How many phenotypes are there in the four pairs of relative traits F2? 5

The trait controlled by two pairs of genes is self-crossed, and the gene AABB of the F1 offspring generation is crossed with AABB due to freedomThe Law of SeparationSixteen plants with nine genotypes were produced, of which AABB accounted for one part, AABB accounted for two copies, AABB accounted for one part, AABB accounted for two copies, AABB accounted for four copies, AABB accounted for two copies, AABB accounted for one, AABB accounted for one, if A, A controlled green or yellow, B, B controlled the wrinkled trait, A was dominant, A was recessive, B was dominant, B was recessive. So there are nine in total.

Know. How many phenotypes of the relative trait F2 are in the four pairs.

Why is it that in the Mendelian cross-breeding experiment of two pairs of relative traits, there are four phenotypes of F2 and 9 genotypes?

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Haven't you seen that picture in your textbook?

Traits controlled by two pairs of genes are self-crossed

The genes of the F1 subgeneration were crossed with AABB, and due to the law of free segregation, 16 plants of 9 genotypes were produced, of which AABB accounted for one

AABB accounts for two copies

AABB accounts for a share

AABB accounts for two copies

AABB accounted for four copies

AABB accounts for two copies

AABB accounts for a share

AABB accounts for two copies

AABB accounts for a share

If a、a controls green or yellow、

b, b control wrinkle traits

A is explicit, A is recessive

b is explicit and b is recessive. So there are nine in total.

In the inheritance of a pair of relative traits, there are 3 genotypes in the F2 generation, with a ratio of 1:2:1.

For example, F1 is AA, and there are two kinds of gametes, A, and A, which are combined and have them. There are two phenotypes, with a ratio of 3:1, and because 1aa, 2aa phenotypes are the same, so 3:1.

1) The first group appears due to the appearance of 9:3:3:

The segregation ratio of 1 is (3:1) (3:1), which means that two pairs of heterozygous are self-inbred, and the parental genotype is AAB AABB The second pair is 1:

The separation ratio of 1:1:1 is (1:.)

1) (1:1), indicating that the two pairs are crossed, then the genotype of the parent is aabb aabb or aabb aabb The third pair appears 3:3:

A separation ratio of 1:1, split is (3:1) (1:.)

1) Indicate that one parental pair is heterozygous self-inbred, and the other pair is assayed, and the parental genotype is AABB AABB or AABB AABB

2) The occurrence of the above segregation ratio must conform to the law of free combination, that is, the free combination of two pairs of genes should be satisfied, and the conditions of equal chance of binding of various male and female gametes and the survival rate of offspring should be the same Therefore, the answer is:

Phenotypic proportion.

Parental genotype.

aabb×aabb

AABB AABB or AABB AABB

AABB AABB or AABB AABB

2) Two pairs of genes are freely combined, the combination of male and female gametes is equal, and the survival rate of offspring is the same.

AABB is self-inbred, looking at it separately, AA generates two phenotypes, and BC is also, so AABB has two times two, which is four phenotypes. AA has three genotypes, BB is the same, so there are three times three, nine genotypes.

Hope it helps!

Haven't you seen that picture in your textbook?

Traits controlled by two pairs of genes are self-crossed

The genes of the F1 subgeneration were crossed with AABB, and due to the law of free segregation, 16 plants of 9 genotypes were produced, of which AABB accounted for one

AABB accounts for two copies

AABB accounts for a share

AABB accounts for two copies

AABB accounted for four copies

AABB accounts for two copies

AABB accounts for a share

AABB accounts for two copies

AABB accounts for a share

If a、a controls green or yellow、

b, b control wrinkle traits

A is explicit, A is recessive

b is explicit and b is recessive.

So the trait expression of the above nine gene types is different

Firstly, biological individuals (homozygous) with two pairs of relative traits controlled by these two pairs of genes (the premise is that these two pairs of genes control two pairs of relative traits respectively) were selected for crossbreeding to obtain F1. For example, aabb aabb, f1 is aabb. It can then be judged in two ways:

1) F1 inbreeding, if the offspring has four phenotypes, and the ratio is 9 3 3 1, then it can be judged that these two pairs of alleles are located on two pairs of homologous chromosomes, and their inheritance follows the law of free combination. Analysis: Because AA and BB are located on two pairs of homologous chromosomes, AABB can produce four gametes, namely AB AB AB AB=1 1 1 1, male and female gametes are randomly combined, and the offspring will appear A-B- A-BB AAB- AABB = 9 3 3 1.

If this is not the case in the offspring, it is known that the two pairs of alleles are located on the same pair of homologous chromosomes, following the law of linkage inheritance. Let a and b be on one chromosome and a and b on one chromosome. There are two more cases:

One is that there is no cross-exchange, then AABB only produces two gametes, that is, ab ab=1 1, male and female gametes are randomly combined, and the offspring will appear a-b- aaabb=3 1, which is a complete linkage. One case is that cross-swapping occurs, then ABB will produce four gametes, namely AB, AB, AB, AB, AB, of which AB and AB account for the majority, and AB and AB account for the minority, so that with the random combination of male and female gametes, the offspring will also appear four phenotypes, but the traits of A-B- and AAB- will be in the majority, and the traits of A-BB and AAB- will be in the minority. This is an incomplete chain.

2) F1 test delivery. From the analysis of root (1), it can be successfully inferred that if there are four phenotypes in the crossed offspring, and the ratio is 1 1 1 1, then it can be known that these two pairs of genes are located on two different pairs of homologous chromosomes, and they follow the law of freedom. If there are two phenotypes in the offspring, then it can be known that the two pairs of genes are located on the same pair of homologous chromosomes, which is fully linked.

If there are four phenotypes in the offspring, but the individuals with the same traits as the parents account for the majority, and the individuals of the recombination type account for the minority, it is incompletely linked.

The most essential reason is actually the essence of the free combination of genes. That is, when F1 produces gametes, alleles are separated, and non-allelic genes on non-homologous chromosomes are freely combined. I hope it helps you, and if you don't understand anything, you can still ask

Two pairs of relative shapes (aabb aabb) f2 represents the present type there are 2 2 kinds 9:3:3:

1. Three pairs of relative traits (AABBCC) AabbCC F2 represents the present type, there are 2 3 types, 27:9:9:

There are 2 n kinds of respect for the relative trait F2 for the relative trait F2 and the result is 3 n:3 (n-1):

Honest judgment ......3^(n-2)：…3^(n-n)