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Survival and Response

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There is always: • a receptor — a structure that detects the stimulus • an effector — a structure, such as a muscle, that produces the response • some kind of linking system or coordinating system — this receives information from the receptor and passes information to the effector • a response — the action that results from the stimulus Many examples of behaviour have evolved to increase the chances of survival of an organism. These include the following: • Plant shoots (stems) grow towards the region of most intense light; this exposes the leaves to the maximum amount of light and increases the rate of photosynthesis.

• Woodlice move more quickly in the light than in the shade; this increases their chances of moving out of the light and into the shade (where it is also likely to be more humid, so the animals will not dehydrate as quickly). Reflex actions in mammals are often protective — for example: – withdrawal reflexes move parts of the body away from potentially damaging stimuli, such as heat or sharp objects – when a bright light is shone into the eye, the iris reflex prevents too much light from entering the eye and damaging the retina Other reflex actions are corrective rather than protective and include: • reducing the heart rate if blood pressure becomes too high • increasing the breathing rate if the partial pressure of oxygen in the blood becomes too low or the partial pressure of carbon dioxide becomes too high We have learned that those organisms with a selective advantage survive and are able to reproduce and pass on their alleles to the next generation.

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The ability to respond to changes in the environment increases the chances of an organism’s survival.

In this section we are going to look at how simple animals respond to their environment and how plants respond to stimuli. Taxis (plural taxes) and Kinesis (plural kineses) Taxes and kineses are simple forms of behaviour that rely almost entirely on simple reflex actions. Taxes and kineses are forms of behaviour involving the locomotion of whole organisms or cells in response to a specific external stimulus. The adaptive value of taxes and kineses is their role in helping to maintain organisms in the most favourable environment for their survival. Taxis Definitions: Taxis: A movement of the entire organism in response to a directional stimulus.

Taxis in which the animal moves along a gradient of intensity of a stimulus either: • towards the greatest intensity of the stimulus (a positive taxis) or • away from the greatest intensity (a negative taxis); There is a directional response to a directional stimulus. Maggots move away from light; many insects are attracted by, and move towards, pheromones (chemicals similar to hormones); Dorsal light reaction in fish, where fish such as plaice maintain their dorsal surface at right angles to the sun. Taxes: are classified as |Stimulus |Taxis |Positive |Negative | |Light |phototaxis |Euglena – swims towards light, |Earthworms, blowfly larvae, woodlice | | | |and cockroaches | | | |Sperms of liverwort, mosses and ferns move |Mosquitoes avoiding repellent | |Chemical |chemotaxis |towards substances released by ovum | | |Air (oxygen) |aerotaxis |Motile aerobic bacteria move towards oxygen | | |resistance |rheotaxis |Planaria (flatworms) move against water | | | | |current, moths and butterflies fly into | | | | |wind | | Activity Some examples of taxes Chlamydomonas is a motile, photosynthetic protoctist that swims directly towards regions of optimum light intensity; this species displays Positive phototaxis Describe the adaptive value of this response.

By moving to an area of high light intensity it can photosynthesise and produce glucose which will allow it to grow and reproduce. Male, silkworm moths display: Positive chemotaxis when they fly directly towards the chemical bombykol, secreted from the abdominal regions of the female moth; the feathery antennae of the males contain millions of bombykol chemoreceptors. Describe the adaptive value of this response. This allows the males to find the females to reproduce and so pass on their alleles. This water flea swims directly towards low-intensity light and towards oxygen-rich environments; it displays positive phototaxis and Positive aerotaxis Describe the adaptive value of this response.

This environment will be where photosynthetic organisms can be found which provide food for the Daphnia but low light intensities ensure lower temperatures and less visibility to predators so more likely to survive and reproduce. KINESES Definitions: Kineses: a non-directional movement response in which the rate of movement is related to the intensity of the stimulus and not the direction of the stimulus. Woodlice display a relatively high surface area to volume ratio and lack a waxy, waterproof cuticle covering their exoskeleton – their primitive respiratory surfaces (a pair of ‘pleopod lungs’) open to the atmosphere through a permanently open pore Wood lice cannot regulate their water loss.

In dry conditions wood lice increase both their rate of movement and their rate of randomly changing direction. As a result they move towards damper conditions where they are less likely to die from water loss. Woodlice display ………….. …………………. behaviour when they encounter atmospheres of different humidity; in drier environments they increase their rate of movement and decrease their rate of turning. Describe the adaptive value of this response. ……………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………………… ………………………………… Look at the following investigation on the PowerPoint ‘Taxes and Kineses’ then discuss the results.

Students investigated rate of movement in woodlice when subjected to environments of different humidity Chambers of differing humidity were prepared, and the distance moved by the woodlice in a given time was determined. Six individual chambers of different humidity were set up by preparing a series of solutions with varying proportions of glycerol and water. A single woodlouse was introduced into each of the humidity chambers and the rate of movement was determined over a period of 60 seconds. Movement was recorded by using a marker pen, on the Petri dish lid, to trace the path of the woodlouse over the 60 second period; the lid was then placed over a sheet of centimetre graph paper and the distance moved in 60 seconds was determined. Results: | | | | | | | |% Relative Humidity |40 |50 |60 |70 |80 |90 | |Mean rate of movement (cm per min) | | | | | | | | |27. 3 |26. 4 |25. 5 |25. 0 |19. 4 |16. 5 | Discuss the findings in terms of the adaptive value of the observed responses to woodlice survival. At low relative humidity the woodlice moved around quickly and had a lot of turns in their path. As the humidity increased their rate of movement slowed down. At the higher humidity the woodlice would be less likely to lose water and more likely to survive. Activity

Now carry out the Practical work ‘Using choice chambers to investigate responses in invertebrates’ Application: Investigating taxes and kineses A student was investigating the hypothesis that woodlice exhibit negative phototaxis – they move away from the stimulus of light. The student covered one half of the choice chamber with light-proof cloth. The other half was illuminated by a light bulb fixed about 20cm above it. 10 woodlice where put into the choice chamber through the holein the centre of the top. 5 minutes later the student counted how many woodlice could be seen in the illuminated side (by subtraction – worked out the number in the dark side). This was repeated 10 times, using different woodlice each time. The results are shown: |Number of woodlice after 5 minutes | |Trial Number | | | |In light |In dark | |1 |3 |7 | |2 |5 |5 | |3 |4 |6 | |4 |4 |6 | |5 |3 |7 | |6 |5 |5 | |7 |6 |4 |8 |3 |7 | |9 |2 |8 | |10 |4 |6 | Looking at these results, the student concluded that woodlice do move away from light and therefore show negative phototaxis. However, let’s do a statistical test to help confirm this conclusion. The chi-squared ((2) test can be used to determine whether the result was statistically significant. Give reasons for your choice of statistical test …………………………………………………………………………………………………………………………………………… State your null hypothesis …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… The equation we need to use is: Remember ‘O’ is the observed results.

We can add the data in each column. From this we can calculate the Expected (E) results Carry out the test and calculate the test statistic. |Category |Observed |Expected (E) |(O – E) |(O – E )2 |(O – E )2 | | |( O) | | | |E | | | | | | | | |Light Side | | | | | | | | | | | | |Dark Side | | | | | | Now, taking the sum of the values for the light and dark sides, we find the (2 value: ……………………………… We now need to look up the values for (2 in the table of critical values. How many degrees of freedom? ……………………. Critical value = ………………………………………………. Interpret the test statistic in relation to the null hypothesis being tested. …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… …………………………………………………………………………………………………………………………………………… Similar experiment but another way to analyse the results: Ten woodlice are placed in each half of the choice chamber.

One half of the chamber is covered with black paper and the apparatus is left for 10 minutes. At the end of this time, the number of woodlice in each environment (dark and light) is noted. To help overcome the small sample size, the experiment is repeated several times and mean numbers for each environment are calculated. The results can then be expressed as a bar chart. To give some idea of the variability of the results, the standard error (SE) of each mean is calculated. Error bars with the value of 1. 96 x SE are shown above and below the mean value. If the error bars for the two values overlap, then we must conclude that there is a greater than 5% probability that the differences are due to chance and have no underlying cause.

If the error bars do not overlap, then we can conclude that there is less than 5% probability that the differences are due to chance and that there is an underlying cause for the differences – the results are significant. Plant Responses – Tropisms Reference Chapter 9. 1 page 142-193 in A2 text book. Flowering plants, like animals, increase their chances of survival by responding to changes in their environment. Examples • They sense the direction of light and grow towards it to maximize light absorption for photosynthesis. • They can sense gravity, so their roots and shoots grow in the right direction. • Climbing plants have a sense of touch, so they can find things to climb and reach the sunlight Definition:

A tropism is the response of a plant to a directional stimulus (a stimulus coming from a particular direction). Plants respond to stimuli by regulating their growth. Tropisms can be positive (growing towards the stimulus) or negative (growing away from the stimulus) and occur in response to a variety of stimuli. Phototropism Phototropism is the growth of a plant in response to light. Shoots are positively phototropic and grow towards light (see below). Roots are negatively phototropic and grow away from light (see below). Phototropism in shootsPhototropism in roots Geotropism Geotropism is the growth of a plant in response to gravity. Shoots are negatively geotropic and grow upwards (see Figure 3).

Roots are positively geotropic and grow downwards (see Figure 4). Geotropism in shootsGeotropism in roots Plants respond to external stimuli by means of plant growth regulators (sometimes called plant hormones). Reference ‘Plant hormones and responses’ Boardworks PowerPoint These growth factors cause the responses in plants by affecting how plant cells grow. Plant growth can be divided into three main types: • cell division • cell elongation / enlargement • cell differentiation (specialization). The main areas of cell division and cell elongation are in the meristems – at the tips of the roots or the shoots. Effect of auxins on shoot growth Plant growth factors are produced in small quantities.

They have their effects close to the tissue that produces them. An example of a plant growth factor is indoleacetic acid (IAA) which acts by making plant cells elongate. IAA is one of a group of plant growth factors known as auxins. Auxins are made continually in the shoot apex and young leaves (apical meristems). They can move by diffusion from cell to cell, or can be transported long distances via the phloem, e. g. from the shoots to the roots. IAA and Phototropism Watch the animation of ‘Plant hormones and responses’ Boardworks PowerPoint Then read Page 158 in the A2 text book and explain why this shoot is growing towards the light. Use the diagram below to help. 1.

Cells in the tip of the shoot produce IAA, which is then transported down the shoot. 2 The IAA is initially transported to all sides as it begins to move down the shoot. 3 Light causes the movement of IAA from the light side to the shaded side of the shoot. 4 A greater concentration of IAA builds up on the shaded side of the shoot than on the light side. 5 As IAA causes elongation of cells and there is a greater concentration of IAA on the shaded side of the shoot, the cells on this side elongate more. 6 The shaded side of the shoot grows faster, causing the shoot to bend towards the light. Activity Discovering the role of IAA in tropisms [pic]

Watch the animation ‘Plant hormones and responses’ Boardworks PowerPoint and Read pages 158 to 159 in the A2 text book then answer the following questions. 1880 Charles and Francis Darwin [pic] 1. Which of these experiments is the control? Explain your answer. Experiment 1- The cover prevents light reaching the tip. There was no response. So the shoot tip must detect the light stimulus. The Darwins concluded that – detection of the stimulus occurs in the tip of the coleoptile – a region some distance behind the tip brings about the response – there was communication between the two, probably chemical in nature 1913 Boysen-Jensen [pic] . Why did Boysen-Jensen use mica in experiments 4 and 5? As mica conducts electricity it will not prevent electrical messages passing from the shoot tip but it will prevent chemical messages passing. 3. Why did Boysen-Jensen use a gelatin block in experiment 6? Gelatin conducts chemicals, but not electricity 4. How do experiments 4 and 5 show that bending only occurs when the IAA passes down the shaded side of the shoot? When mica is inserted on the light side of the shoot there is bending towards the light. When mica is inserted on shaded side of the shoot there is no response, the message must be chemical and must pass down the shaded side. . What do the results of experiment 6 tell us about the cause of phototropism? Gelatin allows chemicals to pass through but not electricity and bending still occurred so the signal must be chemical. Boysen-Jensen concluded that the communication between detection and response was chemical in nature 1919 Arpad Paal [pic] 6. Why did Paal keep his experiments in the dark? To remove the light as a variable. To show that the chemical is still present even in the dark. 7. Explain the results of Paal’s experiments. The experiments shows that the side of the shoot which has the greatest concentration of the chemical signal grows more, causing bending.

Paal concluded that his results provided evidence that curved growth was due to the ‘chemical messenger’ accumulating on one side of the stem. Conclusions to the experiments as shown have to be tentative because: – there is no control to show the effect of not replacing the cut tip – there is no measure of a chemical accumulating 1928 Went Went claimed that his experiments showed that the cause of phototropism was chemical and the concentration of the chemical affected the curvature of the shoot. 8. What control experiments should he have carried out? Place agar blocks, which have not had shoot tips placed on the, on shoots with tips removed.

This will show that the effect is due to the chemical from the shoot tips which had diffused into the agar, and not the agar itself. Went concluded that increasing concentrations of the ‘chemical messenger’ caused increased growth, up to a maximum 1960 Winslow Briggs [pic] Winslow Briggs set out to test three theories a) light inhibits IAA production in the tip and so it is only produced on the shaded side b) Light destroys the IAA as it passes down the light side of the shoot c) IAA is transported from the light side to the shaded side of the shoot. Study the diagrams above and the animation on slide 15 of the PowerPoint before answering the questions. 9.

Why did Winslow Briggs use a glass plate in experiments 9 and 10. It prevents chemicals / IAA, but not light, passing from one side to the other 10. State which of the above theories are supported by the results. Give reasons for your answer. Results support the hypothesis that IAA is transported from the lighter side to the darker side of the shoot. Experiment 8 shows that the total IAA produced and collected is the same whether the shoot is in the light or the dark. This discounts the theory that light destroys IAA or inhibits its production. Experiment 9 shows that the amount of IAA produced at either side of the tip is the same. The glass plate prevents any sideways transfer.

Experiment 10 shows that the IAA is transferred from the light to the dark side of the shoot soon after it is produced because more than twice as much IAA is found on the dark side of the shoot than on the light side. [pic] Interpreting experimental data about auxins In the exam, you could be given some experimental data on auxins and then be asked to interpret the data. You could get something that looks a little like this: Example An experiment was carried out to investigate the role of auxin in shoot growth. Eight shoots, equal in height and mass, had their tips removed. Sponges soaked in glucose and either auxin or water were then placed where the tip should be. Four shoots were then placed in the dark | |Growth | | Shoot A |Shoot B |Shoot C |Shoot D | | |6 mm, right |6 mm, left |6 mm, straight |1 mm, straight | |Experiment A (dark) | | | | | | |8 mm, right |8 mm, right |8 mm, right |3 mm, straight | |Experiment B (light) | | | | | After two days the amount of growth (in mm) and direction of growth was recorded You could be asked to explain the data… The results show how the movement of auxin controls phototropism in plant shoots.

In experiment A shoot A, the auxin diffused straight down from the sponge into the left-hand side of the shoot. This stimulated the cells on this side to elongate, so the shoot grew towards the right. In shoot B, the opposite occurred, making the shoot grow towards the left. In shoot C, equal amounts of auxin diffused down both sides, making all the cells elongate at the same rate. In experiment B, the shoots were exposed to a light source. The auxin diffused into the shoot and accumulated on the shaded side (left-hand side) regardless of where the sponge was placed. All the shoots grew towards the right, because most auxin accumulated on the left, stimulating cell elongation there.

You could be asked to comment on the experimental design… A control (sponge soaked in water) was included to show that it was the auxin having an effect and nothing else. Glucose was included so that the shoots would have energy to grow in the dark (no photosynthesis can take place). Activity Practice Questions —Application Five shoots had their tips removed. Four of the stumps (where the tip should have been) were completely coated with a paste containing IAA and glucose. Each shoot received a different concentration of IAA but the same concentration of glucose. The remaining shoot had its stump completely coated with a paste containing glucose only.

The plants were watered daily and grown in a dark environment for 30 days. Their height was recorded every 5 days. The results are shown in the table below. | |Height (cm) | |Days of growth |Control |Plant 1 |Plant 2 |Plants |Plant 4 | |0 |2. 4 |1. 2 |3. 2 |3. 5 |5. 4 | |5 |2. 5 |1. 8 |4. 9 |4. 1 |7. 1 | |10 |2. 6 |2. 6 |7. 0 |4. 9 |8. 5 | |15 |2. 6 |4. 2 |9. 2 |6. 0 |10. | |20 |2. 7 |5. 9 |11. 6 |6. 9 |12. 3 | |25 |2. 7 |8. 2 |13. 9 |7. 1 |14. 9 | |30 |2. 8 |9. 9 |16. 3 |8. 2 |16. 9 | Q1 What was the mean rate of growth for plant 1? Give your answer in mm/hour. 9. 9 – 1. 2 = 8. 7 cm growth in 30 days 8. 7 cm x 10 = 87mm 30 days x 24 = 720 hrs 87 / 720 = 0. 12 mm/hr Q2 Which plant was given the highest concentration of IAA? Explain your answer. Q3 Evaluate the method that was used in this experiment. (use the headings ‘strengths’ and ‘weaknesses’)

Since 1960’s, the auxin has been isolated and purified and shown to have exactly the same effects as the ‘chemical messenger’ first predicted by the Darwins. Biologists believe that auxins act on growth genes, turning them on (switching genes ‘on’) and stimulating both cell division and cell elongation. Auxin removes the repressor factor. Refer and relate to gene expression. The growth towards light from one side is a result of the auxin being redistributed to the shaded side of the shoot. The phototropic response of a shoot. Auxins move away from the lit side of a shoot and accumulate in the shaded side. This side grows faster as a result, causing a curvature towards the light.

The gravitropic/geotropic (response to gravity) response is also controlled by auxins. However, auxins inhibit growth in roots. Growth is brought about by either the absence, or low concentrations of auxins. If a seedling is placed horizontally and left for a period, the root will grow downwards and the shoot upwards. Investigating how a root and shoot respond to gravity This happens because, in both root and shoot, auxin is redistributed to the lower side. • In the shoot, auxin stimulates growth. Therefore, cells on the lower side grow faster than those on the upper side. This causes an upward curvature. • In the root, auxin inhibits growth.

Therefore, cells on the lower side grow more slowly than those on the upper side. This causes a downward curvature. [pic] Q. At the concentration found in the cells (about 0. 1 ppm), what effect does auxin have on the cells of (i) the shoot cells, and (ii) the root. i) Stimulates growth ii) Inhibits growth Activity Now carry out Practical 6. Investigation into the role of IAA on root growth in mustard seedlings. Activity Practice Questions — Fact Recall Q1 What name is given to the growth of a plant in response to light? Q2 Plant shoots are negatively geotropic. What does this mean? Q3 What parts of a plant produce growth factors? Q4 How do auxins affect plant growth?

Q5 What is indoleacetic acid (IAA) and where is it produced? Q6 How does IAA move around a plant? Q7 Explain how the distribution of IAA affects the growth of: shoots in response to light. b) roots in response to gravity. Activity Exam – Style Question Scientists took three Goosegrass seedlings and planted them in individual pots with soil taken from the same source. They let each seedling grow for 15 days in the conditions shown in the diagram below. [pic] a) Suggest what response the scientists were testing with this experiment(1 mark) b) The scientists didn’t control in their experiment. Describe the conditions that should have been used for a seedling acting as a control(2 marks) ) Describe and explain the pattern of growth in the three plants you would expect to see by the end of the experiment. (6 marks) d) Explain the role of growth factors in controlling the direction of growth in this experiment. (4 marks) Summary Responses in simple animals • Simple animals exhibit taxes and kineses • A taxis involves a directional response to the stimulus e. g. animals may move towards (or away from) the strongest light or the most humid atmosphere. • A kinesis involves a change in the rate of movement as a result of a change in intensity of a stimulus. An increase in the rate of movement when exposed to a stimulus associated with hostile conditions makes it more likely that the animal will move out of the hostile conditions.

Responses in plants • Tropic responses by plants help to maintain different parts of the plant in the most favourable environmental conditions • Plant shoots are positively phototropic (they grow towards the light); this response exposes the leaves to the maximum light intensity, which enhances photosynthesis. • Plant roots are positively geotropic (they grow towards gravity); this response ensures that the roots always grow into the soil from which they absorb water and mineral ions. Plant growth regulators • Auxins are the most widely know plant growth regulators. They are involved in producing the phototropic and geotropic responses of shoots and oots: o In shoots, auxins become concentrated on the side of a shoot away from the light, causing increased growth here and curvature towards the light. o In roots placed horizontally, auxins become concentrated on the lower surface of the root, inhibiting growth here and causing the curvature towards gravity. ———————– Biological Sciences AQA Biology Unit 5: Control in Cell and in Organisms Survival and response (3. 5. 1 and 3. 5. 2)No. 5 Name …………………………………………………………………………………………….. Tutor: ……………………………………………………………………………………. ……… Chambers of differing humidity were prepared, and the distance moved by the woodlice in a given time was determined IAA moves to the shaded parts of the shoots and roots, so there’s uneven growth

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Survival and Response. (2016, Sep 30). Retrieved from https://graduateway.com/survival-and-response/

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