1. About the description of cells, the mistake is that
The plasmodesmata of plant cells has the function of material transport.
B. Adhesion between animal cells is related to glycoprotein on cell membrane.
The energy released by C.C.ATP hydrolysis can be used for energy absorption reaction in cells.
Mammal cells can synthesize sucrose and lactose.
Answer d
Sucrose is synthesized by plant cells, not by animal cells.
2. The nerve cells and muscle cells of the same animal are different in function, and the main reasons for this difference are as follows.
A. They are in different cell cycles. They synthesize different specific protein.
C. They contain different genomes. Their nuclear DNA is replicated in different ways.
Answer b
Analysis of nerve cells and muscle cells are highly differentiated, usually no longer dividing, no cell cycle, no DNA replication; Due to the selective expression of genetic genes and the synthesis of different protein, different kinds of cells have been produced. Somatic cells of the same individual are divided and differentiated from the same fertilized egg, and their genomes are the same.
3. The statement about the generation and reflux of interstitial fluid under normal circumstances is wrong.
A. The oxygen content in the generated interstitial fluid is equal to the oxygen content in the returned interstitial fluid.
B. The tissue fluid continuously generates reflux and keeps dynamic balance.
Some substances in plasma enter interstitial fluid through the arterial end of capillary.
D. Some substances in the interstitial fluid enter the blood through the venous end of capillaries.
Answer a
Analysis Generally speaking, oxygen-enriched plasma oozes from the arterial end of capillary to generate interstitial fluid, and tissue cells living in interstitial fluid consume part of oxygen through aerobic respiration, which leads to the reflux of interstitial fluid from the venous end of capillary, and the oxygen content is usually low.
4. Cut the corolla of the plant into thin strips with the same size and shape, and divide them into groups A, B, C, D, E and F (the number of thin strips in each group is equal). Soak the six groups of thin strips in sucrose solutions with different concentrations for the same time, and then measure the length of each group of thin strips. The result is shown in the figure. If there is only water exchange between sucrose solution and corolla cells, then
A. after the experiment, the solute concentration in the vacuole of group a was higher than that of group B.
B. the vacuole water loss caused by soaking in group f was less than that in group B.
C. the ATP consumption of cells in group a is greater than that in group B.
D. The sucrose concentration of equal-length strips before and after soaking is between 0.4 and 0.5 mol L-1.
Answer d
When the ratio of the length before the experiment to the length after the experiment is 1, water enters and exits the cell in balance; The ratio less than 1 indicates that cells absorb water, and the smaller the ratio, the more water the corolla absorbs. The ratio greater than 1 indicates that the cell loses water, and the greater the ratio, the more the corolla loses water. According to the figure, it can be inferred that group A absorbed more water than group B, so the solute concentration in the cell fluid of group A was lower than that of group B after the experiment. Group f is bigger than group b, so the water loss is bigger than group b; The way of water molecules entering and leaving the cell is free diffusion without consuming energy; Group c absorbed water, while group d lost water, and the concentration of thin cell fluid was 0.4 ~ 0.5mol ~ L- 1.
5. About the description of nucleic acid, the mistake is that
A.RNA polymerase participates in the transcription process in the nucleus.
DNA replication can occur in mitochondria and chloroplasts of plant cells.
C. Phosphoric acid and ribose on one strand of a double-stranded DNA molecule are connected by hydrogen bonds.
D. The distribution of DNA and RNA in cells can be observed by staining with methyl green and Pyrone Red.
Answer c
Analytic transcription mainly occurs in the nucleus and needs the catalysis of RNA polymerase. Both mitochondria and chloroplasts of plant cells contain DNA, which can be replicated. The five-carbon sugar contained in DNA molecule is deoxyribose, and the phosphate and ribose in deoxynucleotide chain are connected by phosphodiester bond; The mixed use of methyl green pyron red can stain cells, methyl green can make DNA appear green, and pyron red can make RNA appear red.
6. The description of photosynthesis and respiration is wrong.
A. phosphoric acid is the reactant needed to synthesize ATP in photoreaction.
B chlorophyll does not need enzymes to absorb light energy in photosynthesis.
C. The energy needed by the human body during strenuous exercise is provided by the decomposition of lactic acid.
D. The replication of viral nucleic acid needs the respiratory function of the host cell to provide energy.
Answer c
Analyze the photoreaction and synthesize ATP； from ADP and phosphoric acid; Chlorophyll absorbs light energy without the participation of enzymes; During strenuous exercise, anaerobic respiration and aerobic respiration are carried out at the same time, and lactic acid produced by anaerobic respiration can no longer be decomposed for energy supply in human body; The virus has no cell structure, and the energy required for its nucleic acid replication comes from the respiration of the host cell.
29.( 10)
The change trend of plant net photosynthetic rate is shown in the figure.
Answer the following questions according to the pictures:
(1) When CO2 concentration is a, the net photosynthetic rate of plants under high light intensity is ①. When the CO2 concentration is between A and B, curve ② shows that the net photosynthetic rate increases with the increase of CO2 concentration.
(2) When the concentration of 2)CO2 is greater than C, the net photosynthetic rate represented by curves B and C will not increase, and the environmental factor limiting its increase is ③.
(3) When the concentration of CO2 in the environment is less than a, the amount of CO2 produced by plant respiration (fill in "greater than", "equal to" or "less than") and the amount of CO2 absorbed by photosynthesis under the three light intensities shown in the figure.
(4) According to the figure, it can be inferred that in the greenhouse, if measures are to be taken to increase the CO2 concentration to improve the yield of this plant, the influence of this factor should also be considered and corresponding measures should be taken.
answer
( 1)①0 ②A、B、C
(2)③ light intensity
③ ④ greater than
④ ⑤ light intensity
analyse
(1) According to the figure, when the CO2 concentration is a, the ordinate (curve A) under high light intensity is 0, that is, the net photosynthetic rate is 0; When the CO2 concentration is between A and B, curves A, B and C all show an increase, that is, the net photosynthetic rate increases with the increase of CO2 concentration.
(2) When the CO2 concentration is greater than C, the net photosynthetic rate (curve A) can still increase with the increase of CO2 concentration under high light intensity. It can be seen that the environmental factor limiting the increase of net photosynthetic rate of B and C is light intensity.
(3) When the concentration of CO2 is less than a, the net photosynthetic rate is less than 0 under the three light intensities, that is, the respiration rate is greater than the photosynthetic rate, which means that the amount of CO2 produced by respiration is greater than that absorbed by photosynthesis.
(4) According to the figure, CO2 concentration and light intensity will affect the net photosynthetic rate, thus affecting the plant yield. Therefore, in order to improve plant yield, the effects of CO2 concentration and light intensity on plants should be considered comprehensively.
30.(9 points)
In order to explore whether the curative effect of compound Chinese herbal medicine on bacterial mastitis is related to the enhancement of immune function, the research group randomly divided the model mice of bacterial mastitis into experimental group (Chinese herbal medicine gavage), blank control group (distilled water gavage) and positive control group (immune enhancer A gavage), and detected the immune indexes.
Answer the following questions:
(1) It was found that the phagocytic capacity of phagocytes in the experimental group was significantly higher than that in the positive control group and significantly higher than that in the blank control group. This result at least shows that this herb enhances the nonspecific immune function of mice. The characteristics of nonspecific immunity are ①.
(2) The study also found that the content of T cells in the experimental group was significantly higher than that in the blank control group, similar to that in the positive control group. This result shows that the drug may enhance the specific immune function of mice by increasing the content of T cells. Usually, in the process of cellular immunity, the role of effector T cells is ②.
(3) In specific immunity, T cells can produce factor ③, and under the action of this factor, cells stimulated by antigens can proliferate and differentiate into plasma cells, producing factor ⑤, and participate in the process of humoral immunity.
answer
(1) (1) Organisms are born with no specific pathogen, but have a certain defense function against many pathogens.
(2) (2) Identify that close contact with host cells invaded by pathogens can cause them to crack and die.
③ Lymphatic 4B ⑤ antibody
analyse
(1) Non-specific immunity is innate. It does not target a specific pathogen, but has a defensive effect on many pathogens.
(2) In the process of cellular immunity, effector T cells can recognize and bind cells (target cells) infected by pathogens, causing them to lyse and apoptosis.
(3) During humoral immunity, B cells stimulated by antigen proliferate and differentiate to produce plasma cells and memory cells under the action of lymphatic factor secreted by T cells, and plasma cells secrete antibodies, which specifically bind to the stimulated antigen.
3 1.(9 points)
In the terrestrial ecosystem, there are only five populations except decomposers: A, B, C, D and E. According to the investigation, the ecosystem has four trophic levels, and the energy transfer efficiency between trophic levels is 10% ~ 20%, and each population is only in one trophic level. The energy values of each group input in a year are shown in the following table, and the units of energy values in the table are the same.
Methyl ethyl propyl butyl pentyl
Energy 3.5612.8010.30 0.48 226.50
Answer the following questions:
(1) Please draw a food web in this ecosystem.
(2) The interspecific relationship between A and B is ①; Population D is composed of organisms in this ecosystem.
(3) Generally speaking, the main functions of an ecosystem include ③ and ④ besides information transmission. Carbon is of great significance to organisms and ecosystems, and the cycle of carbon between ⑤ and ⑤ mainly exists in the form of CO2.
answer
( 1)
(2)① plundering ② consumers
(3) Material circulation, energy flow, biological community and inorganic environment.
analyse
(1) According to the stem, we can know that: 1, the energy flow efficiency between trophic levels is10% ~ 20%; 2. Each population is in only one trophic level. E occupies the most energy and should belong to the first nutrition level; The energy values of B and C are in the same order of magnitude, and their sum (23. 1) is between 10% and 20% (10.20%) of the energy value of E, so B and C should belong to the second trophic level. The energy value of A is between 10% and 20% (15.4%) of the second trophic level, which should belong to the third trophic level. The energy value of D is between 10% and 20% (13.48%) of the third trophic level, which should belong to the fourth trophic level.
(2) According to the food web of (1), it can be inferred that the relationship between species A and B is predatory; It has been explained that the listed data does not include decomposers, while E belongs to the first trophic level and is a producer, so other organisms including D are consumers.
(3) Material circulation, energy flow and information transmission are the three functions of the ecosystem; Carbon circulates between biological community and inorganic environment in the form of CO2.
32.( 1 1)
The goat sex determination method is XY type. The genealogy chart below shows the inheritance of a goat trait, and the dark color in the chart shows the executor of this trait.
It is known that this trait is controlled by a pair of alleles. Regardless of chromosome variation and gene mutation, answer the following questions:
(1) According to the genealogy, this trait is ① (fill in "recessive" or "dominant").
(2) Assuming that the gene controlling the trait is located on the Y chromosome, according to the genetic law of the gene on the Y chromosome, the individual whose third representative type does not conform to the genetic law of the gene is ② (fill in the individual number).
(3) If the gene controlling the trait is only located on the X chromosome, the individual who must be heterozygote in the pedigree map is ③ (fill in the individual number), and the individual who may be heterozygote is ④ (fill in the individual number).
answer
(1)① recessive
(2) ② Ⅲ-1,Ⅲ-3 and Ⅲ-4
(3)③Ⅰ-2、Ⅱ-2、Ⅱ-4 ④Ⅲ-2
analyse
(1)① According to the pedigree chart, II- 1 and II-2 did not show this trait, but their offspring III- 1 did, so it was inferred that this trait was recessive, and II- 1 and II-2 were carriers of this trait.
(2) Assuming that the gene of the trait is located on the Y chromosome and the trait is transmitted by the Y chromosome, then all male offspring and male parents who express the trait will express the trait, while female individuals will not express the trait. According to this theory, the traits of the third generation of individuals should be: III- 1 does not show this trait (because II- 1 does not show this trait), III-2 and III-3 do not show this trait (because they are females), and III-4 shows this trait (because II-3 shows this trait). Combined with the genealogy, it can be seen that the individuals who do not conform to the genetic law of this gene are ⅲ- 1, ⅲ-3 and ⅲ-4.
(3) Assuming that the gene controlling the trait is only located on the X chromosome, (1) shows that the trait is recessive (for convenience of analysis, it is assumed that the trait is controlled by gene A-a); I-2 and II-2 did not show this trait (XAX-), but their male offspring (II-3 and III-1) did. At the same time, considering that the X chromosome of male individuals comes from female parents, it is concluded that I-2 and II-2 must be heterozygotes (XAXa). ⅲ-3 shows this trait (XaXa), and one X chromosome contained in it must be derived from ⅱ-4 (XAX-). At the same time, III-3 does not show this trait, so II-4 is a hybrid (XAXa). ⅲ-2 shows an inappropriate trait (XAX-), and its female parent ⅱ-2 is heterozygote (XAXA), so ⅲ-2 can be heterozygote (XAXa) or homozygote (XAXa).
39. [Biology-Elective 1: Biotechnology Practice] (15 points)
In order to investigate the water quality of a river, a research team determined the content of bacteria in river water samples and carried out the work of bacterial separation. Answer the following questions:
(1) The team used dilution coating plate method to detect the bacterial content in water samples. Before coating and inoculation, some sterilized empty plates were randomly taken and cultured for a period of time. The purpose of this is: then, 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 39, 38 and 37 respectively. Based on this, it can be concluded that the number of live bacteria in each liter of water sample is.
(2) The team separated bacteria from water samples by the method of plate scribing. In operation, the inoculation ring is sterilized, and the second and subsequent scribing always starts from the last end. The purpose of this is.
(3) Schematic diagrams A and B show the results obtained after inoculation and culture by dilution coating plate method.
(4) The team inoculated the obtained strains into the liquid culture medium and mixed them evenly, one part was cultured by standing and the other part was cultured by shaking. The results showed that the bacteria cultured by shaking grew faster than those cultured by standing. The reason is that shake flask culture can increase the content of culture solution, and at the same time make bacteria fully contact with culture solution, thus improving the utilization rate.
answer
(1) Check whether the medium plate sterilization is qualified 3.8× 10.
(2) incineration; Gradually dilute the aggregated bacteria to obtain a single colony.
(3) B (4), dissolved oxygen nutrient
analyse
(1) In order to determine whether the sterilization of the culture medium is qualified, microbiology experiments generally set a blank control: several sterilized empty plates are randomly cultured for a period of time to observe whether there are colonies on the culture medium; Estimating the number of viable bacteria in water samples by the formula of plate colony counting method;
(2) The inoculation ring needs to be burned and disinfected during inoculation; In the second and subsequent streaking, streaking always starts from the last end. The purpose of this is to gradually reduce the number of bacteria in each line by increasing the number of lines, so as to obtain colonies.
(3) As can be seen from the title map, Figure B is the result of plate scribing inoculation (the colony distribution is relatively uniform).
(4) Vibration culture can increase the oxygen content of liquid culture medium and promote the growth of aerobic microorganisms; In addition, shake flask culture can make bacteria fully contact with culture solution and improve the utilization rate of nutrients.
40. [Biology-Elective 3: Special Topics of Modern Biotechnology] (15)
Plant A has strong drought tolerance, and its drought tolerance is related to a certain gene. If the drought-tolerant gene is obtained from this plant and transferred to plant B with low drought tolerance, it is possible to improve the drought tolerance of the latter.
Answer the following questions:
(1) Theoretically, the genome library contains biological genes; CDNA library contains biological genes.
(2) If you want to obtain drought-tolerant genes from plant A, you can first establish the genome library of this plant, and then extract the required drought-tolerant genes from it.
(3) introducing drought-tolerant genes into Agrobacterium, introducing them into plant somatic cells through Agrobacterium transformation, and obtaining regenerated plants through a series of processes. In order to confirm whether the drought-tolerant gene is correctly expressed in the regenerated plant, we should detect the gene in the regenerated plant, and if the detection result is positive, we should detect whether the plant has been improved in the field experiment.
(4) If the obtained diploid transgenic drought-tolerant plants are selfed and the number ratio of drought-tolerant plants to drought-tolerant plants in offspring is 3∶ 1, it can be inferred that drought-tolerant genes are integrated (fill in "one homologous chromosome" or "two homologous chromosomes").
answer
(1) All parts
(2) Screening
(3) Drought tolerance of B expression product
(4) On homologous chromosomes
analyse
The (1) gene library includes a genome library and a cDNA library, wherein the genome library contains all the genes of the biological genome, and the cDNA library is constructed by reverse transcription of mRNA, and only contains the expressed genes (not all genes can be expressed), that is, some genes.
(2) The target gene needs to be screened from the gene library.
(3) In order to improve the drought tolerance of plant B, it is necessary to introduce the drought tolerance gene into the somatic cells of plant B through Agrobacterium transformation. To detect whether the target gene (drought-tolerant gene) is expressed, antigen-antibody hybridization should be used to detect the expression product (drought-tolerant related protein) of the target gene (drought-tolerant gene); The individual level can be tested through field experiments, and its drought tolerance can be observed and tested.
(4) If the drought-tolerant gene is integrated into two homologous chromosomes, the offspring will be drought-tolerant and there will be no character separation. [or "If the drought-tolerant gene is integrated into a homologous chromosome, the genotype of transgenic plants can be represented by A_ (A stands for drought-tolerant gene, and _ means that there is no corresponding gene on another chromosome), and the genotype of A_ selfed offspring is AA: A _: _ = 1: 2: 1, so drought tolerance: drought tolerance = ]
Zhang
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