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2. In order to improve the yield of vegetables in the greenhouse, the temperature should be controlled to ( ) aHigh during the day and low at night.
3. Reasonable dense planting is conducive to the improvement of yield per unit area, and the main reason is ( ) aMake the most of the sun.
7. Filling the warehouse where the grain is stored with carbon dioxide can prolong the time of grain storage, because it is ( ).
b.The high concentration of carbon dioxide inhibits respiration.
8. Among the leaves of leeks and common leeks, the cells with the closest color are ( ) aBoth epidermal cells.
9. The following substances, which generally do not enter and exit from the stomata on plant leaves, are ( ) cInorganic salt.
10. Cut off the veins of the leaves, and the impact is ( ) dAll three of the above statements are true.
11. The main organ-making organ for organic matter in fruits and seeds is ( )cLeaf 12, when planting crops, we should pay attention to reasonable dense planting, the following statement is correct ( ) dCrops can be planted neither too densely nor too thinly, so that the yield of crops will be higher.
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2.High during the day and low at night (why not this?) The temperature is higher during the day, the organic matter produced by net photosynthesis increases, and the lower temperature decreases at night, and the consumption of organic matter by respiration decreases.
It is good for crops to receive enough light, avoid shading each other, and also facilitate air circulation (supply of CO2).
The high concentration of carbon dioxide inhibits respiration and reduces the consumption of organic matter by respiration.
8.I don't know.
Inorganic salts are mainly absorbed by the root system.
Organic matter is a product of photosynthesis and is manufactured from leaves.
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High during the day, which is conducive to photosynthesis, and low at night, can inhibit respiratory consumption.
The main reason for proper dense planting is to make better use of light, remember.
Carbon dioxide is a product of aerobic respiration that inhibits respiration and consumes organic matter.
The epidermal cells are waxy and transparent, and the other three are all green.
Inorganic salts are absorbed by the roots.
The leaf veins have both ducts and sieve tubes, there are no inorganics, and photosynthesis is also affected.
Organic matter is a photosynthetic product, which is made of leaves.
If it is too dense, the covered leaves will not only not be photosynthetic, but will also respire and consume organic matter, so it should be planted reasonably densely.
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2.High during the day and low at night.
I don't know.
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Answer: A
Combined with the title, according to the analysis of the figure, in Figure 1, the pH of the chloroplast matrix = 4, and the ph of the thylakoids = 7 (the pH of the chloroplast matrix is less than the pH of the thylakoids), when both of them become pH = 4, ATP cannot be produced (as shown in Figure 2). Then from Figure 2 to Figure 3, that is, when the chloroplast matrix is pH = 8 and the thylakoids are pH = 4, ATP can be produced. This means that the condition for the production of ATP is that the pH of the chloroplast matrix is greater than that of the thylakoids.
i.e. the thylakoids of chloroplasts should be composed of hydrogen ions higher than the chloroplast matrix.
Light is necessary for photosynthesis, but this experiment is an isolated chloroplast, and there is an original substance, so light is not necessarily needed.
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C, obviously C is not right, they all said that the pH is the same.
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Mr. Wang will explain this problem to you:
This question mainly involves free combinatorial applications.
The scales of this fish had four phenotypes, which were controlled by two pairs of independently inherited alleles, and BB had a lethal effect, indicating that the four phenotypes of this fish population were controlled by four genotypes: A BB, A BB, AABB and AABB. The mating of the single-row scale fish in F1 can produce individuals with four phenotypes, and the genotype of the single-row scale fish in F1 can be deduced as AABB. Crossing a fish with scaleless fish and homozygous wild-type scales, the genotype of a single-row scale fish with AABB can be obtained, first consider the genes b and b, the genotypes of the parents are bb and bb, and the tail of the parental wild-type scales is homozygous, so bb is the genotype of the fish with the parental wild-type scales, and bb is the genotype of the scaleless fish; Considering the genes A and A, there are only two phenotypes in the cross-bred offspring of fish with scaleless fish and homozygous wild scales, and the ratio is 1:
1. Combined with the above analysis, the genotypes of the parents are AA and AA. In this way, there are two ways to combine genes, AABB AABB and AABB AABB, and in the first combination, the individual with the genotype AABB behaves as a single row of scales, so the genotype is AABB AABB.
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ZZ is male and XY is male. The ratio of male to female silkworms has decreased, and the ratio of males to females in humans has decreased.
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Tell me first that ZZ and WW are male or female, respectively. It's been a long time since I've made a creature, and I forgot.
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Investigating species ......Do you think it is necessary to count the trees of tens of thousands of square kilometers of forest one by one?
Of course it's impossible! It must be a circle with a minimum quadrat area to count or investigate the species in this area.
You can understand it as the S3 area, which basically includes all or most of the number of tree species in the area (most of them are defined as at least 84%, and there are more stringent requirements of 95%)
At this time, all the trees in the selected area can already reflect the characteristics of the tree-related species in the forest community, and continuing to expand the area will not help the statistics except for increasing the investment of human and financial resources.
If the area is less than S3, the species will easily be missed and the survey will be inaccurate.
Mr. Wang will explain this problem to you:
This question mainly involves free combinatorial applications. >>>More
Since f(x+1) and f(x-1) are both odd functions, f(x)=f(x+4)t>4 t=4l (l is a positive integer). >>>More
Solution: Using the formula: a 2-b 2 = (a + b) (a - b) a 3-b 3 = (a - b) (a 2 + ab + b 2). >>>More
Method: The test and homozygous were mated.
Objective: To test whether they are homozygous or heterozygous. >>>More
Haha and where pi=
Don't bother me to prove that these two numbers are irrational. >>>More