The characteristics of high school biology genetics and evolution textbooks
The high school biology compulsory textbook," Heredity and Evolution ", had the following characteristics: - ** Organization system **: Different versions have differences. For example, the 2019 edition has a big change in the organization system and content compared to the 2004 edition. The 2019 edition of the second compulsory course was taught in the first year of high school. The overall high school biology textbook arrangement system was different from the previous edition in terms of the curriculum arrangement of each grade. This reflected the need to update the textbooks with the times. - ** In terms of content ** - ** Difficulty **: The knowledge learned in the second compulsory subject of high school biology has always been regarded as the most difficult part. The genetics part is even more difficult in the biology subject. Its academic performance has a greater impact on the overall biology score. - ** Comprehension requirements **: The most important thing in learning this textbook is to understand and not just memorize the knowledge points. After understanding the basic rules thoroughly, the problem can be solved easily. - ** Knowledge Richness **: Covers a wide range of genetics-related knowledge. For example, there are many changes in the content of the 2019 edition of the textbook, including the movement, removal, addition, and modification of the content. In terms of changes, for example, cell cancer was moved to the first section of compulsory chapter 5, etc. In terms of deleting, for example, the content of "Life activities cannot be separated from cells" in chapter 1 was deleted. In terms of adding, for example, the content of "Water is a good solute in cells" was added in chapter 2. In terms of changes, for example,"cyanacea" was changed to "cyanogen" and so on. These changes made the content of the teaching materials continuously optimized and improved, which was more in line with the requirements of modern biology for the construction of genetic and evolutionary knowledge systems. Read more exciting novels for free
What are the high school biology textbooks in Inner Mongolia?
In the autumn of 2023, the biology textbook used by Inner Mongolia High School was the 2019 new edition of the People's Education Department's high school biology textbook. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the methods to solve the high school biology genetics questions?
High School Biology Heredity and Evolution
The thought map of the second compulsory subject of high school biology,"Heredity and Evolution", mainly covered the following points: - ** The discovery of genetic factors **: - The reasons for the success of Mendel's pea crossbreeding experiment included the correct selection of experimental materials (pea self-pollinating, closed pollinating, easily distinguishable relative traits, etc.), the study of relative traits from one pair to many pairs, the analysis of experimental results, the use of hypothesis-deduction method, etc. - The related concepts included relative traits, trait separation, dominant and latent traits, allelic genes, dominant and latent genes, homo and heterozygy, and the combination of genetics and environment to form a trait. - Concepts such as crossing, self-crossing, and test-crossing (F1 hybrids with a recessive-type hybrid can test the F1 gene type). - ** Relationship between genes and genetics **: - Meiosis is related, including the concept of Meiosis (the division in which the number of hormones is halved when mature reproductive cells are produced, and the cell splits twice with each copy of the fetus). - The first division of the Meiosis Division (such as the characteristics of various stages such as the sperm cell in the seminidium, the interphase replication of the sperm, and the syndesis of the prophase) and the second division of the Meiosis Division (such as the characteristics of various stages such as the disorderly arrangement of the prophase). - ** Nature of Genes **:(The relevant content is not described in detail in the materials. It can be further supplemented according to the contents of the teaching materials, such as the exploration of DNA as the main hereditary material, etc.) - ** Gene expression **:(The relevant content is not described in detail in the materials. It can be further supplemented according to the content of the teaching materials, such as the process of transcribing and translation.) - ** Gene mutation and other mutations **:(The relevant content is not described in detail in the materials. It can be further supplemented according to the teaching materials, such as the concept, type, and significance of gene mutation, and other types of mutations such as aberrations, etc.) - ** From crossbreeding to genetic engineering **:(The relevant content is not described in detail in the materials. It can be further supplemented according to the content of the textbook, such as the principle, process, advantages and disadvantages of crossbreeding, and the operation tools and steps of genetic engineering, etc.) - ** Modern Biological Evolution Theory **:(The relevant content is not detailed in the materials. It can be further supplemented according to the teaching materials, such as the change of population gene frequency and biological evolution, isolation and speciation, etc.) <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the scientific experiments in junior high school biology textbooks?
The reference materials mentioned the experiments in junior high school biology textbooks such as " Exploring the Effects of environmental pollution on organisms "," Plant Growth "," Sorting According to the Characteristics of Living Things ", and making and observing the temporary mounting of onion scale leaf inner skin cells. However, it did not explicitly mention all the scientific experiments in junior high school biology textbooks, so it was impossible to answer them completely. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the calculation questions for high school biological genetics and evolution?
High school biological genetics and evolution calculation questions and answers
The following are some common calculation questions and answers in the high school biology book,"Heredity and Evolution." ** 1. Calculation of cell division ** 1. ** Meiosis of the sperm cell ** - For example, an animal's sperm cell can form four quadrants during Meiosis, so the number of embryos, chromatids, and DNA molecules that the secondary sperm cell may have are () - Answer: If the sperm cell can form four quadrants, it means that there are eight embryos in the body. The secondary sperm cell had half of the number of somatocells, namely four. In the prophase and the middle phase of the second division, there were eight Chromatids, which were twice the number of Chromatids. The number of DNA molecules was the same as the number of Chromatids, which was eight. Therefore, the answer was A. 4, 8, 8. 2. ** Calculation of the number of nuclei in different stages ** - For example, it is known that there are 7 embryos in the pea egg cell. Then, the number of embryos in the cell during the anaphase of mitosis, the anaphase of the first division of Meiosis, and the anaphase of the second division of Meiosis is () - Answer: Pea has 14 somacells. In the anaphase of mitosis, there were 28 doubling of the total number of mitotic nuclei. In the anaphase of the first division, there were 14 doubling of the total number of mitotic nuclei. In the anaphase of the second division, there were 14 doubling of the total number of mitotic nuclei. The answers were D. 28, 14, 14. 3. **DNA Marks and Cell Division ** - For example, the double strands of DNA molecules of corn somatic cells (containing 20 embryos) were labeled with 32P, and then these cells were transferred to a medium without 32P for cultivation. At the middle and late stages of the second cell division, the number of embryos in a cell and the number of embryos labeled with 32P were respectively () - Answer: Due to the semi-conservative replication of DNA, in the middle stage of the second cell division, the number of microorganisms was 20, and one strand of the DNA molecules on each of the microorganisms was labeled. Therefore, the number of microorganisms labeled by 32P was 20. In the anaphase, the number of microorganisms became 40, but the number of microorganisms labeled by 32P was still 20. The answer was A. 20 and 20 in the middle stage, 40 and 20 in the anaphase. 4. ** Meiosis with marker DNA ** - For example, if the DNA molecules of a sperm cell containing a pair of homological embryos are labeled with 15N and supplied with 14N-containing materials, the ratio of 15N-labeled sperm among the four sperm produced by the cell during Meiosis is () - Answer: Since DNA replication is semi-conservative replication, a single sperm cell contains a pair of homolog chromomes. After replication, the two sister chromatids of each of the primary sperm cells contain 15N. Among the four sperm formed by Meiosis, each sperm contains 15N. Therefore, the proportion of sperm containing 15N is 100%. The answer is D. 5. ** Type of gametes produced by the sperm cell ** - For example: Droplets have eight embryos, and the type of gametes produced by the Meiosis of a sperm cell is () - Answer: A single sperm cell would produce two types of four gametes during the Meiosis. The answer was A.2. 6. ** Cell division after fusion of neurospora crassa ** - For example, the haploid cell of the neurospora crassa has seven nuclei. Two different types of neurospora crassa become diploids after fusion, and then a typical Meiosis occurs, followed by a mitosis. The final number of daughter cells and the number of embryos in each daughter cell are () - Answer: After the two haploids fuse into a diploid, the number of microorganisms will be 14. After Meiosis, four daughter cells will be formed, and each daughter cell will have seven microorganisms. After another mitosis, the number of daughter cells will become eight, and the number of microorganisms in each daughter cell will still be seven. The answer is A. Eight, seven. ** 2. Calculation of Inheritance Law ** 1. ** Genotype-ratio of offspring of multiple gene hybrids ** - For example, if sunflower hybrids with AaBbCcdd and AabbCcDd are crossed, according to the law of free gene combination, the proportion of individuals with AabBccDd in the offspring should be () - The answer was to calculate the ratio of each pair of genes to produce a specific gene type. The ratio of Aa× AAto AAis 1/2; the ratio of Bb×Bb to BBis 1/4; the ratio of Cc×Cc to CCis 1/4; and the ratio of DD×Dd to Dd is 1/2. Multiplying these proportions, 1/2×1/4 ×1/4×1/2 = 1/64, the answer is D. 1/64. 2. ** Analysis of the proportion of the crossbred offsprings ** - For example, the red fruit of tomato is dominant to the yellow fruit, and the round fruit is dominant to the long fruit, and they are freely combined. Now, the red long fruit and the yellow round fruit tomato are crossed. Theoretically, the proportion of the offspring's genetics that cannot appear is () - Answer: According to the law of free combination of genes, analyze the proportion of the genes in the offspring of different parent combinations. The ratio of 1:0 may appear when a pair of genes are crossed with a pair of genes. 1:1 may appear when a pair of genes are tested. 1:1:1:1 may appear when a pair of genes are tested. 1:2:1 is the ratio of the genes of the self-bred offspring of a pair of genes. The answer is B. 1:2:1. 3. ** Genetic pedigree calculation ** - For example, the picture on the right is the genetic pedigree of an albino family. Please estimate the probability of the couple II-2 and II-3 having an albino child () - Answer: First of all, albinism is an autosomal-type hereditary disease. Suppose the pathogenic gene is A and the normal gene is A. According to the pedigree, the II-2 gene type was 1/3AA or 2/3Aa, and the II-3 gene type was Aa. The probability of them having an albino child (aa) is 2/3×1/4 = 1/6. (The original answer here is wrong. This result is obtained according to the normal calculation process.) 4. ** Calculation of Homozyte Progeny of Pea Self-cross ** - For example, there is a green (yy) round (Rr) pea. Their relative characteristics are yellow and wrinkled. It was known that these two pairs of genes were located on two pairs of homolog genes. The peas were planted and self-pollinated, and the offspring were all planted without selection. They were self-pollinated again and harvested n seeds. It can be inferred that the number of green and round grains in the n grains is about () - Answer: Green (yy) self-bred offspring are all green, round (Rr) self-bred offspring account for 1/4. The first generation of the second generation was obtained by self-crossing. The ratio of green circles (yyRV) was 1×1/4 = 1/4, so there were about n/4 green and round grains in the n grains. The answer was C.n/4. 5. ** Calculating the ratio of the crossbred offspring of two pairs of relative traits ** - For example, the tall and short stems of rice are a pair of relative traits, while the glutinous and non-glutinous traits are a pair of relative traits. Some people let a high-stem non-glutinous rice cross with another short-stem non-glutinous rice, and the resulting offspring are as shown in the right picture (these two pairs of traits are inherited according to the law of free combination). Among the short-stem non-glutinous rice of the offspring, the rice that can be stably inherited accounts for () - Answer: First, determine the parent's gene type according to the ratio of the hybrids, and then analyze the ratio of the dwarf non-glutinous individuals. Assuming tall stem (D), short stem (d), glutinous (R), and non-glutinous (r), the parent genomes were deduced as DdRr and ddRr according to the results of the crossing. The ratio of ddR to ddR was 1/3, and the answer was C. 1/3. 6. ** Calculating the probability of two genetic diseases ** - For example, human albinism is controlled by the autosomal-like gene (a), while hemophobia is controlled by the X-like gene (h). It is known that the parents of a family are normal, and a child is born with albinism and hemophobia. I would like to ask: the probability of only suffering from albinism and the probability of suffering from both diseases. - The answer was to first determine the parents 'genetics. Let albinosis-related genes A and a, hemophilia-related genes H and h. Because the parents were normal and the child was albino and had hemophobia (aaXhY), the parents 'genetics were AaXHY and AaXHXh. The probability of only suffering from albinism: first calculate the probability of suffering from albinism as 1/4, the probability of not suffering from hemophobia as 1 - 1/4 = 3/4, the two multiplied by 1/4×3/4 = 3/16; the probability of suffering from both diseases: the probability of suffering from albinism as 1/4, the probability of suffering from hemophobia as 1/4, the two multiplied by 1/4×1/4 = 1/16. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
High school biological genetics and evolution knowledge experience and harvest perception composition
The characteristics of secondary reaction in high school biology
The secondary immune response could produce a large number of antibodies very quickly (with the presence of memory cells), while the primary immune response was slower. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
What are the characteristics of the satirical novels in junior high school textbooks?
I recommend " Falling in Love Even if I'm Sick in Senior Two ", a novel about urban life written by Stepping on Yan Fei. The male protagonist hated women because of his dark history. The class teacher arranged for two girls to sit at the same table. One was pestering and the other was as cold as ice. It was the beginning of a tragic school life. It was a bit like " There's Something Wrong with My Youth Love Story." The first half was good, but the second half was his own experience. It was considered a light novel with a funny plot. The ending was a little weird with a female teacher. The overall story was refreshing and not melodramatic. It could bring back memories. " What a Secondary Two Teacher Saw and Heard " was not bad either. It was a light novel that sang with the heart. It was super funny that a single middle school student had the ability to become a middle school teacher. There were all kinds of jokes. The author was a newcomer with more than 100 chapters. Although it seemed like he was going to be a eunuch, he had a lot of humor and imagination. The Sunshine of the First Day was a modern romance novel written by Lei Dewen. The female protagonist returned to the 1992 New Year's Day. She was at a loss when faced with algebra textbooks and little wimps. She also had a crush on someone and other problems. It was a new book by Little Poochie. She had just read it, but it felt good. Her soul felt like it had been cleansed. " The Rippling Years of High School " was a young school novel written by Yao Chu. When she returned to high school and looked at her classmates 'young faces, she felt that all her heroic words were empty. Her writing style was good and she felt comfortable reading them. " Myth: The Heavenly Emperor Is Not Working Today " was a fantasy novel written by Death 321. The background of Indian mythology was that the main character had transmigrated to become the Heavenly Emperor, Indra. He did not want to be beaten up, so he cultivated bitterly to change his fate. This was a novel about Indian mythology. It was humorous and popular. The style was relaxed. Those who liked Indian mythology and primitive literature could read it. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>
The Teaching Design Case of Mutation Breeding and Evolution in High School Biology
The following is a teaching design case of high school biological mutation breeding and evolution: ##1. Teaching objectives 1. ** Knowledge target ** - Students will be able to understand the types, structures, and characteristics of genetic mutations, the types of aberrations, and their applications in breeding, as well as the principles and characteristics of various breeding methods. - Master the main points of modern biological evolution theory and understand the comparison between natural selection theory and modern biological evolution theory. 2. ** Ability Target ** - Able to apply the breeding process to the experimental design of biological breeding. - He learned how to build a knowledge network and form his own knowledge system with the knowledge of mutation breeding and evolution. 3. ** Emotional goal ** - Through understanding the application of mutation breeding and evolution knowledge in life, students could experience the practicality of biology and stimulate their interest in biology. ##2. Difficulties in Teaching 1. ** Main point ** - Distinguish three types of hereditary mutations (gene mutation, gene reorganization, and aberrations). - Master the breeding process and be able to carry out relevant experimental design. - Deep understanding of modern biological evolution theory. 2. ** Difficulty ** - Understand the essential differences and connections between the three types of hereditary mutations. - The complicated links in the breeding process and the construction of experimental design ideas. - In modern biological evolution theory, there were abstract concepts such as the calculation of gene frequency. ##3. Teaching Method Teaching method, discussion method, question guidance method, and case analysis method. ##4. Teaching process ###(1) Introduction (5 minutes) By showing some pictures of creatures with obvious differences in traits (such as flowers of different colors, fruits of different shapes, etc.), ask the students how these differences in traits arise, guide the students to think about the concept of mutation, and then lead to the topic of mutation breeding and evolution in this lesson. ###(2) Knowledge explanation (25 minutes) 1. ** Mutated Part ** - gene mutation - Explain the molecular mechanism of genetic mutation, including mitosis, protocaryote fission, and viral DNA replication. It emphasized that gene mutation did not change the number of genes, but only the structure of genes. It explained the impact of gene mutation on traits through examples (such as sickle cell leukemia), including changing biological traits and not changing biological traits. At the same time, the characteristics of gene mutation and the relationship between its advantages and disadvantages and the environment were described. - Illustrate the types of genetic mutations (dominant and latent) and their patterns of expression. - genetic recombination - This paper introduced the concept of gene reorganization from both narrow and broad perspectives, focusing on the free combination of non-allele-like genes on non-homolog and the cross-exchange of non-sister chromatids on homolog during the process of Meiosis. The process of gene reorganization was analyzed through cell division diagrams. Students were asked to discuss the differences between gene reorganization and gene mutation in producing new traits. - chromosome variation - They introduced the types of structural aberrations (loss, repetition, reversal, and translocations) and numerical aberrations (euploids and aneuploids). Through examples (such as the polyploids of some plants), explain the application principle of mutation in breeding. For example, polyploids can increase crop yield and improve quality. 2. ** Breeding Part ** - The more common breeding methods (mutation breeding, cross breeding, haploid breeding, polyploid breeding, genetic engineering breeding, etc.) were explained in terms of breeding principles, methods, advantages, disadvantages, and scope of application. For example, mutation breeding could produce new genes and create new types of mutations, but the frequency of mutation was low and the direction was difficult to control. Crossbreeding could not produce new genes, but it was only a combination of original genes, but the operation was relatively simple. - Give some practical breeding problems, let the students discuss in groups and choose the appropriate breeding method, such as which breeding method should be chosen to obtain a pure spore in a short period of time, etc., to train the students 'ability to use knowledge to solve practical problems. 3. ** Evolution Part ** - This paper introduced the main contents of Darwinian theory of natural selection (over-reproduction, struggle for survival, genetic variation, survival of the fittest), and analyzed its contributions and limitations. - The main points of modern biological evolution theory were elaborated, including population as the basic unit of biological evolution, the concept of gene frequency and its calculation method, the raw materials for evolution produced by mutation and gene reorganization, the direction of biological evolution determined by natural selection, isolation as a necessary condition for speciation, etc. Help students understand modern evolutionary theories through examples, such as the formation of different ground finches on the Galapagos Islands. ###(3) Class interaction (15 minutes) 1. ** Group discussion ** - Give some controversial questions, such as "Is GM food safe?" and let the students discuss in groups. Each group would elect a representative to explain the group's views, and the other groups could ask questions and refute. The teacher patrolled the students during the discussion and guided the students to think about the problem from the perspective of mutation breeding and evolution. 2. ** Question and Answer ** - Students were encouraged to ask questions that they had encountered in the learning process. Teachers would provide detailed answers to students 'questions and focus on some common problems, such as common errors in gene frequency calculation. ###(4) Class summary (10 minutes) 1. Guide the students to review the main knowledge points of mutation breeding and evolution learned in this lesson, including the types of mutation, the comparison of breeding methods, the main points of modern biological evolution theory, etc. 2. Students were required to construct a simple knowledge network, linking the knowledge of mutation, breeding, and evolution. Teachers would select some of the students 'works to display and comment on, further strengthening the students' overall grasp of knowledge. ###(5) Class Practice (10 minutes) Some practice questions related to the content of this lesson were distributed, including multiple-choice questions, fill-in-the-blanks questions, and short-answer questions. Students could complete them independently in class. The practice questions should cover knowledge points such as gene mutation, gene reorganization, aberration variation, breeding method selection, modern biological evolution theory, etc. Through practice, the students should consolidate the knowledge they have learned in time. The teacher should check and feedback the students 'practice situation, and provide individual guidance for the students' problems. <a href="/?from=ask_words" style="color:red" target="_blank">Read more exciting novels for free</a>