Regarding high school biology questions, dear professionals, I ask for a detailed analysis, urgently

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

Progeny DNA refers to two DNA molecules after the first replication.

In this question, one strand of the progeny DNA molecule is the p32 strand and the other strand is the p31 strand. And the two molecules are exactly the same.

The parental DNA molecule has 1000 base pairs. then each chain has 1000 p atoms.

Therefore, the average relative molecular weight decreased by 1000

The answer is D

a 1000 base pairs, i.e. 2000 deoxynucleotides. After two replicates, there were 4 DNA molecules and 8 strands, of which 2 strands contained P32 and 6 strands contained P31, so the average reduction was 2000*3 4=1500 (2 strands in 2 different DNA molecules).

The parental blueflower in the group contains two dominant genes, and the parental blueflower in the group contains one dominant gene, so the f2 in the group is 15:1 and the f2 in the group is 3:1.

From this, it can be seen that all three ABC items are correct.

Item D is wrong, the genotype of the white flower search plant is AAB, the genotype of the group F1 blue flower plant is AAB or AAB, if the white flower plant is crossed with the group F1 blue flower plant, the ratio of progeny to white flower plants should be 1:1.

The first question is indeed B that is wrong

Here's the explanation.

Analogical reasoning is just a kind of reasonable reasoning, to put it bluntly, it is to reasonably infer that genes are on chromosomes based on the parallel relationship between the behavior of contrasting genes and chromosomes, but this is not a rigorous proof, and the proof requires more rigorous experiments. Morgan's Drosophila experiment. It is proved that the white eye gene is located on the X chromosome, which also proves that the gene is located on the chromosome.

The second question is wrong with D

You can't see the order of the base pairs under a light microscope, so you can't see if there's a gene mutation.