A, multiple-choice questions (30)
1. (Hedong District, Tianjin 20 10) The following statements about enzymes. Correct.
Helicase can also play a normal role in PCR amplification of target genes.
B. enzymes synthesized in cytoplasm can also play a role in the nucleus.
C. phosphodiester bonds can only be formed by DNA polymerase catalysis.
D cells that can synthesize enzymes can certainly synthesize hormones.
2 (Taizhou City, Jiangsu Province, 20 10) Some biological variations can be identified by the behavior of cell division in a certain period. Mode A and Mode B represent the phenomenon of "ring" and "cross structure" in the process of cell meiosis, respectively, and the letters in the figure represent genes on chromosomes. The following statement is true: D.
A.a and b belong to chromosome structural variation and gene recombination respectively.
B. A picture is the result that the number of genes on a chromosome changes due to the addition or deletion of individual base pairs.
C. Figure B is the result of cross-exchange of non-sister chromatids of homologous chromosomes in tetrad period.
D. In the prophase of the first meiosis, A and B graphs often appear, and the ratio of chromosome number to DNA number is 1:2.
3. (Hunan Nanxian No.1 Middle School Comprehensive Monthly Exam 20 10) In the following descriptions of organic substances that make up cells and organisms, it is correct.
Once the polypeptide chain is synthesized on ribosomes, it has biological activity.
B The energy released by the oxidative decomposition of sugar, fat and protein with the same mass is the same.
The nucleic acid in the nucleus of C contains only deoxyribose, while the nucleic acid in the cytoplasm contains only ribose.
The chemical essence of all antibodies, most enzymes and some hormones is protein.
4. (Shanghai Pudong New Area 20 10 College Entrance Examination Forecast) The following figure shows the relationship between chromosomes, DNA and genes. The description of their relationship is wrong.
A. each chromosome contains 1 or 2 DNA molecules, and the DNA molecules contain multiple genes.
B can be copied, separated and spread, and the behavior is consistent.
C. all three are genetic materials in biological cells.
The behavior of D chromosome determines the behavior of DNA and genes in biological pedigree.
5. (Senior Three Quality Test in Xiamen, Fujian Province, 20 10) The following figure shows part of the process of intracellular protein synthesis, and the related statement is incorrect.
The synthesis of C may be controlled by more than one gene.
B. There is no transmission of genetic information in the graphic process.
C. process a is only carried out on ribosomes.
Both d a and b contain the start codon.
6. The curve on the right shows the reaction rate catalyzed by enzyme A and enzyme B in a certain pH range. The following options are incorrect.
A. When pH is 5, the reaction rate catalyzed by enzyme A and enzyme B is equal.
B. at pH 4, enzyme a is more active than enzyme B.
C. enzyme a exists in gastric juice.
D. in the experiment of studying the catalytic reaction rate of enzyme a and enzyme b, it is necessary to keep the temperature constant during the reaction.
7. (Ningbo City, Zhejiang Province, 2010) The antigens of Salmonella A and B are different. Mice were injected with two kinds of salmonella at the same time. After a period of time, plasma cells were isolated from rats and each plasma cell was cultured in a culture medium. The serum of rats (antibodies mainly exist in serum in vivo) and the separate culture medium of plasma cells were extracted and preserved respectively. The most likely phenomenon in subsequent experiments is
A. When cultures of different plasma cells are mixed together, a specific immune response will occur.
B. When Salmonella A and Salmonella B are added to the culture medium at the same time, at most only one kind of bacteria will agglutinate.
C. Adding two kinds of Salmonella A and B to rat serum, only one kind of bacteria agglutinates.
D after the rat serum is mixed with the plasma cell separation culture solution, the antibody contained in the former will have an immune reaction with the antigen contained in the latter.
8. (The first comprehensive simulation of No.1 Middle School and No.3 Middle School in Chang 'an, Shaanxi Province 20 10) There are many genes on each chromosome of human beings. For example, the human chromosome 1 includes the following genes:
Alleles of gene control traits and their control traits
Produce amylase A: Produce amylase A: Do not produce amylase.
Rh blood group d: Rh positive D: Rh negative.
Red blood cell shape E: oval cell E: normal cell.
If the genes on the parent chromosome 1 are shown in the following figure respectively, we can't draw any conclusions below without considering chromosome cross-exchange.
A The probability that their children may have oval red blood cells is 1/2.
Their children can't be Rh negative.
The gene composition in the first polar body of normal meiosis of mother is aDe.
Some of their children may not produce amylase.
9. (Wenzhou City, Zhejiang Province, 20 10 Senior Three Comprehensive Model 2) It is known that two pairs of independently inherited genes located on the autosome control the production and deposition of chicken feather pigments. At one site, the dominant gene can produce pigment, while the recessive gene does not produce pigment; At another site, the dominant gene prevents pigment deposition, while the recessive gene can make pigment deposition. The coloring of chicken feathers must be able to produce and deposit pigments, otherwise it is white hair. If a chicken with white phenotype (both loci are recessive homozygotes) crosses with another chicken with white phenotype (both loci are dominant homozygotes), the probability of F2 generation feather coloring is
a . 1/ 16 b . 3/ 16 c . 4/ 16d . 7/ 16
10. (Taizhou City, Jiangsu Province, 20 10 High School Model) Non-waxy rice (Y) is dominant for waxy rice, and disease resistance (R) is dominant for non-disease resistance (R). F 1 is a three-generation cross between two pure-bred rice resistant to non-waxy diseases and non-waxy diseases. Under natural conditions, the change of gene frequency is
A.y gradually increases, r gradually increases, B.y basically remains unchanged, and r gradually increases.
C.y is basically unchanged, R is basically unchanged, D.y is gradually decreasing, and R is gradually decreasing.
1 1. (Hangzhou, Zhejiang Province, 20 10, the first quality inspection of senior three) A, B and C in the following figure represent three regions with different natural conditions, and the thick black lines between the regions indicate certain geographical isolation. Some individuals in Area A began to distribute to Area B and Area C by accident, and gradually formed two new species, in which A, B and C represent three populations respectively. The following statements are correct.
A. The above process shows that geographical isolation is a sign of the formation of new species.
The gene banks of B.A. and B. are very different, so it is impossible to exchange genes.
C.B and C are geographically isolated, but their gene frequencies are the same.
D. There is reproductive isolation between population A and population C, and the gene pool composition of the two populations is completely different.
12. (Shanghai Pudong New Area 20 10 college entrance examination forecast) A mink breeder let his minks mate randomly, and it was found that on average, 9% of the minks had rough skin, so the price of such minks would be low. He expected more smooth-skinned minks, so he decided not to let rough-skinned minks mate. It is the recessive gene located on the autosome that causes rough fur. In the next generation of mink, the theoretical percentage of rough fur individuals is
A 7.3% B 5.3% C 2.5% D 1.2%
Regulation and immunity are closely related to human health. The following statement is correct.
A. The heat dissipation of human body at 30℃ is greater than that at 5℃ to keep the body temperature relatively stable.
B effector T cells can attack cells invaded by type A H 1N 1 by releasing lymphatic factor.
C When the antibody attacks its normal tissues or organs, the corresponding cells will play an immune role by releasing histamine.
The monoclonal antibody synthesized by mouse hybridoma cells may be recognized as antigen by human immune system.
14. (ganxian middle school, Jiangxi, 20 10) The following picture shows the process of eliminating tetanus toxoid by humoral immunity, and the relevant statements are correct.
A. Cells 2 and 3 are B cells and memory cells respectively. B only cell 4 contains the gene for synthesizing substance a.
C. Cell 5 can produce lymphatic factor D2 and ③, which is related to the protein on the cell membrane.
15. (No.14 Middle School, Hangzhou, Zhejiang Province, 20 10 Comprehensive Test of Senior Three) The brewing industry uses yeast fermentation to produce alcohol. The fermentation process is roughly as follows: sterilization and inoculation; Introducing sterile air; Closed fermentation. The following curves describe the changing trend of the number of bacteria and alcohol production in the fermentation process, of which the correct one is
16. (Grade 20 10 in Santai Middle School, Sichuan Province) Which of the following descriptions about carbon cycle in ecosystem and substance exchange in human body fluids is correct?
A. during allergic reaction, the increase of a in figure a leads to tissue edema.
B. the highest oxygen concentration in the human body is b in figure a.
B is the main component of the ecosystem.
D in the food chain established by predator-prey relationship, the trophic level of B in Figure B has the least energy.
17. (Guangzhou City, Guangdong Province, 20 10) The optimum temperatures for plant photosynthesis and cell respiration are 25℃ and 30℃ respectively. Fig. 6 is a graph showing the variation of photosynthesis intensity with light intensity at 25.d egree. C.. The following statement is correct.
The physiological activity of mesophyll cells is only affected by the temperature of point A.A.
Photosynthesis in plants begins at point b.b.
C. If the temperature rises from 25℃ to 30℃, both points A and D will rise.
What is the output of the factory at D.C.
18. (Sichuan Ziyang Senior Three College Entrance Examination Simulation, 20 10) The following are schematic diagrams of four different plasmids, in which ori is the necessary sequence for replication, amp is the ampicillin resistance gene, tet is the tetracycline resistance gene, and the arrow indicates the restriction site of the same restriction enzyme. The following statement is correct.
A gene amp and tet are a pair of alleles, which are often used as marker genes in genetic engineering.
Plasmid B refers to small circular DNA that can replicate itself in bacterial cells and DNA of animal and plant viruses.
The site of action of restriction endonucleases is the hydrogen bond between two specific nucleotides in DNA molecule.
D, introducing the target gene into Escherichia coli with plasmid 4, which can not grow on the medium containing tetracycline.
19. (Dalian, Liaoning, 20 10) According to the latest research results, taking metformin, a diabetes drug, to experimental mice, and chemotherapy (mainly using anticancer drugs to kill cancer cells) can inhibit the spread of breast cancer cells in experimental mice. The following related statements are correct.
A. Chemotherapy will increase the blood sugar concentration of experimental mice, leading to the spread of breast cancer cells.
B metformin can not cause the decrease of glycoprotein on the cell surface of experimental mice.
Proto-oncogene C mutation promotes cell carcinogenesis, while tumor suppressor gene mutation inhibits cell carcinogenesis.
Mice with thymectomy have the same risk of breast cancer as normal mice.
20.(20 10, simulation of senior three in Baotun Senior High School, Henan Province) Gonadal tissue cells were fluorescently labeled. Alleles A and A were both labeled as yellow, and alleles B and B were both labeled as green. The tetrad cells were observed under fluorescence microscope. The following speculation is reasonable.
A. If these two genes are on the 1 homologous chromosome, there are two yellow and two green fluorescent spots in the 1 tetrad.
B If these two pairs of genes are on the 1 homologous chromosome, there are four yellow and four green fluorescent spots in the 1 tetrad.
C If these two pairs of genes are on two homologous chromosomes, there are two yellow and two green fluorescent spots in the 1 tetrad.
D If these two pairs of genes are on two homologous chromosomes, there are four yellow and four green fluorescent spots in the 1 tetrad.
2 1. (Dalian, Liaoning Province, 2010) In 2009, influenza A (H 1N 1) broke out all over the world, and the World Health Organization raised the alert level to level 6. The following related statements are correct.
A. the immunity caused by influenza a vaccine H 1N 1 belongs to humoral immunity in nonspecific immunity.
B. If you are exposed to the pathogen of the disease again after vaccination, the body may react quickly to produce a large number of antibodies.
C the pathogen of the disease is prone to mutation due to gene recombination, and it is necessary to continuously develop new vaccines.
D the pathogen of the disease can be parasitic on human respiratory tract and cultured in nutrient-rich culture solution.
22. (Taizhou City, Jiangsu Province, 20 10 Senior Three and Two Models) The following figure shows some processes of human immune response, and B is correct in judging A, B, C and D cells.
A.a cells have the ability to recognize specific antigens. B. B cells can proliferate and differentiate after being stimulated by antigen.
C.c. Cell lysis and death belong to nonspecific immunity. D.d. cells destroy antigens by producing antibodies.
23. (Yinchuan No.1 Middle School, Ningxia, 20 10) The research shows that there may be two or more transmitters in the same synaptosome, which is called transmitter * * * existence phenomenon. The following statement is incorrect.
A. The transmission of excitation in synapse depends on the fluidity of biofilm, which reflects the information transmission function of cell membrane.
There may be multiple receptors on the same postsynaptic membrane.
There is no transmitter in the neuron where the postsynaptic membrane is located.
D.*** * * transmitters may play a synergistic or antagonistic role.
24. (The first comprehensive simulation of 20 10 in Chang 'an No.1 Middle School and No.3 Middle School in Shaanxi Province) Thymidine deoxynucleoside (thymidine for short) can be converted into thymine deoxynucleotide, which is the raw material for DNA synthesis. The young roots of live kidney beans were treated with nutrient solution containing 3H- thymidine, and the free 3H- thymidine was washed away after a certain period of time. Part A and Part B (A is meristem) in the following figure were detected for 48 hours continuously. With the growth process, the radioactive content of part A and part B changes as follows
25. (Jiangsu yancheng middle school 20 10 Senior Three Model) A biologist investigated the population changes of lemmings and owls in a forest in order to find out the factors controlling the population changes. During the study period, the number of lemmings decreased due to a disease caused by a virus. Under normal circumstances, owls in the forest feed on lemmings, but when the number of lemmings decreases, the number of adult owls (about 20) remains unchanged, while the number of newly hatched owls decreases greatly. The shortage of food supply reduced the number of newborn owls, but did not reduce the number of adult owls. In this regard, some people have made the following explanation. What do you think is right?
① The hatching rate of eggs decreased ② The predation ability of adult owls was stronger than that of newborn owls.
(3) Insufficient nutrition reduces the female owl's egg production; (4) Malnourished newborn owls are more susceptible to diseases and death.
A. only 13B. Only 24C. Only 234D 12344.
26. (Dalian city, Liaoning province, 20 10, senior three, comprehensive model) The following description of the textbook experiment is correct.
A in the experiment of "exploring the influence of temperature on enzyme activity", phenanthrene reagent can be used instead of iodine solution.
B "Observing cell meiosis" can choose blooming pea anthers as experimental materials.
C. In the experiment of "exploring the dynamic change of yeast quantity", the sample solution is usually diluted first and counted by /el cao.
D "Investigating common human genetic diseases" is best to select polygenic genetic diseases with high incidence in the population for investigation.
27. Cell membrane plays an important role in life activities. The following related statements are wrong.
A the fluidity of the cell membrane enables the cell membrane to recognize external information.
B protein on cell membrane will change during cell differentiation.
C. the cell membrane is selective to substances on both sides of the membrane.
Phospholipids and cholesterol on the cell membrane make the cell membrane both fluid and firm.
28. The effect of indoleacetic acid on the rooting of rose cuttings was studied (20 10, Senior Three, One Model, Haidian District, Beijing), and the result is shown in figure 1. Regarding the analysis and evaluation of the experiment, it is correct that
A cuttings should be treated with different concentrations of indoleacetic acid for different time.
In this experiment, it is not necessary to set up a control group to treat drilling cuttings with distilled water.
C, making the cutting branches produce the same amount and concentration of indoleacetic acid.
D when the concentration of indoleacetic acid exceeds a certain range, it will inhibit the rooting of cutting branches.
29. (No.1 Model, Senior Three, Hedong District, Tianjin, 20 10) The following figure is a small fragment of a biofilm. Please choose a reasonable option according to the picture.
A. biofilm must be a fragment of mitochondrial inner membrane.
B.ATP synthase exists on all biofilms.
The interconversion between C.C.ATP and ADP can occur in any living cell.
D. Biofilm must be a segment of chloroplast saccular structure membrane.
30. (Nankai District, Tianjin, 20 10, Grade Three, One Model) The following picture is a picture of animal somatic cell division pattern drawn by a classmate, and two of them are wrong. Please arrange the remaining correct pictures in the order of cell cycle, among which the pictures ranked 1 and 3 are respectively
A.①② B.①⑥ C.③② D.④⑤
I. Non-multiple choice questions (20)
3 1. (Test in the third grade of Foshan No.1 Middle School, Guangdong Province) Asparagus (tender stems commonly known as asparagus) is a rare vegetable, which is a dioecious plant determined by XY sex. Wild asparagus has narrow leaves and low yield. In a wild population, it is found that there are a few broad-leaved asparagus (mutants), both male and female, and the yield of male is higher than that of female.
(1) Some people think that broad-leaved mutant plants have heterosis or polyploid characteristics. Please design a simple experiment to identify whether the mutant is caused by chromosome doubling.
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(2) If it has been proved that broadleaf is caused by gene mutation, there are two possibilities: dominant mutation and recessive mutation. Please design a simple experimental scheme to judge. (It is required to write the hybrid combination, hybrid results and draw a conclusion)
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(3) If it has been proved that it is caused by broad-leaf dominant mutation, the mutant gene may be located on autosome or X chromosome. Please design a simple experimental scheme to judge (it is required to write heterozygous combinations, hybridization results and draw conclusions)
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(4) The yield and quality of male broad-leaved asparagus plants, which are all mutants, are higher than those of female plants. Please consider putting forward two reasonable breeding schemes from the perspective of improving economic benefits (represented by charts, 3 points).
32. (Simulation of Tunxi No.1 Middle School and No.3 High School 20 10 in Anhui Province) The following figure 1 is a dynamic change process of gene pool of two beetle populations. Each beetle in the population has its own genotype. There is no hidden relationship between A and A, and both * * * determine the body color of the beetle. The genotype and phenotype of beetle body color are shown in Figure 2. Please answer the following questions according to the picture.
(1) As can be seen from Figure 2, the relationship between allele A and allele A controlling the body color of beetles is as follows. Beetles with different body colors reflect the diversity of biodiversity. A very reliable way to detect this diversity is to identify different subspecies and different populations.
(2) A beetle with genotype A'A appeared in population I, and the most likely source of A' gene is. The source is biological evolution. The emergence of A'A individuals will make the gene pool of population ⅰ
It has changed.
(3) The arrow in the figure indicates that there are opportunities between the gene pools of two populations through migration. Therefore, there is no relationship between population ⅰ and population ⅱ.
(4) According to the distribution ratio of beetles with different body colors in the two populations in Figure 1, it can be preliminarily inferred that the population is in an environment with serious industrial pollution, which has played a certain role in the survival of beetles.
33. (Comprehensive simulation of senior high school 20 10 attached to Shaanxi Xigong University) Cut off the tip of cultivated oat coleoptile and do the following experiments under suitable conditions. Please answer the questions about IAA.
(1) The experimental phenomena of group A and group B show that group A is the result.
(2) According to the experimental phenomena of group A and group C, the transport direction of auxin was explained.
(3) The growth characteristics of group D indicate that the action characteristics of auxin are as follows.
(4) Group E showed the effects of auxin concentration on the growth of roots, buds and stems. The information given by this chart is.
A. The effects of auxin on three organs are twofold, with low concentration promoting growth and high concentration killing plants.
The auxin concentrations corresponding to points B, A, B and C are the most suitable concentrations to promote the growth of roots, buds and stems, respectively.
The auxin concentration corresponding to point C. D can promote the growth of stem, but inhibit the growth of bud.
D. Young cells are sensitive to auxin, while mature cells are not sensitive to auxin.
(5) How to prove the reason why lateral bud 3 didn't grow into lateral branches? Please use the given materials to complete the experimental steps and answer the relevant questions.
Materials, utensils, reagents and supplies: plants in the same group D, branch shears, lanolin (which can reduce water evaporation), high-concentration 2,4-D lanolin (auxin analogue), distilled water absorbent cotton and absorbent cotton.
Experimental steps:
① Divide 20 plants which are the same as Group D into Group A (control group) and Group B (experimental group), and harmlessly mark the position and length of all lateral buds 3.
(2) In group A, it was cut from a suitable position above the lateral bud 3 with a branch scissors and placed at the fracture; The treatment of group b is.
(3) Under the same suitable conditions, cultivate for a proper time, and observe and record the experimental phenomena.
Predict the experimental results and conclusions:.
34. Scientists found that ultraviolet rays can inhibit the growth of plants. The reason is that ultraviolet light increases the activity of indole acetic acid oxidase (which can only be synthesized by iron) in plants, and indole acetic acid is decomposed into 3- methylene indole oxide under the action of indole acetic acid oxidase (which can not promote cell elongation). In order to verify that the inhibition of plant growth by ultraviolet light is related to the oxidation of indoleacetic acid, please supplement the following experimental design and complete the relevant experimental analysis. (regardless of the influence of iron deficiency on other aspects of plant growth)
A, materials and equipment:
Oat seedlings, complete culture solution, complete culture solution with iron deficiency (appropriate solution concentration), distilled water, agar blocks, leaves, etc.
Second, the experimental steps:
(1) Prepare four beakers with culture scaffolds, which are marked as A, B, C and D respectively;
(2) equal amounts of complete culture solution are added to beakers A and B, and equal amounts of iron deficiency culture solution are added to beakers C and D;
(3) Select and evenly divide into four groups, and put them in beakers from Group A to D4 for a period of time;
(4) Give A and C proper visible light illumination and B and D illumination, and continue to cultivate for a period of time;
(5) Cutting off the top of coleoptile of seedlings, collecting auxin with agar blocks respectively, and marking them as A, C, B and D accordingly;
(6) After a period of culture, observe the curvature of coleoptile.