Shanghai life science examination paper
Last two check codes
Digital code
This paper is divided into two parts: the first volume (1-4 pages) and the second volume (5- 12 pages).
Part. Full volume ***l2 page. Full score 150. Examination time 120 minutes.
The first volume (***60 points)
Note to candidates:
1. Candidates must clearly fill in their name, admission ticket number and check code on the answer sheet with a pen or ballpoint pen, and correctly scribble the admission ticket number and check code with a 2B pencil.
2. Volume 1 (1-3 1) is marked by the machine, and all the answers must be scribbled on the answer sheet. Candidates should use 2B pencil to darken the square representing the correct answer. Pay attention to the one-to-one correspondence between the test question number and the answer sheet number, and do not misplace them. When the answer needs to be changed, the original option must be erased and re-selected. The answer can't be scribbled on the test paper, and you can't get points if you scribble on the test paper.
First, multiple-choice questions (***60 points. There is only one correct option for each small question)
(A) 1 Sub-theme (***6 questions)
1. The picture on the right is a biological model, and the substance that constitutes structure A is most likely.
A. protein B RNA
lipin
2. Transcribe mRNA with the complementary strand of "-GAATTG-", then the sequence of the mRNA is
A.-GAA uug-b .-CTT AAC-c .-cuu AAC-d .-gaat TG-
3. The following curve can best reflect the changing trend of blood glucose concentration in healthy people within 6 hours after meals.
4. If 30% of the bases of a DNA fragment in a cell are A, then this DNA fragment
The content of a.g. is 30%, and that of b.u. is 30%.
C. Purine content is 50%. D. the content of pyrimidine is 40%.
5? The picture on the right shows the chemical reactions in organisms. About this reaction, the following statement is wrong.
We need helicase. B belongs to hydrolysis reaction.
C. there will be energy changes. D. the reaction speed is related to temperature.
6. In knee jerk reflex, the transmission mode of nerve impulse between neurons is
A dendrite → synapse → cell body → axon B axon → cell body → dendrite → synapse.
C dendrite → cell body → axon → synapse D dendrite → synapse → axon → cell body.
(2) Subtitle (***2 1)
7. The substance transformation process that will take place in people and plants is
① glucose is completely oxidized, ② glucose is converted into ethanol, ③ glucose is dehydrated and condensed, and ④ glucose is decomposed into pyruvate.
A.①②③ B.②③④ C.①③④ D.①②④
8. Some kind of parasite is parasitic in human lymphatic vessels, which will cause swelling of lower limbs, which is the cause of swelling.
A. liquid accumulates in intercellular space B. Cells cannot accept amino acids.
C. cells cannot break down fat. D. The concentration of Na+ in extracellular fluid is 12 times that of intracellular fluid.
9. In the following options, a group of compounds containing the same elements are
A. Cellulose and urea B. Fatty acids and phospholipids
C. adenosine triphosphate and ribonucleic acid D. cholesterol and hemoglobin
10. The picture on the right shows the transmembrane transport mode of a substance, and the following statement is correct.
A. The carrier in the membrane can also transport sucrose.
B, this is how iodine enters kelp cells.
C. this method will not be saturated.
D This pattern occurs when the transported substance changes from high concentration to low concentration.
1 1. Among the 9 1 offspring produced by mating of a pair of gray-winged insects, there are 22 black wings, 45 gray wings and 24 white wings. If black wings mate with gray-winged insects, the proportion of black wings in offspring is most likely to be
A.33% B 50% C 67% D 100%
12. The figure on the right is someone's temperature change curve, and the event that leads to the temperature change of ab and bc is the most likely.
A. Fever and chills B. Increase the ambient temperature and chills
C. Decreased cold and sweating D. Increased strenuous exercise and sweating
13. The following statement about human blood pressure regulation is wrong.
A. If the cardiac output remains the same, the diameter of blood vessels becomes smaller, and the systolic blood pressure will increase accordingly.
B. When arterial blood pressure suddenly drops, sympathetic nerve activity will be enhanced and arterial blood pressure will rise again.
C during diastole, when the blood flow to the periphery slows down, the diastolic pressure decreases.
D. Long-term excessive stress will make the balance of the cardiovascular center of the brain out of balance, leading to an increase in blood pressure.
14. The picture on the right shows the changes of the number of four different leaf-eating insects A, B, C and D with the height of the mountain. According to circle analysis, the following statement is correct.
A. The species evenness at 2000m above sea level is higher than that at 3000m above sea level.
The variation of B. b number with altitude can not reflect the genetic diversity of this species.
C is 3,000m above sea level, and the difference between B and C is the result of the interaction between organisms and habitats.
D. genetic diversity of a, b, c and d at an altitude of 4000 meters.
15. The picture on the right shows a cell in the human body and some metabolic processes in it. What is wrong in the following statement is
A cells can not only synthesize cholesterol, but also process excess cholesterol and excrete it.
B. In this cell, glycerol cannot enter glucose metabolism until it is converted into pyruvate.
C in this cell, glycerol can only enter glucose metabolism after being converted into pyruvate.
D. The metabolism of this cell is regulated by glucagon.
16. If the mother's serum does not contain lectin anti-A and anti-B, but the father's serum contains lectin anti-A, then the offspring will be red.
What doesn't happen on the cell membrane
A. lectin a b. lectin b c. lectin a and b d. lectin a and b are deleted.
17. The following statements about enzymes and hormones in human body are correct.
A. both enzymes and hormones are protein B. both enzymes and hormones are related to substance and energy metabolism.
C. Both enzymes and hormones are secreted by endocrine cells. D. both enzymes and hormones must be released into the blood to play a role.
18. The figure on the right shows the structure of mitochondria, in which the impossible reaction is
Tricarboxylic acid cycle occurs in a. ② B. ATP is produced in ①.
C, the dicarbonyl compound is generated in ②, and the binding reaction between H+ and O2 occurs in ③.
19. The following is a description of cell changes during cell division, which can correctly indicate the order of cell division in a cell cycle.
① Two identical DNA molecules are completely separated; ② Radial filaments appear.
③ Centromere multiplication ④ Centromeres are arranged in a plane.
A.②→③→①→④b .②→④→③→①c .③→②→④→①d .②→③→④→①
20. The picture on the right shows a plant cell in a 30% sucrose solution under a microscope. What is wrong in the following statement is
A. if cells are put in clean water, a remains unchanged.
B. if the cells are in 40% sucrose solution, the B/A value will become smaller.
C.b/a value can indicate the degree of water loss of cells.
D.a and b represent the length of cells and vacuoles respectively.
2 1. The picture on the right is the schematic diagram of yeast fermentation experiment.
Where x, y and z respectively represent
A. paraffin oil, carbon dioxide, blue B. Paraffin oil, oxygen, yellow
C. vegetable oil, oxygen, blue D. vegetable oil, carbon dioxide, yellow
22. The picture on the right shows a biochemical reaction chain in a cell. In the figure, E 1~E5 represents different enzymes, and A~E represents different compounds.
According to the picture, judge that the following statement is correct.
A. if the reaction catalyzed by E 1 is inhibited, the consumption rate of a will be accelerated.
B. If the reaction catalyzed by E5 is inhibited, B accumulates to a higher level.
C. If E3 is faster than E4, D will produce more than E.
The catalytic speed of D.E 1 is faster than E5, so the output of b is more than that of a.
23. The following description of Spirogyra and Spirogyra is incorrect.
A. In the living state, oscillatoria is blue-green and Spirogyra is green.
B the number of oscillatoria cells in filaments with the same length is less than that of Spirogyra.
C. There are pigments in the zonular chloroplasts of oscillatoria cells and Spirogyra.
D. After dropping iodine solution, the cells of Spirogyra showed mountain yellow structure, but tremella did not.
24. Rodents on the grassland feed on plants. The figure on the right shows the density of rodent and plant species.
According to the relationship between the number of classes in the picture, the following statement is wrong.
A. the existence of rodents affects the diversity of plants.
B. the diversity of plants depends on the density of rodents.
C. rodents can choose plants.
D. the density of rodents depends on the diversity of plants.
25. If E.coli labeled with 1 35S is infected with phage labeled with 1 32P, all phages will be released after lysis.
A. There must be 35S, maybe 12P B. Only 35S.
C. There must be 32P, maybe 35s d, only 32P.
26.① The right side represents the chromatographic results of photosynthetic pigment paper of fresh spinach leaves, ② The right side is most likely to come from.
A. Onion leaves cultivated in water B. Growing willow leaves
C. Chlamydomonas ginkgo cultivated in autumn and winter
27. The picture on the right shows the investigation results of the incidence of the same genetic disease in twins. A and B represent the percentage of fraternal twins and identical twins who are both sick. According to the picture, judge that the following statement is wrong.
Identical twins are more likely to get sick at the same time than fraternal twins.
B. The probability of simultaneous onset of identical twins is influenced by non-genetic factors.
When one fraternal twin gets sick, the other may get sick.
D. When one identical twin gets sick, the other will get sick.
(3) Subtitle (***4 questions)
28. In the population of white-flowered pollinators, 30% of parents are aa genotype individuals and 20% are AA genotype individuals, so parents
The frequencies of A-generation gene and AA genotype of F 1 are respectively
A.55% and 32.5% B.55% and 42.5%
C.45% and 42.5%
29. A stands for genetic engineering, C stands for fermentation engineering, D stands for cloning technology, and E and B stand for respectively.
A. Embryo transfer and cell culture B. Microinjection technology and cell culture
C. cell culture and pollen culture in vitro D. microinjection technology and pollen culture in vitro
30. The safflower of pea is dominant for white flowers, and the long pollen is dominant for round pollen. Existing safflower long pollen and white flower round pollen
F 1 of plant hybridization is safflower long pollen. If F 1 selfed, 200 F2 plants were obtained, including 10 white flowers and round pollen.
32 plants, the proportion of hybrid safflower round pollen plants in F2 is
A.7% B.8% C.9% D. 10%
3 1. The three pairs of alleles A/a, B/b and C/c, which control fruit weight, have equal effects on fruit weight, and they are located on three pairs of homologous chromosomes respectively. It is known that the fruit weight of genotype aabbcc is 120g, and the fruit weight of AABBCC is 2 10g. At present, fruit trees A and B are crossed, the genotype of A is AAbbcc, and the single fruit weight of F 1 35-165g. So the genotype of B is
A.aaBBcc B. AaBBcc C. AaBbCc D. aaBbCc
[Source: Z. xx .k.Com]
20 10 national unified entrance examination for colleges and universities
Shanghai life science examination paper
Volume II (***90 points)
Second, the comprehensive analysis questions (***90 points)
(four) to answer questions about the growth and development of valuable things and cross breeding. (9 points)
Take some high-quality and high-yield oat seedlings and insert mica sheets under the top of the coleoptile, as shown in the following figure.
32. Light shines on the left side of the seedling in the picture, resulting in the seedling. Explain the reason why seedlings produce this phenomenon.
33. After illumination, several tissue blocks were cut from seedling A, disinfected and then inoculated into the medium for inducing redifferentiation, and the concentration of two phytohormones in the medium was the same. Theoretically, the result of differentiation is:
The reason is.
34. Now, if you want to study the hybrid between oats and corn with high quality and high yield, you can use technology, because this technology can be solved and it is possible to cultivate new ones.
(5) Answer the following questions about human immunity. (10)
35. The human immune organs mainly include (at least two).
The following figure is a schematic diagram of three immune defense lines of human body against external pathogen infection, in which ① ~ ⑦ represent immune cells, and ⑧ and ⑨ represent immune molecules.
36. The following do not belong to 1 immune defense line.
A. lysosomes B. sweat C. oral mucosa D. gastric acid
37. Bacteria ① in the picture can devour many pathogens, so the characteristic of this immune process is.
38. The names of ② and ④ in the figure are and respectively; ⑧ and ⑨ stand for and respectively.
39. After vaccination, if the corresponding pathogen invades the human body again, the human body will quickly produce an immune response. The reasons for this phenomenon are analyzed with charts.
40. Vaccination is an immunization method.
(6) Analyze the information about plant photosynthesis and answer the questions. ( 1 1)
The photosynthetic rates of plants A and B under different light conditions were measured under certain concentration of CO2 and suitable temperature conditions. The results are shown in the following table, and the questions are answered according to the data in the table.
Photosynthetic rate and respiratory rate
Equal illumination
(kilolux) When the light is saturated,
intensity of illumination
CO2 absorption at light saturation (kilolux)
(mg/ 100 cm2 leaf? Hours) CO2 emission under dark conditions.
(mg/ 100 cm2 leaf? Hours)
A plant 1 3 1 1 5.5
Factory B 3 9 30 15
4 1. Compared with plant B, plant A grows under light conditions, and the judgment is based on
.
42. When the light intensity exceeds 9 kilojoules, the photosynthetic rate of plant B can't keep up with the reaction.
43. When the light intensity is 9 kilolux, the total photosynthetic rate of plant B is (mg CO2/ 100 cm2 leaf? Hours). When the light intensity is 3000 lux, the difference between the fixed CO2 amount of plant A and plant B is as follows
(mg CO2/ 100 cm2 leaf? Hours)
The photosynthetic rate is also affected by the output rate of photosynthetic products in leaves.
A plant is in the fruiting stage, as shown on the right.
44. If only one leaf is left, the other leaves are removed, as shown in Figure 2 on the right, and the photosynthetic rate of the leaves is left, because.
(7) Analyze the information about gene expression and answer questions. (9 points)
Take different types of cells from the same organism and detect their gene expression. The result is shown on the right.
45. If one of the genes 1 ~ 8 is a gene that controls ribosomal protein synthesis, it is most likely to be a gene.
46. Among the cells shown in the figure, the function of cells is the most similar.
A. 1 and 6 B.2 and 5
C.2 and 3 D
47. The basis for judging the approximate degree of cell function in the figure is.
48. Now we want to study the function of the new gene formed by the connection of gene 1 and gene 7. Before introducing plasmid, the correct position to restrict enzyme cleavage is.
49. The upper right picture is a schematic diagram of a plasmid used to express a new gene. If a new gene is to be inserted into the A position of the plasmid, the restriction endonuclease available for cutting the gene is.
A.hindⅲb . EcoRⅰ
C.Eco B D. Pst I
50. The probability of introducing new genes and DNA molecules after plasmid recombination into recipient cells is very small, so it is necessary to carry out tests to determine that recipient cells already contain the target genes.
5 1. If the plasmid's anti-penicillin gene lacks two bases in the recipient cell with the target gene, the possible results in this cell are as follows: (1) Inoculate such recipient cells into a culture medium containing penicillin.
A. penicillin resistance genes cannot be transcribed. B. Genes 1 and 7 were not expressed.
C. Genes 1 and 7 cannot be linked. Restriction sites on the plasmid have changed.
(8) The following figure is a schematic diagram of the partial regulation mechanism of osmotic pressure balance of human extracellular fluid. Answer the questions according to the pictures. (9 points)
52. Write the names of A and B in the picture: A and B.
When the ambient temperature of a healthy human body is 38℃, the internal environment can still maintain a relatively stable temperature and osmotic pressure.
53. At this point, the main reflex arc of temperature regulation in human body is
.
54. Write the nerve-body fluid regulation mechanism of human osmotic pressure at this time:
.
55. When the osmotic pressure of extracellular fluid decreases, the regulating function of A is the activity of A, which widely exists in the regulation of environmental balance in human body.
(9) Analyze the information about biological evolution and answer questions. (9 points)
56. The number of individuals of any living thing in nature cannot increase indefinitely. According to Darwin's theory of natural selection, this is because.
57. The right picture shows three natural selection types of the population, and Figure ② shows the selection types of the giraffe population. The medium-sized sparrows are selected to stay, and this selection type can be represented by a graph. Among the three selection types, graph is the most likely to produce new species.
The picture on the right shows the distribution of three populations of an amphibian in a mountain range. In summer, members of populations A and B and populations A and C can migrate across mountains. Through research, 1990 to 2000, mines were built between the three habitats, and 1920, roads were built between the habitats of population A and population C. During the period of 100, the temperature gradually increased and the rainfall gradually decreased.
58. After mining, population B may disappear or become a new species different from population A and population C. ..
Analyze the reasons why population B may form new species;
.
The following table shows the size of populations A and C and the frequencies of alleles 1(T/t) and 2(W/w), and the frequencies of recessive genes are shown in the table.
Population A[ population c]
Scale t(%) w(%) scale t(%) w(%)
1900 46000 5 1 1000 5 1
1920 45000 5.5 1 850 7 1
1940 48000 7 1 850 9 0.8
1960 44000 8 1 800 12 0.6
1980 42000 6 1 600 10 0.8
2000 400 00 5 1 550 1 1 1
59. According to the data in the table and the above information, the description of group C is more accurate.
A. The heterozygosity of allele 1 increased gradually. B there is no gene exchange with population a.
C. is experiencing adaptive radiation. Washington is more affected by the climate.
60. According to the data analysis in the table, the gene pool of population C is higher than that of population A; Population size is closely related to the frequency change of genes.
(10) Analyze information about microorganisms and answer questions. (10)
1982, an Australian scholar isolated Helicobacter pylori from gastric biopsy tissue.
6 1. The area where the genetic material of Helicobacter pylori is concentrated is called.
62. The four test tubes in the above figure respectively represent the growth state of four kinds of microorganisms in semi-solid medium (agar content 3.5g/L), among which test tube ② represents the growth state of Helicobacter pylori. Judging from the picture, this kind of bacteria can't grow under this condition. The growth state of methanogenic bacteria is best represented by test tubes.
63. The following table shows the formula of culture medium.
Component glucose KH2PO4 MgSO4 NaCl CaSO4 CaCO3 agar distilled water
Content10g0.2g0.2g3.5g1l
The number of Helicobacter pylori in the culture medium is higher than that in the inoculation after inoculating Helicobacter pylori into the culture medium with suitable pH and culturing at 37℃ for a period of time, mainly because:
The form of Helicobacter pylori, as shown on the right, can cause stomach diseases such as gastric ulcer.
64. The optimum pH for the growth of Helicobacter pylori is 6 ~ 7, and the pH in human stomach cavity is 1 ~ 2, but the pH of mucous layer of gastric mucosa near epithelial cells is about 7.4. If Helicobacter pylori enters the stomach cavity with food, how can it survive in the stomach in combination with its structural characteristics and the characteristics that can cause gastric ulcer?
65. According to the information in question 10, does Helicobacter pylori belong to archaea? .
The reason is.
(eleven) analyze the information about genetic diseases and answer questions. ( 1 1)
Figure 1 shows two genetic diseases in a family, which are controlled by gene A/a on autosome and gene B/b on sex chromosome respectively.
66. The X chromosome of Ⅲ-1 4 comes from1generation.
67. The pathogenic genes of onychomycosis are located on chromosomes. This is a genetic disease.
68. If III- 14 marries a woman with the same genotype III- 15 in Figure 2, the probability of her offspring suffering from nail infection is.
69. Assuming that III- 1 1 marries III- 15, if an egg is fertilized with e sperm, the developed IV- 16 suffers from two diseases, and its genotype is. If egg A is fertilized with sperm B, the genotype and phenotype of Ⅳ-17 will be.
If Ⅳ-17 marries a normal person whose parents are normal, but whose brothers and sisters have onychomycosis, then the probability that their offspring will not get sick is.
70. By taking measures, the recurrence risk rate of genetic diseases can be estimated and preventive measures can be put forward.
(twelve) to analyze the information about scientific inquiry and answer questions. (12)
Rhizobia of leguminous crops can fix nitrogen, while Gramineae plants can't. Therefore, in agricultural practice, leguminous plants and gramineous plants are intercropped to improve the yield of gramineous plants. It is found that the increase of yield is directly related to the absorption of hydrogen in soil by bacteria. In order to explore the specific mechanism, the following three experiments were carried out.
[Experiment 1]
The nitrogen fixation reaction of leguminous plants can produce hydrogen, which is absorbed by soil.
Alternative materials: leguminous alfalfa seedlings and gramineous wheat seedlings; Sterilized sand, ordinary soil.
Optional instrument: equipment for collecting hydrogen.
Experimental scheme:
7 1. If the hypothesis holds, complete the table on the right.
Plant name: experimental results of substrate planting (with or without hydrogen)
Experimental group
No soil
control group
Experimental results: no hydrogen was detected in the soil of the experimental group, and the rest are shown in the above table.
[Experiment 2] In order to explore the way of hydrogen absorption in soil, the following assumptions were made.
Hypothesis: Hydrogen is absorbed by bacteria in the soil.
Optional materials: alfalfa seedlings, common soil, antibiotics (rhizobia are insensitive), fungicides, 2,4-D, and extracted acetic acid.
Optional instrument: equipment for collecting hydrogen.
Experimental scheme:
72. According to the hypothesis, in addition to selecting and cultivating alfalfa seedlings after soil treatment, alfalfa seedlings cultivated in soil should also be used as control.
73. If the hypothesis holds, describe the experimental results according to the experimental scheme.
[Experiment 3] Whether hydrogen-absorbing bacteria (oxyhydrogen bacteria) in soil can promote plant growth, continue to explore.
Hypothesis: Oxyhydrogen bacteria can promote plant growth.
Materials for selection: 1.2m×2m experimental field, wheat seeds, hydrogen-oxidizing bacteria A 1, B 1, C 1, D 1, E1; Non-oxidizing bacterial strains A2, B2, C2, D2, E2; E. coli.
Experimental scheme: seed dressing of different strains, planting in experimental field, and recording the phase data of primary wheat strains after a period of time.
The results showed that the average radicle length (mm) and relative root growth (%).
A 1: the average radicle length is 13 and the relative root length is163; E2: The average radicle length is 8, and the relative root length is100;
D2: the average radicle length is 8 and the relative root length is100; B 1: the average radicle length is 30, and the relative root length is 375;
C2: The average radicle length is 8 and the relative root length is100; C 1: the average radicle length is 12 and the relative root length is150;
D 1: the average radicle length is 33, and the relative root length is 4.63; E 1: the average radicle length is 20 and the relative root length is 250;
A2: The average radicle length is 8 and the relative root length is100; B2: the average radicle length is 3, and the relative root length is 38;
Escherichia coli: the average radicle length is 8, and the relative root length is 100.
74. According to the hypothesis, the above data are statistically processed and expressed in an appropriate table.
Conclusion According to the above three experimental results, it can be seen that hydrogen-oxidizing bacteria in soil play an important role in promoting plant growth.
20 10 national unified entrance examination for colleges and universities
Shanghai life science examination paper
Answers and scoring criteria
Note: 1. This answer is marked and graded. Candidates can refer to the grading criteria if they write other correct answers.
2. Write typos for scientific nouns with birth names, and deduct points appropriately.
This paper consists of two parts, the total score is 150.
The first volume (***60 points)
Volume II (***90 points)
2. Comprehensive analysis questions (***90 points)
32. Bend to the light (bend to the left)
Auxin is synthesized at the top of coleoptile and transported to the lower part of coleoptile to promote the elongation of coleoptile cells; Light will affect the distribution of auxin, making the auxin on the backlight side more than that on the light side; At the same time, the mica sheet inserted on the left side of the seedling prevented the auxin from being transported downward to the light side, resulting in uneven elongation of cells on the backlight side, so the seedling bent toward the light direction.
33. Take root
The concentrations of cytokinin and auxin in the original culture medium are equal, but the irradiated tissue block A contains more auxin, so the ratio of cytokinin and auxin concentration in the tissue block is less than 1, thus inducing rooting.
34. Cell fusion (cell hybridization) distant hybridization incompatibility (reproductive isolation)
Species (species or hybrids)
(5) (10)
35. Thymus and spleen bone marrow lymph nodes (at least 2).
36.A
(8) (9)
52. Hypothalamic pituitary gland
53. Skin temperature receptors afferent nerves, hypothalamus temperature regulation center efferent nerve sweat glands.
54. At this time, sweating increases, blood volume decreases, and osmotic pressure of extracellular fluid increases, which stimulates hypothalamus and promotes antidiuretic shock.
Increased hormone secretion; Through pituitary release, it acts on renal tubules, enhances reabsorption and reduces urine output; At the same time, down
Thalamic thirst center is excited, causing thirst, taking the initiative to drink water and increasing body water content.
55. Inhibition (negative feedback) maintains the relative stability of hormone levels (maintaining the relative stability of internal environment/system/organism).
(9) (9 points)
56. Survival competition (survival competition, survival of the fittest)
57.① ③
58.
Because of the geographical isolation from population A, the gene communication between population B and population A, population B and population C was blocked. Therefore, the genetic mutation of population B began to accumulate, and the mutation was preserved through environmental selection. The population gradually adapted to the living environment, and the population size began to expand, forming reproductive isolation and new species.
59.A
60. Small
6 1. Quasi-nucleus (nuclear region)
62. Oxygen concentration is too high or too low ③
63. Lack of nitrogen sources (lack of nitrogen sources and growth factors)
( 12)
7 1.
Plant name: experimental results of substrate planting (with or without hydrogen)
The sand amount of alfalfa seedlings in the experimental group is as follows
In the control group, wheat seedlings have no sand.
No soil
72. Antibiotics (fungicides) Fungicides (antibiotics) will not be treated.
73. Planting alfalfa seedlings in soil treated with antibiotics can collect hydrogen, but planting alfalfa seedlings in soil treated with fungicides and untreated soil cannot collect hydrogen.
74.
Average radicle length of plant type (mm) and relative root growth (%)
Average distance
Escherichia coli 8.0 100.0
Non-oxyhydrogen bacteria strain 3-8 7.0 38- 100 87.6
Hydrophilic bacterial strain12-3321.6150-413270.2