Biology questions, urgent answers, please point

This is a biology mock question for the college entrance exam, and I did it, and your answer is correct.

We assume that the genotype of the parent is AABB [黄甜] (paternal parent) aabb [白非甜] (female parent).

Then the genotype of the child generation is AABB [non-sweet white seed coat].

The offspring is then inbred to F2 and the genotype of F2 is a*b* a*bb aab* aabb, but the seed coat of F2 is yellow [same as F1 genotype].

Then F2 is inbred again The probability of AA in F2 accounts for 1 4 (i.e., the white seed coat, and the offspring produced by its mother must be a white seed coat).

We then consider traits determined by embryonic genes in F2 BB 1 4 BB 1 4 BB 1 2

The probability of non-sweetness in the seeds formed by f2 is (1 2*1 4+1 4=3 8).

That is, 1 4 * 3 8 = 3 32

In fact, the genotype of the seed coat should consider the genotype of F2.

The sweet and non-sweet kernels are based on the normal algorithm of taking into account the genotype of the seeds produced by F2.

The first empty in your answer is wrong, the dominant trait in the color of the seed coat is white seed coat, because the first generation of seed coats are white seed coats, assuming that the gene of homozygous yellow seed coat sweet corn is yyxx, and the homozygous white seed coat non-sweet corn is yyxx, then there are 16 gene combinations of F2 (can the combination be formed?). All genes with y in the gene are white seed coats, but the other part must be two xx is sweet corn, so it accounts for three-sixteenths, or the two traits are separated, the genes of the F2 generation must be yyxx, the F2 generation is inbred yy yy, the offspring have y accounted for three-quarters, the xx xx offspring have xx accounted for a quarter, 3 4 * 1 4 is 3 16, I say this I don't know if you understand, hehe, my expression ability is not very good.

The dominant trait is yellow seed coat, because F1 all bears yellow seed coat, which is the genotype of F1. F1 itself is non-sweet (when they are still seeds), so non-sweet is the dominant trait.

Therefore, the color of the seed coat of the seeds produced by F2 is determined by F2's own genes, and the white seed coat is only 1 4.

The sweet and non-sweet traits of seeds produced by F2 are determined by the genes of F3 (so even if the two pairs of genes are not related to each other on the homologous chromosomes, they are still independent of each other), non-sweet is the dominant trait, corn and peas are pollinated in different ways, generally not self-pollinated like peas, and the distribution of traits is similar to that of F2. Sweet only 1 4.

b.Methylene blue stain.

Methylene blue, also known as methylene blue, blue, is a non-toxic dye, its oxidation type is blue, reduced colorless, when dyeing the living cells of yeast with melan, due to the metabolism of cells, the cell has a strong reducing ability, which can make the blue from the blue oxidized type to the colorless reduced form. As a result, the reducible yeast cells are colorless, while dead or senescent cells with weak metabolism are blue or light blue, allowing the yeast to distinguish between dead and live yeast cells. It is often combined with eosin acid dyes to form Wright's dye.

Select D for supplementary questions

Because the yeast will convert glucose into water and carbon dioxide, if you use sucrose, the yeast also needs to use sucrase to turn sucrose into glucose for reuse, one more step, it is definitely not as good as adding glucose directly, and a large amount can ensure that it is sufficient.

b Methylene blue should be a stain for tissue sections and bacteria. d

Ribosomes are mitochondria.

A has to pass through the membrane but wears two layers of nuclear membrane, and then it enters the ribosome, and the mitochondria enters the pigment C

Ribosomes are mitochondria.

A Upstairs, you don't need to wear a membrane to enter the mitochondria?

1) Elogeneity (genetically between the offspring of the individual and the parent.

Similar. The phenomenon is called heredity; Variation refers to the existence between offspring and parents, as well as between offspring individuals.

Difference. phenomenon).

aa aaaa can be known from the second set of results.

Tongue curling. It is a dominant trait and cannot be tongue curled.

Recessive traits. So it can be judged. Xiao Ming. Target.

Genotype. is AA, and one of his two genes comes from his father and the other from his mother, so his parents each have a gene A, and because his parents both show dominant traits, all have A genes, so the genotypes of Xiao Ming's father and mother are all AA; Xiaoshan can't roll his tongue, so Xiaoshan's genotype is AA, and one of the A genes comes from his mother, so Xiaoshan's mother must also have an A gene, which is manifested as being able to roll her tongue, and she must also have a gene).

When a student was doing an experiment on the digestion of starch by salivary amylase, the steps were as follows:

Inject 2 ml of 10% starch paste into one tube, add 2

Immediately after ml of saliva, drop the iodine solution, put the test tube into a water bath in warm water at 37 degrees Celsius for 10min, and then observe the phenomenon. The observation recorded by the student was that the starch paste did not change blue, and the experimental conclusion was that salivary amylase had a digestive effect on starch.

1) The student's shortcomings in experimental design are 1No controlled trials were done2The time for adding iodine solution in the experiment was not suitable.

2) The errors in the experimental protocol are ---1No controlled trials were done2The time for adding iodine solution in the experiment was not suitable.

3) The experimental phenomenon observed by the student should be --- turning blue.

4) Therefore, the conclusion reached by the student should be --- wrong.

(1) The student's shortcomings in experimental design are that there is no control ---.

2) The error in the experimental procedure is --- to add the enzyme solution first, then heat, then add reactants, and finally test and observe.

3) The experimental phenomenon observed by the student should be --- test tube A turns blue, and test tube B does not change blue.

4) Therefore, the student should conclude that --- salivary amylase has a digestive effect on starch.

P.S. The correct step is 1).

Inject 2 ml of saliva into one tube A

2) Inject 2 ml of distilled water into another tube

3) Put the test tube into a water bath in warm water at 37 degrees Celsius for 10min.

Inject 2 ml of 10% starch paste into both tubes.

5) Drop in iodine solution and observe the phenomenon.