Questions about high school biology calculations, a high school biology calculation question. Hurry

DNA is double-stranded, RNA is single-stranded, and if there are six bases on one DNA, then each single strand is three. The mRNA is complemented by only one of the bases, so the RNA is three bases. Then three codons correspond to one amino acid, so three bases on mRNA correspond to one amino acid.

So the ratio is 6:3:1

The actual DNA (base number): mRNA (base number) should be 2:1 mRNA (base number): protein (number of amino acids) should be 3:1 Multiply the first formula by three, and add the second formula to 6:3:1

DNA is a two-strand structure, mRNA is a strand, then first, DNA (base number): mRNA (base number) = 2:1

The 3 bases (also called codons) on the mRNA strand determine an amino acid, so mRNA (base number): protein (number of amino acids) = 3:1

At the same time, the multiplier is 6:3:1

I've just finished my studies, so you should be deserting in class, right? This one is simple.

Three codons encode an amino acid, so it takes three bases on the mRNA to code for an amino acid.

DNA is double-stranded, so one codon corresponds to the DNA, which is two strands and 6 bases.

But that's not the case, one amino acid corresponds to more bases on the DNA. Because DNA is divided into coding and non-coding regions, only the bases of the coding region can encode proteins.

In addition, the coding region of eukaryotes is also divided into exons and introns, and introns are bases that are in the coding region but do not code for proteins. Sometimes the number of introns is still very large.

But that's true in high school. Or rather, the ratio is at least 6:3:1

The transcription of DNA into mRNA is a one-to-one relationship, but DNA is two strands, and only one of them carries out the transcription process, so the relationship between DNA and mRNA is 6:3 - > 2:1.

The three codons in mRNA are translated into one amino acid, which is a 3:1 relationship.

Then the relationship between the three can be expressed as 6:3:1

When the temperature is 30 degrees, the total photosynthesis amount is 640 + 220 * 2 44 * 32 = 960mg per ten hours, so it is 96mg per hour.

The amount of respiration at fifteen degrees is 110 5

Therefore, 64 32 6 * 180 * 15-9 * 110 44 5 6 * 180 = 765mg

Why didn't I practice this kind of question that year?

C6Hi2O6 + 6H2O+ 6O2 = 6CO2 + 12H2O + energy, (= should be an arrow, it says 'enzyme'), blood sugar refers to glucose C6H12O6, with the ratio relationship C6H12O6: O2 = 1:6 can be calculated 40mol After the complete oxidation and decomposition of 1mol of glucose in the cell, 1161kJ of energy can be stored in ATP, so 1161x 40=46440kJ

Blood sugar is calculated using the formula of glucose breaking down into carbon dioxide and water, not to mention that you can't calculate moles. If you want to store ATP, you need to have a conversion rate, which can be seen in the compulsory textbook Respiration section of the reasoning on the efficiency of converting glucose into ATP. I don't know how to type the chemical formula in detail, so pass ......

Total Reaction. C6H12O6 + 6H2O + 6O2 6CO2 + 12H2O + Massive Energy (38ATP).

When ATP is hydrolyzed, it can release about energy.

So according to the reaction formula, N(C6H12O6)=N(O2) 6=40molQ=38·

aa*aa=1aa+2aa+1aa three genotypes, two phenotypes; The probability of AA occurring is 2 4;b Same thing.

cc*cc=1cc+1cc digenotype biphenotype.

dd*dd=dd one genotype one phenotype.

The four alleles occur at the same time, and a certain result needs to be multiplied, so it is 3*3*2*1 with a total of 18 genotypes, and 2*2*2*1 with a total of 8 phenotypes.

There are 4*4*2*1=32 genotypes, and 2*2*2*1=8 phenotypes.

The chance of generating AABBCCDD is 1 2*1 2*1 2*1 2*1=1 8

1.Phenotypes: 8.

A, B, C, and shape were all 2 species, and D was 1 2x2x2x1=8 genotypes: 18 species.

aShape genes: aa aa aa 3 b: bb bb bb 3 c: cc cc 2 d: dd 1.

3x3x2x1=18

2.The probability is 1 8

Explained in detail hi said above.

The first pair of offspring phenotype 2 genotype 3 2 table 2 base 3 3 table 2 base 2 4 table 1 base 1 then phenotype = 2 2 2 1 = 8 genotype = 3 3 2 1 = 18 (2) that is, the probability that the offspring of aa aa is aa is 1 2, bb1 2, cc1 2, dd1, then the answer = 1 2 1 2 1 2 1 = 1 8

Green plants account for 1 2, that is, from plants, divided by 10% equals 5; 1 4 from small carnivores, small carnivores belong to the third trophic level, there is from it, converted into the weight of the plant, that is, the consumption divided by 10% to the third is equal to 250, sheep are herbivores, belong to the second trophic level, divided by 10% of the quadratic is equal to 25kg, the sum of the three equals 280kg.

5kg from green plants, from meat, and carnivores are herbivores, from 250 plants, the same goes for. The herbivores come from herbivores, while the herbivores come from 25kg of greenery, which adds up to 280kgSmall carnivores belong to the third trophic level, and sheep are herbivores and belong to the second trophic level.

Important knowledge points in the senior one biology exam.

Chapter 1: Approaching the Cell.

The first section goes from the biosphere to the cell.

Knowledge combing:1 Viruses do not have a cellular structure, but they must rely on (living cells) to survive.

2 Life activities are inseparable from cells, which are the (basic units) of the structure and function of living organisms.

3. Structural levels of living systems: (cells), (tissues), (organs), (systems), (individuals), (populations) (communities), (ecosystems), (biosphere).

4 Blood belongs to the (tissue) level, ** belongs to the (organ) level.

5 Plants do not have (systemic) levels, and single-celled organisms can be transformed into both (individual) and (cellular) levels.

6 The most basic living system on Earth is (cells).

7 populations: the sum of individuals of the same species in a certain area. Example: All carp in a pond.

8 Community: The sum of all living things in a given area. Example: All the creatures in a pond. (Not all fish).

9 Ecosystem: A unified whole formed by the interaction between a biological community and the inorganic environment in which it lives.

10. The exchange of matter and energy between organisms and the environment based on cellular metabolism; growth and development based on cell proliferation and differentiation; Inheritance and variation based on the transmission and change of genes within cells.

Section II: Diversity and Uniformity of Cells.

Knowledge combing: 1. Steps for the use of high-magnification lenses (especially pay attention to steps 1 and 4).

1 Find the object under the low magnification, move the object to (field of view**), 2 turn the ** changer), and replace it with a high magnification.

3 Adjust (aperture) and (reflector) to make the field of view brighter appropriate.

4. Adjust (fine focus spiral) to make the image clear.

Second, the use of microscope common sense.

1Two ways to brighten the field of view (magnifying the aperture) and (using a concave mirror).

2. High magnification: object image (large), field of view (dark), number of cells seen (small).

Low magnification: object image (small), field of view (bright), number of cells seen (large).

3 Objective: (with) thread, the longer the barrel, the greater the magnification.

Eyepiece: (none) thread, the shorter the barrel, the greater the magnification.

The greater the magnification, the smaller the field of view, the darker the field of view, the smaller the number of cells in the field of view, the larger each cell.

The smaller the magnification, the larger the field of view, the brighter the field of view, the greater the number of cells in the field of view, the smaller each cell.

4 magnification = magnification of the objective lens magnification of the eyepiece.

5. The number of cells in a row can be inversely proportional to the magnification according to the field of view.

Calculation method: Number of cells Reciprocal of magnification = last number of cells seen.

For example, there is a row of cells in the field of view of the eyepiece 10 objective lens 10, and the number is 20, and the objective lens is replaced by 40 without changing the eyepiece, then how many cells can be seen in the field of view? 20×1/4=5

Since it is a free mating in a closed environment, the gene frequency does not change, so just calculate the gene frequency of the parent v.

The frequency of the v gene of the parent (20000*15%+20000*55%*1 2+2000) (20000+2000)=, so choose b

First, find the gene frequency after the introduction of 2000 animals; If the number of Drosophila in VV is 20000x15%=3000+2000=5000, and VV=20000x55% 11000, then the V gene frequency (5000x2+11000x1) 22000x2=21 44; Because Drosophila mates randomly, the law of genetic equilibrium, which states that the gene frequency of v in the F1 generation is equal to the gene frequency of V in the parent 21 44, the answer should be B 48%.

After introduction, the v and v genes are (20000+2000)*2=44000, where v is (20000*15%*2+20000*55%+2000*2) 44000=, because the frequency of the mutated gene does not change, so v in F1 is selected