At present, the answers to the 2023 biology test questions of the third year of high school have not been announced yet. Please continue to pay attention to the relevant websites to obtain the latest information. Read more exciting novels for free
The following are some of the answers and analysis of the biology test questions of the first mock exam of the third year of high school in 2023 in Shandong Province: 1. Single choice questions 1. Answer: B. - Analysis: During the synthesis of cell membrane shaping protein, the location is provided by the Ribosome, and the power is provided by the mitochondria.(The main energy supplier in the cell, the plastids are mainly found in plant cells and are mainly related to photosynthesis, not involved here), A is wrong; the "molecular garbage bag" is a vesicle, mainly composed of lysomes and protein, B is correct; the "recycling factory" may be a lysoma, and the "component" may be an amine acid.(Because it is a product of protein decomposition, not a primer), C error; The organelle that can form vesicles in human cells includes the plasmatic reticule and the Golgi apparatus, but the centric body cannot form vesicles, D error. 2. Answer: A. - Analysis: The V-disorderly transport of H+ from the cellular matrix into the vesicle was a process that consumed ATP. It was an active transport, not an auxiliary spread. A was wrong. Inhibition of the activity of Cys transporter on the tonoelle would affect the process illustrated in the diagram, resulting in abnormal function of the mitochondria. B was correct. Cys entered the vesicle against the concentration slope by using the potential energy of H+. It was an active transport. C was correct. The process illustrated in the diagram indicated that there was a connection between the vesicle and the mitochondria. There was both division of labor and cooperation. D was correct. 3. Answer: D. - Analysis: The cells in the root meristematic zone of onions had no fat, so orange granules could not be observed with Sultan III staining. A was wrong; The cells in the root meristematic zone of onions became square and tightly arranged after being dissociated, rinsed, stained, and made into slices. B was wrong; The widest color of the green tube-shaped leaf on the filter paper was the cyanophile a (blue-green), C was wrong; The color of the purple onion scale leaf was cyanophile, which could not be extracted with absolute alcohol (absolute alcohol was used to extract the color in the plastids). D was correct. 4. (As the question did not give all the questions about this question, the answer and analysis could not be given accurately.) 2. Multiple-choice questions (As the questions did not give all the questions in this part, all the answers and analysis could not be given accurately) Third, non-multiple choice questions 21. - (1) Answer: Light intensity and CO2 concentration;Gs. - Analysis: According to the experimental content, the influencing factors are light intensity and CO2 concentration, and Gs is the relevant measurement index. - (2) Answer: Shade. Compared with natural light, the stomatal conductivity of leaves under shade increased slowly, and the Ci value decreased very little in the initial stage after light. - [Analysis: Obtain the relevant characteristics under the shade condition from the comparison of experimental data.] - (3) Answer: [H] and ATP-produced by light reaction increased, and dark reaction strengthened, but stomatal conductivity was lower, and intercellular CO2 concentration decreased. As stomatal conductivity increased, intercellular CO2 concentration gradually increased. - Analysis: This phenomenon was explained by the relationship between light and dark reactions and the effect of stomatal conductivity on intercellular CO2 concentration. 22. - (1) Answer: Genetic mutation; a trait can be affected by multiple genes. - [Analysis: Genetic changes may be caused by genetic mutation. From the content of the question, it can be analyzed that one trait is related to multiple genes.] - (2) Answer: 3. - (The analysis needs to be based on the complete information related to the genes and traits of the fruit fly. Due to partial missing, it is impossible to give details.) - (3) Answer: In mutant 1, choose the double-balanced lethal line fruit fly to cross with the wild type non-lethal line fruit fly, and let the male and female fruit flies in F1 mate freely. If F2 wild type: mutant = 6:1, then the mutant gene is located on the 2nd; if F2 wild type: mutant =3:1, then the mutant gene is located on other genes. - [Analysis: Through crossbreeding experiments, according to the inheritance law of the genes on different microorganisms, the corresponding proportion of offspring is obtained to determine the genetics of the genes.] - (4) Answer: 4, 2/3. - (The analysis needs to be based on the complete information related to the genes and traits of the fruit fly. Due to partial missing, it is impossible to give details.) 23. - (1) Answer: Assist in spreading;Na+ channel is open, Na+ flows in; cerebral hemisphere; efferent nerve endings and the diaphragm and abdominal muscles they inhabit. - [Analysis: According to the characteristics of the material transport method, it is determined to assist in the spread. The nerve center of the vomiting reflex is in the cerebral hemisphere, and the relevant muscles are indominated by the efferent nerve endings.] - (2) Answer: Choose healthy mice with the same physiological state and divide them into two groups, A and B. Group A used chemical genetics to specifically suppress M neurons, while Group B was used as a control. Stimulate the two groups of mice with toxins and observe the vomiting behavior of the mice. Inhibition of Ca2 + channel activity, reduction of Ca2 + influx, inhibition of the expression of the Tachkinin gene in the M-type neurons. - Analysis: Investigate the role of M neurons in vomiting behavior by setting up a control experiment. From the perspective of affecting the physiological process of the neurons, propose measures to suppress the relevant channels and gene expression. 24. - (1) Answer: Species composition; dominant species in the community. - [Analysis: This is a concept related to community. Species composition and dominant species are important aspects of community research.] - (2) Answer: Yunnan pine forest. The survival rate and mortality rate of Pseudotacia glazei in the Yunnan pine forest community reached a balance at a lower age class. - [Analysis: According to the relevant data of Yunnan pine forest in the map, the survival characteristics of Lancang yellow cedar can be obtained.] - (3) Answer: No. The data above does not show the proportion of individuals of different ages in the population. - [Analysis: According to the knowledge of the age structure of the population, the data in the picture cannot reflect the age structure.] - (4) Answer: The young individuals of the population of Pseudotacia glazei in Fig. 2 and Fig. 3 compete more fiercely with other species for survival resources and space. - [Analysis: Obtain the competition status of young individuals from the comparison of the data in the map.] 25. - (1) Answer: Denature the protein and separate it from the DNA; separate the DNA from the protein. - [Analysis: In related experimental operations, the purpose of this operation is to deal with the relationship between protein and DNA.] - (2) Answer: B. The primer is too short and has a weak specialty. The two primer will complement each other. - Analysis: Explain the reason for choosing B according to the characteristics of the primer and the requirements in the experiment. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
First, the answer to the multiple-choice question: 1. C 2. D 3. C 4. C 5. B 6. D 7. D 8. D 9. B 10. D 11. B 12. C 13. C 14. A 15. A Second, the answer to the multiple-choice question: 16. D 17. ABD 18. CD 19. BC 20. AC Third, non-multiple choice questions: 21. (9 points) - (1) Light intensity and CO2 concentration (2 points);Gs (2 points). - (2) Shady. Compared with natural light, under shade, the stomatal conductivity of the leaves increased slower, and Ci value decreased slightly (2 points) at the initial stage after illumination. - (3) The light reaction produced (H) and ATPincreased, and the dark reaction strengthened, but the stomatal conductivity was lower, and the intercellular CO2 concentration decreased; the stomatal conductivity increased, and the intercellular CO2 concentration gradually increased (2 points). 22. (15 points) - (1) Genetic mutation (2 points), one trait can be affected by multiple genes (2 points). - (2) 3 (2 points), 5 (2 points) - (3) In mutant 1, select the double balanced lethal line fruit fly and the wild type non-lethal line fruit fly to cross, and let the male and female fruit flies in F1 mate freely. (2 points) If F2 wild type: mutant = 6:1, then the mutant gene is located on the 2nd (1 point); if F2 wild type: mutant = 3:1, then the mutant gene is located on the other (1 point). <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Here are some questions and answers that might be involved in junior high school biology research: ** 1. Regarding nutrition ** 1. [Question: How do common meat dishes affect the development of brain intelligence?] - Answer: Different meat dishes contained different nutrients. For example, fish was rich in fatty acid, which was beneficial to the development and maintenance of brain cells and had a positive impact on the development of brain intelligence. Meat was rich in protein, iron, and other nutrients. The protein was the basic material that constituted the brain cells, and iron was an important raw material for the synthesis of hemoglobin. It ensured the oxygen supply of the brain and indirectly played a role in the development of the brain's intelligence. However, excessive consumption of meat dishes could lead to problems such as excessive fat intake, affecting health and possibly affecting brain function. 2. [Question: What is the relationship between nutrition and health for middle school students?] - ** Answer **: Middle school students are in a critical period of growth and development and need adequate nutrition. A reasonable intake of nutrients, such as sufficient calories to provide energy, protein to build body tissues, vitamins and minerals to maintain physiological functions, could ensure the normal development of the body, improve immunity, and enhance learning ability. If the nutrition is not balanced, there may be health problems such as leukemia (lack of iron, etc.), growth delay (lack of protein, etc.), and vision loss (lack of vitamins A, etc.). ** 2. About the environment and biology ** 1. [Question: How do common plants react to environmental changes?] - [Answer: Some plants are more sensitive to environmental changes.] For example, when the temperature rises, the growth cycle of some plants may be shortened, and the flowering and fruiting of some plants may be accelerated. In dry environments, plants will reduce water loss by closing their stomata, and their roots will grow deeper to find water. In terms of countermeasures, the environment could be improved by planting plants with strong adaptability to the environment. For example, plants that could absorb harmful gases could be planted in heavily polluted areas, and plants with developed roots could be planted in areas prone to soil erosion. 2. [Question: What are the effects of air pollution on crops?] - ** Answer **: Sulfur dioxide, nitrogen dioxide, and other gases in the air pollution may damage the leaf tissues of crops and affect photosynthesis. For example, after sulfur dioxide enters the leaves of plants, it will form sulfurous acid, destroy the photosynthesis, cause the leaves to lose their green color and turn yellow, and in serious cases, cause the leaves to die. In addition, air pollution could also affect the growth and development of crops, resulting in lower yield and quality. ** 3. Regarding physiological structure and function ** 1. [Question: Which layers of cells does oxygen pass through when it enters the blood?] - ** Answer **: It passed through the two layers of cells in the walls of the lungs and capillaries. 2. [Question: Which component in the blood will decrease after gas exchange?] - ** Answer **: Less carbon dioxide. 3. [Question: What kind of blood does the blood that flows through the capillaries around the lung become?] - [Answer: Arterial blood.] 4. [Question: Which substance does oxygen combine with after entering the blood?] - ** Answer **: Binding with hemogloblin. 5. [Question: What are the two parts of a gene?] - ** Answer **: It consists of two parts, DNA and protein. 6. [Question: What is the structure of DNA?] - ** Answer **: DNA has a double spiral structure. 7. [Question: How many genes are there in a strand of DNA?] - ** Answer **: There are many genes. 8. [Question: What does the gene control?] - ** Answer **: Genes control the traits of living things, but the traits of living things are also affected by the environment. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
There was not much information about the biology teachers of the No.1 Middle School. They only knew that Yuan Bo was the deputy director of the office of the No.1 Middle School. He had a bachelor's degree and was a first-level biology teacher of the high school. He was rated as "Excellent Tutor","Excellent Class Teacher of the Shan Tou City Tou", and "Expert of the Youth Position of the Shan Tou City Tou" in the Guangdong Province Middle School Biology League. There was also a recruitment information that showed that the school's recruitment of biology teachers required a master's degree and other conditions. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Biology answer for the 2022 Guangdong Province School Entrance Examination: First, multiple-choice questions 1. Fish will die soon after leaving the water. The reason is B. They can't breathe. 2. The following are mollusks (incomplete answers). As the reference materials did not give all the answers to the 2022 Guangdong junior high school geography and biology test paper, only part of the answer information could be provided. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
1. ** What is the structure of the glomerulus? ** - The glomerulus was formed by dozens of capillaries that were twisted and coiled by the glomerular artery. The other end gathered to form the glomerular artery, which was surrounded by the glomerular sac. It was a part of the nephron, and the nephron was the basic unit that formed urine. 2. ** What is the role of glomeruli in the formation of urine? ** - In the process of urine formation, the glomeruli and the walls of the kidney capsule have a filtering effect. Other than blood cells and large molecules of protein, all the plasma components could be filtered and formed into urine in the kidney vesicle. 3. ** How to evaluate the glomerular filtering capacity? ** - Clinically, the glomerular filtering rate was commonly used to evaluate the filtering capacity of the glomeruli. The normal value was 90, and the lower the value, the worse the kidney function. Genders, age, weight, and so on all affect glomerular filtration rate. For example, if a man's blood biochemistry was also 200 and his weight was also 70 kilograms, his glomerular filtration rate would be 47 for a 30-year-old, 43 for a 40-year-old, 39 for a 50-year-old, 34 for a 60-year-old, and 29.9 for a 70-year-old. When the glomerular filtration rate was <30, it was severe kidney failure. However, it was more accurate to measure the glomerular filtering rate using the electron capture electron microscope. 4. ** What's with glomerulonepathy? ** - Glomerulonepathy was a common kidney disease caused by damage to the glomeruli due to various reasons. It was a common kidney disease that showed partial or complete symptoms such as hematuria, proteinuria, edema, high blood pressure, decreased urine output or anuria, and abnormal kidney function. The most commonly seen in clinical practice was chronic glomerulonepathy, which was based on edema, hematuria, proteinuria, and high blood pressure. The disease progressed slowly, and there might be varying degrees of kidney function damage. Some patients might eventually develop end-stage kidney failure. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The following are some of the PEP edition high school biology elective one exercise questions and answers: - When making fruit wine, fruit vinegar, pickled vegetables, and fermented bean curd, the metabolism types of microorganisms used were facultative-anaerobic, aerobic, anaerobic, and aerobic. - The following experiment on DNA extraction and identification: - Yeast, cauliflower, and pig red blood cells could all be used to extract DNA. - DNA was easily dissolved in a solution of NaCl3, but not in an alcohol solution. - A solution containing a large amount of DNA could be obtained by adding an appropriate amount of distilled water to the chopped plant tissue. - After dissolving the extracted silk-like substance, it turned blue in the water bath after adding the diphenylethanamine reagent. - Description of practical applications of biotechnology: - Immobilized immobilized immobilized - Beef extract peptone medium was a commonly used basic culture medium for bacteria, which could be sterilized by high-pressure steam sterilization. - At the same pH. The washing effect of the washing powder with an enzyme was better than that of ordinary washing powder. - By adding cooled 95% alcohol to the chicken blood cells, one could obtain impurity-free DNA. - Regarding the production of fruit wine, vinegar, and fermented bean curd: - Fruits with high sugar content can be used to make fruit wine. - The fermented fruit vinegar was an airborne one. - When making fermented bean curd, it was necessary to prevent the growth of other microorganisms other than the fungus. - They were all final products that were obtained through the use of microorganisms. - Pectinase could catalyze pectins, making it easier to extract juice and increasing the juice yield of fruits. In order to investigate the effect of temperature on pectinase activity, a student added the same amount of pectinase into the same amount of apple puree at different temperatures. After the reaction time was the same, the reaction liquid was filtered for the same time. The volume of apple juice filtered out was measured with a measuring cylinder. (The specific curve that reflects the experimental results is not given here.) - The process of two types of protein with different relative molecular masses moving through the gel column (specific options are not given here). - The objective and result analysis of the immobilized yeast cell experiment: - The dry yeast cells were in a dormant state and needed to be activated in distilled water. - The preparation of the solution of alginic acid required a small fire or intermittent heating to prevent the alginic acid from being burnt. - If the concentration of the seaweed was too low, the gel beads would be white, and the number of fixed cells would be too small. - The length of time that the gel beads were soaked in CaCl2 solution affected the fermentation effect. - In the process of making fruit wine and fruit vinegar: - The grape juice could not be filled to the brim of the fermentation flask. There had to be some space left, or else the fermentation liquid would overflow. - During the wine fermentation process, there was no need to open the bottle cap to deflate in the early stage. If a large amount of gas was produced in the later stage, the bottle cap could be opened appropriately to deflate. - The temperature of the fruit wine fermentation process was controlled at 18 - 25°C, and the temperature of the fruit vinegar fermentation process was controlled at 30 - 35°C. - In the process of fruit vinegar fermentation, it was necessary to inflate the air through the inflating port in time to facilitate the metabolism of the vinegar bacteria. - Galactoside was not part of the cellulase component. The cellulase component included C-, C-, and beta-glucose. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
You can download the 15 sets of real questions and answers for the Teacher Qualification Examination Junior High School Biology (subject) on the Star Talent Network (<anno data-annotation-id ="0000004 - 4444 - 4000 - 8000 - 9000 - 80000000000"> Constellation Talent Network </anno>(<anno data-annotation-id ="3c0000000 - 4c64 - 4c00-a777777777766678"></anno>), which includes the real questions and answers for the Teacher Qualification Examination Junior High School Biology in the second half of 2020. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
There were seven types of error-prone questions in high school biology: 1. ** Concept error **: For example, an inaccurate understanding of biological concepts. 2. ** Comprehension error **: There is a deviation in the understanding of biological knowledge. 3. [Judgement error: A mistake in judging a biological phenomenon or principle.] 4. [Spatial position error: For example, misjudgment of the spatial position of the cell structure and organs in the body.] 5. ** Mistakes in experimental methods **: For example, the experimental design was unreasonable, the experimental steps were wrong, and the analysis of the experimental results was inaccurate. 6. ** Words, writing errors **: Mistakes in writing biological terms, inaccurate expressions, etc. when answering questions. 7. ** Question review error **: You did not accurately understand the requirements of the question and answered incorrectly. However, because there was no detailed information on the specific question types and answers, it was impossible to provide examples of answers for each error-prone question type. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
Here are some common questions and answers to them: ** 1. Physical basis of cells ** - Error-prone problem: Believes that all sugar is an energy source. - Answer: Wrong. Not all sugar was an energy substance. For example, Ribose and Deoxy Ribose were the components of DNA, and Cellulose was a component of plant cell walls. They could not be used as energy substances to supply energy to cells. - Error-prone question: Believing that macro elements are important in the body and micro elements are not. - Answer: Wrong. Although there were many large amounts of elements in the body, trace elements also played an irreplaceable role in the body. For example, iron was an important component of hemoglobin. A lack of iron would lead to iron-deficiency leukemia. ** 2. In terms of the protein and the protein ** - Error-prone problem: Believing that the protein only works inside the cell. - Answer: Wrong. The digestive system could be activated in the digestive tract. - Error-prone problem: It is believed that the relationship between the activity of the protein and temperature and pH. - Answer: Wrong. The relationship between the activity of the protein, temperature, and pH. The activity of the protein was the highest at the optimal temperature and pH. The activity of the protein would decrease when it was lower or higher than this value. However, the high temperature and the overly acidic or overly basic pH. ** 3. Photosynthesis and cell breathing ** - Wrong question: Think that plants only photosynthesize when there is light. - Answer: Wrong. Plants carried out photosynthesis and respiration at the same time when there was light, but the oxygen and organic matter produced by photosynthesis might be more than the oxygen and organic matter consumed by respiration. The overall performance was the release of oxygen and the accumulation of organic matter. - Wrong question: Think that only the mitochondria are the places of oxygen. - Answer: Wrong. The first stage of oxygen breathing was carried out in the cellular matrix. The glucose was broken down into gly ruvate and a small amount of [H], releasing a small amount of energy. The second stage was carried out in the mitochondria matrix. The reaction of gly ruvate and water produced carbon dioxide and a large amount of [H], releasing a small amount of energy. The third stage was carried out on the inner membrane of the mitochondria.[H] combined with oxygen to produce water, releasing a large amount of energy. ** 4. Meiosis and fertilisation ** - Error-prone problem: think that the same type of DNA only exists during the process of Meiosis. - Answer: Wrong. Homoeotic embryos also existed in the body cells. During the process of Meiosis, the homological embryos underwent special behaviors such as association and separation. - Wrong question: Think that fertilisation is just a simple combination of sperm and egg. - Answer: Wrong. During the process of fertilisation, the head of the sperm entered the egg cell while the tail remained outside. At the same time, the nucleus of the sperm fused with the nucleus of the egg cell. Half of the embryos in the fertilized egg came from the father and the other half came from the mother. In addition, almost all the genetic material in the fertilized egg came from the mother. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>