Senior 1 Compulsory 1 Biology Supplementary Study Case Question Help!!

The first one: 1is a biofilm.

2.Water molecules, gas molecules, non-polar fat-soluble molecules can pass through. Ions and macromolecular substances cannot pass through.

3.Small intestinal epithelial cells have carrier proteins that selectively absorb glucose molecules.

Similarity: both are transported across membranes.

Differences: whether carrier proteins are needed, whether energy is needed, and whether the pro-concentration gradient or the inverse concentration gradient is required.

Second sheet: Free diffusion: high to low concentration Carrier: None No energy required Oxygen.

Assisted diffusion: high to low concentration Carrier: Yes No energy required.

Active transport: low to high concentration Carrier: There is energy required for sodium and potassium ions.

This diagram is not a biofilm, which is a membrane structure based on a lipid bilayer and embedded with proteins and so on.

gases such as carbon dioxide, oxygen; Water; There are also fat-soluble substances that can freely pass through the lipid bilayer. Some ions, macromolecular substances cannot enter and exit freely.

The way in which glucose enters the epithelial cells of the small intestine is active transport. Glucose cannot pass through the protein-free lipid bilayer and can enter small intestinal epithelial cells mainly because of carrier proteins on the cell membrane of small intestinal epithelial cells.

The main difference between free diffusion and assisted diffusion (facilitated diffusion) is that assisted diffusion requires carrier proteins.

Direction of transport Carrier Energy Examples.

Free diffusion High to low concentration Not required Not required Water, gases, fat-soluble substances in and out of the cell.

Assists in diffusion of high to low concentrations that require and do not require glucose to enter red blood cells.

Active transport is not limited by the concentration required for glucose to enter the small intestinal epithelial cells.

The most soluble carotene on the filter paper is the least, and the solubility and pigment content are not proportional.

This is because in chromatography experiments, a pigment with high solubility only means that it moves the fastest, and it has nothing to do with the content. When we draw a fine line on the filter paper, the amount of each pigment is fixed, and it does not increase because of its high solubility.

The large degree of dissolution here just means that the pigment has a large solubility in the chromatography solution, and it has nothing to do with our extraction.

If the pigment is soluble in the extract at the time of extraction, perhaps its content can be large.

There is no relationship between solubility and pigment content, solubility is related to the speed of diffusion.

The amount is determined by the leaf itself, as for why it is not positively correlated, you can change the chromatography, maybe it is positively correlated.

What about water? b Some enzymes are RNA and also elements.

c Low temperature will cause the protein to lose its biological activity, but the temperature will be normal when the temperature is restored, c indicates the selective permeability of the cell membrane.

dRibosomes have no membrane structure.

The other options secrete proteins that are connected and can be converted into each other.

1, C is not necessary, you look at the compound of water, which does not contain element C, as we all know, water is the most important compound in the human body.

2. After the change of the spatial structure, there are two situations, one is reversible and the other is irreversible, which is closely related to the selected chemical decomposition reagent. Some RNAs have catalytic functions.

3. Option A, the university biology textbook proposes that the extensive pseudopodia of amoeba is related to the cytoplasmic flow in its cells, and option C is related to the selectivity of its osmosis, that is, selective absorption.

4,D cell membrane is connected to the Golgi membrane, Golgi membrane is connected to the endoplasmic reticulum membrane, and the endoplasmic reticulum membrane is connected to the nuclear membrane, and there is material transport between the three, so the boundary structure is similar.

1.The basic elements can be understood as the components and specific gravity of each element that makes up the cell. So b2BC is wrong.

B: Some RNAs also have catalytic activity and become ribozymes.

C: There is nothing to say, not all protein inactivation is irreversible.

3.The reason for choosing Ab,C is the selective permeability of the membrane, due to the sodium-potassium ion channels on the membrane.

D, the protein enters the nucleus because there are nuclear pores on the nuclear membrane, and proteins enter the nucleus through the nuclear pores, and Cd has nothing to do with fluidity.

4.Pay attention to the C upstairs, this question is: structure.

AB is a monolayer membrane structure.

d are all double-layer membrane structures.

C: The endoplasmic reticulum is a monolayer and the nuclear membrane is a bilayer.

1.Does the stem talk about the "most basic elements"?

It could also be RNA

c is the indication of the choice of permeability.

The other options secrete proteins that are connected and can be converted into each other.

a. According to the analysis of the figure, the chromosome of Figure B has been translocated, which belongs to the variation of chromosome structure, and a is correct;

b. Screening is the screening of chromosomal mutation of individuals, which can be used with a light microscope, while screening is the offspring produced by genetic recombination, which cannot be observed under the light microscope, b is correct;

c. According to the chromosome composition of Figure B, it can be seen that the proportion of gamete abnormalities produced by it is 3 4, so the cells of 3 4 Drosophila in F1 contain abnormal chromosomes, and c is wrong;

D. It is known that individuals without this allele cannot develop, so the ratio of gray and black fruit flies in F2 is 2:1, and D is correct

Therefore, c

Seed C is developed from ovules, and one seed corresponds to one ovule, so there are 2 ovules.

The fruit is developed from the ovary, and it can be considered that the peanut shell part is developed from the ovary, so only one coax letter ovary (one ovary can hold several ovules).

Since peanuts are dicots, dicots are double-fertilized, i.e., two sperm cells are required to form a fertilized egg, one sperm cell combines with the egg to form a fertilized egg, and the other is combined with the polar nucleus to close the wheel to form the fertilized polar nucleus. To form two seeds, four sperm cells are required. One pollen can produce two sperm cells, so two pollen are needed.

The answer is 1,2,2,

I choose B, where A is the legacy of the previous year, not the total energy inflow per year, where N1 is greater than N2 and W1 is greater than W2

c b and d can represent the energy used by the decomposer and the energy that is not utilized, respectively, and d is similar to c.

b, excluding a, because a is the previous year.