A photosynthesizing mouse?

A photosynthesizing mouse?

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N. Shubin's Your Inner Fish makes the point several times that there is a lot of functional similarity between some seemingly remote gene cousins. If that needed reinforcing we have the spider-goat, whose milk contains spider's silk, and a knock-in mouse whose vision resembles that of humans.

The last two examples are interesting because, as I recall, nothing beyond the gene insertion (already a feat) had to be done to confer the extra/new capability. The machinery at the cellular level (for the mouse to perceive a new color or for the goat to somehow process the milk) already existed.

My question is whether it might not be possible in theory to create a mammal with the ability to photosynthesize? If this is a polygenic trait perhaps it would be accomplished in multiple stages. I realize an answer here would be highly speculative but a careful answer might cast some light on the process of conferring new traits/abilities in this way.

While I see no obvious benefit of creating a photosynthesizing mouse, at least the food bills for their maintenance might be low. This sounds like a joke but it's not a trivial benefit.

Thanks for any insights.

It is almost impossible for following reasons:

  1. For photosynthesis you need chloroplasts
  2. To maintain chloroplasts you need many genes in the nucleus that will support its endosymbiosis

I said almost impossible because there are some natural examples of what you are asking. A sea slug called Elysia acquires choloplasts from green algae on which it feeds. However, it cannot maintain the chloroplasts and pass them on to the next generation but it acquires enough to appear green and survive on photosynthesis when there is no food.

Another example is that of the protist, Rhopalodia gibba, which acquired a cyanobacteria like organism. This protist already had a red alga derived secondary plastid before it acquired a "green" cyanobacteria. This acquisition gave the host the ability to fix nitrogen in the presence of light.

The case of Paulinella chromatophora is very interesting because the acquisition of an endosymbiont happened very recently. The endosymbiont is close to Synechococcus clade of cyanobacteria.

Intuitively, it can be understood that these kinds of acquisitions would be quite difficult for a multicellular organism.


Photosynthesis is a complex reaction which requires a dedicated compartment which not only harvests light and produces ATP but also has enzyme complexes required for anabolism (carbon fixation etc). For now we can accept the hypothesis that it would not be possible for a huge eukaryotic cell to perform these functions in absence of a dedicated organelle. For a discussion on why this is so, you can refer this post. Having said that, there is an easy way to impart partial photosynthetic ability to a eukaryotic cell. Some archaea and bacteria employ rhodopsin to pump out protons against its gradient, in the presence of light; this is coupled with ATP-synthase just like the complexes of ETC in mitochondria. In this study, Hara et al have expressed delta-rhodopsin in mammalian mitochondia which now makes the mitochondria generate ATP in presence of light. Furthermore, these cells were immune to mitochondrial toxins that affect complex-I activity.

PS: thanks biogirl. I almost forgot about this and I just remembered when I was reading about something else

Summarize the experimental results that revealed details about the process of photosynthesis.

The history of the studies done on photosynthesis dates back into the 17th century with Jan Baptist van Helmont. He rejected the ancient idea that plants take most of their biomass from the soil. For the proof, he performed an experiment using a willow tree. He started with a willow tree with a mass of 2.27 kg. Over 5 years, it grew to 67.7 kg. However, the mass of the soil only decreased by 57 grams. Van Helmont came to the conclusion that plants must obtain most of their mass from water. He did not know about gases.

Joseph Priestley ran a series of experiments in 1772 (Figure (PageIndex<1>)). He tested a mouse, a candle, and a sprig of mint under hermetically sealed (no air can go in or out) jar. He first observed that a mouse and a candle behave very similarly when covered, in that they both &ldquospend&rdquo the air. However, when a plant is placed with either the candle or mouse, the plant &ldquorevives&rdquo the air for both.

Figure (PageIndex<1>): Experiments of J. Priestley (1772) demonstrating that plants provided "air" needed for a candle to burn or a mouse to breath.

Further ideas were brought about in the late 1700&rsquos. Jan Ingenhousz and Jean Senebier found that the air is only reviving in the day time and that CO(_2) is assembled by plants. Antoin-Laurent Lavoiser found that &ldquorevived air&rdquo is a separate gas, oxygen.

But what is the oxygen &ldquomaker&rdquo? There are many pigments in plants, and all accept and reflect some parts of rainbow. To identify the culprit, Thomas Engelmann ran an experiment (Figure (PageIndex<2>)) using a crystal prism that shine different wavelengths (colors) of visible light on the algae Spirogyra algae. He then measured oxygen production with aerotactic bacteria, which move towards areas of high oxygen concentration. A high density of bacteria cells accumulated in the blue and red parts of the spectrum, indicating this was where the most oxygen was produced and the most photosynthesis was conducted. This was a huge find. It tells that the key photosynthetic pigment should accept blue and red rays, and thus reflect green rays. The photosynthetic pigment chlorophyll a best fits this description.

Figure (PageIndex<2>): The experiment of Thomas Engelmann (1881) involved separating the wavelengths of visible light using a prism. The rainbow of visible light ranges from approximately 400 nm (violet) to 700 nm (red). The algae Spirogyra produced the most oxygen bubbles in blue and red parts of the visible light spectrum, and bacterial cells thus accumulated in these regions.

Another important fact was discovered by Frederick Blackman in 1905. He found that if light intensity is low, the increase of temperature actually has very little effect on the rate of photosynthesis. However, the reverse is not exactly true, and light is able to intensify photosynthesis even when it is cold.

This could not happen if light and temperature are absolutely independent factors. If temperature and light are components of the chain, light was first (&ldquoignition&rdquo) and temperature was second. This ultimately shows that photosynthesis has two stages (now called the light-dependent and light-independent stages). The light-dependent stage relates the to intensity of the light. The light-independent stage relates more with the temperature as it involves many enzymes.

Methods of Producing Transgenic Mouse (With Diagram)

It is designed to support investigators doing biology of aging research by creating mice that have been genetically altered by either inserting a new gene or removing a normal gene.

This method has become one of the most exciting approaches of discovering the functions and interactions of genes in mammals. At the University Of Washington, Nathan Shock Center, this transgenic technology is used to develop new animal models for studying genetic mechanisms of the aging process (Fig. 18.4).

During the previous year, transgenic mouse production has focused on constructs with enhanced defense against free radical injury in aging (e.g., catalase, superoxide dismutase, glutathione S-transferase), Werner Syndrome, adult onset diabetes, Alzheimer’s disease, thrombospondin, and rheumatoid arthritis in aging. Almost 4000 embryos, mainly of the C57BL/6 ingred strain, have been transferred, 498 pups analyzed and at least 40 contained the integrated construct.

In addition, this core concentrated an appreciable portion of effort into embryonal stem line (ES) methodologies for generation of knockouts and targeted ES transgenic. This included work to generate mouse models of Werner’s Syndrome, models for study of presenillin genes related to Alzheimer’s Disease and study of models of thrombospondin in aging. In the past year, a total of 396 embryos were transferred, and 79 pups were born, of which 37 were chimeric.

The isolation of mammalian genes is of utmost importance to the biology and medicine of aging because of the contributions these studies can make to the understanding of physiology and development. Techniques for introducing foreign genes into the mouse germ line provide novel approaches for modeling human genetic and chronic degenerative diseases. Since the initial report in 1980 describing transgenic mice, methods for the direct microinjection of DNA into the pro-nuclei of fertilized embryos have become established. Foreign genes can be incorporated into somatic germ-line tissues, with expression of these elements in the progeny of founder mice.

The creation of “transgenic” animals that make a specified gene product presents a spectrum of opportunities for basic studies in molecular pathogenesis and pre-clinical investigations applicable to a wide variety of medical problems of aging. An additional gene transfer technology developed in the 1980’s involved the use of stem cells from the early embryo, so-called embryonic stem (ES) cells. The capacity of ES cells to undergo differentiation makes them useful for investigating the effects of genetic modifications of either the gain of function or loss of function.

These pluripotent genetically modified ES cells can then be used to make mice with deleted genes (gene knockout) or targeted mutagenesis of genes thought to be involved in the aging process. It is also possible to develop lines of transgenic mice carrying very large DNA constructs (>600 kb) transected into ES cells. The University of Washington Aging Program provides a focused level of expertise and resources to enhance and facilitate the development of transgenic animal models using ES cell transfer technology.

Two methods of producing transgenic mice are widely used:

(1) Transforming embryonic stem cells (ES cells) growing in tissue culture with the desired DNA

(2) injecting the desired gene into the pro-nucleus of a fertilized mouse egg.

Method 1- The Embryonic Stem Cell Method:

Embryonic stem cells (ES cells) are harvested from the inner cell mass (ICM) of mouse blastocysts. They can be grown in culture and retain their full potential to produce all the cells of the mature animal, including its gametes.

1. Make your DNA-Using recombinant DNA methods, build molecules of DNA containing the structural gene you desire (e.g., the insulin gene), vector DNA to enable the molecules to be inserted into host DNA molecules, promoter and enhancer sequences to enable the gene to be expressed by host cells.

2. Transform ES cells in culture- Expose the cultured cells to the DNA so that some will incorporate it.

3. Select for successfully transformed cells.

4. Inject these cells into the inner cell mass (ICM) of mouse blastocysts.

5. Embryo transfer- Prepare a pseudo pregnant mouse (by mating a female mouse with a vasectomized male). The stimulus of mating elicits the hormonal changes needed to make her uterus receptive. Transfer the embryos into her uterus.

6. Test her offspring – Remove a small piece of tissue from the tail and examine its DNA for the desired gene. No more than 10-20% will have it, and they will be heterozygous for the gene.

7. Establish a transgenic strain – Mate two heterozygous mice and screen their offspring for the 1:4 that will be homozygous for the transgene. Mating these will found the transgenic strain.

Method 2 -The Pro-nucleus Method:

1. DNA is prepared as in Method 1.

2. Transform fertilized eggs – Freshly fertilized eggs are harvested before the sperm head has become a pro-nucleus. The male pro-nucleus is injected with DNA. When the pro-nuclei have fused to form the diploid zygote nucleus, the zygote is allowed to divide by mitosis to form a 2-cell embryo. The embryos is implanted in a pseudo pregnant foster mother and preceded as in Method 1.

The Figure 18.5 shows a transgenic mouse (right) with a normal littermate (left). The giant mouse developed from a fertilized egg transformed with a recombinant DNA molecule containing the structural gene for human growth hormone and a strong mouse gene promoter.

The levels of growth hormone in the serum of some of the transgenic mice were several hundred times higher than in control mice.

Random vs. Targeted Gene Insertion:

The early vectors used for gene insertion could, and did, place the gene (from one to 200 copies of it) anywhere in the genome. However, if you know some of the DNA sequence flanking a particular gene, it is possible to design vectors that replace that gene. The replacement gene can be one that (a) restores function in a mutant animal or (b) knocks out the function of a particular locus.

In either case, targeted gene insertion requires:

(2) Neor, a gene that encodes an enzyme that inactivates the antibiotic neomycin and its relatives, like the drug G418, which is lethal to mammalian, cells

(3) Tk, a gene that encodes thymidine kinase, an enzyme that phosphorylates the nucleoside analog gancyclovir.

DNA polymerase fails to discriminate against the resulting nucleotide and inserts this nonfunctional nucleotide into freshly-replicating DNA. So ganciclovir kills cells that contain the tk gene. Knockout mice are valuable tools for discovering the function(s) of genes for which mutant strains were not previously available.

Two generalizations have emerged from examining knockout mice: Knockout mice are often surprisingly unaffected by their deficiency. Many genes turn out not to be indispensable. The mouse genome appears to have sufficient redundancy to compensate for a single missing pair of alleles. Most genes are pleiotropic. They are expressed in different tissues in different ways and at different times in development.

The Cre/loxP System:

One of the bacteriophages that infects E. coli, called PI, produces an enzyme — designated Cre — that cuts its DNA into lengths suitable for packaging into fresh virus particles. Cre cuts the viral DNA wherever it encounters a pair of sequences designated loxP. All the DNA between the two loxP sites is removed and the remaining DNA ligated together again (so the enzyme is a recombinase).

Using “Method 1” (above), mice can be made transgenic for the gene encoding Cre attached to a promoter that will be activated only when it is bound by the same transcription factors that urn on the other genes required for the unique function(s) of that type of cell a “target” gene, the one whose function is to be studied, flanked by loxP sequences (Fig. 18.6).

In the adult animal, those cells that receive signals (e.g., the arrival of a hormone or cytokine) to turn on production of the transcription factors needed to activate the promoters of the genes whose products are needed by that particular kind of cell will also turn on transcription of the Cre gene. Its protein will then remove the “target” gene under study. All other cells will lack the transcription factors needed to bind to the Cre promoter (and/or any enhancers) so the target gene remains intact.

The result: a mouse with a particular gene knocked out in only certain cells. The CrdloxP system can also be used to remove DNA sequences that block gene transcription. In such a “knockin” mouse, the “target” gene is turned on in only certain cells.

Priestly was quite intrigued by the air that floated over the fermenting grain in the brewery next to his church. From his experiments, he was able to deduce that this brewery gas extinguished a lighted wood chip and was heavier than normal air. He later designed an experiment to produce the same gas in his lab. He also noted that the gas produced a tangy taste when dissolved in water which led to his invention of the first drinkable glass of carbonated water (soda water).

In 1794, Priestley immigrated to the United States with his family and settled in Northumberland, Pennsylvania. He later retired to a peaceful life with his writings. He died quietly in his home on February 6, 1804.

Chapter 10 – Photosynthesis Objectives

1. Distinguish between autotrophic and heterotrophic nutrition.

2. Distinguish between photoautotrophs and chemoautotrophs.

3. Describe the structure of a chloroplast, listing all membranes and compartments.

The Pathways of Photosynthesis

4. Write a summary equation for photosynthesis.

5. Explain van Niel’s hypothesis and describe how it contributed to our current understanding of photosynthesis. Explain the evidence that supported his hypothesis.

6. In general terms, explain the role of redox reactions in photosynthesis.

7. Describe the two main stages of photosynthesis in general terms.

8. Describe the relationship between an action spectrum and an absorption spectrum. Explain why the action spectrum for photosynthesis differs from the absorption spectrum for chlorophyll a.

9. Explain how carotenoids protect the cell from damage by light.

10. List the wavelengths of light that are most effective for photosynthesis.

11. Explain what happens when a solution of chlorophyll a absorbs photons. Explain what happens when chlorophyll a in an intact chloroplast absorbs photons.

12. List the components of a photosystem and explain the function of each component.

13. Trace the movement of electrons in noncyclic electron flow. Trace the movement of electrons in cyclic electron flow.

14. Explain the functions of cyclic and noncyclic electron flow.

15. Describe the similarities and differences in chemiosmosis between oxidative phosphorylation in mitochondria and photophosphorylation in chloroplasts.

16. State the function of each of the three phases of the Calvin cycle.

17. Describe the role of ATP and NADPH in the Calvin cycle.

18. Describe what happens to rubisco when O2 concentration is much higher than CO2 concentration.

19. Describe the major consequences of photorespiration. Explain why it is thought to be an evolutionary relict.

20. Describe two important photosynthetic adaptations that minimize photorespiration.

F. F. Blackman

The above equation shows the relationship between the substances used in and produced by the process. It tells us nothing about the intermediate steps. That photosynthesis does involve at least two quite distinct processes became apparent from the experiments of the British plant physiologist F. F. Blackman. His results can easily be duplicated by using the setup on the left. The green water plant Elodea (available wherever aquarium supplies are sold) is the test organism. When a sprig is placed upside down in a dilute solution of NaHCO3 (which serves as a source of CO2) and illuminated with a flood lamp, oxygen bubbles are soon given off from the cut portion of the stem. One then counts the number of bubbles given off in a fixed interval of time at each of several light intensities. Plotting these data produces a graph like the one below.

Since the rate of photosynthesis does not continue to increase indefinitely with increased illumination, Blackman concluded that at least two distinct processes are involved: one, a reaction that requires light and the other, a reaction that does not. This latter is called a "dark" reaction although it can go on in the light. Blackman theorized that at moderate light intensities, the "light" reaction limits or "paces" the entire process. In other words, at these intensities the dark reaction is capable of handling all the intermediate substances produced by the light reaction. With increasing light intensities, however, a point is eventually reached when the dark reaction is working at maximum capacity. Any further illumination is ineffective, and the process reaches a steady rate.

This interpretation is strengthened by repeating the experiment as a somewhat higher temperature. Most chemical reactions proceed more rapidly at higher temperatures (up to a point). At 35°C, the rate of photosynthesis does not level off until greater light intensities are present. This suggest that the dark reaction is now working faster. The fact that at low light intensities the rate of photosynthesis is no greater at 35°C than at 20°C also supports the idea that it is a light reaction that is limiting the process in this range. Light reactions depend, not on temperature, but simply on the intensity of illumination.

The increased rate of photosynthesis with increased temperature does not occur if the supply of CO2 is limited. As the figure shows, the overall rate of photosynthesis reaches a steady value at lower light intensities if the amount of CO2 available is limited. Thus CO2 concentration must be added as a third factor regulating the rate at which photosynthesis occurs. As a practical matter, however, the concentration available to terrestrial plants is simply that found in the atmosphere: 0.035%.

Researchers Image Photosynthesis in Action

Using the SLAC National Accelerator Laboratory’s Linac Coherent Light Source – the world’s most powerful X-ray laser, a multinational group of scientists was able to take detailed snapshots of photosynthesis in action as it splits water into oxygen, protons and electrons.

The oxygen-evolving complex (OEC) of photosystem II cycles through five states, where four electrons are sequentially extracted from the OEC in four light–driven, charge-separation events. The ellipses show snapshots of the metal cluster observed in the study. Image credit: Mary Zhu.

Photosynthesis, a process catalyzed by plants, algae and cyanobacteria, is one of the fundamental processes of life on Earth. It converts sunlight to energy thus sustaining all higher life on Earth.

Two large protein complexes, called photosystem I and II, act in series to catalyze the light-driven reactions in photosynthesis.

Photosystem II produces the oxygen we breathe, which ultimately keeps us alive.

The revealing of the mechanism of this process is essential for the development of artificial systems that mimic and surpass the efficiency of natural systems.

“Photosynthetic organisms already know how to do this, and we need to know the details of how photosynthesis carries out the process using abundant manganese and calcium,” said Prof Devens Gust of Arizona State University, who was not involved in the research.

In photosynthesis, oxygen is produced at a special metal site containing four manganese atoms and one calcium atom, connected together as a metal cluster.

This oxygen-evolving cluster is bound to photosystem II that catalyzes the light-driven process of water splitting. It requires four light flashes to extract one molecule of oxygen from two water molecules bound to the metal cluster.

There are two major drawbacks to obtaining structural and dynamical information on this process by a traditional technique called X-ray crystallography: (i) the pictures one can obtain with standard structural determination methods are static (ii) the quality of the structural information is adversely affected by X-ray damage.

“The trick is to use the world’s most powerful X-ray laser, SLAC’s Linac Coherent Light Source. Extremely fast femtosecond pulses record snapshots of the photosystem II crystals before they explode in the X-ray beam, a principle called diffraction before destruction. In this way, snapshots of the process of water splitting are obtained damage-free. The ultimate goal of the work is to record molecular movies of water splitting,” explained Prof Petra Fromme of Arizona State University, the senior author of the paper published in the journal Nature.

Prof Fromme and her colleagues performed the time-resolved femtosecond crystallography experiments on photosystem II nanocrystals, which are so small that you can hardly see them, even under a microscope.

The crystals are hit with two green laser flashes before the structural changes are elucidated by the femtosecond X-ray pulses.

The team discovered large structural changes of the protein and the metal cluster that catalyzes the reaction.

The cluster significantly elongates, thereby making room for a water molecule to move in.

Christopher Kupitz et al. Serial time-resolved crystallography of photosystem II using a femtosecond X-ray laser. Nature, published online July 09, 2014 doi: 10.1038/nature13453

Photosynthesis ICSE Class-10 Concise Selina Biology Solutions Chapter-6

A. MULTIPLE CHOICE TYPE Photosynthesis ICSE Class-10 Concise Selina Biology

The production of starch, and not glucose, is often used as a measure of photosynthesis in leaves because

(a) starch is immediate product of photosynthesis

(b) glucose formed in photosynthesis soon gets converted into starch

(c) starch is soluble in water

Answer 1

(b) glucose formed in photosynthesis soon gets converted into starch

Question 2

The number of water molecules required in the chemical reactions to produce one molecule of glucose during photosynthesis is

Answer 2

Question 3

The rate of photosynthesis is not affected by

Answer 3

Question 4

Chlorophyll in a leaf is required for

(a) breaking down water into hydrogen and oxygen

(d) storing starch in the leaves

Answer 4

Question 5

If the rate of respiration becomes more than the rate of photosynthesis, plants will:

(a) continue to live, but will not be able to store food

(c) grow more vigorously because more energy will be available

(d) stop growing and die gradually of starvation

Answer 5

(a) continue to live, but will not be able to store food

Question 6

Which one chemical reaction occurs during photosynthesis?

(a) Carbon dioxide is reduced and water is oxidised

(b) Water is reduced and carbon dioxide is oxidised

(c) Both carbon dioxide and water are oxidised

(d) Both carbon dioxide and water are reduced

Answer 6

(a) Carbon dioxide is reduced and water is oxidised

Question 7

The specific function of light energy in the process of photosynthesis is to

Answer 7

Question 8

A plant is kept in a dark cupboard for 48 hours before conducting any experiment on photosynthesis in order to

(a) remove chlorophyll from leaves

(b) remove starch from the leaves

(c) ensure that no photosynthesis occurred

(d) ensure that the leaves are free from starch

Answer 8

(d) ensure that the leaves are free from starch

Question 9

During photosynthesis, the oxygen in glucose comes from

Answer 9

B. VERY SHORT ANSWER TYPE Photosynthesis ICSE Class-10 Concise Selina Biology

Question 1

(a) The category of organisms that prepare their own food from basic raw materials.

(b) The kind of plastids found in the mesophyll cells of the leaf.

(c) The compound which stores energy in the cells.

(d) The first form of food substance produced during photosynthesis.

(e) The organisms that can be called “natural purifiers” of the air.

(f) The source of carbon dioxide for aquatic plants.

(g) The part of chloroplast where the dark reaction of photosynthesis takes place.

(h) The tissue that transports manufactured type of starch from leaves to all parts of the plants.

Answer 1

(c) ATP (Adenosine triphosphate)

(f) Carbon dioxide dissolved in water

C. SHORT ANSWER TYPE for ICSE Class-10 Chapter 6 Photosynthesis

Question 1

Mention one difference between the following on the basis of what is given in brackets.

(a) Respiration and photosynthesis (gas released)

(b) Light and dark reactions (products formed)

(c) Producers and consumers (mode of nutrition)

(d) Grass and grasshopper (mode of nutrition)

(e) Chlorophyll and chloroplast (part of plant cell)

Answer 1

Respiration Photosynthesis
The gas released during respiration is carbon dioxide. The gas released during photosynthesis is oxygen.

Light Reaction Dark Reaction
Hydrogen and oxygen are produced here, along with release of electrons, which converts ADP into ATP. Glucose is the main product formed during dark reaction.

Grass Grasshopper
Green grass being a producer is capable of producing its own food by photosynthesis. Grasshopper is a primary consumer (herbivore) and directly feeds on producers like grass.

Chlorophyll Chloroplast
Chlorophyll is the green pigment present in cell organelles called chloroplasts. Chloroplasts are cell organelles, situated in the cytoplasm of plant cells. They are present mainly in the mesophyll cells and in the guard cells of stomata.

Question 2

Identify the false statements and rewrite them correctly by changing the first or last word only.

(a) Dark reaction of photosynthesis occurs during night time.

(b) Immediate product of photosynthesis is glucose.

(c) Starch produced in a leaf remains stored in it for 2-3 weeks before it is used by other parts of the plant.

(d) Photosynthesis requires enzymes.

(e) Green plants are consumers.

(f) Photosynthesis results in loss of dry weight of the plants.

(g) Photosynthesis stops at a temperature of about 35 o C.

(h) Photosynthesis occurs only in cells containing chloroplasts.

(i) Green plants perform photosynthesis.

Answer 2

Correct Statement: Dark reaction of photosynthesis is independent of light and occurs simultaneously with light reaction.

Correct Statement: Starch produced in a leaf is stored temporarily in the leaf until the process of photosynthesis. At night it is converted back into soluble sugar and translocated to different part of the body either for the utilization or for the storage.

Correct Statement: Green plants are producers.

Correct Statement: Respiration results in loss of dry weight of the plants.

Correct Statement: Photosynthesis stops at a temperature of above 40 o C.

Question 3

Fill in the blanks with the appropriate answer from the choices given in the brackets.

(a) The site of light reaction in the cells of a leaf is ………… (cytoplasm, stroma, grana)

(b) The chemical substance used to test the presence of starch in the cell of a leaf is …………. (CaCl2, iodine solution, Benedict solution)

(c) Stroma is ground substance in ……………. (cytoplasm, chloroplast, ribosomes)

(d) The dark reaction of photosynthesis is known as ………….. (Hill reaction, cyclic phosphorylation, Calvin cycle)

(e) In the flowering plants, food is transported in the form of ……………… (sucrose, glucose, starch)

Answer 3

(a) The site of light reaction in the cells of a leaf grana

(b) The chemical substance used to test the presence of starch in the cell of a leaf is iodine solution

(c) Stroma is ground substance in chloroplast.

(d) The dark reaction of photosynthesis is known as Calvin cycle

(e) In the flowering plants, food is transported in the form of sucrose

Question 4

Are the following statements true or false? Give reason in support of your answer.

(a) The rate of photosynthesis continues to rise as long as the intensity of light rises.

(b) The outside atmospheric temperature has no effect on the rate of photosynthesis.

(c) If you immerse a leaf intact on the plant in ice cold water, it will continue to photosynthesise in bright sunshine.

(d) Destarching of the leaves of a potted plant can occur only at night.

(e) The starting point of carbon cycle is the release of carbon dioxide by animals during respiration.

(f) If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will fail to occur.

(g) Photosynthesis is considered as a process supporting all life on earth.

Answer 4

Photosynthesis increases with the light intensity up to a certain limit only and then it gets stabilized.

The atmospheric temperature is an important external factor affecting photosynthesis. The rate of photosynthesis increases up to the temperature 35 o C after which the rate falls and the photosynthesis stops after 40 o C.

Ice cold water will hamper the process of photosynthesis in the immersed leaf, even if there is sufficient sunshine because the temperature is an important factor for the rate of photosynthesis.

For destarching, the potted plant can kept in a dark room for 24-48 hours.

There is no start point or end point in the carbon cycle, the carbon is constantly circulated between the atmosphere and the living organisms.

If a plant is kept in bright light all the 24 hours for a few days, the dark reaction (biosynthetic phase) will continue to occur because the dark reaction is independent of light and it occurs simultaneously with the light dependent reaction.

Question 5

Given below are five terms. Rewrite the terms in the correct order so as to be in logical sequence with regard to photosynthesis: (i) water molecules, (ii) oxygen, (iii) grana, (iv) hydrogen and hydroxyl ions, (v) photons.

Answer 5

Photons, grana, water molecules, hydrogen and hydroxyl ions, oxygen

Question 6

State any four differences between photosynthesis and respiration.

Answer 6

Photosynthesis Respiration
Carbon dioxide is used up and oxygen is released. Oxygen is used up and carbon dioxide is released.
Photosynthesis occurs in plants and some bacteria. Respiration occurs in all living organisms.
Photosynthesis results in gain of dry weight of the plants. Respiration results in loss of dry weight of the plants.
Glucose is produced which is utilized by the plants. Glucose is broken down to obtain energy.
The raw materials for the photosynthesis are water, carbon dioxide and sunlight. The raw material for respiration is glucose.

Question 7

“Oxygen is a waste product of photosynthesis.” Comment.

Answer 7

Oxygen is released during photosynthesis. Some of this oxygen may be used in respiration in the leaf cells, but the major portion of it is not required and it diffuses out into the atmosphere through the stomata. However, in a sense, even this oxygen is not a waste because all organisms require it for their existence including the plants.

Question 8

Why is it necessary to place a plant in the dark before starting an experiment on photosynthesis? Explain.

Answer 8

The presence of starch is regarded as evidence of photosynthesis. Hence before starting an experiment on photosynthesis, the plant should be placed in the dark for 24-48 hours to destarch the leaves. During this period, all the starch from the leaves will be sent to the storage organs and the leaves will not show the presence of starch. So the various experiments on photosynthesis can be carried out effectively.

Question 9

Why is it not possible to demonstrate respiration in a green plant kept in sunlight?

Answer 9

If a green plant is kept in bright light, it tends to use up all the CO2 produced during respiration, for photosynthesis. Thus, the release of CO2 cannot be demonstrated. Hence, it is difficult to demonstrate respiration as these two processes occur simultaneously.

Question 10

Most leaves have the upper surface more green and shiny than the lower one. Why?

Answer 10

The chloroplasts are concentrated in the upper layers of the leaf which helps cells to trap the sunlight quickly. Also the epidermis is covered by a waxy, waterproof layer of cuticle. This layer is thicker on the upper surface than the lower one. Hence most leaves have the upper surface more green and shiny than the lower one.

Question 11

How would you demonstrate that green plants release oxygen when exposed to light?

Answer 11

  • Place hydrilla plant (a water plant) in a beaker containing pond water and cover it by a short-stemmed funnel. (Make sure the level of water in the beaker is above the level of the stem of the funnel)
  • Invert a test tube full of water over the stem of the funnel.
  • Place the set up in the sun light for a few hours

Bubbles appear in the stem which rise and are collected in the test tube. When sufficient gas gets collected, a glowing splinter will be introduced in the test tube, which will burst into flames.

The splinter glows due the presence of oxygen in the test tube which proves that the gas collected in the test is released by hydrilla during photosynthesis.

Question 12

Describe the main chemical changes which occur during photosynthesis in:

Answer 12

(i) Light Reaction:

The light reaction occurs in two main steps:

(1) Activation of chlorophyll – On exposure to light energy, chlorophyll becomes activated by absorbing photons.

(2) Splitting of water – The absorbed energy is used in splitting the water molecule into hydrogen and oxygen, releasing energy. This reaction is known as photolysis of water.

2H2O 4H + + 4e – + O2

The fate of H + , e – and (O) component are as follows:

The hydrogen ions (H + ) obtained from above are picked up by a compound NADP (Nicotinamide adenine dinucleotide phosphate) to form NADPH.

The oxygen (O) component is given out as molecular oxygen (O2).

The electrons (e – ) are used in converting ADP into energy rich ATP by adding one inorganic phosphate group Pi.ADP + Pi ATP. This process is called photophosphorylation. (ii) Dark reaction: The reactions in this phase does not require light energy and occur simultaneously with the light reaction. The time gap between the light and dark reaction is less than one thousandth of a second. In the dark reaction, ATP and NADPH molecules (produced during light reaction) are used to produce glucose (C6H12O6) from carbon dioxide. Fixation and reduction of carbon dioxide occurs in the stroma of the chloroplast through a series of reactions. The glucose produced is either immediately used up by the cells or stored in the form of starch.

Question 13

Complete the following food chains by writing the names of appropriate organisms in the blanks:

Answer 13

Complete the following food chains by writing the names of appropriate organisms in the blanks:

(i) Grass → Rabbit. → Snake → Hawk

(ii) Grass/Corn → Mouse → Snake → Peacock

Question 14

How do non-green plants such as fungi and bacteria obtain their nourishment?

Answer 14

Non-green plants such as fungi and bacteria obtain their nourishment from decaying organic matter in their environment. This matter comes from dead animals and plants. Fungi and bacteria break down the organic matter to obtain the nourishment and they release carbon dioxide back in the atmosphere.

Question 15

All life owes its existence to chlorophyll. Give reason.

Answer 15

Chlorophyll is the foundation site for the photosynthesis in green plants. The initiation of photosynthesis takes place when the chlorophyll molecule traps the light energy. The light energy is then converted into chemical energy in the form of glucose using carbon dioxide (CO2) from the atmosphere, and water (H2O) from the soil. All other organisms, directly or indirectly depend on this food for their survival. The starting point of any food chain is always a plant. If green plants were to suddenly disappear, then so would virtually all life on Earth. Thus, we can say that all life owes its existence to chlorophyll.

Question 16

Complete the following by filling the blanks 1 to 5 with appropriate words/ terms/ phrases:

To test the leaf for starch, the leaf is boiled in water to …………… (1). It is next boiled in methylated spirit to ……………(2). The leaf is placed in warm water to soften it. It is then placed in a dish and ………….(3) solution in added. The region, which contains starch, turns ……………. (4) and the region, which dose not contain starch, turns ………………(5)

Answer 16

To test the leaf for starch, the leaf is boiled in water to kill the cells. It is next boiled in methylated spirit to remove chlorophyll. The leaf is placed in warm water to soften it. It is then placed in a dish and iodine solution in added. The region, which contains starch, turns blue-blackand the region, which does not contain starch, turns brown.

D. STRUCTURED / APPLICATION / SKILL TYPE for ICSE Class-10 Chapter 6 Photosynthesis Concise Selina Biology

Question 1

A candidate studied the importance of certain factors in photosynthesis. He took a potted plant and kept in the dark for over 24 hours. In the early hours of the next morning, he covered one of the leaves with black paper in the centre only. Then he placed the plant in sunlight for a few hours and tested the leaf which was covered with black paper for starch.

a. What aspect of photosynthesis was being investigated?

b. Is there any control in this experiment? If so, state it.

c. Why was the plant kept in the dark before the experiment?

d. Describe step by step, how the candidate proceeded to test the leaf for the presence of starch?

Answer 1

(a) The student wanted to show that sunlight is necessary for photosynthesis. / The role of sunlight in photosynthesis is being investigated.

b. Yes. The other uncovered leave of the potted plant act as a control.

c. Destarching ensures that any starch present after the experiment has been formed under experimental conditions. Therefore, the plant was kept in the dark before the experiment.

d.The student dipped the leaf in boiling water for a minute to kill the cells.

Then he boiled the leaf in alcohol / methylated spirit over a water bath to remove chlorophyll. The leaf becomes hard and brittle.

He then places the leaf in hot water to soften it.

Next the student spreads the leaf in a dish and pours iodine solution on it. The presence of starch is indicated by a blue-black colour.

The uncovered portion (exposed to sunlight) turned blue-black colour and the covered portion showed brown colour. The difference in the colours of covered and uncovered part of leaves indicates the importance of sunlight in photosynthesis.

Question 2

Photosynthesis in green plants is directly and indirectly dependent on so many plant structures. Explain briefly the role of the following structures in this process.

(d) Xylem tissue in the leaf veins

(e) Phloem tissue in the leaf veins

Answer 2

(a) Guard cells: They regulate the opening and closing of stomata and thus regulate the entry of carbon dioxide through the stomata.

(b) Cuticle: Cuticle is transparent and water proof due to which light can penetrate this later easily.

(c) Mesophyll cells: Mesophyll cells are the main sites for photosynthesis. Chloroplasts are mainly contained in the mesophyll cells. When sunlight falls on the leaf, the light energy is trapped by the chlorophyll of the upper layers of mesophyll, especially the palisade cells.

(d) Xylem Tissue in the Leaf Veins: Water is essential for photosynthesis to occur. Water is taken up by the roots from the soil, sent up through the stem and finally brought to the leaves (site of photosynthesis) through the xylem tissue. The water is then distributed in the mesophyll tissue.

(e) Phloem Tissue in the Leaf Veins: The prepared food is transported from leaves to all parts of the plant by the phloem tissue. The glucose is converted into insoluble starch and later into soluble sugar i.e. sucrose, which is transported in solution through the phloem in the veins of the leaf and down through the phloem of the stem.

(f) Stomata: The main function of stoma is to let in carbon dioxide from the atmosphere for photosynthesis. Also most of the oxygen produced during photosynthesis diffuses out into the atmosphere through the stomata.

Question 3

Given below is a schematic diagram to illustrate some aspects of photosynthesis.

(a) Fill up the gaps, in blank spaces (1-4), by writing the names of the correct items.

(b) What phenomenon do the thick arrows A and B indicate?

Answer 3 (a)

(b) A – Transpiration

Question 4

Given below is the representation of a certain phenomenon in nature. With four organisms 1-4.

(a) Name the phenomenon represented.

(b) Name any one organism that could be shown at No .5

(c) Name the biological process which was the starting point of the whole chain.

(d) Name one natural element which all the organisms 2-4 and even 5 are getting from No. 1 for their survival.

Answer 4

Question 5

Enumerate the steps involved in testing a green leaf for the presence of starch.

Answer 5

Test to determine the presence of starch in a leaf:

  • Dip a leaf in boiling water for a minute to kill the cells.
  • Boil the leaf in methylated spirit in a water bath to remove the chlorophyll, till the leaf turns pale blue and becomes hard and brittle.
  • Now place the leaf in hot water to soften it.
  • Place the leaf in a Petri dish and pour iodine solution over it.
  • The appearance of a blue-black colour on the leaf is indicative of the presence of starch.
  • The absence of starch is indicated by a brown colouration.

Question 6

Given alongside is the diagram of an experimental set-up:

(a) What is the objective of this experiment?

(b) Will it work satisfactorily? Given reason.

(c) What alteration (s) will you make in it for obtaining expected result?

(d) Would you take any step before starting the experiment? Describe this step and explain its necessity.

Answer 6

(a) To demonstrate the importance of carbon dioxide in photosynthesis

(b) No, the experiment will not work satisfactorily, as the beaker contains lime water and not potassium hydroxide to absorb CO­2.

(c) Place potassium hydroxide in the beaker instead of lime water

(d) Before starting the experiment, it is necessary to destarch the leaves of the plant by keeping the plant in complete darkness for 48 hours. This is because if the plant is not destarched, then the experiment will give false results because starch stored previously may be detected in the leaf placed in the beaker even if no starch is produced during the experiment.

Question 7

Draw a neat diagram of the stomatal apparatus found in the epidermis of leaves and label the Stoma, Guard cells, Chloroplast, Epidermal cells, Cell wall and Nucleus.

Answer 7

Question 8

A potted plant was taken in order to prove a factor necessary for photosynthesis. The potted plant was kept in the dark for 24 hours. One of the leaves was covered with black paper in the centre. The potted plant was then placed in sunlight for a few hours.

(a) What aspect of photosynthesis was being tested?

(b) Why was the plant placed in the dark before beginning the experiment?

(c) During the starch test, why was the leaf

(1) boiled in water (2) boiled in methylated spirit

(d) Write a balanced chemical equation to represent the process of photosynthesis.

(e) Draw a neat diagram of a chloroplast and label its parts.

Answer 8

(a) Light is required for photosynthesis.

(b) Before beginning the experiment, the plant was kept in dark in order to destarch it, i.e. to remove the pre-existing starch from the storage organs.

(1) The leaf was boiled in water to destroy enzymes so that further chemical changes do not take place in the leaf.

(2) The leaf was boiled in methylated spirit to dissolve chlorophyll.

(3) Chemical equation for the process of photosynthesis:

Question 9

The diagram below shows two test-tubes A and B. Test-tube A contains a green water plant. Test-tube B contains both a green water plant and a snail. Both test-tubes are kept in sunlight. Answer the questions that follow:

(a) Name the physiological process that releases the bubbles of oxygen.

(b) Explain the physiological process as mentioned above in (a).

(c) What is the purpose of keeping a snail in test-tube B?

(d) Why does test-tube B have more bubbles of oxygen?

(e) Give an example of a water plant that can be used in the above experiment.

(f) Write the overall chemical equation for the above process.

Answer 9

(a) Photosynthesis releases bubbles of oxygen.

(b) Photosynthesis is a physiological process by which plant cells containing chlorophyll produce food in the form of carbohydrates by using carbon dioxide, water and light energy. Oxygen is released as a by-product.

(c) Carbon dioxide released by the snail during respiration is used by the plant for photosynthesis. This increases the rate of photosynthesis in the plant placed in test tube B. This also suggests that both respiration and photosynthesis are complementary processes to maintain the concentration of oxygen and carbon dioxide in the atmosphere.

(d) A plant and a snail are kept in test tube B. The plant in test tube B has more concentration of CO2 available because the snail releases CO2 during respiration. This increases the rate of photosynthesis in the plant placed in test tube B which leads to the release of more amount of oxygen.

Stages in the Process of Photosynthesis

Light-dependent Reactions

This is the first stage of the photosynthetic process. These reactions take place in the presence of sunlight, and use light energy from the sun to produce ATP molecules and other molecules known as NADPH. These molecules are used as the energy source to carry out the chemical changes in the next stage of photosynthesis.

Light-independent Reactions (Calvin Cycle)

In this stage of photosynthesis, energy-containing sugar molecules are synthesized. The ATP and NADPH produced are used to fuel the reactions in this stage. Here, CO2 molecules are broken down and converted into sugars and other compounds. The Calvin Cycle is repeated twice in order to yield one molecule of glucose.

Cellular Respiration

Cellular respiration takes place in the same way in both plants and animals. Living cells obtain the products of photosynthesis (sugar molecules) and undergo cellular respiration to produce ATP molecules. Some cells respire aerobically, using oxygen, while others undergo anaerobic respiration, without using oxygen. The process involves a set of chemical reactions to convert chemical energy from the glucose molecules into ATP molecules.

Chemical reaction in Cellular Respiration

Glucose + Oxygen → Carbon dioxide + Water + Energy (ATP)

Tips for Teaching Photosynthesis

Photosynthesis… there was something about it that stumped me as a teacher. It was important to me that my kids understood the big picture and that they were able to answer their questions, but teaching it didn’t come naturally, because, well… photosynthesis. For you life science and bio teachers out there my BIGGEST piece of advice would be to focus on the “big picture”. Do not teach light reactions, followed by dark reactions. Teach them together, and then go back and fill in the details where you can.

I’ve compiled 5 helpful tips for teaching photosynthesis:

1. Photosynthesis can be a complex subject for students to learn. Scaffolding will provide optimal success: start off with the big picture and then work your way into the nooks and crannies.

2. Students should be able to comprehend energy flow in photosynthesis, from sunlight to sugar, and everything in between. Use my Mouse Trap game analogy to help!

3. Dark reactions and light reactions are dependent on each other through the ATP –> ADP and NADPH –> NADP pathway.

4. Teach the basic steps of the electron transport chain. There are two separate ones, each associated with a photosystem: one for ATP (indirectly through hydrogen ions) and one for NADPH.

5. Visuals. Visuals. Visuals. Keep the diagrams simple at first! Labs will help all learner types, too!

To help with this, I’m offering a free Photosynthesis Graphic Organizer!

What is the mouse trap analogy referenced in #2? Just like in the game “Mouse Trap”, each step depends directly on the previous step. If one of the processes stopped, the next step would not be able to go on (energy flow). Ask your students what they think the marble and the foot that kicks the marble out of the bucket represent!

For more photosynthesis resources that I offer, check out my photosynthesis interactive notebook, or other types of materials you’ll find in my store (scroll down to see all the resources I have to offer in the photosynthesis category).

Watch the video: What Does a Giant Monster Neodymium Magnet do to a Mouse? (August 2022).