Senior high school biology photosynthesis breathing foundation question and the answer detailed explanation
The following are some basic questions and answers for high school biology:
** I. Comprehension of the concepts of photosynthesis and respiratory rate **
1. ** Title *
- The following statement about net photosynthesis rate is correct ()
- A. The net photosynthesis rate refers to the amount of carbon dioxide released by plants in the dark
- B. The net photosynthesis rate can be expressed by the amount of O
Senior High School Biology Experiment on Photosynthesis and Breathing
The following are some of the experimental designs for photosynthesizing breathing in high school biology: ** 1. Using droplet movement to measure photosynthesis and respiratory rate ** 1. ** Device Setting ** - For the measurement of breathing rate: - Use a sealed container to place plant materials (such as leaves or germinated seeds). The container was filled with a solution that could absorb carbon dioxide (such as a solution of NaOx). - The container was connected to a graduated glass tube (such as a U-shaped tube) with a drop of colored liquid in it. - "Principle: Under dark conditions (ensure that only breathing is carried out), the plant absorbs oxygen and releases carbon dioxide. As carbon dioxide is absorbed by the solution, the gas pressure in the container will decrease, and the droplets will move towards the container. The distance (or volume) that the droplet moved in a unit of time could represent the breathing rate (commonly used to measure the breathing rate by the amount of oxygen absorbed in a unit of time). - As for the measurement of the net photosynthesis rate: - A similar device, but the container is filled with a substance that can provide a stable concentration of carbon dioxide (such as a solution of NaHCO3, which can decompose to produce carbon dioxide). - When the device was placed under light, the plants would undergo photosynthesis and breathing. The oxygen produced by photosynthesis would increase the pressure in the container, and the droplets would move away from the container. The distance (or volume) that the droplets moved in a unit of time could represent the net photosynthesis rate (usually expressed as the amount of oxygen released or the amount of carbon dioxide absorbed). - The calculation of the true photosynthesis rate: true photosynthesis rate = net photosynthesis rate + respiratory rate. 2. ** Experiment Notes ** - Variant Control: - Light intensity: If you want to explore the effect of light intensity on the rate of photosynthesis, you can use different power bulbs or change the distance between the bulb and the plant to control the light intensity. - [Temperatures: Different thermostats can be used to control the experiment temperature.] - [Concentration of carbon dioxide: For example, carbon dioxide buffer solutions of different concentration can be used for adjustment.] - Principle of comparison: You can't experiment with just one set of equipment by gradually changing its conditions. Instead, you should use a series of equipment to compare with each other. ** 2. Black-and-white bottle method to measure the photosynthesis and respiratory rate ** 1. ** Operation method ** - Prepare several sets of black and white bottles. The black bottle was wrapped in a black material (so that it was in a dark environment), while the white bottle was normally transparent. - Put an equal amount of plants (aquatic plants such as algae are more convenient) and an equal amount of water in each bottle. 2. ** Principle ** - The plant in the black bottle only breathed, and the measured value was the intensity of the breathing (such as the reduction of oxygen or the increase of carbon dioxide). - The plants in the white bottle were photosynthesizing and breathing under light, and the measured value was the net photosynthesis intensity (such as the increase in oxygen or the decrease in carbon dioxide). - Combining the data of the black bottle and the white bottle, the true photosynthesis intensity value could be obtained (true photosynthesis rate = net photosynthesis rate + respiratory rate. The net photosynthesis rate here is the white bottle data, and the respiratory rate is the black bottle data). ** 3. Half-leaf method to measure the photosynthesis and respiratory rate (half-leaf weighing method)** 1. ** Operation process ** - Choose the leaves of the plant and cover half of them from the light (such as covering them with black paper), while the other half is under normal light. - After a period of time (usually a few hours to a day, depending on the plant species and experimental conditions), the leaves were taken down, and a hole cutter was used to punch out leaf discs of the same area on the light-blocking part and the non-light-blocking part of the leaves. - The leaf discs were dried to a constant weight and weighed. 2. ** Principle ** - The light-blocking part was only used for breathing, and the weight reduction was due to the consumption of organic matter by breathing. - The unshielded part underwent photosynthesis and respiration, and its weight change was the difference between the organic matter produced by photosynthesis and the organic matter consumed by respiration. By comparing the weight changes of the two parts of the leaf discs, the total amount of organic matter produced by photosynthesis could be calculated, and thus the photosynthesis rate (commonly used to measure the photosynthesis rate of field crops) could be obtained. ** IV. Studying the influencing factors of photosynthesis and respiratory rate with the gradient-based method ** 1. ** Device Setting ** - Prepare a series of devices with the same plants in each device. - Different conditions such as light intensity, temperature, or carbon dioxide concentration were set in different devices. 2. ** Purpose * - The effects of light intensity, temperature, or carbon dioxide concentration on the intensity of photosynthesis could be investigated. For example, by setting different light intensity (such as using different power bulbs or changing the distance between the bulb and the plant to set low, medium, and high light intensity), the plant's photosynthesis rate under each light intensity can be measured (the change in oxygen or carbon dioxide can be measured by the droplet movement method mentioned above or other methods), and the relationship curve between light intensity and photosynthesis rate can be obtained. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Basic Questions and Analysis of Photosynthesis and Breathing in High School Biology
The following are some basic questions and analysis of high school biology's photosynthesis and breathing: * * 1. Multiple choice questions ** 1. As shown in Figure A and B, they are part of the structural diagrams of the two types of organelle. A and B represent the gas exchange between the two types of organelle in the mesophysical cells under light. The correct description is (). - A. The nail cell organ can carry out complete cell breathing. - B. In the second picture, the movement direction of the ADC is from 5 to 4. - C. If all the O <2> is used by the A structure, the photosynthesis rate is the same as the respiratory rate. - D. Under suitable conditions, there is an electron transfer process at both 2 and 4. - Analysis: - Option A: The alpha cell organ is a mitochondria. Mitrons can only carry out the second and third stages of oxygen breathing. Complete cell breathing also includes the first stage of oxygen breathing (carried out in the cellular matrix), so Option A is wrong. - Option B: B is the plastids, and the movement direction of the light reaction produced by the light reaction is from the thylakoid membrane (light reaction site) to the plastids stroma (dark reaction site), that is, from 4 to 5, so B is wrong. - Option C: If all the O <anno data-annotation-id ="00000000 - 4150 - 4410-a160 - 999b6111000"> 2 </anno> is used by the A structure (mitochondria), it means that the plant's photosynthesis rate is equal to the respiratory rate, because the O <anno data-annotation-id ="2c3cd75 - 4c60 - 4c60 - 4c60 - 99999999999998"> 2 </anno> produced by photosynthesis is consumed by the respiratory process. C is correct. - Option D: Under suitable conditions, there is an electron transfer process in both the mitochondria and the thylakoid membrane. D is correct. 2. Mitochondria and plastids were important places for energy conversion in plant cells, and they had certain similarities. The following statement is wrong (). - A. Both of them have a matrix, and both of them can synthesize bound molecules. - B. Both contain DNA, which controls protein synthesis. - C. Both are covered by two layers of membrane, and the role of the inner and outer membranes is different. - D. In the process of energy conversion, both are accompanied by gas exchange. - Analysis: - [Option A: The second stage of oxygen metabolism can be carried out in the mitochondrion stroma, and it can be used to synthesize ATP. However, the plastids stroma is a dark reaction site, and it consumes ATP. It cannot be used to synthesize ATP. Option A is wrong.] - Option B: Both mitochondria and plastids contain a small amount of DNA, which can control the synthesis of a part of the protein they need. - Option C: Both mitochondria and plastids have two membranes. The inner membrane and outer membrane are different in structure and function. For example, the inner membrane of the mitochondria is the location of the third stage of oxygen metabolism. The inner membrane and outer membrane of the plastids have the function of selectively permeating substances. Option C is correct. - "Option D: The mitochondria have gas exchange during their respiratory process (absorbing O <2> and releasing O <2>), while the plastids have gas exchange during their photosynthesis (absorbing O <2> and releasing O <2>). D is correct. 3. The following statement about the process of studying the human body's cell oxygen demand breathing by the radioactive label method is wrong (). - A. Using Thousand-O to label glucose, the product water can be detected to be radioactive. - B. Use Thousand-C to label glucose, and the carbon dioxide product can be detected as radioactive. - C. Labelling oxygen with ¹ O, the product can be detected as H ^¹ O and C ¹ O ^ - D. Using H to label glucose, radioactive activity can be detected in both the cell-sol and mitochondria. - Analysis: - Option A: In the process of oxygen breathing, the oxygen in the glucose is eventually transferred to carbon dioxide. The oxygen in the water comes from oxygen, so the glucose is labeled with monarch O. The product water cannot be detected as radioactive. - [Option B: Labeled glucose with 1 C. The decomposition of glucose produces carbon dioxide, so the product carbon dioxide can be detected as radioactive. Option B is correct.] - Option C: Use Thousand-O to mark oxygen. The oxygen is involved in the third stage of oxygen breathing to produce water, and the water can participate in the second stage of oxygen breathing to produce carbon dioxide. Therefore, the product can detect H Chi Thousand-O and C Thousand-O Chi. C is correct. - "Option D: Dextrose is decomposed into methyruvate in the cell-colloid, and methyruvate enters the mitochondria for further decomposition. Therefore, when the hydrogen is used to label the glucose, the radiation can be detected in both the cell-colloid and the mitochondria. D is correct. 4. The long-leaved thorny sunflower is a kind of palm plant. The following picture shows the curve of the photosynthesis intensity of the long-leaved thorny sunflower measured by a research team under sufficient water conditions within 24 hours. The following description of the curve is wrong (). - A. The curve a represents the total photosynthesis intensity, and the curve b represents the net photosynthesis intensity. - B. After 10:00, the reason for the decrease of curve b is due to the insensitivity of the thermo-responsive photosynthesis related to the temperature. - C. After 14:00, both a and b decreased because the light intensity weakened. - D. Maximum accumulation of organic matter around 18:00 - Analysis: - Option A: Total photosynthesis intensity is equal to the net photosynthesis intensity plus the respiratory intensity. The curve a includes both respiratory and net photosynthesis, so curve a represents the total photosynthesis intensity, and curve b represents the net photosynthesis intensity. A is correct. - Option B: The reason why curve B drops after 10:00 is mainly because of the increase in temperature, the partial closure of the stomata, and the decrease in carbon dioxide supply, rather than the fact that the protein is not sensitive to temperature. - Option C: After 14:00, the light intensity weakens, resulting in a decrease in total photosynthesis intensity (a) and net photosynthesis intensity (b). C is correct. - Option D: When the net photosynthesis intensity is 0, the accumulation of organic matter reaches the maximum. At about 18:00, the net photosynthesis rate is 0. At this time, the accumulation of organic matter is the maximum. D is correct. 5. The picture below is a diagram of the photosynthesis and respiratory processes of a plant's mesophysical cells (A-D represents the relevant processes, a-e represents the related substances). The following analysis is wrong (). (Since there is no picture, the assumption here is an analysis of the relationship between the substances and processes of photosynthesis and respiration.) - [Analysis: It needs to be analyzed according to the specific map. If it is about the material connection, for example, a is carbon dioxide, which is the raw material of photosynthesis and the product of respiration. If it is about the process connection, the light reaction of photosynthesis (process A) provides the dark reaction (process B) with [H] and [Ang]. Respocation provides carbon dioxide and other substances for photosynthesis.] The wrong choice needed to be judged based on the wrong relationship between the material and the process marked on the specific map. * * 2. Non-multiple choice questions ** 1. The meaning and relationship between net and total photosynthesizing rate: - (1) What are the methods to express the net photosynthesis rate? - Analysis: The net photosynthesis rate can be expressed in the following ways: (1)"measure" the amount of carbon dioxide absorbed by the plant (leaf) or the reduction of carbon dioxide in the "experimental container";(2) the amount of oxygen dioxide released by the plant (leaf) or the increase of oxygen dioxide in the "container";(3) the amount of glucose accumulated by the plant (leaf) or the increase in the mass (organic matter) of the plant (leaf). - (2) What are the methods to express the total photosynthesis rate? - Analysis: The total photosynthesis rate is expressed in the following ways: (1) the amount of carbon dioxide absorbed by the plastids;(2) the amount of carbon dioxide assimilated;(3) the amount of oxygen dioxide released by the plastids;(4) the total amount of oxygen dioxide produced;(5) the amount of glucose produced by plants or plastids;(6) the amount of organic matter produced by plants. - (3) If a plant is under light, the increase in O <anno data-annotation-id ="00000000 - 4110 - 4000 - 8000 - 9000 - 80000000000"> 2 </anno> mole/h in the container and the absorption of O </anno> in the dark is 1 mole/h, then calculate the total and net photosynthesizing rate. - Analysis: The net photosynthesis rate can be expressed by the increase in O <anno data-annotation-id ="0000004 - 4445 - 4445-a110 - 99999999999"> 2 </anno> in the container, so the net photosynthesis rate is 2 mole/h. The total photosynthesis rate is equal to the net photosynthesis rate plus the respiratory rate. The respiratory rate can be expressed as the amount of O <2> absorbed in the dark, so the total photosynthesis rate = 2 mole/h +1 mole/h = 3 mole/h. 2. Experimental measurement related questions: - (1) How do you measure the rate of a plant's breathing? - [Analysis: You can use the following devices to measure the plant's breathing rate.] There was a small bottle in the device. Plant tissue was placed in the bottle. Under the condition of no light, the tissue cells absorbed O <2> and released CO <2>. The CO <2> was absorbed by the solution of NaH <2> in the bottle, reducing the pressure of the gas in the whole container, and the water droplets in the scale tube would move to the left. The volume of the water droplet moving to the left in a unit of time represented the breathing rate. However, if the experimental material is a green plant, the entire device should be shielded from light, otherwise the photosynthesis rate will interfere with the detection of the respiratory rate. If the seeds are selected as experimental materials, the existence of microorganisms must be considered to interfere with the detection of the experiment. Therefore, the device and the tested seeds should be disinfected. In order to avoid errors caused by physical factors such as air pressure and temperature during the experiment, a control experiment should be set up, and the tested biological materials should be deactivated (such as cooking the seeds). Other conditions remain unchanged. When using seeds with high fat (or fat) content, such as oil seeds, the water droplet movement is more obvious (because it consumes more O <2>). - (2) How to measure the net photosynthesis rate of plants? - "Analysis: A device can be used, in which the decomposition of hydrogen carbonates produces carbon dioxide to ensure the constant carbon dioxide in the container. The increase in oxygen in the container can be measured to determine the net photosynthesis rate of the plant. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
College entrance examination biology big question photosynthesis and breathing
光合作用是植物通过叶绿体利用光能把二氧化碳和水转化成储存能量的有机物，并且释放氧气的过程。呼吸作用是细胞利用氧将有机物分解成二氧化碳和水，并且将储存在有机物中的能量释放出来供生命活动需要的过程。 在高考生物大题中，光合作用与呼吸作用相关题型属于高频考点，主要涉及以下几个方面： **一、光合呼吸曲线类（包括影响因素）** 1. **曲线分析位点移动** - **A点的移动**：A点代表呼吸速率，若细胞呼吸增强，A点下移;反之，A点上移。 2. **影响因素** - **光照强度**：影响光合作用的光反应阶段。在一定范围内，光照强度增加，光合作用速率增大。当光照强度达到光饱和点后，光照强度再增加，光合速率不再增加。而呼吸作用有光无光均可进行。 - **二氧化碳浓度**：是光合作用暗反应的原料。二氧化碳浓度增加，光合速率增大，当达到一定浓度后，光合速率不再增加。 - **温度**：通过影响酶的活性来影响光合作用和呼吸作用。光合作用和呼吸作用都有最适温度，在最适温度之前，温度升高，速率增大;超过最适温度，速率下降。 **二、光合作用过程的判断** 1. **物质转化方面** - 把简单的无机物（二氧化碳和水）转化成复杂的有机物并释放氧气。 2. **能量转化方面** - 把光能转变为贮存在有机物中的化学能。 **三、C3、C4、CAM植物及光呼吸** 1. **C3植物** - 二氧化碳固定后的最初产物是三碳化合物（3 - 磷酸甘油酸）。其光合作用的暗反应途径为C3途径，这类植物如大豆、小麦等。 2. **C4植物** - 二氧化碳固定后的最初产物是四碳化合物（草酰乙酸）。C4植物具有特殊的结构（如维管束鞘细胞）和生理机制，能够在低二氧化碳浓度下进行光合作用，如玉米、甘蔗等。 3. **CAM植物** - 这类植物的气孔在夜间开放，吸收二氧化碳并固定成有机酸，白天有机酸脱羧释放二氧化碳用于光合作用，主要是为了适应干旱环境，如仙人掌等。 **四、呼吸作用过程及方式判断** 1. **有氧呼吸** - 反应式为：\(C_{6}H_{12}O_{6}+6O_{2}+6H_{2}O \xrightarrow[]{酶} 6CO_{2}+12H_{2}O + 能量\)，主要场所是线粒体，分三个阶段进行，包括糖酵解（细胞质基质）、柠檬酸循环（线粒体基质）和电子传递链（线粒体内膜），释放大量能量。 2. **无氧呼吸** - 在无氧条件下进行，产生酒精和二氧化碳（如酵母菌）或者乳酸（如乳酸菌），释放少量能量。反应式如产生酒精时：\(C_{6}H_{12}O_{6}\xrightarrow[]{酶} 2C_{2}H_{5}OH + 2CO_{2}+ 少量能量\);产生乳酸时：\(C_{6}H_{12}O_{6}\xrightarrow[]{酶} 2C_{3}H_{6}O_{3}(乳酸)+ 少量能量\)。 在解答高考生物光合作用和呼吸作用大题时，要熟悉这些基础知识点，同时具备识图、表格分析、图像解读、情境信息理解及实验设计等综合能力。例如，根据给出的图像判断植物的光补偿点、光饱和点，分析不同植物对光能的利用率;根据实验数据判断植物的呼吸方式等。 <a href="/?from=ask_words" style="color:red" target="_blank">点击前往免费阅读更多精彩小说</a>
Junior high school biology breathing and photosynthesis knowledge points summary
Photosynthesis: - [Location: Phosphor.] The chlorites in the plastids could absorb light energy and provide energy for photosynthesis, so the plastids were the place where green plants carried out photosynthesis. - "Description: Green plants use light energy to convert carbon dioxide and water into organic matter that stores energy and releases oxygen." - Reaction formula: carbon dioxide + water $/stack {light energy}{→}$organic matter (stored energy)+ oxygen. - The conversion in the process: one was material conversion, which converted simple minerals (carbon dioxide and water) into complex organic matter and released oxygen; the other was energy conversion, which converted light energy into chemical energy stored in organic matter. - "The influencing factors include the intensity of light, because light is the source of energy for plants to carry out photosynthesis. Only under a certain intensity of light can they carry out photosynthesis to produce organic matter. It can only be formed under light. Breathing effect: - [Description: The process in which cells use oxygen to decompose organic matter into carbon dioxide and water, and release the energy stored in the organic matter for life activities.] <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
A question on the topic of oxygen and anoxia breathing in high school biology.
以下是一些关于高一生物有氧呼吸和无氧呼吸的常见题型及解答： **一、概念辨析类** 1. **判断对错** - 有氧呼吸和无氧呼吸的第一阶段完全相同。（正确）。有氧呼吸和无氧呼吸的第一阶段都是葡萄糖分解为丙酮酸和[H]，该过程在细胞质基质中进行。 - 无氧呼吸的全过程都在细胞质基质中进行。（正确）。无氧呼吸分为两个阶段，两个阶段都在细胞质基质中完成，而有氧呼吸的第二阶段在线粒体基质，第三阶段在线粒体内膜。 - 所有生物都能进行有氧呼吸。（错误）。例如乳酸菌等微生物只能进行无氧呼吸，蛔虫是动物但也只进行无氧呼吸。 2. **选择题** - 下列关于有氧呼吸的叙述，正确的是（） A. 有氧呼吸产生的能量全部储存在ATP中。（错误，有氧呼吸产生的能量一部分以热能形式散失，一部分储存在ATP中） B. 有氧呼吸过程中，水既是反应物又是生成物。（正确，有氧呼吸第二阶段有水参与反应，第三阶段有水生成） C. 有氧呼吸的场所只有线粒体。（错误，有氧呼吸第一阶段在细胞质基质，第二阶段在线粒体基质，第三阶段在线粒体内膜） - 无氧呼吸产生酒精的生物有（） A. 乳酸菌（错误，乳酸菌无氧呼吸产生乳酸） B. 酵母菌（正确，酵母菌无氧呼吸可产生酒精和二氧化碳） C. 人（错误，人无氧呼吸产生乳酸） **二、反应式书写类** 1. **写出有氧呼吸的总反应式** - 正确反应式：\(C_{6}H_{12}O_{6}+6O_{2}+6H_{2}O \xrightarrow[]{酶} 6CO_{2}+12H_{2}O + 能量\)。需要注意反应条件是酶，不能把箭头写成等号，反应物和生成物两边的水不能随意消去，能量不能用ATP代替。 2. **写出产生酒精的无氧呼吸反应式** - \(C_{6}H_{12}O_{6} \xrightarrow[]{酶} 2C_{2}H_{5}OH + 2CO_{2}+少量能量\) 3. **写出产生乳酸的无氧呼吸反应式** - \(C_{6}H_{12}O_{6} \xrightarrow[]{酶} 2C_{3}H_{6}O_{3}(乳酸)+少量能量\) **三、计算类** 1. **根据反应式计算能量或物质的量** - 例如：已知1mol葡萄糖进行有氧呼吸，求产生二氧化碳的物质的量。 - 根据有氧呼吸反应式\(C_{6}H_{12}O_{6}+6O_{2}+6H_{2}O \xrightarrow[]{酶} 6CO_{2}+12H_{2}O + 能量\)，1mol葡萄糖进行有氧呼吸会产生6mol二氧化碳。 - 又如：1mol葡萄糖进行无氧呼吸产生酒精，求产生的能量（已知每摩尔葡萄糖生成酒精释放的能量为225.94kJ）。 - 答案为225.94kJ，因为1mol葡萄糖进行产生酒精的无氧呼吸释放能量为225.94kJ。 **四、实验探究类** 1. **探究有氧呼吸和无氧呼吸产物的实验** - 实验装置：为了探究酵母菌的呼吸方式，可以设置两组实验装置。 - 有氧呼吸装置：将酵母菌培养液加入到带有通气管（通气管先经过氢氧化钠溶液以除去空气中的二氧化碳）的锥形瓶中，再连接澄清石灰水的锥形瓶。 - 无氧呼吸装置：将酵母菌培养液加入到密封的锥形瓶中，连接澄清石灰水的锥形瓶。 - 实验现象及结论： - 如果有氧呼吸装置中澄清石灰水变混浊，说明有氧呼吸产生了二氧化碳;无氧呼吸装置中澄清石灰水也变混浊，说明无氧呼吸也产生二氧化碳。 - 若要进一步探究无氧呼吸是否产生酒精，可以从无氧呼吸装置中取培养液，加入酸性重铬酸钾溶液，若溶液由橙色变为灰绿色，则说明产生了酒精。 <a href="/?from=ask_words" style="color:red" target="_blank">点击前往免费阅读更多精彩小说</a>
The answer to the compulsory textbook of biology in senior high school
The following are the answers to some of the questions in the Jiangsu version of the high school biology compulsory textbook: Chapter 1: The molecular composition of cells The Elements and Inorganic Compound in the Cell - ** Speculation Questions ** - Among the chemical elements that made up living things, C, H, O, and N were the most abundant. - Free water was a good solution for many ions, not bound water. - If muscle weakness occurs, the substance that may be lacking is calcium. - ** Usage Questions ** - The trace elements were iron, mn, zn, copper, b, al, co, and se. - The carbon exists in the form of carbon dioxide, oxygen in the form of water, hydrogen in the form of water, calcium in the form of carbon dioxide, phosphorus in the form of dipotassium hydrogen orthhosphate, and potassium in the form of kci. They are involved in the composition of substances in the body; Na ions maintain the stability of the osmonic pressure of the extra-cellular fluid; K ions participate in the activation of more than 60 kinds of membranes during plant growth and development; Ferrous ions are the components of many membranes; and Magnesium ions are an important component of the phylls. - Legume plants such as alfalfas and soybeans had a nitrogen fixing bacteria at the roots, which could help them survive the winter. - Free water was a good solute. The ions and molecules needed for cell life could be dissolved in it and flow with the water to participate in the reaction. Metabolic waste could also be discharged from the body through water. ** Section Two: Carboids and Lipids in the Cell ** - ** Speculation Questions ** - Glycogen and glucose were the main energy substances, fat was an important energy storage substance, and Cellulose was a plant cell structure substance that could not be used as an energy substance. - Lipids were divided into fats, lysomes, and steroids, but lysomes did not belong to steroids. - It was not advisable to reduce overweight weight only by reducing the intake of sugar and fatty foods. It was also necessary to combine it with moderate physical exercise. - ** Usage Questions ** - Plant cells contained starch and starch, and the process of making syrup was similar to the biological method. - The second biological method was used to utilize the raw material of the Cellulosic. - Celluloses and other glycans were important structural substances in plant cells, while glucose was the main energy source for life activities. - Ferlin's reagent could only identify reducing sugar. Sweet potatoes mainly contained non-reducing sugar such as starch, so it could not be used to identify them. However, Sultan III staining solution could be used to detect the fat in sweet potatoes. ** Section Three: The Nucleic Acid and Nucleic Acid in the Cell ** - ** Speculation Questions ** - Nucleic acid in the human body was composed of eight kinds of nuclei (adenine deoxy, guanine deoxy, cystine deoxy, thymine deoxy, adenine Ribo, guanine Ribo, cystine Ribo, and uracil Ribo). Uridine Ribo was composed of uracil, Ribo, and Phosphoryl. - The seeds of wheat, sorghums, and other plants are rich in starch, while animals such as cattle and sheep are rich in protein and fat. - ** Usage Questions ** - In contrast to the "food pyramid", if you eat too much dessert and meat, you may be obese, and if you eat too little vegetables and fruits, you may lack vitamins. - Mung beans may contain more starch; it is not possible to judge whether they are nutritious based on the content of protein; the cell composition of beans cannot be separated from protein. Cell structure and life activities (Part) - In the content related to organic matter in cells, there were various statements: - Not all kinds of sugar could be used as a plant energy material. Ribose and deoxy sugar formed the genetic material, and the plant cell wall made of Cellulose could not be used as an energy material. Therefore, it was wrong to say that all kinds of sugar could be used as a plant energy material. - The basic framework of the cell membrane was composed of a double-layer consisting of lysomes. Choline was an important component of animal cell membranes and was involved in the transport of fats in the blood. Therefore, it was correct to say that both lysomes and Choline were components of animal cell membranes. - The main reason for protein degeneration was the destruction of the spatial structure rather than the cleavage of the bond. Deoxy acids were connected to each other by the bond of the ester. Therefore, it was wrong to say that the high temperature degeneration of the protein was due to the cleavage of the bond, and the single chain was formed by the bond of the deoxy acids. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The Analysis of the Questions and the Answer in the Biology Course of Senior High School
The following are some questions and answers that might be involved in the high school biology seminar: ** I. Problems related to the metabolism of cells ** 1. ** Question **: briefly describe the process of scientific discovery of the essence of the enzyme and the characteristics of scientific exploration. - Before Pasteur, he believed that fermentation was a pure chemical reaction and had nothing to do with life. Pasteur and Liebig respectively proposed that the fermentation in brewing was related to living yeast cells. Without living cells, sugar could not be turned into alcohol, and that the fermentation was caused by certain substances in the yeast cells, and these substances could only play a role after the yeast cells died and cracked. Through experiments, Bichner proved Liebig's point of view was correct, saying that the substance that caused fermentation was a zymase, but what exactly was the zymase was still unclear. Later, Sumner extracted urease crystals with high purity from sword bean seeds for the first time and proved that urease was a protein. This process showed that scientific discoveries were not smooth sailing. Many scientists needed to repeatedly demonstrate, debate, and experiment to reach scientific conclusions. 2. [Question: What is the relationship between the metabolism of the cell and the metabolism of the cell?] - Cell metabolism is composed of a series of chemical reactions in the cell. These reactions need to be carried out at high speed under normal temperature and pressure, and they require the catalyze of an enzyme. Because the reaction environment inside the cell was at normal temperature and pressure, in order to make the chemical reaction proceed quickly, the reaction required the activation energy to be reduced. 3. [Question: How do you determine if an protein is an protein?] - [Answer analysis: Biuret reagent can be used to determine whether this type of protein is a protein.] ** 2. Questions related to the ability to explore biological experiments ** 1. [Question: In the college entrance examination, how do you test your ability to design an experiment plan?] - For example, the 30th question of National Volume A required students to design an experimental plan to verify that protein fragments could appear in the lysomes of engulfing cells. The 32nd question of the new curriculum standard required students to design an experimental plan to test the AIDS virus in blood samples. This tested the students 'ability to use basic experimental methods and techniques to solve new problems. 2. [Question: Using genetic engineering experiments as an example, how does the biology college entrance examination test the understanding of the principles and objectives of the key links of the experiment?] - Question 38 of National Volume A used genetic engineering to express the target protein as the background. It examined the principles and purposes of the key links such as the double digestion of the plasmid-carrier and the preparation of competent cells. It allowed students to feel the rigorous and realistic characteristics of scientific research in the process of solving common problems in scientific research, so as to examine their understanding of the principles and purposes of the key links of the experiment. ** 3. Questions related to the application of biological knowledge in production and life ** 1. [Question: Give an example of how the college entrance examination biology questions reflect the application of biological knowledge in solving practical production and life problems.] - ** Answer Analysis **: Question 34 of the new curriculum standard volume mentioned the practical problem of the high yield of the female plants of the white thorn melon being favored by consumers. Students were required to use genetic crossing methods to screen out the pure hybrids of the female plants of the white thorn melon to solve the problem of high yield. Question 37 of National Volume A focused on the theme of rational use of disinfectant to reduce infectious diseases. It examined the key points of the exploration process such as microorganisms cultivation, bacteria counting, coating plates, and analysis of experimental results. It guided students to use the basic methods of scientific inquiry to solve practical problems in production and life, reflecting the application of biological knowledge in solving practical production and life problems. ** IV. Core literacy and teaching-related issues in biology ** 1. [Question: How to implement the core knowledge of biology in biology class?] - ** Answer Analysis **: Taking the classroom teaching " Passive Transportation " as an example, at the beginning of the class, the students were used to think about life phenomena, stimulate the curiosity of students to explore new knowledge, arouse the enthusiasm of students in the form of question strings, set up independent inquiry and group discussion sessions, and let students feel a sense of participation. In the teaching process, they combined life examples to understand the knowledge points, such as explaining the principle of water entering and leaving the cell. This could well fulfill the requirements of developing students 'scientific thinking and social responsibility in the core accomplishment of biology. In addition, the course " Gene Mutation " focused on the occurrence and mechanism of cancer. It guided the students to analyze and explore the concept, essence, characteristics, significance, etc. of gene mutation independently. It was also helpful to improve the students 'core biological quality. 2. ** Question **: How does the inter-school biology exchange activity promote teachers 'teaching? - Take the inter-school biology exchange between Jigang High School and Laiwu No.1 Middle School as an example. The inter-school exchange provided a platform for teachers to exchange teaching models and study school-based curriculum. Through the exchange, teachers could learn the experience of developing and implementing the biology school-based curriculum in other schools. For example, the teachers of Jigang High School realized that the content of the biology school-based curriculum in Laiwu No.1 Middle School was rich and layered. They could also learn different classroom teaching methods, such as teacher guidance and student inquiry in the classroom of Laiwu No.1 Middle School. They could truly realize the student-centered teaching mode, thus promoting the updating of teachers 'teaching ideas and the reform of teaching methods, and improving the teaching level of teachers. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The Concept of Aerobic Breathing in High School Biology
Aerobic breathing referred to the process in which cells, with the participation of oxygen, completely oxided and decomposed organic substances such as glucose into carbon dioxide and water, releasing energy and generating a large amount of Ang. Aerobic breathing was the main form of breathing for higher animals and plants. Usually, the so-called breathing referred to the use of oxygen. Aerobic breathing was carried out in the cellular matrix and the mitochondria, and the mitochondria was the main place for cells to carry out oxygen breathing. It could be divided into three stages. The first stage was the initial decomposition of glucose in the cellular matrix. One molecular sugar was decomposed into two molecules of methyruvate, four activated hydrogen (H), and a small amount of ATP. The second stage was the complete decomposition of the Pyruvate in the mitochondria matrix. The hydrogen in the two molecules of Pyruvate and the six water molecules was completely removed. The Pyruvate was then oxided and decomposed into carbon dioxide, producing a small amount of energy. The third stage was the combination of the hydrogen and oxygen removed in the previous two stages into water on the membrane of the mitochondria, releasing a large amount of energy. The fate of the energy includes storage in the form of active chemical energy in the ATPs and dissipation in the form of heat. The complete decomposition of 1 mole of glucose in the body can release 2870kJ of energy, of which about 1161kJ of energy is stored in the ATP-bound substance (about 38 moles of ATP-bound substance). The energy utilization rate is about 40.45%, and the rest of the energy is lost in the form of heat energy. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
High school biology plant photosynthesis knowledge framework map
Photosynthesis was an important knowledge in high school biology. The following is the framework of its knowledge: ** 1. Concept ** 1. [Range: Green plants] 2. [Location: Phosphor.] 3. [Energy source: Light energy.] 4. [Materials: carbon dioxide and water.] 5. [Production: Energy-storing organic matter and oxygen.] ** 2. The process ** 1. Light reaction stage - [Location: On the thylakoid membrane of the plastids.] - material changes - Photolyzation of water: Water is decomposed into the reduced form of hydrogen and oxygen under light. - The synthesis of ATP-binding reagents: AAD and Phosphorous Acid synthesize ATP-binding reagents. - Energy change: The conversion of light energy into chemical energy in the form of bound bound 2. Dark Reaction Stage (Calvin Cycle) - [Location: Protoplasts stroma.] - material changes - The fixing of carbon dioxide: 1-molecular five-carbon compound and carbon dioxide form 2-molecular three-carbon compound. - The reduction of three-carbon compounds: When two molecules of three-carbon compounds are supplied with energy by ATP-and hydrogen by ADMPH, part of them will be reduced to form five-carbon compounds, and the other part will form organic substances such as sugar. - Energy change: The chemical energy in the molecules of ATPs and ADPs is converted into the chemical energy in organic matter. ** 3. Impact factors ** 1. internal factor - Plant species (Different plants have different photosynthesis abilities, such as C3 plants and C4 plants). - Leaf growth condition (such as leaf age, content of photosynthesis, etc.) 2. external factor - Light intensity: Affects the light reaction stage. Within a certain range, the increase in light intensity will increase the rate of photosynthesis. When it reaches the light saturation point, the increase in light intensity will no longer increase the rate of photosynthesis. - [Temperatures: By affecting the activity of the microorganisms, it affects photosynthesis. Different plants have different optimal temperatures for photosynthesis.] - [Concentration of carbon dioxide: Affects the dark reaction stage. Within a certain range, the concentration of carbon dioxide increases, and the rate of photosynthesis increases. When it reaches the carbon dioxide saturation point, the concentration increases, and the rate of photosynthesis no longer increases.] - Water: Water is one of the raw materials for photosynthesis. A lack of water will cause the stomata to close, affecting the entry of carbon dioxide and thus affecting photosynthesis. ** 4. Experiment-related ** 1. Extraction and Separation of Pigments from Green Leaves - extracting the coloring matter: dissolving the coloring matter with absolute alcohol. - Separation of Pigments: Separation of Pigments by using a separation solution according to the different dissolutes of different Pigments in the separation solution. The Pigments with high dissolutes will spread quickly on the filter paper (such as beta-Carotenin), and the Pigments with low dissolutes will spread slowly (such as Chlorella b). 2. Research on the conditions and products of photosynthesis (such as the use of radioactive labels to investigate the source of oxygen in photosynthesis). <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Zero foundation high school biology and geography
For high school biology students with zero basics: - Biology was one of the liberal arts in science. There was little calculation content and most of it relied on memory, but memorization could easily confuse the knowledge. First of all, he had to spend about two weeks to make up for the basics. He had to go back to the textbook and focus on the large black words. He had to pay attention to the dialog boxes and small tips beside him. Then, he had to construct a complete knowledge system from the cell to the biosphere through the mind map. After completing the foundation, he would carry out special training. He would accurately brush the questions and do them in the order of knowledge points and difficulty. He would pay attention to the required knowledge points, question types, and corresponding solution methods. There was no need to repeatedly brush the questions. For high school geography students with zero foundation: - The difficulty level of Year One was similar to Biology, while Year Two's Human Geography was more difficult. The basic knowledge of geography was scattered and messy. He had to sort out the basics first because the basics accounted for more than half of it. As for the big questions, because there was an answer template, he could focus on solving the big questions. If you want to get more than 70 points in a short period of time, you can follow the steps above. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>