Biogenetics ExercisesHelp !!。。。

eg1.(1) RR RR F1: all red fruits, genotype is RR, RR, the ratio is 1:1

2) RR RR F1: The genotype is RR, RR, RR. Ratio 1:2:1; Phenotype red fruit: yellow fruit = 3:1

3) RR RR F1: All red fruits, genotype RR.

eg2.The female genotype is IAIB, the male genotype is II, and the child genotype is IAI or IBI. Blood type A or B respectively.

eg3.(1) wwdd wwdd f1: wwdd, wwdd, wwdd, wwdd ratio 1:

1:1:1, the phenotype is white cone, white cone, yellow cone, yellow cone, ratio 1:

2) wwdd wwdd f1: wwdd, wwdd, wwdd, wwdd, wwdd, wwdd, wwdd, wwdd, wwdd ratio 1:2:

1. The phenotype is white cone, white cone, yellow cone, yellow cone, and the ratio is 9:3:3:

1z b z b w f1 genotype: z bz b, z b z b, z bw, z b w. The phenotypes were reed, non-reed, reed, and non-reed

wz^b w、

Non-reed flowers

Red fruits; Huang Guo = 1;0

Red fruits; Yellow fruit = 3;1

Red fruits; Yellow fruit = 1;1

Type A or B. IAI or IBI

1)wwdd

wwddwwdd

wwdd

1.Monosomy: If an individual is missing a chromosome in a chromosome set, and the number of chromosomes in the somatic cell is 2n-1, the organism is monosomy and the karyotype is (ABCD)(ABC)====-

2.Missing body: If a pair of homologous chromosomes are missing in the chromosome set of a diploid, the number of chromosomes in the somatic cell is 2n-(2).This organism is deficient and has a karyotype of (ABC) (ABC) ===

3.Trisomy: An individual with an additional chromosome in the chromosome group of a diploid and an individual with a chromosome number of 2n+1 in the somatic cell is called a trisomy with a karyotype of (ABCD) (ABCD) A===

4.Tetrasomies: Organisms with two identical chromosomes added to the chromosome set of a diploid and a chromosome number of 2n+2(1) in a somatic cell are called tetrasomies. The karyotype is (ABCD) (ABCD) AA

5.Ditrisomy: If two pairs of homologous chromosomes and one member is added, an organism with a number of chromosomes in a somatic cell of 2N+1+1 is a ditrisomy with a karyotype of (ABCD) (ABCD) AB

6.Dimonosome: If both pairs of homologous chromosomes have one chromosome missing each, a monomer is called a dimonosome. The karyotype is 2n-1-1

It is the variation in the number of chromosomes tested.

Fill in the results as described above.

Fill in what, I don't understand, can you be specific?

1, P is single-petaled, but the self-crossed offspring has a double petal, which is a phenomenon of trait separation, so following the law of separation, the single valve is dominant.

2.It is hypothesized that the univalve genotype in F1 may be BB or BB. The pollen was B or B, haploid in vitro culture, and colchicine treatment was obtained as BB (single petal) or BB (double petal).

However, the title says that the offspring only have double lobes (bb), indicating that the single lobes (bb) are dead. So bpollen is sterile.

3.It is known from 2 that pollen is 50% of male plants, sterile. The female is fertile, and it accounts for 50%.

4.Known by 2, B is sterile, so the short is B

The law of gene segregation reads: in the somatic cells of organisms, the genetic factors that control the same trait exist in pairs and do not fuse; During the formation of gametes, pairs of genetic factors are separated, and the isolated genetic factors enter different gametes and are passed on to offspring along with the gametes. Obviously, this question is talking about a pair of relative traits, so it is the law of separation.

In general, the law of free combination of genes is answered only when non-allelic genes on two or more pairs of non-homologous chromosomes are involved. In the question, the trait separation occurred in the self-inbred of the unipetal violet, so the unipetal is the dominant trait.

The genotype of the unipetal violet in F1 should theoretically be bb bb, and there are two kinds of gametes produced: b and b, so the violet obtained by the anther in vitro culture should theoretically be an individual with the genotype bb and bb, from the title, only the double violet, from the first question, the genotype of the unipetal violet is bb, bb, and the genotype of the double violet is bb, that is to say, only the individual with the genotype bb exists after the anther is cultured in vitro. From this, it can be concluded that the pollen of B is sterile and the pollen of B is fertile.

Because the anther in vitro culture is actually the culture of male gametes, obtained from (2), male gametes chromosome deletion and sterility, and then from the title, single-petal violet self-breeding, theoretically speaking, the male gametes produced are b, b, female gametes are also b, b, but male gametes b sterility, so there will be a trait segregation ratio of 1:1. If we look at it this way, what we just said in (2) is not accurate, because the male gamete B is sterile, so we can think that he only produces the male gamete of B, then B must be from the female gamete, because there is no male gamete to provide B, so the genotype of the unipetal violet can only be BB.

Therefore, the types of female gametes produced by single-petal violets are b and b, and the ratio is 1:1, and the male gametes produced are only b

After selecting the positive and negative crosses of the genotype as the answer, the results obtained are not the same, so it can be proven.

Because any deoxynucleotide contains one base, one deoxyribose and one phosphate molecule, therefore, the number of deoxynucleotides = base number = number of deoxyribose = number of phosphate, this question is a typical high starting point and low point. It may seem complicated, but it's actually very simple.

b.To AFor single-stranded DNA, the purine base number is not necessarily equal to the pyrimidine base number.

and DNA polymerase, both catalytic have specificity, and the substrates are different.

d.3 of a deoxyribose'bits and 5'Two phosphoric acids are attached to the bit.

Phosphoric acid - deoxyribose - base - base - deoxyribose - phosphate base complementary arrangement, phosphoric acid and deoxyribose alternately linked to form a skeleton.

The structure of the nucleotide is.

Phospho--deoxyribose--base.

See the textbook Structural Plan of DNA.

In a quarter, ABB can form two genotypes AB and AB, and AABB can form four types of AB, AB, AB, AB. Individuals that can be stably inherited are homozygous, i.e., AABB and AABB, 2 2 * 4 = 1 4

Another method: aa aa=aa:aa:aa=1:2:1, in which the one that can be stably inherited accounts for 1 2, bb bb = bb:bb = 1:1, and the one that can be stably inherited accounts for 1 2, and the sum is 1 2 * 1 2 = 1 4

A stably inherited individual is one in which both pairs of genes are homozygous, and there are two types of them: AABB and AABB, where the probability of AABB = 1 4 1 2 = 1 8 and the probability of AABB = 1 4 1 2 = 1 8

In addition, there is a 1 4 probability of AAB, a 1 4 probability of AAB, a 1 8 probability of AAB, and a 1 8 probability of AABB, all of which are unstable inheritance.

So, a stable inherited individual = 1 8 + 1 8 = 1 4

Stable inheritance is homozygous, AABB and AABB

Probability of AABB = 1 4 1 2 = 1 8

Probability of AABB = 1 4 1 2 = 1 8

3 (provided that neither of the parents has albinism).

2. aa=1 10000, then a=1 100 a=99 100 So: aa:aa=99 100 x 99 100: 2 x 1 100 x 99 100=99:2

That is, the probability that the husband is a carrier is 2 101

3 x 2 101 x 1 4=1 3034, no condition.

5. **: Genetics**, prevention: prohibition of inbreeding, genetic counseling, prenatal diagnosis.

1 aa aa aa aa

The wife is normal, the meaning is aa or aa or aa 2 32, the carrier is twice the probability, 1 50003, the probability that both are carriers, multiplied by 2 (5000*3*4).

4. Calculate the probability of the father's carrier, which cannot be calculated, and it is not stated whether the grandmother, that is, the maternal grandmother, is the sick person5, and it is forbidden for close relatives to marry.

1 To find out the probability of carrying a wife, you need to ask about the characteristics of his parents.

If both parents analyze the genotype normally and the wife's younger brother is a patient, then both parents are carriers.

aa x aa||aa aa aa aa because the wife is behaving normally, the probability that the wife is a carrier is 2 3 If one of the parents is a patient then.

aa x aa|aa aa so the wife must be the carrier.

2. The incidence of albinism (AA) in the population is 1 in 10,000, so according to the law of genetic equilibrium, the gene frequency of a is the quadratic root of the frequency of the aa genotype, which is 1 100, and the probability that the husband is a carrier is (2 * 99 100 *1 100) 9999 10000 = 2 101 (I was negligent just now, and I miscalculated here, and it is now corrected.) Excuse me! ）

3. If the parents are normal people. The probability of the couple having a child with albinism is 2 3*2 101*1 4=1 303 (I was negligent just now, and I miscalculated here, and I am correcting it now.) Excuse me! ）

4. If you are not sure about the performance type of the couple's previous generation, there is really no way to calculate this.

5. There is no way to deal with albinism at this time. Prevention can only be to avoid consanguineous marriages, genetic counseling, prenatal testing, and selective termination of pregnancy.