. The name of the plant comes from the first product, never from the acceptor.

This comparison answers chapter-end exercise 9 (a).

Question 45

Q. Give a comparison between the anatomy of the leaf in $\mathrm{C_3}$ and $\mathrm{C_4}$ plants.

Answer. Take vertical sections of the two leaves and the difference shows up around the vascular bundle.

Feature of the leaf	$\mathrm{C_3}$ plant	$\mathrm{C_4}$ plant
Kranz anatomy	Absent	Present - Kranz means wreath, a reflection of the arrangement of cells
Bundle sheath cells	Not large and not specialised in this way	Particularly large cells around the vascular bundles, and they may form several layers
Chloroplasts in the bundle sheath	Few or none	A large number
Walls of the bundle sheath cells	Not specialised	Thick walls impervious to gaseous exchange
Intercellular spaces in the bundle sheath	Not a distinguishing feature	No intercellular spaces
Mesophyll arrangement	Ordinary mesophyll, not organised in a wreath around the bundle	Mesophyll arranged around the bundle sheath, which in turn wraps the vascular bundle
Enzyme in the mesophyll cells	RuBisCO	PEPcase; the mesophyll cells lack RuBisCO
Enzyme in the bundle sheath cells	Not the site of fixation	Rich in RuBisCO, but lack PEPcase
Where the Calvin cycle runs	In all the mesophyll cells	Only in the bundle sheath cells
Examples	Wheat, rice	Maize, sorghum

Why each anatomical feature is there. The many chloroplasts are needed because the bundle sheath is the only place the Calvin cycle runs in a $\mathrm{C_4}$ plant. The thick walls impervious to gaseous exchange and the absence of intercellular spaces keep the $\mathrm{CO_2}$ released from the $\mathrm{C_4}$ acids from leaking back out, which is what increases the concentration of $\mathrm{CO_2}$ at the enzyme site and so ensures that RuBisCO functions as a carboxylase, minimising the oxygenase activity. The anatomy and the biochemistry are one design.

How to use it in the laboratory. Cut vertical sections of leaves of diverse species and observe them under the microscope, looking for the bundle sheath around the vascular bundles - the presence of the bundle sheath would help you identify the $\mathrm{C_4}$ plants. Maize or sorghum are the leaves to cut to see Kranz anatomy.

This comparison answers chapter-end exercise 9 (c).

Question 46

Q. What is photorespiration, what does it cost the plant, and why do $\mathrm{C_4}$ plants escape it?

Answer. Photorespiration is one more process that creates an important difference between $\mathrm{C_3}$ and $\mathrm{C_4}$ plants, and it begins in the first step of the Calvin cycle.

Where it comes from. RuBisCO is characterised by the fact that its active site can bind to both $\mathrm{CO_2}$ and $\mathrm{O_2}$ - hence the name carboxylase-oxygenase. RuBisCO has a much greater affinity for $\mathrm{CO_2}$ when the $\mathrm{CO_2}$ to $\mathrm{O_2}$ ratio is nearly equal, this binding is competitive, and it is the relative concentration of $\mathrm{O_2}$ and $\mathrm{CO_2}$ that determines which of the two will bind to the enzyme.

What happens in a $\mathrm{C_3}$ plant. In $\mathrm{C_3}$ plants some $\mathrm{O_2}$ does bind to RuBisCO, and hence $\mathrm{CO_2}$ fixation is decreased. Here the RuBP, instead of being converted to 2 molecules of PGA, binds with $\mathrm{O_2}$ to form one molecule of phosphoglycerate and one molecule of phosphoglycolate, which is a 2-carbon compound. This pathway is called photorespiration.

What it costs.

In the photorespiratory pathway there is neither synthesis of sugars, nor of ATP.
Rather, it results in the release of $\mathrm{CO_2}$ with the utilisation of ATP.
In the photorespiratory pathway there is no synthesis of ATP or NADPH.
The biological function of photorespiration is not known yet.

So the plant loses carbon it had already fixed, spends ATP doing it, and gains no sugar and no reducing power in return.

Why $\mathrm{C_4}$ plants escape it. In $\mathrm{C_4}$ plants photorespiration does not occur, and the reason is entirely a matter of concentration at the enzyme site, not of a different enzyme.

The $\mathrm{C_4}$ acid from the mesophyll is broken down in the bundle sheath cells to release $\mathrm{CO_2}$ .
This results in increasing the intracellular concentration of $\mathrm{CO_2}$ in exactly the cells where RuBisCO sits, and the thick walls impervious to gaseous exchange with no intercellular spaces hold it there.
In turn, this ensures that the RuBisCO functions as a carboxylase, minimising the oxygenase activity.

Because the relative concentration of the two gases decides which one binds, flooding the bundle sheath with $\mathrm{CO_2}$ effectively shuts oxygen out of the active site. $\mathrm{C_4}$ plants have the same RuBisCO - the bundle sheath cells are rich in RuBisCO but lack PEPcase; what differs is the gas mixture the enzyme is sitting in. The consequences follow: since $\mathrm{C_4}$ plants lack photorespiration, productivity and yields are better in these plants, and in addition these plants show tolerance to higher temperatures.

Question 47

Q. RuBisCO acts both as a carboxylase and as an oxygenase. Compare the two activities and say what decides which one the enzyme performs.

Answer. RuBisCO is the most abundant enzyme in the world, and its full name, ribulose bisphosphate carboxylase-oxygenase, records both jobs. Its active site can bind to both $\mathrm{CO_2}$ and $\mathrm{O_2}$ - hence the name.

Point of comparison	Carboxylase activity	Oxygenase activity
Which gas binds the active site	$\mathrm{CO_2}$	$\mathrm{O_2}$
What RuBP is converted to	Two molecules of 3PGA	One molecule of phosphoglycerate and one molecule of phosphoglycolate, which is a 2-carbon compound
The pathway it starts	The Calvin cycle	Photorespiration
Where it is the first step	Carboxylation, the most crucial step of the Calvin cycle	The photorespiratory pathway
Sugar formed	Yes, through the Calvin cycle	No - there is neither synthesis of sugars, nor of ATP
ATP and NADPH	Spent productively - 3 ATP and 2 NADPH per $\mathrm{CO_2}$ fixed	ATP is spent with nothing gained - it results in the release of $\mathrm{CO_2}$ with the utilisation of ATP, and there is no synthesis of ATP or NADPH
Net effect on the plant's carbon	Carbon is gained	Carbon already fixed is lost
Known biological function	Carbon fixation, the basis of all the plant's sugar	The biological function of photorespiration is not known yet

What decides which activity runs. It is the relative concentration of $\mathrm{O_2}$ and $\mathrm{CO_2}$ that determines which of the two will bind to the enzyme. The binding is competitive - the two gases compete for the same active site - and RuBisCO has a much greater affinity for $\mathrm{CO_2}$ when the $\mathrm{CO_2}$ to $\mathrm{O_2}$ ratio is nearly equal. It is not the absolute amount of oxygen, not the amount of enzyme, and not a second, different enzyme.

Why RuBisCO carries out more carboxylation in a $\mathrm{C_4}$ plant follows from that one sentence. $\mathrm{C_4}$ plants have a mechanism that increases the concentration of $\mathrm{CO_2}$ at the enzyme site: the $\mathrm{C_4}$ acid from the mesophyll is broken down in the bundle sheath cells to release $\mathrm{CO_2}$ , this results in increasing the intracellular concentration of $\mathrm{CO_2}$ , and this in turn ensures that the RuBisCO functions as a carboxylase, minimising the oxygenase activity.

Question 48

Q. State Blackman's Law of Limiting Factors and illustrate it with a worked example, showing how it explains the shape of the light curve.

Answer. Photosynthesis is under the influence of several factors, both internal (plant) and external. The plant or internal factors are the number, size, age and orientation of leaves, mesophyll cells and chloroplasts, internal $\mathrm{CO_2}$ concentration and the amount of chlorophyll, and they are dependent on the genetic predisposition and the growth of the plant. The external factors are the availability of sunlight, temperature, $\mathrm{CO_2}$ concentration and water. As a plant photosynthesises, all these factors will simultaneously affect its rate, but usually one factor is the major cause, or is the one that limits the rate, and hence, at any point the rate will be determined by the factor available at sub-optimal levels.

The law, stated in full. Blackman's (1905) Law of Limiting Factors: if a chemical process is affected by more than one factor, then its rate will be determined by the factor which is nearest to its minimal value: it is the factor which directly affects the process if its quantity is changed.

Read the second half as a practical test. Change the limiting factor and the rate moves; change anything else and nothing happens.

Worked example 1 - the chapter's own. Despite the presence of a green leaf and optimal light and $\mathrm{CO_2}$ conditions, the plant may not photosynthesise if the temperature is very low. This leaf, if given the optimal temperature, will start photosynthesising. Light and $\mathrm{CO_2}$ were already optimal, so raising them further changes nothing. Only the factor nearest its minimum - here the temperature - moves the rate, and temperature bites because the dark reactions, being enzymatic, are temperature controlled.

Worked example 2 - the shape of the light curve.

Part of the light curve	What is happening	Which factor is limiting
The rising, straight portion at low light	There is a linear relationship between incident light and $\mathrm{CO_2}$ fixation rates at low light intensities	Light itself - it is nearest its minimal value
The bending region	The increase slows	Light, and increasingly other factors
The flat plateau	At higher light intensities, gradually the rate does not show further increase, as other factors become limiting	Something other than light - typically $\mathrm{CO_2}$ concentration or temperature
Beyond a point	The rate falls	Increase in incident light beyond a point causes the breakdown of chlorophyll and a decrease in photosynthesis

The plateau is the law made visible: once light is no longer the factor nearest its minimum, adding light does nothing. Two facts complete this: light saturation occurs at 10 per cent of the full sunlight, and hence, except for plants in shade or in dense forests, light is rarely a limiting factor in nature.

Worked example 3 - which factor to add. On the plateau of the light curve, the factor to add is carbon dioxide, because carbon dioxide is the major limiting factor for photosynthesis - its concentration is very low in the atmosphere, between 0.03 and 0.04 per cent. But the answer depends on the plant. $\mathrm{C_4}$ plants show saturation at about $360\ \mu\mathrm{L\,L^{-1}}$ , so extra $\mathrm{CO_2}$ does little for them, while $\mathrm{C_3}$ plants respond to increased $\mathrm{CO_2}$ concentration and saturation is seen only beyond $450\ \mu\mathrm{L\,L^{-1}}$ . Thus, current availability of $\mathrm{CO_2}$ levels is limiting to the $\mathrm{C_3}$ plants, which is why greenhouse crops such as tomatoes and bell pepper are allowed to grow in a carbon dioxide enriched atmosphere, which leads to higher yields.

The Chapter-End Exercises

There are nine exercises at the end of this chapter, and because exercise 9 asks for three separate comparisons, that comes to eleven questions to write out. Every one of them is already answered in full inside the twelve teaching sections of this chapter - there is nothing left over to work out here. The table below tells you exactly where each answer sits, so you can attempt the exercise yourself first and then check your version against a complete one.

Exercise	Answered as
1. By looking at a plant externally, can you tell whether a plant is $\mathrm{C_3}$ or $\mathrm{C_4}$	Question 4 of the The C4 Pathway section
2. By looking at which internal structure of a plant you can tell whether a plant is $\mathrm{C_3}$ or $\mathrm{C_4}$	Question 5 of the The C4 Pathway section
3. Even though a very few cells in a $\mathrm{C_4}$ plant carry out the biosynthetic Calvin pathway, yet they are highly productive	Question 14 of the The C4 Pathway section
4. RuBisCO acts both as a carboxylase and oxygenase - why does it carry out more carboxylation in $\mathrm{C_4}$ plants	Question 12 of the Photorespiration section
5. Plants with a high concentration of chlorophyll b but lacking chlorophyll a - would they photosynthesise, and why do plants have accessory pigments	Question 11 of the Where Photosynthesis Happens, and the Pigments Involved section
6. Why does a leaf kept in the dark frequently become yellow or pale green, and which pigment is more stable	Question 12 of the Where Photosynthesis Happens, and the Pigments Involved section
7. Leaves on the shady side compared with leaves on the sunny side - which are darker green and why	Question 13 of the Where Photosynthesis Happens, and the Pigments Involved section
8. Figure 11.10, the effect of light on the rate of photosynthesis - parts (a), (b) and (c)	Question 9 of the Factors Affecting Photosynthesis section
9 (a). Comparison between the $\mathrm{C_3}$ and $\mathrm{C_4}$ pathways	Question 14 of the C3 Plants and C4 Plants Compared section, and again as Question 44 above
9 (b). Comparison between cyclic and non-cyclic photophosphorylation	Question 12 of the Cyclic and Non-cyclic Photophosphorylation section, and again as Question 40 above
9 (c). Comparison between the anatomy of the leaf in C3 and C4 plants	Question 6 of the The C4 Pathway section, and again as Question 45 above

Work all eleven out on paper before the exam, in your own words and with the tables drawn out, rather than reading the answers and moving on. And notice how the set is built: exercises 1, 2, 3 and 9 are all the same $\mathrm{C_3}$ against $\mathrm{C_4}$ material asked four different ways - the external appearance of the plant, the internal anatomy of the leaf, the productivity of the bundle sheath cells, and the two pathways side by side. Learn the Kranz anatomy of the bundle sheath, the PEP-PEPcase-OAA route in the mesophyll, the release of $\mathrm{CO_2}$ in the bundle sheath that shuts out photorespiration, and the full comparison table once, properly, and four of the nine exercises answer themselves.

How to Use This Section

The Facts These Questions Draw On

Tier 1 - Concept Checks

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Tier 2 - Application and Identification

Question 23

Question 24

Question 25

Question 26

Question 27

Question 28

Question 29

Question 30

Question 31

Question 32

Question 33

Question 34

Question 35

Question 36

Tier 3 - Comparisons and Long Answers

Question 37

Question 38

Question 39

Question 40

Question 41

Question 42

Question 43

Question 44

Question 45

Question 46

Question 47

Question 48

The Chapter-End Exercises

Great job!