Seek the classic practice of high school biology compulsory 2

In a genetic experiment with a pair of relative traits, the following description is incorrect (b).

a.If the offspring are crossed with different shapes, and the offspring have two shapes, it is impossible to judge the dominant recessive trait and the genetic factor composition of the two parents.

b.If two parents with the same trait are crossed, and the offspring only appear in the same trait of the parent, then both the parent and the offspring are homozygous.

c.If two parents with the same trait are crossed, and the offspring has a different trait from the parent, the parent is heterozygous.

d.If a parent is self-bred, and the offspring have two traits, and the trait segregation ratio is 3:1, the offspring that account for 3 4 show dominant traits.

Such as bb x bb

A couple with double eyelids had a total of four children, three single eyelids and one double eyelid, and the best explanation for this phenomenon is (c).

1 Conform to the law of segregation of genes bThis inheritance does not conform to the law of segregation of genes.

c.The couple had the possibility of developing a single eyelid in each fetus dThe single eyelid gene was swapped with the double eyelid gene.

Just like the gender, the first birth of a boy and the second birth of a boy and a girl are possible.

The gray (g) to the white (g) of the pea seed coat is dominant, and the f1 (heterozygous) is now planted and continuously self-crossed. The statement in question is incorrect (d).

The seed coat of the seeds on the plant is gray.

There are three types of cotyledon genetic components of seeds produced on plants.

The seed coat of the seeds on the plant is gray:white = 3:1

The probability that the cotyledon genetic composition of the seeds produced on the plant is homozygous is 1 2

Should be greater than 1 2

In order to observe the trait segregation of pea plants, it is necessary to observe the second generation of children at least until the first year (c.). ）

a.A bII cIII DFour.

The seeds of the second generation of children only grow into plants in the third year.

1 option b If the parents are all heterozygous, e.g. bbThen the offspring will appear bb, bb, bb 3 genotypes, of which the first 2 traits are the same as the parents, but the second is not homozygous, and the parents are not homozygous.

2 choose c, single eyelid and double eyelid are a pair of relative traits, so the gene that controls their traits is an allele. If the gene for a single eyelid is AA, and the gene for both eyelids is AA, then the genotype of the parents may be AA, or it may be AA and AAThe former has a separation ratio of 3 to 1 and the latter has a 1 to 1 ratio

So A is wrong.

3.The cotyledons are developed from the zygote, that is, the cotyledons of F1 are the genetic components of F2, because we can judge by biological totipotency that the genes in each cell are the same. So the title is not wrong, so B is right, and GG is heterozygous.

3 genotypes. So the cotyledons on F2 is the genotype of F3, because the pea is a self-inbred plant, so we can calculate the ratio of F3 through the gene ratio of F2, that is, 1 4 AA + AA + AA + have, AA has, AA has.

4 I broke down, there is a problem with the question of this question. Pick B. First of all, in the first year, we crossed the heterozygous F1AA from the female parent P, (because of the trait separation experiment, so we need to separate different traits from the unified trait in order to achieve the experimental purpose), and then in the second year, we self-crossed the first generation F2 from F1, at this time the shape separation has occurred, so it only takes 2 years.

Why the question still asks the child 2 generations for this key hint, despise the teacher.

1. Choose B, in this case, it is possible that one is zygosis bai and the other is heterozygous.

2. Select DUC, from the question condition

See, the probability of double eyelids in offspring is dao3 4 and single eyelid is 1 4, but each birth is random, and there is a possibility of both single eyelids and double eyelids, and the proportion cannot be obtained in the case of too few samples.

3. Choose B, because the condition given in the title is to study the seed coat, and the genetic composition of the cotyledons is not clear.

4. Choose B, and the second year is the seed produced on the F1 generation.

1 b 。If genes a and a are appointed, then item a corresponds to aa aa aa, aa or aa aa aa, aa; then item b is inconsistent with aa aa aa aa and aa. Then the c and d terms symbolize the concept of round trait separation. Answer.

a.The simplest is to assume that AA seeds are 1 and AA seeds are 2. The seed setting rate is the same, and we assume that 1 seed produces 50 offspring.

According to Mendodel's law of heredity, the ratio of aa, aa, and aa to aa seed is 1:2:1, so the ratio of aa, aa, and aa to the number of aa, aa, and aa in the offspring of 2 aa seeds is 25:

50:25, plus 1 AA's 50 descendants are all AA, so AA has 75. So aa:

aa:aa=75:50:

aa: aa = 1:2, aa accounts for 1 3, aa accounts for 2 3, peas are inbred, 1 3aa is inbred, and the offspring aa accounts for 1 3;2 3AA is inbred, the offspring AA accounts for 1 6, AA accounts for 1 3, and AA accounts for 1 6.

In summary, aa==1 3+1 6==3 6,aa==2 6,aa==1 6.

So, aa:aa:aa==3:2:1.

The state of nature is both self-inbred.

The parental is aa aa, so the probability of all aa of the offspring is one-third, the parent is aa aa, so the offspring aa is one-sixth, aa is one-third, and aa is one-sixth.

In summary: aa:aa:aa=3:2:1

Human I, IA ib genes can control blood type, in general, genotype II is blood type O, IA IA or IAI is blood type A, IB IB or IBI is blood type B, IAIB is blood type AB, because the husband's blood type A, the wife's blood type B, gave birth to a son with blood type O, so the husband's blood type is IAI, the wife is Ibi, and their child's genotype may be iaib, iai, ibi, ii, all 1:1:1: 1: 1: 1:

1, then the probability of having another child with the same blood type as the husband is 1 4, and because it is a girl, it is multiplied by 1 2, that is, the probability is 1 8

The genotype of the father is AI, and the genotype of the mother is BI;

The husband has the same blood type AI, and the probability is 1 4