High School Biology Heredity and Variation, Biology High School Heredity and Variation in High Schoo

Because there are non-sweet corn seeds on the ears of sweet corn, and the sweet corn plant itself is purebred, the gene of non-sweet corn is dominant, and the dominant gene is A, and because the sweet corn plant itself is purebred, the gene is recessive, and it can only be Aa

Sweet corn seeds are not found on the ears of non-sweet corn, so the non-sweet corn plant itself is also purebred, which is AASweet corn plants (AA) were crossed with non-sweet corn plants (AA), and the sweet corn plants themselves were hybridized, and the ears of sweet corn plants had non-sweet corn seeds (AA) and sweet corn seeds (AA), while non-sweet corn plants had only non-sweet corn seeds (AA).

Your question is why it's not co-manifested.

If it is co-manifestation, then it should be both sweet and non-sweet on the same corn cob, and there are non-sweet corn kernels on the ears of sweet corn", the sweet corn in this sentence refers to the genotype AA where the parents are sweet, and the kernels are offspring, which are of different generations, so they are not co-manifest.

Mating is defined between different species.

Let the sweet gene be S and the non-sweet gene be S. Purebred sweet corn (SS) and non-sweet corn (SS) are interplanted with a recessive gene that will only be sweet in the case of SS, note: the prerequisite here is"Purebred"

Because in the case of non-sweetness, the SS gene will not be produced, and in the ears of sweet corn, if the S gene is found, it will be non-sweet corn.

Non-sweet is dominant to sweet

Hello, it's a pleasure to serve you and give you the following answer: High School Genetics and Variation in the Third Year of High School. Heredity and variation are important knowledge in biology, and they are the basis of biological evolution.

Heredity and variation refer to changes in the genetic makeup of an organism that can affect the traits of an organism. Reasons for the problem:1

Genetic variation may be due to environmental factors, such as radioactive materials, drugs, contaminants, etc., that affect the genome of an organism. 2.Genetic variation can also be due to genetic mutations within the organism, which can be due to empty residues due to genetic recombination, deletion, or insertion.

Workaround and practice steps:1First of all, it is necessary to understand the causes of genetic variation so that effective measures can be taken to prevent it from occurring.

2.Secondly, it is necessary to take effective measures to control environmental factors and reduce the impact on living organisms. 3.

Thirdly, effective measures should be taken to control gene mutations, such as using genetic engineering technology to change the structure of the genome to reduce the occurrence of gene mutations. 4.Finally, effective measures should be taken to control genetic variation, such as the use of genetic regulation technology to control gene expression, so as to reduce the occurrence of genetic variation.

Teaching & Personal Tips:1When teaching, students should understand the causes of genetic variation, so that they can take effective measures to prevent the occurrence of variation.

2.When teaching, students should understand the solutions and practical steps of genetic variation, so that effective measures can be taken to control the occurrence of genetic variation. 3.

Personal tip: Learn more about genetic variation so that you can better understand the causes and solutions of genetic variation, so as to better control the occurrence of genetic variation.

From the meaning of the title, the gray body is explicit. Let the control gene be a,a, then the genome of F2 becomes 1 3aa, 2 3aa

To find the ratio of gray fruit fly and black fruit fly, it is simple to find the ratio of black fruit fly AA.

AA can only be produced by 2 3 aa self-inbred, and its ratio is 2 3 * (2 3) * (1 4) = 1 9

Then Drosophila gray: Drosophila melanosophila = 8:1

Pure gray and pure black are hybridized to get gray, so the gray is black and hidden, and F1 is all BB. then bb x bb gets a quarter bb two-quarters bb a quarter bb. Only BB black. So gray is 3 to 1 compared to black

Ash is a dominant trait and is denoted as a. f1 is aa:aa:

aa=1：2：1。

The gray body is aa and aa, and the ratio is 1::a=2:1.

Self-inbreeding produces aa and aa and aa. In the offspring, the square of (a+a) = :aa:

aa=4：4：1。

Therefore the gray body is more black than the black body::1. Pick D.

1.Rice is the least suitable (monoecious, small flowers. Yuan Longping was able to successfully engage in hybridization because he found male sterile rice. ）

A Drosophila: reproduces quickly, is abundant, and is a good hybrid material.

c Pea: a classic hybrid experimental material in textbooks, with large flowers, and self-pollination and closed-flower pollination.

d Corn: monoecious but heterogeneous, it is also an ideal hybrid material.

Trait segregation generally refers to the isolation of offspring with recessive traits from parents with dominant traits. The appearance of recessive traits must have recessive homozygous and will not separate traits. ）

3.Not entirely true: the number of male and female gametes in living organisms is not equal, and generally the number of male gametes is greater than that of female gametes, but the combination of female and male gametes is random.

1b Because the flowers of rice are too small to be manipulated2b In the hybrid offspring, only the appearance of recessive traits is homozygous.3 Fault In nature, the ratio of male and female gametes is more female than male.

White spherical: AABB

Yellow disc: AABB

f1:aabb

F2 has four traits – white disk: white disk: yellow disk:

Yellow Ball = 9:3:3:

1. Among them, there are two genotypes of white balls - aabb: aabb = 1:2, and yellow disk also has two genotypes - aabb:

AABB = 1:2, and the total number of white and yellow discs in F2 is equal.

The title says that there are 1001 white spheroids that can be stably inherited in F2, which tells you that there are 1001 ABBs, so AABB also has 1001, so the yellow discs that cannot be stably inherited in F2 are 1001*2=2002.

b option.

Process: Because the white spherical pumpkin is crossed with the yellow disc-shaped pumpkin, F1 is all white disc-shaped pumpkin, and it can be seen that white is dominant and represented by a gene, yellow is A, discoid is dominant and represented by B, and the ball is B.

Think of two pairs of relative traits separately.

1. F1 is all white, the genotype is AA, the genotype of F1 is self-inbred to F2, 2, F2 is all disc-shaped, the genotype is BB, and the genotype of F1 self-inbred F2 is ,, comprehensive thinking:

There are 1001 white spherical pumpkins in F2 that can be stably inherited, and they are homozygous for their genotype ABB, and the proportion of such individuals in the total number of individuals is: . Theoretically, the genotype of the yellow discoid pumpkin in F2 that cannot be stably inherited is ABB, and the proportion of such individuals to the total number of all individuals is: . It is exactly twice as large as the individual white spherical pumpkin that can be stably inherited.

There are 1001 white spherical pumpkins that can be stably inherited, and the yellow discoid pumpkins that cannot be stably inherited in F2 are 1001*2 2002.

Other ratios can also be calculated in the following way:

If calculated in F2, the white spherical pumpkin that cannot be stably inherited is, also 2002.

Simple thinking.

The white spherical and yellow discoid were hybridized, and F1 was all white disc, indicating that the white discoid was the dominant trait, the F1 genotype was inbred, the genotype of the white spherical individual that could be stably inherited in F2 was AAB, and the proportion of the yellow discoid individual (ABB) that could not be stably inherited in F2 was 2 16, so the yellow disc was 1001*2=2002

The offspring of the BB*BB cross with the F1 gene are 1 2 BB and 1 2 BB, and the result is in three cases:

1 2bb*1 2bb, then the offspring is 1 4bb;

2*(1 2bb*1 2bb), then the offspring is 1 4bb, 1 4bb; 1 2bb*1 2bb, offspring 1 16bb, 1 8bb and 1 16bbSince BB is dead, so subtract its proportion to calculate that the proportion of BB alive is 6 15, then the proportion of B is 3 15, that is, 20%. Then the proportion of b is 80% hehe.

BB and BB plants were crossed to obtain F1, F1 was 1 2BB, 1 2BB, B gametes produced in F1 generation accounted for 3 4, B gametes accounted for 1 4, if the F2 obtained by random mating did not die, then BB accounted for 9 16, BB accounted for 6 16, BB accounted for 1 16, but due to BB death, so in the mature population, BB accounted for 9 15, BB accounted for 6 15, and the gene frequency of B gene was 9 15 + 6 15 * 1 2 = 12 15 = 4 5

f1bb1 2bb1 2, the two mate randomly, there are bb*bb, bb*bb, bb*bb. Let's do the math again, using the genotype method.

The F1 gene of BB*BB cross is 1 2 BB and 1 2 BB, and the result of F1 random assignment has three cases, 1 2BB*1 2BB, then the offspring is 1 4BB; 2*(1 2bb*1 2bb), then the offspring is 1 4bb, 1 4bb; 1 2bb*1 2bb, offspring 1 16bb, 1 8bb and 1 16bbSince BB is dead, so subtract its proportion to calculate that the proportion of BB alive is 6 15, then the proportion of B is 3 15, that is, 20%. Then the proportion of b is 80%.

In the F1 generation, there are only BB, BB

In the F2 generation, it is also BB, BB

In the F2 generation, it is a random mating, and only when BB*BB has 1 4 BBs killed. Thus, BB is left with 1 2 and BB 3 8.

Note that the frequency of b is being sought. b accounts for only 7 8 1 5

How do I think it's 75% that, there is no way that there will be a BB situation at all.