1.(20 19 Ningxia Yinchuan Quality Inspection) The following description of some life phenomena and their biological significance is correct ()
A. Cyanobacteria can carry out photosynthesis, which is related to their chloroplasts.
B. Active transportation makes the concentration of substances inside and outside the cell membrane tend to be consistent, and maintains the normal metabolism of cells.
C. The vertical structure of plants in the forest creates habitat space and food conditions for animals.
D glucagon secreted by islet B cells can promote the decomposition of hepatic glycogen and maintain blood sugar stability.
Answer c
Analysis of cyanobacteria is prokaryote, with no chloroplast and errors; Active transportation is the transportation against the concentration gradient, which is generally used to maintain the concentration difference of substances inside and outside the cell membrane, and B error; The vertical structure of plants creates habitat space and food conditions for animals, C is correct; Glucagon is secreted by islet A cells.
2. During aerobic respiration, 2,4-dinitrophenol (DNP) can inhibit ATP synthesis, but has no effect on water production. The following statement is true ()
A.DNP will not affect the first stage of aerobic respiration.
B.DNP mainly plays a role in mitochondrial matrix.
C. When Clostridium DNP acts on muscle cells, the heat energy lost on the inner membrane of mitochondria will increase.
D.DNP can inhibit the glucose in human blood from entering red blood cells, thus inhibiting the process of aerobic respiration of cells.
Answer c
Analysis of the first stage of aerobic respiration can produce ATP, so DNP will affect the first stage of aerobic respiration, which is wrong; According to the meaning of the question, DNP mainly acts on the inner membrane of mitochondria, inhibiting ATP synthesis in the third stage of aerobic respiration, B error; According to the meaning of the question, when DNP acts on tissue cells, the enzymes in the mitochondrial inner membrane can not catalyze the formation of ATP, resulting in most energy loss in the form of heat energy, so the heat energy lost on the mitochondrial inner membrane will increase, and C is correct; The way glucose enters red blood cells is to assist diffusion without consuming ATP, so DNP has no effect on the process of glucose entering red blood cells. Red blood cells just breathe anaerobically, and D is wrong.
3.(20 19 Ermo, Loudi, Hunan) The defense mechanism of plants against herbivores and pathogenic microorganisms has been paid more and more attention. People found a tropical shrub. The caterpillar eats one of its leaves. Instead of eating nearby leaves, it bites leaves at a certain distance. Some people speculate that "deeply damaged blades may send some kind of chemical signal to nearby blades". The following statement is wrong ()
A. Pathogenic microorganisms can be viruses, prokaryotes and eukaryotes.
B. If this conjecture is true, the leaves that received the signal may have synthesized some chemical substance.
The defense mechanism of this plant is the result of gene directed mutation.
D. Removing newly bitten leaves and observing whether caterpillars "stay near and seek far" can be used to test this conjecture.
Answer c
The analysis of pathogenic microorganisms includes viruses, bacteria (prokaryotes) and fungi (eukaryotes). A is correct; The plant cells receiving the information have corresponding physiological reactions, which may be the synthesis of a chemical substance, and B is correct; Gene mutation has no direction, and the adaptive characteristics of organisms are the result of natural selection, C error; Removing the newly bitten leaves and cutting off the contact between the bitten leaves and the adjacent leaves can be used to verify that D is correct.
4. Some aphids living in the north of China live by sucking the sap of young branches and leaves of deciduous trees, as shown in the figure. Syrphids and ladybugs feed on aphids, and ants get honey dew from aphids to drive them away. The following statement is wrong ()
A. data sampling method of mathematical model in the figure
B before June, the aphid population had little resistance to the living environment.
C. The rapid decline of aphid population is mainly due to the increase of natural enemies.
D. There are four species and interspecific relationships among organisms.
Answer c
The activity range of aphids is very small, so the population density can be investigated by sampling method, so the data of mathematical model in the title map comes from sampling method, and A is correct; Before June, the number of aphids increased sharply, indicating that the environmental resistance of their population was very small, and B was correct; It is understood that aphids living in the north of China live by sucking the sap of young leaves of deciduous trees. After June, due to the lack of young branches and leaves of trees, the aphid population decreased rapidly due to the lack of food sources. C is wrong. According to the meaning of the question, aphids and trees are parasitic, syrphids and ladybugs are competitive, syrphids and aphids are predatory, and ants and aphids are mutually beneficial. D is correct.
5.(20 19 Xiamen Quality Inspection) The following picture shows the experiment of exploring DNA replication mode by using Escherichia coli, and the following statement is correct ().
Phage can be used instead of E.coli to carry out the above experiment.
B.(NH4)SO4 and (NH4)SO4 can be used to replace 15NH4Cl and 14NH4Cl respectively in the above experiments.
C. both DNA with b in test tube ③ contain 14N.
D. Comparing the results of test tubes ② and ③ alone cannot prove that the DNA replication pattern is semi-conservative.
Answer d
Analysis phage is a virus, which can't be cultured in ordinary culture medium, so it can't be used to replace Escherichia coli in the above experiment. A is wrong. S is not a characteristic element of DNA, so it is wrong to use compounds containing S instead of 15NH4Cl and14n4cl to carry out the above experiments. In test tube ③, both DNA in band A contain 14N, while one DNA in band B contains 14N and the other contains 15N, which is a C error. Comparing the results of tube ② and ③ only, it can't prove that DNA replication mode is semi-conservative replication and D is correct.
6. The genotype of animal (2n=6) is AaBbRrXTY, in which A and B genes are located on the same autosome and R and R genes are located on another pair of autosomes. One spermatogonia of this animal produces four spermatocytes through meiosis: A, B, C and D. A and B come from one secondary spermatocyte, and C and D come from another secondary spermatocyte. Among them, the genotype of A is AbRXTY, regardless of gene mutation and chromosome structure variation, the following judgment is correct ()
A.a contains five chromosomes.
The genotype of B. B. is ambiguous.
C c contains two autosomes.
The genotype of D.D. is either ABr or aBr.
Answer c
The genotype of animal (2n=6) is AaBbRrXTY. Because genes A and B are located on the same autosome, and the genotype of A is AbRXTY, cross-exchange of non-sister chromatids between homologous chromosomes will occur during meiosis of animals. If A and A are cross-exchanged, the genotypes of the two secondary spermatocytes are AabbRRXTXTYY and AaBBrr respectively, and the second meiosis continues. The former produces AbRXTY and b, while the latter produces c and d, namely ABr and aBr. If B and B are cross-exchanged, the genotypes of the two secondary spermatocytes are AABbRRXTXTYY and aaBbrr, and the second meiosis continues. The former produces A (AbXTY) and B (AbXTY), while the latter produces c and d, namely abr and aBr. Because the genes A and B are located on the same autosome, the genotype of A is AbRXTY, so the number of chromosomes changes during the formation of A, A contains four chromosomes, and A is wrong. The genotype of b is AbXTY or AbXTY, and b is wrong; The genotype of C is ABr or abr or aBr, which contains two autosomes, and C is correct; The genotype of d is ABr or abr or aBr, and D is wrong.
7.(20 19 Luoyang, Henan) CO2 is an important compound in biological and ecological environment. Answer the following related questions:
(1) The place where CO2 is produced in human cells is _ _ _ _ _ _ _ _.
(2) The CO2 compensation point of plants refers to the CO2 concentration in the environment when the photosynthetic rate is equal to the respiration rate due to the limitation of CO2. As we all know, the CO2 compensation point of Class A plants is larger than that of Class B plants. Normal A- plants and B- plants are placed in the same closed room and cultivated under suitable lighting conditions. When the net photosynthetic rate of plant A is 0, the net photosynthetic rate of plant B is 0. If the light intensity is properly reduced, the CO2 compensation point of a plant will become _ _ _ _ _ _ (fill in "big" or "small"). Because _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
_______________________________________________________________________________。
(3)CO2 is a regulator of life activities. Compared with neuroregulation, the regulation mode of CO2 participation has the characteristics of _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
(4)CO2 is the main form of carbon cycle between biological community and inorganic environment, but it is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Answer: (1) The compensation point that the mitochondrial matrix (2) is larger than the original CO2 is measured under suitable light intensity, so the photosynthetic rate decreases after the light intensity is appropriately reduced, and the CO2 concentration in the environment needs to be increased to make the photosynthetic rate equal to the respiratory rate.
(3) The route of action requires transport of body fluids, and the action time is relatively long; (4) Fossil fuels such as coal and oil.
Analysis (1) shows that the products of aerobic respiration of human cells are CO2 and H2O, and the products of anaerobic respiration are lactic acid, so only aerobic respiration can produce CO2. The physiological process of CO2 production is the second stage of aerobic respiration, and the place is mitochondrial matrix.
(2) The normally growing plants A and B are placed in the same closed room and cultured under suitable lighting conditions. It is known that the CO2 compensation point of A plant is greater than that of B plant, so when the net photosynthetic rate of A plant is 0 (that is, when it reaches the CO2 compensation point), B plant has exceeded the CO2 compensation point, so its net photosynthetic rate is greater than 0. If the light intensity is properly reduced, the photosynthetic rate of plants will decrease, and it is necessary to increase the CO2 concentration in the environment to make the photosynthetic rate equal to the respiration rate, so the CO2 compensation point of plants will become larger.
(3)CO2 participates in the regulation through body fluid regulation. Compared with neuromodulation, humoral regulation requires humoral transport and takes a long time.
(4) The main form of carbon cycle between biological community and inorganic environment is CO2. However, the excessive exploitation and utilization of fossil fuels such as coal and oil will release the carbon accumulated in the stratum in a short time and disrupt the carbon cycle balance in the biosphere.
8.(2065 438+9, Eight Cities in Guangxi in April) 20 18 Nobel Prize in Physiology or Medicine was awarded to James Allison and Tasuku Honjo who discovered immunotherapy for cancer cells. James Ellison found in mouse experiments that CTLA-4 protein on T cells can prevent T cells from attacking cancer cells. Therefore, this kind of protein is called "brake molecule". As long as CTLA-4 protein is inhibited by CTLA-4 antibody, T cells can be activated and continue to attack cancer cells. Answer the following questions:
In the process of (1)T cells attacking cancer cells, antigens on the surface of cancer cells stimulate the proliferation and differentiation of T cells, forming _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(2)CTLA-4 antibody inhibits CTLA-4 protein, which belongs to _ _ _ _ _ immune process in specific immune response, and it is _ _ _ _ cells that can produce this antibody.
(3) When the body fights against cancer cells, immune cells can't exert their maximum fighting power due to the "braking" effect of CTLA-4. If drugs release the "braking" effect of CTLA-4, immune cells can attack cancer cells with all their strength, but it may cause autoimmune diseases in the body. The reason is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
________________________________________________________________________________
_______________________________________________________________________________。
Answer (1) Effect T cell monitoring and clearance (2) Body fluid plasma (3) If drugs release the "braking" effect of CTLA- 4, immune cells will become overactive, which may attack normal somatic cells and cause autoimmune diseases.
Analysis showed that (1)T cells were stimulated by antigen, which made them proliferate and differentiate into effector T cells. Effector T cells are in close contact with cancer cells, causing them to crack and die, which embodies the monitoring and clearing function of immune system. (2) The process of antibody action belongs to humoral immunity, and antibody is synthesized and secreted by plasma cells. (3) If drugs release the "braking" effect of CTLA- 4, immune cells will become too active, which may attack normal somatic cells and cause autoimmune diseases.
9.(20 19 Shaanxi Senior Three Quality Inspection) Zhu was known as the "Oriental Pearl" and was once on the verge of extinction. In order to save wild crested ibis, governments and institutions at all levels in China have taken a series of measures to increase the number of wild crested ibis from 7 in 198 1 to more than 2,200 in 20 16. Answer the following questions:
(1) Under the ideal conditions of sufficient food and space and no natural enemies, the growth curve of crested ibis population is "_ _". And _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ and so on. The most effective protection measures for crested ibis are _ _ _ _ _.
(2) In order to investigate the population density of crested ibis and golden pheasant in a certain area, a classmate randomly set up several bird nets in this area. The statistics of capture results are as follows:
There are about _ _ _ _ crested ibis in this area, and the calculation result is _ _ _ _ _ _ (fill in "too big", "too small" or "unchanged"), because the probability that the marked individual will be captured again becomes smaller.
(3) The crested ibis is parasitized by some bacteria, and the researchers found that this kind of bacteria reproduces every 20 minutes. If the initial number of bacteria is N0, the number of bacteria can be expressed as _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
The answer is (1)J Environmental pollution, food shortage and habitat loss. In-situ protection (or establishment of nature reserves) (2)322 is too large (3)N9=N0×29.
Analysis (1) Under the ideal conditions of no natural enemies and sufficient food and space conditions, the population growth rate of crested ibis has remained unchanged, and the population number has increased in a "J" shape; Environmental pollution, food shortage and habitat loss are the reasons leading to a large number of wild crested ibis. The most effective measure to protect it is to protect it in situ, that is, to establish a nature reserve.
(2) To investigate the population density of crested ibis and golden pheasant, it is necessary to use the marked recapture method. Total population = (number of marked individuals × number of recaptured individuals) ÷ number of marked individuals in recapture, so there are about (46×42)÷6=322 crested ibis in this area. Because the marked individual is recaptured,
(3) It is known that this kind of bacteria produces a generation every 20 minutes, and the initial number of bacteria is N0, so a * * * produces 60÷20×3=9 (times) within 3 hours, so the number of bacteria after 3 hours can be expressed as N9=N0×29 (times) in the mathematical model.
10.(20 19 Chongqing Nankai middle school senior three April simulation) In the process of cultivating a certain crop (2n=42), haploid plants (2n- 1) are sometimes found, for example, a haploid plant is called haploid plant 6 because it lacks a chromosome 6 than a normal plant.
The variation type of haploid plant (1) No.6 is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.
(2) The researchers conducted a hybridization experiment with the haploid plant No.6, and the results are shown in the following table:
Monomer ♀ forms n- 1 gamete _ _ _ _ _ _ (fill in "greater than", "equal to" or "less than") N gamete during meiosis, because chromosome 6 cannot be lost during the first meiosis.
There are two varieties of this crop. A variety is disease-resistant but poor in other characters (the disease-resistant gene R is located on chromosome 6), while B variety is not disease-resistant but excellent in other characters. In order to obtain varieties with excellent disease resistance, the ideal breeding scheme is to cross the haploid plant No.6 of variety B with variety A, and select the haploid from its progeny. Then cross with _ _ _ _ _ _ _ _ _ _ for several generations in succession, and select a single plant with excellent disease resistance and other properties at a time, and finally make this single plant _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Answer (1) Chromosome variation (or chromosome number variation) Homologous chromosome or sister chromatid of chromosome 6 (2) Multigamete pairing (forming tetrad) (3) Monomer selfing of female parent variety B 6.
The analysis showed that (1) haploid plant No.6 lacked a chromosome 6 compared with normal plants, so the variation type of haploid plant No.6 was chromosome number variation, which was formed because the homologous chromosome or sister chromatid of chromosome 6 did not separate during meiosis of a parent.
(2) According to the table, among the progeny of haploid ♀ 6 and normal diploid ♀, haploid accounts for 75% and normal diploid accounts for 25%. Therefore, we know that in the process of meiosis, haploid ♀ forms more n- 1 gametes than N-type gametes. This is because chromosome 6 was lost during the first meiosis, because it cannot be associated and paired.
(3) Through cross breeding, excellent characters can be concentrated on one individual. According to the analysis in (2), the ideal breeding scheme is that more n- 1 gametes are formed in the process of meiosis of monomer, and the fertility of n- 1 male gametes is very low, so the monomer plant No.6 of variety B should be used as the female parent to cross with variety A and in its offspring.
Answer any questions of 1 1 and 12.
1 1. [Biotechnology Practice] (20 19 Quality Inspection in Qingdao, Shandong Province) The sewage pool of a chemical plant contains a harmful and refractory organic compound A, and the researchers successfully screened out the bacteria (target bacteria) that can degrade compound A efficiently by using the culture medium prepared from compound A, phosphate and magnesium salt. Please analyze and answer the following questions:
(1) This medium only allows the growth of bacteria that decompose compound A. This medium is called _ _ _ _ _ _ _ _. The culture medium should be sterilized by _ _ _ _ _ _ _ _ method before inoculation.
(2) Purifying and counting the target bacteria obtained by preliminary screening. Commonly used inoculation method is _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(3) dilute 1 mL water sample by 100 times, and apply 0. 1 mL diluent to three plates by coating method; After proper culture, the number of colonies on the three plates was 5 1, 50 and 49, respectively, from which it can be concluded that the number of live bacteria in each liter of water sample was _ _ _ _ _ _ _ _ _. In fact, the number of live bacteria per liter of solution is more than that obtained by statistics, because _ _ _ _ _ _ _ _.
_______________________________________________________________________________。
(4) Compared with common enzyme preparation, the advantages of immobilized enzyme technology are: _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
Answer (1) Choose high-pressure steam sterilization (2) Dilution coating plate method to check whether the medium plate sterilization is qualified (3) The statistical result of 5.0× 107 is expressed by the number of colonies. When two or more kinds of bacteria are connected together, only one colony is observed on the reusable embedding plate (4).
The analysis (1) medium is a selective medium, which only allows the bacteria that decompose compound A to survive by adding specific components. The culture medium is usually sterilized by high pressure steam sterilization.
(2) Countable inoculation method is dilution coating plate method. Before inoculation, several sterilized blank plates were randomly selected and cultured for a period of time to test whether the sterilization of culture medium plates was qualified.
(3) The number of live bacteria per liter of water = (51+50+49) ÷ 3 ÷ 0.100 ×103 = 5.0 ×107 (. When two or more bacteria are connected together, a colony is observed on the plate, which will lead to a lower statistical result than the actual one.
(4) Common enzyme preparations are easily inactivated by environmental influences and cannot be recycled. Mixing into the product after the reaction will also affect the quality of the product; The immobilized enzyme can be reused, which is beneficial to the purification of products. Cells are large and usually fixed by embedding.
12. [Special Topics of Modern Biotechnology] (20 19 Quality Inspection in Qingdao, Shandong Province) Catharanthus roseus is a wild flower native to the east coast of Africa, and its vinblastine has a good anti-tumor effect. The encoded product of gene tms can promote the synthesis of auxin. Researchers used gene tms to construct recombinant Ti plasmid and genetically modified callus, thus solving the problem that vinblastine was in short supply. The operation process is as follows. Please answer the following questions:
(1) Process ① refers to _ _ _ _ _ _ _, and Process ② adopts _ _ _ _ _ _ _.
(2) If the gene tms is obtained from a gene library, then the gene obtained from _ _ _ _ _ _ _ _ (fill in "genome library" or "cDNA library") contains a promoter whose function is _ _ _ _ _ _ _ _ _ _ _ _ _ _. As a vector, the Ti plasmid should contain _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(3) PCR was used to amplify the target gene. After heating denaturation, the DNA containing the target gene unwinds into a single strand, and _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _
(4) Vinblastine not only kills cancer cells, but also has toxic effects on normal cells. In order to reduce the toxicity of vinblastine to normal cells, it can be made into a "biological missile" by using the preparation technology of _ _ _ _ _ _ _ _ _, which can be used for targeted therapy.
Answer (1) Dedifferentiated Agrobacterium transformation method (2) Genome library drives target gene to transcribe mRNA marker gene (3) Primer (4) Monoclonal antibody.
Analysis (1) explants dedifferentiated to form callus, and the target gene was introduced into plant cells through Agrobacterium transformation.
(2) 2) The cDNA library is formed by RNA reverse transcription, and the genes obtained from it do not contain promoters, while the genes obtained from the genome library contain promoters. Promoter is the binding site of RNA polymerase and starts the transcription process. The vector needs to contain marker genes, which can be used to screen cells containing target genes.
(3) (3) The process of PCR amplification is as follows: the first step is to denature at high temperature to open the double strand; The second step is annealing, so that the primer is combined with the template chain; The third step is to extend the sub-chain. This process needs the participation of thermostable Taq enzyme.
(4) Monoclonal antibodies have strong specificity and can recognize specific antigens, so they can be made into "biological missiles" to directionally transport vinblastine to tumor sites to kill cancer cells, which can reduce the toxicity of vinblastine to normal cells.
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