The phenotype and ratio of F1 offspring mainly depend on

If F1 is heterozygous, then the phenotype of the offspring is part of the dominant trait, part of the recessive trait, if F1 is homozygous, then the phenotype of the offspring must be the dominant trait.

Yes, question A asks about phenotype, not genotype. The phenotype and ratio of F1 offspring are mainly influenced by environmental conditions.

For example, the F1 generation is a homozygous rice with a tall stem, and the final mature rice is similar to a dwarf stalk due to the poor land.

Crossing refers to the crossing of a heterozygous and a recessive homozygous. So the genotype and proportion produced are certain! (I'm a pair of genes, you can also think about it one-on-one).

So I think it should be the influence of the environment!

The type and proportion of gametes produced by Cf1 should be chosen.

It is better to believe in the book than to have no book.

Agree that the upstairs phenotype has a lot to do with the environment.

Summary. If AB and AB are linked, and are not completely linked, that is, the genotype of F1 is AB AB, when producing gametes, they will be swapped, that is, there will be four genotypes of gametes: AB, AB, AB, AB, the proportion of these four gametes is unequal, of which AB and AB account for the majority (these two proportions are equal), and AB and AB account for a minority (these two proportions are equal).

F1 assays, that is, AB AB and AB AB types mate, so that they can produce offspring of four genotypes, namely AB AB, AB AB, AB AB, AB AB, and the first two proportions are equal and the majority; The latter two are in equal proportions and account for a minority.

If AB and AB are linked, and are not completely linked, that is, the genotype of F1 is AB AB, when producing gametes, there will be interchange between them, that is, there will be four genotypes of gametes: AB, AB, AB, AB, the proportion of these four search gametes is unequal, of which AB and AB account for the majority (these two proportions are equal), and AB and AB account for a minority (these two proportions are equal). F1 assays, that is, AB AB and AB AB types mate, so that they can produce offspring of four genotypes, namely AB AB, AB AB, AB AB, AB AB, and the first two proportions are equal and the majority; The latter two are equal in proportion and account for a minority.

Fellow, I really didn't understand, I can be more specific.

If AB and AB are linked, and are fully linked, i.e., the genotype of F1 is AB AB, there will be no interchange between them when gametes are produced, that is, only two genotypes of gametes can be produced: AB and AB, with a ratio of 1:1.

The offspring will produce two types of genotypes, ab ab and ab ab, with a ratio of 1:1.

There are four genotypes of BABB, AABB, AABB, and AABB, and now the present type of DU is zhi1:3, so it can be guessed that this 1 part is double DAO dominant or double rectile or single gene single manifest (no matter why the relative 3 copies will show the same traits, it does not affect the solution).

Assuming that 1 is a double display, F1 inbred descendants are double display: single hidden + double hidden = 9:7, a is possible.

Hypothesis 1 is a double hidden, F1 inbred double hidden: dominant = 1:15, b is possible.

Hypothesis 1 is a single gene single display (assuming that it is A), only A single display: double display + double hidden + B single display = 3:13, C pair.

Hence the choice D. The most difficult thing about this question is what to assume what it is, especially the third hypothesis, some students may think, how can it be aabb: aabb+aabb+aabb=1:3, how can the phenotypes of the last 3 genotypes be the same, because the current knowledge points cannot be explained, but it does not prevent us from making bold assumptions.

a. The genotype of F1 can be detected by the crossing test, but it cannot directly reflect the essence of the law of gene segregation, and A is wrong;

The phenotypic ratio of B, F1 inbred offspring is 3:1, which cannot directly reflect the essence of the law of gene segregation, and B is wrong.

c. When producing gametes, genes that control the same trait are separated from each other, c is wrong;

d. The essence of the law of gene segregation is that alleles are separated with the separation of homologous chromosomes, that is, when gametes are produced, genes that control different phenotypes of the same trait are separated from each other, d is correct

Therefore, d

a、f1

The phenotype and ratio of the inbred offspring mainly depend on the type and proportion of gametes produced by F1, but due to the influence of environmental conditions, a is wrong.

b. The genotype of the other parent intersecting with F1 should be recessive homozygous, B is wrong;

c. Since the crossing is a hybridization between F1 and recessive homozygous, the type and proportion of gametes produced by F1 determine the phenotype and ratio of the offspring, and C is correct;

d. The other parent is recessive homozygous and can only produce one type of gamete, so c

Detailed travel dissipation:

Since it's 10 16 and 6 16, then obviously F1 is AABB

Since the ratio of inbred offspring is 10:6, i.e., double display (aabb, aabb, aabb, aabb) and double hidden (aabb) are a phenotype of the bridge book, accounting for 10 16. Single display (aabb, aabb, aabb, aabb) is a phenotype that accounts for 6 10.

A pair of heterozygous dominant genes and a pair of homozygous recessive genes were crossed, and the offspring F1 gene type was two types with a ratio of 1:1