1. The following belongs to a pair of relative figures ()

A. Long hair and white hair of rabbits B. Yellow round grains and green wrinkled grains of pea seeds

C. tall stems of wheat and short stems of barley D. rough and smooth hair of guinea pigs

2. The recessive trait refers to ()

A. Unknown traits of hybrid offspring B. Unknown traits of hybrid F 1

C. Unexplained inbreeding offspring characteristics D. Unexplained offspring characteristics

3. Test refers to ()

A.f 1 intersects with one of the parents. b.f 1 intersects with the dominant parent.

C. crossing with recessive homozygote d for self-crossing.

4. The father has six fingers (pathogenic genetic factor A), the mother is normal, and the fingers of the first child are normal. The probability of having another sick boy is ()

12.5% 25% 50% d . 75%

5. Let the heterozygote Aa self-cross for three generations in a row, then the proportion of the fourth generation heterozygote is ()

A.B. C. D。

6. The relationship between genotype and phenotype is ()

① The genotype is the same, and the phenotype is definitely the same; ② The genotype is the same, and the phenotype is roughly the same.

③ The phenotype is the same, and the genotype is roughly the same; ④ The phenotype is the same, but the genotype is not necessarily the same.

A.①② B.②④ C.②③ D.③④

7. In the following groups, the phenotype is the same (the external conditions are the same) ().

A.BBDD and BBDD B. aaCC and aaCC

C.BBCC and BBCC Didi and ddee

8. The white disc-shaped pumpkin is crossed with the yellow spherical pumpkin, and the Fl is all white disc-shaped. There are 200 heterozygous white spherical pumpkins in F2, so the homozygous yellow disc pumpkin should be ().

100

9. If the tall rice stalk (D) is dominant to the short rice stalk (D), the rice blast resistance (R) is dominant to the susceptible rice blast (R), and two pairs of related traits are independently inherited. A homozygous susceptible dwarf variety (lodging resistance) crossed with a homozygous disease-resistant tall variety (lodging resistance), and F 1 selfed. There are both disease-resistant genotypes and lodging-resistant genotypes in F2 generation, and the ratio is ().

A.Rrdd Bay RRdd,

C.Rrdd and rrDD, D.rrDD and RRdd,

10. The table below shows the phenotype and plant number of three different wheat hybrid combinations and their offspring.

combine

Phenotype and plant number of offspring of serial number hybrid combination type

Disease-resistant red seed coat disease-resistant white seed coat disease-resistant red seed coat disease-resistant white seed coat

A disease-resistant, red seed coat × susceptible, red seed coat 416138 410135.

2. Disease resistance, red seed coat × susceptible, white seed coat180184178182.

Three diseases, red seed coat × disease, white seed coat140136 420 414.

According to the table analysis, the following inference is wrong ()

A. all six parents are heterozygotes. B. for susceptible diseases, disease resistance is dominant.

C. the red seed coat is dominant to the white seed coat. D. these two pairs of characters are freely combined.

1 1. Tetrad means ()

A. There are four chromosomes in the nucleus.

B. The nucleus contains four homologous chromosomes.

C. each homologous chromosome pair contains four chromosomes.

D. a general term for four chromatids in postsynaptic homologous chromosomes

12. Among the following phenomena, () is unique to splinter cell.

A. The appearance of the spindle

B. Chromosome appearance

C. the chromosome moves to the two poles of the cell.

D. homologous chromosomes are separated and move to the two poles respectively.

13. Sex chromosomes exist in ()

A. somatic cells B. sperm C. egg cells D. or above

14. Human sex can be determined by sperm type, because it contains ().

A.x chromosome

B.y chromosome

C. One X chromosome and one Y chromosome

D.X chromosome or Y chromosome

15. During the spermatogenesis of Bb genotype animals, B and B, B and B and B and B genes were separated at () respectively.

① Spermatogonial cells form primary spermatocytes.

② Primary spermatocytes form secondary spermatocytes.

③ Secondary spermatocytes form spermatocytes.

④ Sperm cells are transformed into sperm.

A.①②③b .③②c .②②②d .②③④

16.A and A are a pair of alleles that control the body color of Drosophila melanogaster, and only exist on the X chromosome. In the process of cell division, the cell with allelic segregation is ()

A. Primary spermatocytes

C. Primary oocytes

17. About human genetic diseases, the correct statement is ().

A. if the parents are normal, the child will not have genetic diseases.

B. The mother suffers from albinism and the son suffers from albinism.

C. The mother is color-blind, and so are her sons.

D. The father is color blind and the son is color blind.

18. The picture below is a genetic map of a family hereditary disease, which can't be ().

A. autosomal dominant genetic diseases B. autosomal recessive genetic diseases

Chromosome recessive genetic disease

19. The following statement about the X chromosome is wrong ().

A. Normal female somatic cells have two X chromosomes.

B. There is an X chromosome in normal male somatic cells.

Genes on chromosome C. X are all related to sex determination.

D. Chimpanzees and other mammals also have X chromosomes.

20. Human curly hair is dominant for straight hair, and the gene is located on the autosome. Hereditary chronic nephritis is an X chromosome dominant genetic disease. A woman with curly hair and hereditary chronic nephritis married a man with straight hair and gave birth to a son with straight hair and no nephritis. The probability that the couple will have another child with curly hair and hereditary chronic nephritis is ()

A. 1/4

Second, non-multiple choice questions

1. The following figure is the pedigree of a genetic disease in a family (A stands for dominant gene), please analyze and answer:

(1) The disease is hereditary.

(The genotype of Ii8 is;

(3) 3) The probability that III10 is a heterozygote is.

(4) If Ii8 and Ii 10 are illegal marriages, then the probability that they give birth to sick children is.

2. Endosperm yellow (a) is dominant to white (a), and non-waxy (b) is dominant to waxy (b). Two pairs of characters can be combined freely. Now there are two genotypes of homozygous corn kernels, the phenotypes are: yellow waxy and white waxy.

(1) Please take the above two kinds of corn seeds as parents, and get 4 seeds through hybridization test. Phenotypes are yellow glutinous rice, yellow glutinous rice, white glutinous rice and white glutinous rice, and the ratio is close to 1: 1: 1 (genetic map answer).

(2) If the parents are the same, the above four seeds are obtained, but the ratio is close to 9∶3∶3∶ 1. What is the main difference between this hybrid experiment and the previous hybrid experiment? (Answer in words)

_______________________________________________________________________________________________________________________________________________________。

3. The picture below is a schematic diagram of cell division. Please analyze and answer:

(1) The number _ _ _ _ _ _ is in the metaphase of the second meiosis. With homologous chromosomes is figure _ _ _ _ _ _ _ _.

(2) In the tetrad stage, Figure _ _ _ _ _ _ _ _.

(3) In Figure A, there are _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _.

4. The picture below shows the chromosome and gene composition of spermatogonia of a higher male animal. Please analyze and answer:

(1) A and A in the diagram are called.

(2) ① and ② in the figure are called.

(3) There are many kinds of gametophytes produced by this cell, and the ratio is.

(4) There are two phenotypes in the progeny of biological test cross, among which there are two different types of biological test cross, and the phenotypic ratio of the progeny of biological test cross is.

(5) The offspring of self-pollinated organisms have two genotypes and two phenotypes. The phenotypic ratio is. The proportion of individuals who can inherit stably in offspring is.

(6) The proportion of representative species after the existence of organisms intersecting with aaBB individuals is.

Reference answer

First, multiple choice questions

1.D 2。 B 3。 C 4 explosive B

5.B

Analysis: The heterozygote Aa continues to self-pollinate, and its trend is gradually homozygous (the proportion of heterozygotes is getting smaller and smaller). According to the segregation phenomenon, the first generation is self-crossing, in which heterozygote (Aa) accounts; Self-cross second generation, heterozygote accounts for × =; After three generations of continuous self-crossing, the proportion of heterozygotes in the fourth generation is 1/8.

6.B 7。 C 8。 B 9。 C

10.B

Analysis: according to the meaning of the question, the white seed coat appears in the offspring of one or two red seed coats, indicating that the red seed coat is a dominant trait and both parents are heterozygotes. The offspring of three or two susceptible parents showed disease resistance, indicating that the susceptible parents were dominant and both parents were heterozygotes. Combination 2 belongs to test crossing, and there are four phenotypes in the offspring with the same proportion, which shows that these two pairs of genes are freely combined. 1 1.D 12。 D 13。 D 14。 D 15。 B

16.C

Analysis: Alleles are located on homologous chromosomes, and the sex chromosome of male Drosophila is XY, so there is only one X chromosome in somatic cells, and it is impossible to contain alleles A and A. Allele segregation occurs during the first meiosis. Oocytes undergo the first meiosis after DNA replication, and the cells at this time are primary oocytes.

17.C 18。 C

19.C

Analysis: The sex determination of our common animals (including chimpanzees) is mostly XY type, that is, males have two heterozygous XY chromosomes and females have two identical X chromosomes. Sex chromosomes exist in pairs in somatic cells, and they exist alone in meiotic cells. Some genes on sex chromosomes are related to sex, while others have nothing to do with sex determination (such as color blindness and hemophilia). ).

20.D

Analysis: assume that curly hair is dominant over straight hair, represented by a and a; Hereditary chronic nephritis is dominant to normal, represented by B and B. According to the meaning of the question, the genotype of the straight-haired son without nephritis is aaXBY, so the genotypes of this couple are AaXBXb and aaXbY respectively. Then the probability of a child with hereditary chronic nephritis (AaXB__) regenerating a curly hair is:

Aa× XB__=3/8 AaXB__ .

Second, non-multiple choice questions

1.( 1) Hide (2)AA or Aa (3) (4)

2.( 1)P AABB×AaBb f 1 AaBb×AaBb

f 1 AaBb AaBb∶AaBb∶AaBb∶AaBb

1 ∶ 1 ∶ 1 ∶ 1

(2) F 1 was used for test crossing in the first experiment and F 1 was used for selfing in the second experiment.

Analysis: mainly investigate the application of the law of free combination, and pay attention to the correct writing of gene map. With homozygotes as parents, four phenotypes will appear in offspring, and the ratio is close to 1: 1: 1, which can only be achieved through the test crossing of F 1. So AABB× AABB → F 1: AABB, let F 1 cross with recessive homozygote, and four phenotypes will appear in the offspring, and the ratio is close to the four ratios of 1. If there are four phenotypes, and the ratio is close to 9∶3∶3∶ 1, F 1 selfing is needed.

3.( 1)C； A and b (2) a (3) 4; 8; eight

4.( 1) allele (2) homologous chromosome (3)2 1: 1

(4) 4, 3, 1∶ 1∶ 1∶ 1 (5)9, 4,9∶3∶3∶ 1

(6) 2, 1∶ 1