Reference answer to a compulsory subject: 1. 1. 1. (1) Living cells: A, D, G, I; (2) Dead cells: B, E; (3) cell products: c, f, h. (1) cell level (also individual level, because Escherichia coli is a unicellular organism); (2) population level; (3) Community level. 1.2 1 .b. Self-test: 1. 2.×。 3.√ Multiple choice questions 1. C 2。 D.3.b. The conjunctions represented by the three question marks in the concept map are: no, yes, yes. 2. 1 1.( 1)? √; (2)×? . 3. B.2.21.(1) √; (2)√。 2. A.3.B.2.31.(1) √; (2)√? ; (3)√? . 2. C.3.C.2.41.(1) √; (2)√。 2. C.3.C.4.C.5.C.2.5 1 c.2.A.3.D. Self-test:1. √.2.×。 3.×。 4.√。 5.×。 6.×. Multiple choice questions 1. A 2。 B.3.D.4. Answer. Draw the concept map II. Knowledge transfer is free water, combined water and free water. 3. 1 1. C.2.A.3.C.3.2 1. In the figure1,the "endoplasmic reticulum" in the annotation should be Golgi apparatus and Golgi apparatus should be endoplasmic reticulum. The position of chromatin marker line is wrong. The centrosome should also include centriole below the indicator line. In Figure 2, nucleoli should be chloroplasts, chloroplasts should be mitochondria, and ribosomes should be centrosomes. 2.c.3.b.4.c.3.31.(1) √; (2)×。 2. C.3.C. Self-test judgment question 1. 2.×. Multiple choice questions C. Connection questions 2. Knowledge transfer tips: related to the role of lysosomes. After cell death, the lysosomal membrane breaks, releasing various hydrolases and decomposing substances such as protein in cells. At this time, the meat of livestock and poultry is more tender after cooking. This process will take some time. Third, there are two ways to darken the field of vision: one is to turn the mirror to reduce the amount of light entering; The second is to choose a small aperture to reduce the amount of light entering. 4. 1 1.√。 2.√。 3.×。 4.2 1. Hint: The cell membrane is too thin to be seen under the optical microscope, and there was no electron microscope in19th century, so scholars had to start with the physiological function of the cell membrane. High school each grade courseware teaching plan exercises summary Chinese mathematics English physical chemistry II. Lipids and protein. 3. Hint: Both structural models think that the main substances that make up the cell membrane are lipids and protein, which is their similarity. The differences are as follows: (1) The fluid mosaic model suggests that the distribution of protein in the membrane is uneven, some span the whole lipid bilayer, some are partially or completely embedded in the lipid bilayer, and some are embedded in the inner and outer surfaces of the lipid bilayer; However, the three-layer structure model thinks that protein is evenly distributed on both sides of lipid bilayer. (2) The fluid mosaic model emphasizes that the molecules that make up the membrane are moving; However, the three-layer structure model thinks that biofilm is static structure. 4.d.4.3 1 .d.2. Answer. Answers and skills of self-test1. Concept test judgment 1. 2.×。 3.×。 4.×。 5.×. Multiple choice questions 1. D 2。 C. Draw the concept map 5. 1 (5) Practice the basic questions in the first section 1. Pasteur: Fermentation is related to living cells. Fermentation is the whole cell rather than some substances in the cell. Justus von Liebig: Fermentation is caused by some substances in cells, but these substances can only play a role after yeast cells die and rupture. Buchner: Some substances in yeast cells can continue to play a catalytic role after yeast cells are broken, just like in living yeast cells. Sumner: The enzyme is protein. 2. Hint: (1) There are thousands of chemical reactions in cells all the time. These chemical reactions need to be carried out efficiently, and the catalytic efficiency of enzymes is much higher than that of inorganic catalysts. (2) Chemical reactions in cells need to be carried out under mild conditions such as normal temperature, normal pressure and moderate pH, while inorganic catalysts often need the assistance of severe conditions such as high temperature, high pressure, strong acid and strong alkali to have high catalytic efficiency. 3.d. Exercise 1. In the second part. 2.b.3. Hint: In this model, A stands for an enzyme, B stands for reaction substrate, and C and D stand for reaction products. The significance of this model is that enzyme A and substrate B specifically combine to catalyze the reaction to produce products C and D. This model reveals the specificity of enzyme. Extended problem 1. (1) point a: with the increase of the concentration of the reaction substrate, the reaction speed is accelerated. Point B: At this time, the reaction rate reaches the highest. Point C: The reaction rate no longer increases with the increase of the concentration of the reaction substrate, but remains at a relatively stable level. (2) If the temperature at point A rises 10℃, the rising range of the curve becomes smaller. Because the original curve in the figure shows the change of catalytic rate with substrate concentration at the optimal temperature. When the temperature is higher or lower than the optimum temperature, the reaction speed will slow down. (3) The curve shows that the concentration of reaction substrate at point B is large enough, and the improvement of reaction rate is limited by the number of enzymes. At this time, adding a small amount of enzyme will speed up the reaction (figure omitted). 5.2 1 .b.5.31.c.2.b.3. Tip: The first stage of aerobic respiration and anaerobic respiration is exactly the same: no oxygen is needed; Has nothing to do with mitochondria. According to the structure of early earth and prokaryotic cells, it can be inferred that anaerobic respiration appeared first and then aerobic respiration appeared in the history of biological evolution, that is, aerobic respiration was formed by the development and change of anaerobic respiration. Prokaryotic cells appeared first, then eukaryotic cells appeared, that is, eukaryotic cells evolved from prokaryotic cells. 4. no. Because green plants can also form ATP in photosynthesis. 5.4 The first paragraph is 1. ( 1) ×; (2)√。 2. B.3. The conclusion is that chloroplasts mainly absorb red light and blue light for photosynthesis and release oxygen. Exercise 1 in the second quarter. ( 1) √; (2)×。 2. B.3.D.4.C.5.D.6.B.7. In photosynthesis, the energy source of light reaction stage is light energy, and that of dark reaction stage is ATP. 8. If the supply of carbon dioxide is suddenly interrupted during the day, the first substance accumulated in chloroplasts is pentacarbon compounds. Answers and skills of self-test 1. Concept test judgment 1. √.2.×。 3.√ Multiple choice questions 1. D 2。 D. Draw the concept map 6.1.1.c.2. A.3.D.4. Chinese, English.5.B.6.21.The number increases, but the chromosome number and stability are different. 2. C.6.31.(1) ×; (2)√; (3)×。 2. C.6.41.(1) √; (2)√。 2. The characteristics of cancer cells: cell proliferation is out of control and can proliferate indefinitely; The morphological structure of cells changed significantly; Easy to disperse and transfer in the body. Answers and skills of self-test 1. Concept test judgment 1. 2.√ Multiple choice questions 1. C 2。 D.3.Answer.4.Answer.5.C.6.Answer.2. Skill application 1. Prompt: (1) 2 ~ 21.3; (2)2 1.3~40.6。 2. Cell cycle:19.3h; ; Interval:17.3h; ; Division period: 2 h 3. Thinking development tips: Embryonic cell culture and adult nuclear transfer experiments were carried out on mammalian mice. The experimental design and expected experimental results are as follows, which proves that the higher the degree of differentiation of animal cells, the more limited their totipotency, but the nucleus is still totipotency. Step 1: isolate one cell of mouse 8-cell embryo, culture it to blastocyst stage, transplant it into the uterus of surrogate mother, and finally develop into a mouse; Step 2: isolate blastocyst cells and transplant them into the uterus of surrogate mothers. As a result, they cannot develop into mice. Step 3: isolate blastocyst cells, transplant their nuclei into enucleated eggs, and develop into mice; Step 4: Isolate mouse intestinal epithelial cells and transplant their nuclei into enucleated eggs to develop into mice. Compulsory 2 Chapter 4 4.11.tgcctagaa; UGCCUAGAA3; 3; Cysteine, leucine and glutamic acid. 2. C. Extension problem 1. Tip: You can write the changed codons separately, and then look up the codon table to see which amino acids the changed codons correspond to. This example shows that the degeneracy of password can prevent the change of genetic information caused by the change of base to some extent. 2. Hint: Because several codons may encode the same amino acid, some base sequences do not encode amino acids, such as termination codes, so we can only write a definite amino acid sequence according to the base sequence, but not according to the amino acid sequence. The transmission of genetic information was lost in this process. 4.2 (5) Skills training 1. It is pointed out that the development of wings needs an enzyme-catalyzed reaction, and the enzyme is synthesized under the guidance of genes, and the activity of the enzyme is affected by temperature, pH and other conditions. 2. Genes control biological traits, and the formation of traits is also influenced by the environment. (6) Exercise 1. A 2。 ( 1)×; (2)×; (3)√ Extended problem 1. Normal red-eye gene is a necessary but not sufficient condition for red-eye formation. When the red eye gene is normal and other genes involved in red eye formation are normal, the red eye of Drosophila can be formed; If the red eye gene is abnormal, even if all other genes involved in red eye formation are normal, the red eye of Drosophila can not be formed. Chapter IV Answers and Tips for Self-test 1. Concept test fill-in-the-blank questions DNA double strand 1 cgt2gcamrna gcatrna, amino acid alanine (codon GCA) multiple-choice questions1. D 2。 D.3. Answer.4. C. The hydrogen bond is broken (1); Helicase; Energy. (2)acu GAA； Transcription. (3)2。 (4) base complementary pairing. Draw a concept map 2. The combination of knowledge transfer ribosome, tRNA and mRNA is indispensable for the synthesis of protein. Antibiotics interfere with the formation of bacterial ribosomes or prevent the combination of tRNA and mRNA, interfere with the synthesis of bacterial protein, and inhibit the growth of bacteria. Therefore, antibiotics can be used to treat diseases caused by bacterial infections. Third, skill application 1. Tip: You can write the possible base code of each amino acid by consulting the codon table. 2. This method can only infer the possible base sequence, but can't write the definite base sequence. This method is simple and fast, and no experiment is needed. 3. It is presumed that gene sequencing cannot be replaced by sequencer. Because only a few possible base sequences can be inferred, but no definite base sequence can be obtained. Fourth, thinking development1.C. Chapter 5. 1 1. ( 1) √; (2)×; (3)×。 2. C.3.B.4. Answer. Extended question 1. The function of radiotherapy or chemotherapy is to interfere with the replication of DNA molecules in tumor cells and cancer cells through a certain amount of radiation or chemical agents, so as to cause gene mutation, thereby inhibiting their division ability or killing cancer cells. Radiotherapy rays or chemotherapeutic agents have effects on both cancer cells and normal somatic cells, so patients are very weak after radiotherapy or chemotherapy. 2. Sickle cell anemia patients have strong resistance to malaria, indicating that sickle cell mutation is beneficial to the survival of local people in malaria-prone areas. Although the homozygote of this mutant is not conducive to survival, its heterozygote is beneficial to the survival of local people. 5.2 1.( 1)×; (2)×。 2. B.3. The chromatid number of gametes in somatic cells belongs to octoploid extension of ploidy pea 14 7 2 1 diploid common wheat 422 163 hexaploid triticale 56 28 8 4.