Find 50 in high school biology to add

There are only a few differences, the development of plant seeds, which involves the operation of the chromosomes, and the new syllabus of C3 and C4 plants has already said that it is not examined, you don't have to worry.

It's okay to go back and review the junior high school biology book, the physiological processes on it must be familiar, and now we must focus on the existing textbooks, which is the point.

I wish you excellent results.

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If you ask me to say, I'll have to say it for half a day. Let's do it: you can go to the old bookshelves in the school library and find the [Textbook Analysis] from the early years, which has some content on it. If you can't find it, you can go to some old teachers and ask them to help!

I think the Century Gold List is very easy to use, I used that in high school, and the biology score has always been very good, as for C3, C4, plants, and the like, I will still take the test, and it's a little difficult, just do more questions, it's not too difficult to understand, and the content of many reference books is still complete, but it is very detailed, and many test points are easy to ignore Small details.

It is known that II2 does not carry the causative gene and that the disease is recessive with the X chromosome. ii3 and ii4 about the disease.

We count one disease at a time:

1. For the disease: III4 The possible genotypes are: 1 2xbxb, 1 2xbxb and their spouses are xby

Therefore, the probability of offspring getting this disease is 1 2 * 1 4 = 1 8, and the probability of the disease is 7 8

2. For albinism, the genotype of the couple is AA, and the probability of the offspring getting the disease is 1 4 and the probability of not albinism is 3 4

Therefore, there are two situations in which the probability of a child getting a disease is that there are no albinism, and if you have albinism, you cannot get nail disease.

So, the probability is: 1 8 * 3 4 + 7 8 * 1 4 = 5 16

Have you become a person with disease A + disease B? That's wrong.

If II-2 does not carry the causative gene of the disease, it can be judged that the disease is recessive on the X chromosome, so the genotypes II-1, 2, 3 and 4 are XBXB, XBY, XBXB, XBYGenotypes III-1, 2, 3, and 4 are XBY, XBXB, XBY, 1 2XBXB or 1 2XBXB, respectively, and if the III-4 spouse does not have the disease, the genotype should be XBY. Therefore, the probability of giving birth to a child with this disease is 1 2 * 1 4 = 1 8, that is, when the mother is xbxb and the father is xby, the probability of giving birth to a child of xby.

III-4 and their spouses are all phenotypically normal albinism carriers, so their albinism genotype is AA, AA, and the probability of having an albinistic child is 1 4, and having only one disease means that the probability of having A and not B plus the probability of having B and not getting A, so it is 1 8*3 4+1 4*7 8=10 32=5 16. Note that the probability of replacing a child with one disorder needs to be determined that the child will not have another disorder.

After 20min, I found that I have been making -3 and -4 have children, which is really a dog.

II-2 does not carry the causative gene of the genetic disease, indicating that the disease is recessive on the X chromosome. If the probability of genotype III-4 is XBXB is 1 2, then the probability of children born to III-4 and their spouses is 1 2*1 4=1 8, the probability of albinism is 1 4, and the probability of only one disease is 1 8*3 4+7 8*1 4=5 16

If you only suffer from one disease, do you count both diseases?

The components of enzymes are proteins or RMAs, and their components are amino acids, nucleotides, respectively.

It is conceivable that aerobic respiration produces the same amount of carbon dioxide as it consumes, while anaerobic respiration does not absorb oxygen.

Hypoxia, anaerobic respiration produces alcohol, poisoning and rotten roots

Carbon dioxide inhibits aerobic respiration, thereby reducing the consumption of organic matter.

The cycle of matter and the reversal of energy are a single flow.

Same as the first three questions.

Two glucose should be 12(CH2O).

12CO2 + 12C5 compound 24C3 compound 24C3 compound 12(CH2O) + 12C5 compound.

The essence of most enzymes is protein, a few are Zhizhiling RNA, and the amount of carbon oxide released by aerobic respiration is equal to the absorption of oxygen.

Anaerobic respiration in most plants produces alcohol, which has a toxic effect.

Inhibits respiration.

The change of matter is reversible, the change of energy is irreversible, the conversion of ADP to ATP is the conversion of light energy into chemical energy, and the conversion of ATP to ADP is the conversion of active chemical energy to stable chemical energy.

Same suspicion car 3CO2+C5=2C3 2C3-->CH2O ; i.e. CO2, C5, CH2O1, 6 (C6H12O6); 2C6H12O6 has 12 C's and requires 12 CO2 and C5

It's that simple, find your own book!

Homework Help, can help you solve.

4. There are many kinds of cells in the root, and the long roots of plants must have a differentiation process. b The callus is produced in the phloem of carrots, which is a process of dedifferentiation (the process of differentiation from high to low differentiation). The production of plasma cells by C B cells is the process by which cells produce differences in stability and are differentiated.

d The loss of nucleus during maturation of human red blood cells is the continued differentiation of red blood cells.

15,d should be a variation in the number of chromosomes, and the number of chromosomes in the daughter cell has changed. The haploid karyotype is also incorrect, and haploid refers to an individual with a somatic chromosome count of the gamete chromosomes of the species.

1. A cell growth, that is, the process of increasing the volume of cells. As the cell volume increases, the ratio of membrane area to volume decreases, and the rate of material transport decreases (see the experiment on the rate of material transport in the textbook).

The first diagram is to choose C. Item A is formed by the change of hormone distribution and dedifferentiation after plant injury. As for b, dedifferentiation. As for D, there is no differentiation at all. It's just a structural change to adapt to its function.

In the second figure, the first half of option d is correct, and it is an addition to the chromosomal variation. The second half is wrong, and haploid has only one set of chromosomes.

In the third diagram, option A is wrong. The transport efficiency of substances is positively correlated with the ratio of membrane area to volume, and the transport efficiency decreases as the cell grows.