A biological question about the calculation of genes

The ratio of homozygotes to heterozygotes in the second generation of medium-dwarf disease resistance is 1:2, that is, heterozygous accounts for two-thirds. The offspring of these heterozygous self-inbred offspring are one-quarter disease-resistant homozygous and one-half heterozygous.

Therefore, the total number of heterozygotes is two-thirds multiplied by one-half equals one-third. Whereas the total number of disease-resistant homozygous is one-third plus (two-thirds multiplied by one-quarter) equals one-half. Therefore, the pure sum ratio is heterozygous as one-half to one-third, that is, three to two.

This is because the original DDRR is 2:1 compared to DDR

There are two possibilities for them to self-bring, one is DDRR and DDRR, and the other is DDRR and DDRR

Then in the first case, the DDR: the DDRR is 2 3 *1 4:2 3 *2 4, the front is the probability of the DDR, the second is the probability of the result of this situation, and in the second case, the probability of the occurrence of the DDRR is 2 3 *1 2

Then the total ratio is 2 3 *1 4 +2 3 *1 2 :2 3 *2 4 =1 2 :1 3=3:2

Analyze from generation to generation.

The F1 genotype was all DDRR;

F2 is obtained from self-breeding, in which the short-rod resistance genotype is DDR, and it is easy to obtain 1 3DDR, 2 3DDRR (DDRR must be removed and then the proportion is calculated);

This is simple, F2 is self-inbred and the DDRR offspring is still DDRR, accounting for 1 3 of F3;DDRR self-inbred has 1 4DDR, 1 2DDR, multiplied by 2 3 to get 1 6DDR, 1 3DDRR

Then DDRR accounts for 1 3+1 6=1 2 of F3, and DDRR accounts for 1 3, so the ratio is 3:2

I wanted to use the ...... of the gene frequencyBut since it's self-breeding, there's no need for this It's a little confusing to write, and I don't know how to ask hi

Concept: DNA double helix diameter 2 nm; The helix week contains 10 base teams; The pitch is; The spacing of adjacent base pairs in planes.

Find the logarithm of deoxynucleotides first: 3 105 618 = pair DNA length: logarithm spacing of deoxynucleotides.

Volume: Base area multiplied by height.

Base area: squared] = square nanometer height.

Volume: cubic nanometers.

Number of turns: Number of nucleotide logs pitch turns.

Select C Analysis:

As you know, the genotype of d is AAAA and the pollen genotype produced is: 1AA : 4AA : 1AA

Therefore, the phenotype and its proportion of plants directly cultivated with pollen of d are 5a: the proportion of pollen aa of 1 aa is 1 6 so the proportion of inbred aaa is (1 6) 2 i.e. 1 36 . Then a accounts for 35 36 so the phenotype of d inbred offspring and its proportion is 35a:

1aaaa

F1 AA was obtained by hybridization with AA, and D AAAA, a was obtained after colchicine treatment;

b.Haploid individuals usually develop directly from unfertilized egg cells, b ;

c.From A, it can be seen that the genotype of D is AAAA, then the pollen genotype produced is AA :AA :AA=1:4:1, so the phenotype and its proportion are 5a :1AA, C;

d.The genotype of pollen produced by C is AA:AA:

AA=1:4:1 shows that AA accounts for 1 6 in gametes, so AAAA accounts for (1 6), that is, 1 36, so the phenotype and proportion of D inbred offspring are 35a:

1aaaa。∴d√.

f1: aa, the processed d d is aaaa, so a is correct;

Haploid is an individual developed from the gametes of this species, so b is correct;

The pollen of d has 1aa, 2aa, 1aa, so the phenotype of the pollen-bred plant is 3:1 correctly;

For example, the gametes and ratios of d are the upper three, so the phenotype of the offspring after fertilization is clear with a checkerboard, and it should be 15:1, so d is wrong.

d should be wrong because d is aaaa, and a is a percentage, so aaaa should be one-sixteenth.

Upstairs person's point of view, I don't agree that the genotype d of d is aaaa and the pollen genotype produced should be: 1aa : 2aa : 1aa

Gray straight hair, gray body forked hair, black body straight hair, black body split hair.

Female flies 3 4 0 1 4 0

Male flies 3 8 3 8 1 8 1 8

f and f are superscripts, the same below. ）

The title itself does not indicate who is autosomal and who is cooperable. It cannot be directly inferred. ）

Among the offspring females, the genotype and proportion of grey body were as follows: bb bb = 1 2

The genotype and proportion of straight hairs is: xfxf xfxf = 1 1

In the offspring of female flies, the probability of homozygous bbxfxf is; 1/3×1/2=1/6。Heterozygous is 5 6.

So among the offspring female flies, homozygous heterozygous = 1 5

1. The straight hair and bifurcated hair of the male and female are sex-separated, and the gray body and black body are still 3:1 Therefore, the parent is bbx(f)x(f) bbx(f)y, and the female fly with straight gray body is b?x(f)x(？）

For x(f)x(?) The probability of x(f) x(f) is 1 2 for b?The probability of :bb is 1 3

So the probability of homozygous is 1 6

Homozygous: Heterozygous 1:5

First of all, according to the title, it can be known that the genotype of the parent Drosophila is ffxbxb ffxbyb. In the offspring of female Drosophila, the genotype of XBXB accounted for 1 2, and the genotype of FF accounted for 1 3, and the proportion of pure and somatic in the offspring was 16, so the ratio of the two was 15

Combining all the experiments, it was concluded that the suit color was controlled by two pairs of alleles (set to aa, bb), because two or four is a deformation of 9331.

In experiment 1, purple to red is dominant. Comparing 24, it can be seen that purple needs AB, and red needs only A or B

In the second experiment, white nails and red obtained purple F1, indicating that they carried different dominant genes.

In experiment 3, white B is a true recessive homozygous AAB, because the offspring of A and B only have the dominant gene from A.

Experiment 4, purple is AABB

In summary, purple is dominant and requires ab to be present at the same time, and recessive homozygous is white. This is typical recessive epitaxis.

We can assume that AA is the epistatic gene, and when present, bb, bb, bb all appear white. And when there is only a present, a bb is red.

According to this, the white nail is AABB, and the red flower is AABB

Purple is AABB, red is AABB, and white nail is AABB

According to the characteristics of the segregation ratio of the two pairs of relative traits, it is not difficult to see that the flower color of the plant is controlled by two pairs of alleles (set to Ab two genes), and the normal gene segregation ratio of two pairs of traits should be 9:3:3:

1, except that one of the manifestations of the single manifestation of the question is the same as the manifestation of the double implicit. According to the data of experiment 2 and experiment 4, it is not difficult to see that purple is a double display, a red single display, and white is another single display and double concealment. According to the results of experiment 2, red (AAB) is crossed with white nail (AA, F1 is purple, so white nail must have B gene, so white nail is AAB; According to the experiment 4, purple (AAB) is crossed with white B (AA), no matter what genotype B is F1 must be purple, but if B is AAB, then F1 is AABB, and the genotype of F2 has AAB (purple):

AABB (purple): AABB (white) = 1:2:

1. It does not fit the topic, if B is AAB, F1 is AAB, and the phenotype of F2 is purple; Red; White = 9:3:4 fits the topic.

Purple aabb, red aabb, white nail aabb, white B aabb

F2 showed 9 purple:3 red:4 white, indicating that the trait was controlled by two pairs of alleles. From the conditions in the question, it can be seen that there should be a double dominant gene in purple, red is a dominant and a recessive, and white nail is a dominant and a recessive; White B is double hidden.

Experiment 2: Red and white nails, F1 is purple, F2 is 9 purple (double display): 3 red (one dominant, one recessive): 4 white (3 parts one dominant, one recessive; 1 part double recessive).

I learned this thing in college.,Is it so hard in high school now?。。。

In my impression, it is called the superposition effect, and the small B gene has a superposition to the A and A genes, that is, as long as the B gene is pure and BB, then no matter what the A gene is, the final result will show the trait of B, white. The composition of the A gene is considered only when the B gene is pure and BB or BB, AA, AA is purple, and AA is red.

So purple pure and aabb

Red pure and aabb

White armor aabb

White B AABB

Maybe so, I didn't have time to think about it in the middle of the experiment, so I'm sorry.