Help, high school biology, ask a high school biology question

The idea of option d is the same as that of the experiment done by Saxophone in the book, if the whole leaf should be blue according to the operation method of the option, the first thing to do in this kind of experiment is to starve and consume the accumulated starch in the experiment.

Option b here wants to emphasize that the site of photosynthesis is in the chloroplast, and the place of starch production is in the chloroplast, if we follow your idea we also have to distinguish whether this is a C3 plant or a C4 plant It is too complicated to consider.

The answer is d

a Chlorophyll absorbs most of the red and violet light and reflects green light. A is correct b chloroplast is the place where plants carry out photosynthesis, plant photosynthesis produces starch, starch is blue when exposed to iodine so b is correct.

c Undoubtedly, correct.

dNote that the leaf was not removed for this experiment around 10 a.m., but in the afternoon. Nutrients are still present in the area covered by tin and platinum foil, so it is heated with alcohol and decolorized, and after being treated with iodine solution, the entire leaf is blue.

I think b is wrong The starch produced by chloroplasts is also transported to other parts of the leaf, and other parts will also appear blue.

1.This problem is mainly considered from the two steps of light and action, light reaction and dark reaction.

The light reaction mainly requires light, while the dark reaction is mainly related to the concentration of CO2.

In the XY segment, the diagram shows that the light intensity is relatively low during this period, and the plants need to absorb oxygen from the outside for aerobic respiration. This is a low-light environment, and photosynthesis is divided into two steps, light reaction and dark reaction. At this time, the light intensity is low, which affects the progress of the light reaction.

In the YZ segment, the light and action intensity on the graph are gradually increasing, and we need to consider the light and two steps, which affect both the light intensity and the CO2 concentration.

In the Z segment and beyond, the photosynthetic rate does not increase with the increase of light intensity, and the CO2 concentration related to the dark reaction is considered to be the most likely influencing factor, that is, the upper limit of CO2 uptake by plants is reached, and the light saturation point is reached. You said why you didn't fill in the temperature, because you can't see the relationship with temperature in the description of the question, and you asked about the main limiting factor.

2.The experimental setup depends on the difference between the light and the effect and the respiration, which is the amount of net photosynthesis. What is called net photosynthesis, that is, black algae will breathe and light, and respiration consumes oxygen, but light and sum can be produced, so black algae respiration will use the oxygen produced by light and action, and the oxygen discharged into the external environment is the amount of oxygen released.

Of course, the CO2 produced by the respiration of black algae will also be used by light, and the amount of oxygen produced by plants is generally more than CO2, (otherwise there will be no oxygen used by animals), so after this experimental setup, the experimental results will be seen from the tube of the colored droplets, that is, the released oxygen will gradually accumulate, and the pressure generated by the oxygen will press the small droplets to the right side.

Black algae itself is an aquatic algae, so it can breathe in water.

(1) The first void: the light intensity is not enough, resulting in the decrease of photosynthesis intensity, which can be seen from the chart: the horizontal axis represents the increase of the amount of light intensity, and the photosynthetic intensity gradually increases with the increase of it; When the amount of oxygen absorbed is 0, it is a critical point.

Plants need carbon dioxide to release oxygen, so the second void is light intensity and CO2 concentration; The third null has no temperature because the question sets the temperature to be constant at 30.

2) It can be respirated, but it is very weak and can be disregarded. It's like measuring the photosynthesis of plants at noon.

As far as the first question is concerned, the main factors affecting respiration are temperature, oxygen, carbon dioxide, light intensity and photosynthesis, but not respiration, and all temperatures are more important than CO2 concentration.

The second question is that black algae contains chlorophyll, which can be photosynthesized by light irradiation, produce oxygen, and can be respirated, and the precipitation of oxygen in CO2 buffer will cause small colored droplets to move.

(1) Because it has been indicated that it is at the optimal temperature.

2) This is an experimental device that measures the net photosynthetic rate, which can be used for respiration and photosynthesis, so it involves the change of two gases, oxygen and carbon dioxide, but he uses carbon dioxide buffer, which means that carbon dioxide will not cause a change in the volume of gas in the bottle, so the movement of the droplets is the oxygen produced by photosynthesis minus the oxygen absorbed by breathing, which is the net photosynthetic amount, understand?

Let A be the normal gene and A be the galactosemia gene.

The wife's father is homozygous and does not have the disease, so the wife's father is AA and the wife's maternal grandmother is AA, so the wife's mother is a carrier AA so the probability that the wife is a carrier AA is 1 2

The husband's brother fell ill, which proved that the husband's parents were both carrying AA

So the probability that the husband is a carrier AA is 2 3

Therefore, the probability of having a child is 1 2 2 3 1 4 = 1 12, and a should be chosen

The probability of husband 2 3 is AA, the probability of wife 1 2 is AA, and the probability of son disease is 2 3x1 2x1 4=1 12

By doing this with my students, I think your question appears on (2).

According to Crick's theory, if ribosomal RNA is a messenger, then heavy ribosomes are formed before the phage does not infect the bacteria, and should carry the genetic information of the bacteria, while the light ribosomes are synthesized after the phage infects the bacteria, according to the title, "After the phage infects the bacteria, the protein synthesis of the bacteria immediately stops, and the protein of the bacteriophage is synthesized instead." "The light ribosomes carry the genetic information of the bacteriophage. Light and heavy glycosomes are different because of the n content, one is 14N, and the heavy is 15N, but they are both non-radioactive.

We distinguish them by their position in the centrifuge tube.

Whereas amino acids are labeled with 35-s to detect who will have protein synthesis on light and heavy ribosomes.

According to the initial assumption, if ribosomal RNA is a messenger, the ribosome synthesized only phage proteins, and only light ribosomes should be radioactive; If both light and heavy ribosomes are reflexive, it means that both ribosomes can synthesize phage proteins, which means that ribosomal RNA is not a messenger.

I don't know if I explain it this way, but you understand

The option for this question is correct, because the ribosome does not contain DNA (contains rRNA and tRNA at translation), so there will be no thymine deoxynucleotide (T in DNA, U in RNA). In option D, O2 is inhaled by mice and used in the body for the third stage of aerobic respiration, combined with [H] to form H2O, and the resulting water is then used for the second stage of aerobic respiration, and the O in H2O will be in CO2, so D is wrong.

B is wrong because the label with h is deoxynucleotide, it includes bases, deoxyribose and phosphoric acid, the main structure of ribosomes is RNA, it and deoxyribonucleotides have phosphoric acid, H can label phosphoric acid, of course, you can detect radioactivity.

b, because it is labeled with hydrogen. Deoxynucleotides dehydrate and condense when they form DNA. With radioactive hydrogen in the water, there is no guarantee that radioactivity will not be detectable elsewhere.

Who fooled you into saying that you were bwrong??? B: Because plants can't directly use organic matter. Ribosomes are DNA-free.

A: Wheat is a C3 plant, so that's right.

D: After the rat inhales 18O2, it has to perform aerobic respiration, the equation: C6H12O6 + 6H2O + 6O2---6CO2 + 12H20 + energy.

That must be 18o d wrong.

B phages infect bacteria, that's certainly right, don't explain...

I'm too little... The boys in front of me are less misguided, so don't answer if you don't...