In the first year of high school, the master will help analyze it

The first problem: the hydrolysis of proteins basically belongs to the category of organic chemistry, and there is no principle required in biology. So the landlord just needs to memorize the examples listed in the book.

There is no need to forcibly resolve the hydrolysis of arbitrary proteins. (The college entrance examination is only in the book or supplemented by the teacher).

The second question: First of all, let's talk about the answer a, after heating on the alcohol lamp, the plant cells will be colored, and it is impossible to recover, so it is excluded.

B answer, a certain concentration of sucrose solution is not clear, and the concentration of plant vacuoles is not clear, this answer is not considered for the time being.

C Answer, nitric acid can indeed observe plasmo-wall separation, but this conclusion is not the effect of the cell wall, but another, and the plant has died.

The last answer is the conclusion. So choose D.

Problem 1: The basic unit of protein is amino acids, and generally speaking, protein hydrolysates are -amino acids.

Problem 2: After the isolation of the cytoplasmic wall, it still has normal physiological functions because it can be restored.

b is the reason that the concentration of the sucrose solution = = has not yet been stated

Question 1: Natural amino acids are amino and carboxyl groups attached to the same carbon, and there is also a hydrogen atom and an R group on the carbon.

Question 2: Item B does not state the concentration... If the sucrose solution is the same as the intracellular sucrose concentration, there will be no plasmo-wall separation.

LZ Biology class slept?

1. I remember that natural protein hydrolysis produces a-amino acids, that is, amino acids with amino and carboxyl groups attached to the same carbon, and there is no limit to the artificial synthesis of amino acids.

Second, let's not talk about the problem of a certain concentration, does B say that it is a living cell?

Choose A. Antibodies are specific.

B, lysozyme in tears is non-specific immune (first line of defense). Humoral immunity is specific immunity dominated by B cells.

c, it is the plasma cells that produce antibodies. Secondary immunity is mainly caused by the rapid proliferation and differentiation of memory cells into plasma cells, and then the production of antibodies.

d, AIDS patients, will eventually lose all immunity. Moreover, in the process of humoral immunity, T cells are involved (releasing lymphokines and promoting the proliferation and differentiation of lymphocytes).

D T cells are transported through humoral fluids, so there's humoral immunity wrong.

c The secondary immunization process is certainly not this one process, the most important thing is that the antibody recognizes the antigen, which is a bit of a word-picking mistake.

A T cell that recognizes antigens is not phagocytic cells.

Antibodies secreted by plasma cells in AA are selected to be specific.

Lysozyme in B is involved in non-specific immunity, at least in the high school stage, memory cells in C do not produce antibodies, and the first differentiation effect should be B cells, T cells in D can present antigen information in the induction stage of humoral immunity, and can also secrete lymphokines.

Plasma cells secrete antibodies that recognize specific antigens.

The mode of immunity in which antibodies are produced by B lymphocytes is called humoral immunity, and secondary immunity is not caused by the rapid production of antibodies by memory cells, but plasma cell HIV also affects humoral immunity.

a, the antibodies secreted by plasma cells have specific immunity, and plasma cells do not have the function of recognition.

a。Antibodies secreted by plasma cells can specifically recognize antigens. b does not belong to enable non-specific immunity. The c-reflection stage is the specific recognition of antigens. d, T cells secrete lymphokines to help humoral immunity, so they indirectly affect humoral immunity.

C, don't blame me if you're wrong, the college entrance examination has been away from me for three years, you can take a look at the process of immunization in the book, if the book is still there, it may be able to help you.

aThe answer should be the original words in the textbook.

dThe answer is incorrect. A system that can complete life activities independently is called a living system, and a cell must be a single-celled organism in order to be a living system.

Living system: cell-tissue-organ-system-individual-population-community-ecosystemCells are living systems, and it's true that cells are the most basic living systems, but this is a data question.

The topic is only written about the photosynthesis (function) of mesophyll cells, and the beating of the animal heart is through the contraction and relaxation of cardiomyocytes (also function).

It has nothing to do with the living system, so D is not selected

Such data questions must start with the data, otherwise it is easy to make mistakes.

The mesophyll cells of green plants can carry out photosynthesis, and the beating of the animal heart is done by the contraction and relaxation of cardiomyocytes. Illustration of the above example (a.).The cell is the basic unit of structure and function of living organisms).

It's not a living system, it's a unit.

A mesophyll cell or cardiomyocyte cannot survive on its own, so it is not a living system.

A system that can complete life activities independently is called a living system. Whereas, life activities have several basic characteristics, namely metabolism, ability to grow and develop, ability to reproduce and genetics, and stress.

Obviously, mesophyll cells and cardiomyocytes cannot meet the characteristics of independent reproduction and heredity, so they cannot be said to be living systems.

D, for single-celled organisms, is correct. However, the example in question does not lead to this conclusion.

The structure and function of the organism are described in the structure and function of the organism by performing light and action and completing the beating of the heart through the contraction and relaxation of new myocardiomyocytes.

So choose A

Because this question is about function, photosynthesis, and the beating of the heart, all of them are talking about the function of cells. But what he asked to do the question depended.

For example, the example "most animals and plants are multicellular animals" can illustrate d.

That's it. Landlord,Crack down on the copying party.。。

Of course, it is A, this question does not explain BCD, and the A question reflects the function.

Select CE analysis: This topic is a two-factor variable type, and a single variable should be used for analysis when solving the problem.

According to the figure, growth regulator X has the effect of promoting rooting on leafy shoots (compared with nutrient treatment), item A is correct;

After nutrient treatment, the experimental results of leafless branches and leafy branches were the same, and b was correct.

After the treatment of growth regulator X, the promotion effect of leafy shoots was significant, but the results of leafless shoots did not change (compared with nutrient treatment), C was wrong.

In the presence of leaves, the rooting effect of growth regulator X alone was similar to that of nutrients and growth regulator X in the absence of leaves, indicating that the leaves may have produced substances similar to the effects of nutrients, and item d was correct.

B says that nutrients have no obvious effect on root formation, so it is of course wrong to say that nutrients and growth regulator X are beneficial to root formation, and E is wrong;

For leafless shoots, the addition of growth regulator X or nutrients alone had little effect on rooting, while the presence of nutrients and growth regulator X had a promoting effect on rooting, and item f was correct.

Hope it helps

b False. Has an effect on leafless branches. and growth regulator synergist. Compare the first and second sets of histograms.

The first group of histograms showed that somatomodulin alone promoted rooting of leafy shoots.

The second group further showed that the histogram showed that, along with nutrients, somatomodulin promoted rooting of leafless shoots.

The third group of nutrients alone did not promote the rooting of branches.

Answer C, Description: A, from the first comparison; b, this conclusion can be reached in the third place;

c, this experiment only made the conclusion of the branches of one plant, if it is a high concentration is also promoted?

d, the first white and the second white contrast can be launched;

e, the first place has less rooting without leaves, and the second place can be rooting when there is no leaves, indicating that the nutrients are beneficial to rooting;

f, the second place is also available.

b Illustrated by the two sets of histograms.

The only variable in the 1 and 2 groups showed that the number of roots increased significantly after the addition of nutrients to leafless shoots. The number of roots in leafy branches is almost constant, indicating that the leaves produce some kind of nutrient-like substance.

Three kinds of processing, comparison between two and two, grasp the variable analysis, and you can do it.

bcd b: the comparison under the leafless condition can be known to have an effect c: this question cannot be deduced from the result d: the number of roots does not change when there is a leaf condition, d is wrong.

The man's younger brother does not have polydactyly, so both parents are AA, so the man's AA probability is 1 3, and the AA probability is 2 3;

The man's younger brother is BB, so the man's parents are BB, so the man's BB probability is 1 3, and the BB probability is 2 3;

The woman is normal, so she is AA and has a brother with albinism, so the woman's BB probability is 1 3, BB probability is 2 3;

To sum up the childNopeProbability of polydactyly: paa=2 3*1 2=1 3;YesProbability of albinism: PBB = 2 3 * 2 3 * 1 4 = 1 9

Therefore, the probability that the child is not sick is p=1 3*(1-1 9)=8 27, and the probability that the child is sick is 1-8 27=19 27

1.The peptide is composed of several amino acids, of which how many glutamic acids are there?

10 amino acids.

Because there are 10 n

Glutamic acid can be listed as an equation.

4x+2(10-x)-9=19

Solve according to the number of o.

x=42.It is known that the glutamate molecule contains two carboxyl groups, so how many amino groups and how many carboxyl groups should this peptide have?

There is only 1 amino group.

at the n-terminus of the peptide chain.

5 carboxyl groups. 1 on the C-side.

4 on glutamic acid R group.

3.How many water molecules are needed to hydrolyze this peptide into amino acids?

9 because it's a 9 peptide.

If the light intensity is equal to 3, it only ensures that the carbon dioxide absorption is zero during the day, and the rate of respiration is equal to the rate of photosynthesis, but what about at night (12 hours of day and 12 hours of night in a day)?

It is known that the release of carbon dioxide per unit time of respiration is 4, and a day is 4*24=96. 12 hours of day, i.e. the actual rate of photosynthesis is 96 12 = 8. The origin coordinates are (0,-4), so you can find (x,4) points. The corresponding x is 6

Breathing is also performed during the day.

Do you see the "12 hours of day, 12 hours of night" given?

The one on the first floor has a problem, and the absorption rate in the figure is positive and negative, which means that it is the total rate (absorption-release). 12 hours at night 12*(-4)=-48 12 hours during the day then 48 48 12=4 at 6 kg of Les is absorbed