遗传学 双语课 练习题
Medialian genetics
1 Shown are F
2 results of two of Mendel's monohybrid crosses. State a null hypothesis that you will test using chi-square analysis. Calculate the x2 value and determine the p value for both crosses, then interpret the p values. Which cross shows a greater amount of deviation?
(a)Full pods 882
Constricted pods 299
(b)Violet flowers 705
White flowers224
2 In Dmsophihla, gay body color is dominant over ebony body color, while long wings are dominant over vestigial wings. Work the following crosses through thee F2 generation, and determine the genotypic and phenotypic ratios for each generation. Assume that the Pl individuals are homozygous:
(a)gray, long ×ebony, vestigial (b) gray, vestigial ×ebony, long (c) gray, long ×gray, vestigial
3.How many different types of gametes can be formed by individuals of the following genotypes? What are they in each case?
(a) AaBb,(b) AaBB, (c)AaBbCc, (d)AaBBcc, (e)AaBbcc and (f) AaBbCcDdEe?
https://www.wendangku.net/doc/a65315038.html,ing the forked-line method, detennine thee genotypic and phenotypic ratios of these trihybrid crosses:
(a)AaBbCc×AaBBCC, (b)AaBBCc×aaBBCc, and (c) AaBbCc ×AaBbCc.
5. Mendel crossed peas with round, green seeds with peas having wrinkled, yellow seeds. All Fl plants had seeds that were round and yellow-Predict the results of test crossing these F1 plants.
6. In one of Mendel's dihybrid crosses, he observed 315 round, yel low; 108 round,green; 101 wrinkled, yellow; and 32wrinkled, green F2 plants. Analyze these data using chi--square analysis to see whether (a) they fit a 9:3:3:1 ratio; (b)the round, wrinkled trait fit for a 3:1 ratio; or (c) the yellow, green traits fit a 3:1ratio.
7. A geneticist, in assessing data that fell into two phenotypic classes, observed values of 250:150. He decided to perform chi-square analysis using two different null hypotheses: (a) The data fit a 3:1 ratio; and (b) the data fit a l:1 ratio.Calculate the X2 values for each hypothesis-What can you conclude about each hypothesis?
8. The basis for rejecting any null hypothesis is arbitrary. The researcher can set more or less stringent standards by deciding to raise or lower the critical p value. Would the use of a standard of p=0.10 be more or less stringent in failing to reject the null hypothesis? Explain.
9. Consider three independently assorting gene pairs, A/a, B/b, and C/c, where each demonstrates typical dominance (A-, B-, C- ), and recessiveness (aa, bb, cc). What is the probability of obtaining offspring that is AABbCc from parents that are AaBbCC and AABbCc?
10. What is the probability of obtaining a triply recessive individual from thee parent as shown in Problem 9?
11. Of all offspring of the Parents in Problem 9, what proportion will express all three
dominant traits?
12. For the following pedigree, predict the mode of inheritance and the resulting genotypes of each individual. Assume that the alleles A and a control the expression of the trait.
13. Which of Mendel's postulates are demonstrated by the pedigree in Problem 12? List and define these postulates.
14. The following pedigree follows the inheritance of myopia (near-sightedness) in humans. Predict whether the disorder is identified as a dominant or a recessive trait. Based on your predic6on, indicate the most probable genotype for each individual.
15. Two true-breeding pea plants are crossed. One parent is round, terminal, violet, constricted, while the other expresses the contrasting phenotypes of wdnkled, axial, white, full. The four pairs of contrasting traits are controlled by four genes, each located on a separate chromosome. In the F1 generation only round, axial, violet, and full are expressed. In the F2 generation, all possible combinations of these traits are expressed in ratios consistent with Mendelian inheritance.
(a) What conclusion can you draw about the inheritance of these traits based on the Fl results?
(b) Which phenotype appears most frequently in the F2 results? Write a mathematical expression that predicts the frequency o f occurrence of this phenotype.
(c) Which F2 phenotype is expected to occur 1east frequently? Write a mathematical expression that predicts this frequency.
(d) How often is either Pl phenotype likely to occur in the F2 generation?
(e) If the Fl plant is testcrossed, how many different phenotypes will be produced, and how does this number compare to the number of different phenotypes in the F2 generation discussed in part (b)?
16. Tay-Sachs disease (TSD) is an inborn error of metabolism that results in death,
usual1y before the age of five. You are a genetic counselor, and you interview a phenotypically normal couple who consult you because the man had a female first cousin (on Ms father's side) who died tom TSD, and the woman had a maternal uncle with TSD. There are no other known cases in either family, and none of the matings were/are between related individuals. Assume that this trait is rare in this population.
(a)Using standard pedigree symbols, draw a pedigree of these individuals, families, showing the relevant individuals.
(b)The couple asks you to calculate the probability that they both are heterozygous for the TSD allele.
(c)They also want to know the probability that neither of them is heterozygous.
(d)They also ask you for the probability that one of them is heterozygous but the other is not.
[Hint: thee answers to (b), (c), and(d) should add up to1.0.]
17 Chlamydomonas, a eukaryotic green alga, is sensitive to the antibiotic erythromycin, which inhibits protein synthesis in prokaryotes. There are two mating types in this alga, mt+ and mt-, If an mt+ cell sensitive to the antibiotic is crossed with an mt- cell that is resistant, all progeny cells are sensitive. The reciprocal cross (mt+ resistant and mt-sensitive) yields all resistant progeny cells. Assuming that the mutation for resistance is in the chloroplast DNA, what can you conclude from the results from these crosses?
sex determination
1. As related to sex determination, what is meant by (a) homomorphic and
heteromorphic chromosomes; (b) isogamous and heterogamous organisms?
2. Contrast the life cycle of a plant such as zea mays with an animal such as
C.elegans-.
3. Discuss the role of sexual differentiation in the life cycles of chlamydomonas, ZE
mays, and C .elegans.
4. Distinguish between the concepts of sexual differentiation and sex detem1ination.
5. Contrast the Protenor and Lygaeus modes of sex determination.
6. Describe the major difference between sex determination in Drosophila and in
humans.
7. What specific observations (evidence) support the conclusions you have drawn
about sex determination in Drosophila and humans?
8. Describe how nondisjunction in human female gametes can give rise to Klinefelter
and Turner syndrome offspring following fertilization by a normal male gamete.
9. An insect species is discovered in which the heterogametic sex is unknown. An
X-linked recessive mutation for reduced wing (rw) is discovered. Contrast the F1 and F2 generations from a cross between a female with reduced wings and a male with normal sized wings when (a)the female is the heterogametic sex; (b) the male is the heterogametic sex.
10. Based on your answers in boblem9, is it possible to distinguish between Protenor
and Lygaeus mode of sex determination based on the outcome of these crosses? 11.When COWS have twin calves of unlike sex (fraternal twins), the female twin is
usually sterile and has masculinind reproductive organs. This calf is referred to as
a freemartin .In cows, twins may share a common placenta and thus fetal
circulation. Predict why a freemartin develops.
12. It has been suggested that any male-determining genes contained on the Y
chromosome in humans cannot be located in the limited region that synapses with the X chromosome during meiosis. What might be the outcome if such genes were located in this region?
13. What is a Barr body, and where is it found in a cell?
14. Indicate the expect ed number of Barr bodies in interphase cell of the following
individuals; KIinefelter syndrome; Tumer syndrome; and karyotypes 47,XYY, 47,XXX, and 48,XXXX.
15. Define the Lyon hypothesis.
16 .Can the Lyon hypothesis be tested in a human female who is homozygous for one
allele of the X-linked G6PD gene? Why, or why not?
17. Predict the potential effect of the Lyon hypothesis on the region of a human
female heterozygous for the X-linked red-green color blindness trait.
18. Cat breeders are aware that kittens expressing the X-linked calico coat pattern and
tortoise shell pattern are almost invariable females. Why?
19. What does the apparent need for dosage compensation mechanisms suggest about
the expression of genetic information in normal diploid individuals?
Linkage and Chromosome Mapping in eukaryotyes
1. What is the significance of crossing over (which leads to genetic recombination to the process of evolution?
2. Describe the cytological observation that suggests that crossing over occurs during the first meiotic prophase.
3. Why does more crossing over occur between two distantly linked genes than between two genes that aree very close together on the same chromosome?
4. Why is a 50 percent recovery of single-crossover products the upper limit, even when crossing over always occurs between two linked genes?
5. Why are double-crossover events expected less frequently than single-crossover events?
6. What is the proposed basis for positive interference?
7. What three essential criteria must be met in order to execute a successful mapping cross?
8. The genes dumpy wings ( dp ),clot eyes (cl), and apterous wings (ap) are linked on chromosome II of Drosophilia .in a series of two-point mapping crosses, the genetic distances shown below were determined-What is the sequence of the three genes?
dp-ap 42 dp-cl 3 ap-cl 39
9. Consider two hypothetical recessive autosomal genes a and b, where a heterozygote is testcrossed to a double-homozygous mutant. Predict the phenotypic ratios under the following conditions: (a) a and b are located on separate autosomes; (b) aand b are linked on the same autosome but are so far that a crossover always occurs between them; (c) a and b are linked on the same autosome but are so close together that a crossover never occurs (d) a and b are linked on the same autosome about l0mu apart.
10.Colored aleurone in the kernels of corn is due to the dominant allele R. The recessive allele r when homozygous, produces colorless aleurone. The plant color (not kernel color) is controlled by another gene with two alleles, Y and y. The dominant Y allele results in green color, whereas the homologous presence of recessive allele causes the plant to appear yellow. I a testcross between a plant of unknown genotype and phenotype and a plant that is homozygous recessive for both traits, the following progeny were obtained:
CY 88; Cy12; cY8; cy92
Explain how these results were obtained by determining exact genotype and phenotype of the unknown plant, including the precise association of the two genes on the hormologs (i.e. the arrangement).
11. In the cross shown here, involving two linked genes, ebony (e) and claret (ca), in Drosophila, where crossing over does not occur in males, offspring were produced in a (2+:1ca:1e) ratio:
These genes are 30 mu apart on chromosome III. What did crossing over in the female
contribute to these phenotypes?
12. With two pairs of genes involved (P,p and Z,z), a testeross (to ppzz) with m organism of unknown genotype indicated that the gametes were produced m these proportions :PZ=42.4%; Pz=6.9%; pZ=7.1%; and pz=43.6%. Draw all possible conclusions tom these data.
13 in a serious of two-point map crosses involving five genes located on chromosome II in Drosophila, the following recombinant (single-crossover) frequencies were observed
Pr-adp 29 Pr-vg 13 Pr-c 21 Pr-b 6 Adp-b 35 Adp-c 8 Adp-vg 16 Vg-b 19 Vg-c 8 c-b 27
(a) If the adp gene is present near the end of chromosome II, (locus83), construct a map of these genes.
(b) In another set of experiments, a sixth gene (d) was tested. against b and pr, and the results were d-b =17% and d-Pr =23%. Preedict the results of two-point maps between d and c, d and vg, and d and adp.
14 Two different female Drosophila were isolated, each heterozygous for the autosomally linked genes black body (b), dachs tarsus (d) ,and curved wings (c).These genes are in the order d-b-c with b closer to d than to c. Shown below is the genotypic arrangement for each female, along with the various gametes formed by both. Identify which categories are noncrossovers (NCO),single crossovers (SCO), and double crossovers (DCO) in each case. Then, indicate the relative frequency in which each will be produced.
15 In Drosoophilia, a cross was made between females expressing the three X-linked recessive traits, scute bristles (sc),sable body (s), and vermilion eyes (v) and wild-type males. All females were wild type in the F1, while all males expressed all three mutant traits.The cross was carried to the F2 generation, and 1000 offsprings were counted, with the results shown in the table. No determination of sex was made in the F2 data. (a0) Using proper nomenclature, determine the genotypes of the Pl and F1 parents, (b) Determine the sequence of the three genes and the map distance between them. (c) Are there more or fewer double crossovers than expected? (d)Calculate the coefficient of coincidence; does this represent positive or negative interference?
16. A cross in Drosophila involved the recessive, X-linked genes yellow body (y), white eyes(w), and cut wings (ct). A yellow-bodied, white-eyed female with normal wings was crossed to a male whose eyes and body were normal, but whose wings
were cut. The F1 females were wild type for al1 three traits, while the Fl males expressed the yellow-body, white-eye traits. The cross was carried to F2 progeny, and only male offspring were tallied. On the basis of the data shown here, a genetic map was constructed. (a)Diagram the genotypes of the F1 parents; (b)Construct a map, assuming that w is at locus 1.5 on the x chromosome. (c) Were any double-crossover offspring expected? (d) Could the F2 female offspring be used to construct the map? Why or why not?
17.In Drosophila, Dichaete (D) is a mutation on chromosome III with a dominant effect on wing shape. It is lethal when homozygous. The genes ebony body (e) and pink eye(p) are recessive mutations on chromosome III. Flies from a Diet1aete stock were crossed to homozygous ebony, pink flies, and the Fl progeny with a Diet1aete phenotype were backcrossed to the ebony, pink homozygotes. Using the results of this backcross shown in the table, (a)diagram the cross, showing the genotypes of the parents and offspring of both crosses. (b) What are sequence and interlocus distance between these three genes?
18. Drosophila females homozygous for the third chromosomal genes pink eye (p) and ebony body (e) were crossed with males homozygous for the second chromosomal gene dumpy wings (dp). Because these genes are recessive, all offsprings were wild type (normal). Fl females were testcrossed to triply recessive males. We assume that the two linked genes (p and e) are 20 mu apart, predict the results of this cross. If the reciprocal cross were made (F1 males-where no crossing
over occurs-with triply recessive females), how would the results vary, if at all?
19.In Drosophila, the two mutations Stubble bristles (Sb) and curled wings (cu) are linked on chromosome E. Sb is a dominant gene that is lethal in a homozygous state, and Cu is a recessive gene. If a female of the genotype
Sb Cu
+ +
is to be mated to detect recombinants among her offspring, what male genotype would you choose as her mate?
20.In Drosophila, a heterozygous female for the X-linked recessive traits a, b, and c was crossed to a male that phenotypically expressed a, b, and c. The offspring occurred in the phenotypic ratios in the following table, and no other phenotypes were observed. (a)What is the genotypic arrangement of the alleles of these genes on the X chromosome of the female? (b)Determine the correct sequence, and construct a map of these genes on the x chromosome. (c)What progeny phenotypes are missing, and why?
21. Are sister chromatid exchanges effective in producing genetic variability in an individual? In the offspring of individuals?
22.What conclusion can be drawn from the observations that male Dromphila, no crossing over occurs and that during meiosis, synaptonemal complexes (ultrastuctured components found between synapsed homologs in meiosis) are not seen in males but are observed in females where crossing over occurs?
23. An organism of the genotype AaBbCc was testcrossed to a triply recessive organism (aabbcc). The genotypes of the progeny are in the following table.
(a) Assuming simple dominance and recessiveness in each gene pair, if these three genes were all assorting independently, how many genotypic and phenotypic classes would result in the offspring and in what proportion?
(b) Answer part (a) again, assuming the three genes are so tightly linked on a single chromosome that no crossover gametes were recovered in the sample of offspring? (c)What can you conclude tom the actual data about the location of the three genes
relation to each other?
24 In a plant, fruit color is either red or yellow, and fruit shape either oval or long.
Red and oval are the dominant traits. Two plants, both heterozygous for these traits ,were testcrossed, the results shown below. .Determine the location of the genes relative to one another and the genotypes of the two parental plants.
25. In a cross in Neurospora involving two alleles, B and b, the tetrad patterns in the following table were observed. Calculate the distance between the gene and centromere.
Tetrad pattern Number Tetrad pattern Number Tetrad pattern Number Tetrad pattern Number Tetrad pattern Number
BBbb 36 bbBB 4 BbBb 4 BBbB 6 BbbB 3 bBBb 7
26. D.melanogaster has one pair of sex chromosomes (XX or XY) and three autosomes (chromosomes II, III ,and IV). A genetics student discovered a male fly with very short (sh) legs-Using this male, the student was able to establish a pure-breeding stock of this mutant and found that it was recessive. She then incorporated the mutant into a stock containing the recessive gene black (b, body color, located on chromosome II) and the recessive gene pink (p, eye color, located on chromosome III). A female from the homozygous black, pink, short stock was then mated to a wild-type male. The F1 males of this cross were all wild type and were then backcrossed to the homozygous b, p, sh females. The F2 results appeared as shown in the table that follows, and no other phenotypes were observed .(a)Based on these results, the student was able to assign sh to a linkage group (a chromosome). Determine which chromosome, and include step-by-step reasoning. (b)The student repeated the experiment, making the reciprocal cross: F1 females backcrossed to homozygous b, p, sh males. She observed that 85 percent of the offspring fell into the given classes, but that15 percent of the offspring were equally divided among b+p, b++, +shp, and +sh+ phenotypic males and females. How can these results be explained, and what information can be derived tom these data?
Mapping in Bacteria and Bacteriphages
1. Distinguish among the three modes of recombination in bacteria.
2. With respect to F+ and F- bacterial matings, (a)How was it established that physical contact was necessary? (b)How was it established that chromosome transfer was unidirectional?(c) What is the genetic basis of a bacterium being F+?
3.List all of the differences between F+ ×F- and Hfr ×F-bacterial crosses.
4 Bacterial conjugation, mediated mainly by conjugative plasmids like F, represents a potential health threat through the sharing of genes for pathogenicity or antibiotic resistance. Given that more than 400 different species of bacteria co-inhabit a healthy human gut and more than 200 co-inhabit human skin, Dionisio and colleagues investigated the ability of plasmids to undergo between species Conjugal transfer. Data (Dionisio et al.2002. Genetics 162:1525-1532) involving various species of the enterobacterial genus, Escherichia, are presented in the fo11owing table. The data are presented as "log base l0 "values, where for example, -2.0 would be equivalent to10-2 as a rate of transfer. Assume that all differences between values are statistically significant.
(a) What general conclusion(s) can be drawn from these data? (b) In what species is with in-species transfer most 1ikely? In what species pair is between-species transfer most likely?
(c)What is the significance of these findings in terms of human health?
5 Describe the origin of F` bacteria and merozygotes.
6. Describe the mechanism of transformation.
7. In a transformation experiment involving a recipient bacterial strain of genotype a-b- , the results below were obtained. What can you conclude about the location of the a and b genes relative to each other?
8. In a transformation experiment, donor DNA was obtained from a prototroph bacterial strain (a+b+c+), and the recipient was a triple auxotroph (a-b- c-). What general conclusions can you draw about the linkage relationships among the three genes from the following transformant classes that were recovered?
9. The bacteriophage genome consists primarily of genes encoding proteins that make up the head, collar and tail, and tail fibers. When these genes are transcribed following phage infection, how are these proteins synthesized, since the phage genome lacks genes essential to ribosome structure.
10. Describe the temporal sequence of the bacteriophage life cycle.
11. In the plaque assay, what is the precise origin of a single plaque?
12. In the p1aque assay, exactly what makes up a single plaque?
13. A plaque assay is performed beginning with 1.0 mL of a solution containing bactedophages. This solution is serially diluted three times by taking 0.1 mL and adding it to 9.9mL of liquid medium. The End dilution is plated and yields l7 plaques. What is the initial density of bacteriophages in the original 1.0 mL?
14.Describe the diffemnce between the lytic cycle and lysogeny when bacteriophage infection occurs.
15. .Define prophage.
16. Explain the observations that led Zinder and Lederberg to conclude that the prototophs recovered in their transduction experiments were not the results of Hfr mediated conjugation.
17.Describe generalized transduction and distinguish between abortive and complete transduction.
18. Describe how generalized transduction can be used to map bacterial chromosome. How does cotransduction play a role in mapping?
19. Two theoretical genetic strains of a virus (a-b-c- and a+b+c+) are used to simultaneously infect
a culture of host bacteria. Of the plaques scored, the genotypes in the following table were observed. Determine the genetic map of these three genes on viral chromosome..
20. Describe the conditions under which genetic recombinations occur in bacteriophages.
21. If a single bacteriophage infects one E coli cell present in a culture of bacteria and, upon lysis, yields 200 viable viruses, how many phages will exist in a single plaque if three more lytic cycles occur?
22. A phage-infected bacterial culture was subjected to a serial dilutions, and a plaque assay was performed in each case with the following results. What conclusion cm be drawn in the case of each dilution?
23. When the interrupted mating technique was used with five different strains of Hfr bacteria, the order of gene entry during recombination shown in the following table was observed. On the basis of these data, draw a map of the bacterial chromosome. Do the data support the concept of curcularity?
24. In B.subtilis, linkage analysis of two mutant genes affecting the synthesis of the two amino
acids, typtophan (typ2), and tyrosine (tyr1), was performed using transformation. Examine the data
in the table that follows, and draw all possible conclusions regarding linkage.
25. An Hfr strain is used to map three genes in an interrupted mating experiment. The cross is Hfr/a+b+c+rif s×F-/a-b-c-rif r(no map order is implied in the listing of the deles; rif= the antibiotic rifampicin). The a十gene is required for biosynthesis of nutrient A, the b+ gene for nutrient B, and c+ for nutrient C. The minus alleles are auxotrophs for these nutrients. The cross is initiated at time=0, and , at variuous intervals, the mating mixture is plated on different types of medium. Each plate contains minimal medium (MM) plus rifampicin plus the specific supplements indicated in the fo11owing table (results for each time point are shown as number of colonies growing on each plate). (a)What is the purpose of rifampicin in the experiment? (b) Based on these data, determine the approximate location on the chromosome of the a, b, and c genes relative to one another and to the F factor. (c) Can the location of the rif gene be determined in this experiment? If not, design experiment to determine the location of rif relative to the F factor and to gene b.
26 In a cotransformation experiment using various combinations of genes, two at a time, the following data were produced. Determine which genes are linked and to which others.
Quantitative Genetics
1 What is the difference between continuous and discontinuous variation? Which of the two is most likely to be the result of polygenic inheritance?
2 Define the following: (a) polygene, (b) additive alleles, (c) correlation, (d) heritability.
3 A homozygous plant with 20-cm diameter flowers is crossed with a homozygous plant of the same species that has 40-cm diameter flowers. The Fl plants all have flowers 30cm in diameter. In the F2 generation of 512 plants, 2 plants have flowers 20 cm in diameter, 2 plants have flowers 40cm in diameter, and the remaining 510 plants have flowers of a range of sizes in between. (a) Assuming that a1l alleles involved act additively, how many g enes control flower size in this plant?
(b) What frequency distribution of flower diameter would you expect to see in the progeny of a backcross between m Fl plant and the large-flowered parent?
(c)Calculate the mean, variance, and standard deviation for flower diameter in the backcross progeny from (b).
4.A dark-red strain and a white strain of wheat are crossed and produce intermediate, medium-red F1. When the Fl plants are interbred, an F2 generation is produced in a ratio of 1 dark-red: 4 medium-dark-red; 6 medium-red: 4 light-red; 1whte. Further crosses reveal that the dark- red and white F2 plant are true breeding.
(a)Based on the ratio of offspring in the F2, how many genes are involved in the production of color?
(b)How many additive alleles are needed to produce each possible phenotype?
(c)Assign symbols to these alleles and list possible genotypes that give rise to the medium-red and the light-red phenotypes
(d)Predict the outcome of the F1 and F2 generations m a cross between a true-breeding medium- red plant and a white plant.
5.Height in humans depends on the additive action of genes. Assume that this trait is controlled by the four loci R, S, T, and U and that environmental effects are negligible. Instead of additive versus nondditive alleles, assume that additive and partially additive alleles exist. Additive alleles contribute two units, and partially additive deles contribute one unit to height.
(a)Can two individuals of moderate height produce offspring that are much taller or shorter than either parent? If so, how?
(b) If an individual with the minimum height specified by these genes marries an individual of intermediate or moderate height, will any of their children be taller than the tall parent? Why or why not?
6. An inbred strain of plants has a mean height of 24cm. A second strain of the same species from
a different geographical region also has a mean height of 24cm. When plants from the two strains are crossed together, the Fl plants are the same height as the parent plants. However, the F2 generation shows a wide range of heights; the majority are like the Pl and Fl plants, but approximately 4 of 1000 are only 12cm high and about 4 of 1000 are 36cm high.
(a) What mode of inheritance is occurring here?
(b) How many gene pairs are involved?
(c) How much does each gene contribute to plant height?
(d) Indicate one possible set of genotypes for the original Pl parents and the Fl plants that could
account for these results.
(e)Indicate three possible genotypes that could account for F2 plants that are 18 cm high and three that account for F2 plants that are 33cm high
7. In a series of crosses between two true-breeding strains of peaches the Fl generation was uniform, producing 30-g peaches.The F2 fruit mass ranges from 38 to 22g at intervals of 2g.
(a)Using these data, determine the number of polygenic loci involved in the inheritance of peach mlass. (b) Using gene symbols of your choice, give the genotypes of the parents and the F1.
29. Consider a true-breeding plant, AABBCC, crossed with another true-breeding plant, aabbcc, whose resulting offspring AaBbCc. If you cross the Fl where independent assortment operational, the expected fraction of offspring in each phenotypic class is given by the expression N!/[M!(N-M!)], where N is the total number of alleles (six in this example) aIl number of upperuse alleles. In a cross of AaBbCc ×AaBMti what proportion of the offspring would be expected to contain two uppercase alleles?
7.Erma and Harvey were a compatible barnyard pair but a cunous sight.H 町vey's tail was only6cm,while EEII1a's was3Ocm. Their F1piglet offspring allgrew tails th at were 18cm.When inbred,an F2generation resulted in many piglets (Erma and H 缸- veys grandpigs),whose tails ranged in 4-cm intervals from 6t0 3Ocm (6,10,14,18,22,26,and 30).Most had 18-cm taih, while 1/64had 6-cmtails and 1/64had 30-cmtails-
(a)Explain how taillength is inherited by describing the mode
of inheritance,indicating how many gene pairs are at work and designating the genotypes of H 缸vey,Ermaand their 18- cm o 旺spring-
(b)If one of the 18-cm Fl pigs were mated with one of the 6-cm F2pigs,what phenotypic ratio would be predicted if mmy o 旺spring resulted?Diagram the cross.
8.h the following table,average diEerences of height,weight,and Erlgefpdnt ddge countbetween monozygoUc twins (Ieared toge 也er and apart),dizygodc twins,and nontwin siblings are compared:
Based on th e dauin this table,which of these qumtitdve traits have the highest heritability values?
9.What kind of heritability estimates (broad sense or narrow
sense)are obtained from human twin studies?
10.List as many human traits as you cm that 缸e likely to be under
th e controIof a polygenic mode of inhedance-
11.Corn plants from a test plot are measured,and the distribution of heights at 10-cm intervals is recorded in the following table
Calculate (a)the mean height,(b) th e v 缸imce,(c)the standard deviadOIL and (d)the standard error of the mean.Plot a rough graph of plant height against frequency.Do the values represent anormal distribution?B ωed on your calCElla6ons,how would you assess d1e variation within th is population?
12.The following variances were calculated for two traits in a herd
of hogs.
ι
13.The mem and variance of plant height of two highly iIlM
(a)Calculate broad-sense (H2)and narrow-sense d)impitabili-
ties for each trait in this herd.
strains(Pl and p2)md their progeny mand Fd are shownheft
(a)Calculate the naITOW-sense heritability(h2)for both trau (b)Which trait,if either,is likely to respond to selecUon?
15.The following table shows measurements for Eber lengths and
fleece weight in a small nock of eight sheep
(a)What are the mem,variance,and standard deviation foread
trait in th is flock? (b)What is the covariance of the two traits? (c)What is th e cOETelation coemdent for Eber length mdh ω
weight?
(d)Do you think m increase inEber length causes greaterfleect
weight?Why or why not?
16.In a herd of dairy cows,the narrow-sense hedtability formilt protein content is0.76,and for mik buttedat it is0.82.ThecOI- relation coemcient between milk protein content and blltidalil 091.If the farmer selects for cows producing mom bullerfaliR their milk,what will be the most likely eEect on mlk prom content in the next generation?
17.h 阻挡sesSIIlent of lean1ing m Dmsophi 缸,ni 臼may be 往回ned to avoid certain olfactory cum-h one poPulaHOIL a mem of 85 回als was required.A subgoup of 出is paren 臼1popuhtion th at was tr 垣ned most quicMy (mean =6.O)was in 田bred,md 血ekprogeny was examined.ITmse 血es demonstra 能 d a mm training value of 75. Calculate rea1ized hedtabi1ity for olfactory kaming in Dmsophih1.
l8.Suppose you want to develop a population of Drosophila th at W0111d rapidly le 缸卫to avoid certain substances the flies could detect by smell.Based on the heritability estimate you obtained
in Problem 17,would it be wodh doing th is by aruncialselec-' tioMWhy or why not?
l9.In a population of tomato plants,mean fmit weight is60g and dis03.Predict the mem weight
ofthe progeny iftomato plants whose fmit averaged 80g were selected from the original popu- lation and interbFed.
20.In a population of 100inbred genotypically identical rice plants, variance for grain yield is4.67.What is th e heEttability for yield? Would you advise a rice breeder to improve yield in th is strain of rice plants by seleCHon?
2l.A mutant s 位ain ofDrosophila was isolated and shown to be resis- 国lt to m experimental insecticide,whereas normal (wild-type) flies were sensitive to the chemical.Following a cross between resistant Hies and sensitive flies,isolated populations were derived th at had various combinations of chromosomes from the two strains.Each was tested for resistance,as shown in the Eg- lim.Analyze the data and draw any appropriate conclusion about which chromosome(s)contain a gene responsible for idledunce ofresistance to the insecticide.
Examine Wilkens's results and respond to the followmg questIons:
(a)Based sMctly on th e Fl and F2results of Wilkens's initial crosses,what possible explanation concerning the inheritance of eye size seems most feasible?
(b) Based on the results of the Fl backcross with cavefish, is your explanation supported? Explanation.
(c) Based on the results ofthe Fl backcross with lakefish,is your e xplanation supported?Explain. (d) Wilkens examined about 1000F2 progeny and estimated that six to seven genes are involved in determining eye size.Is the samples inadequate to judge this conclusion? Propose an expedmental protocol to test the hypothesis.
(e) A comparison of the embryonic eye in cavefish md1abash revealed that both reach approximately 4mm in diameter. However,lakeEsh eyes continue to grow,while cavefish eye size is greatly reduced.Speculate on the role of the genes involved in th is problem-[R 写 fkrenceJWilkens,H.(1988)- Ecol.Biol.23:271-367.]
23.ALinch plant was crossed with a 15inch plant and 41Fl plants were 9-inches-h the F2,plants exhibited a "normal dis 位 ibuHon" with heights of 3,4,5,6,7,8,9,10,11,12,13,14,md15inches. (a)What ratio will constitute the "normaldis 位 ibution"M the
F2?
(b)What outcome will occur if th e Fl plants are testcrossed with
plants th at are homozygous for a11nonadditive alleles?
24.The following table gives the percentage of twin pairs studied in which both twins expressed the same phenotype for a trait (con- cordance).Percentages listed are for diEerent trats m monozy- gotic(MZ)and dizygotic(DD wins.Assuming that both twins in each p 组 r were raised together in the same environment,what do you conclude about the relative importance of genetic versus environmental factors for each trait?
25.In a cross between a strain of large guinea pigs and a strain of small guinea pigs, 也 e F1 缸 e phenotypicalIy uniform with m average size about intermediate between that of the two parentaI Trait DZ%
Blood types 100
66 Eye co1or 99
28 Mental retardation 97
37 h4easles 95
87 Hair color 89
22 Hande 伽 ess 79
77 Idiopathic ep 丑 epsy 72
15 Schizophrenia 69
10 Diabetes 65
18 Identical allergy
5 59
Clefk lip 42 5
32 Club foot
3 6 Mammazy cancer
3
专升本医学遗传学练习题(A)
专升本《医学遗传学》练习题(A) 班别：姓名：学号：成绩： 一．选择题 1. 最常见的染色体三体综合征是_______________________； A．18号三体 B. 13号三体 C. 9号三体 D. 21号三体 2. 200个初级母细胞最终形成的卵子数是___________; A. 800; B. 600; C. 400; D. 200; 3. 常染色体隐性遗传病家系中，患者双亲__________； A．都是携带者；B. 都是患者; C. 有一个患者; D. 没有患者; 4. 镰状贫血是由于血红蛋白β链第6位谷氨酸被____________所取代; A. 胱氨酸; B. 缬氨酸; C. 亮氨酸; D. 赖氨酸; 5. 下面__________疾病不属于多基因疾病; A. 高血压; B. 糖尿病; C. 先天性幽门狭窄; D. 毛细管扩张性共济失调; 6. 嵌合型克氏综合征的核型为_____________; A. 46, XY/47, XXY; B. 46, XX/47, XXX; C. 46, XY/47, XYY; D. 46, XX/47, XYY; 7. 如果一种多基因病,其男性发病率高于女性,则其后代复发风险是 A. 男性高于女性; B. 男女相同; C. 女性高于男性; D. 与双亲发病无关; 8. 一患者核型为难47,XXY, 在细胞分裂间期,其性染色质组成为:_______ A. 1个X染色质,1个Y染色质; B. 2个X染色质,1个Y染色质; C. 1个X染色质, 无Y染色质; D. 2个X染色质,1个Y染色质; 10. 一个个体核型为: 46,XY,-14,+t(14q21q), 该个体是___________; A. 正常人; B. 先天愚型患者; C. 平衡易位携带者 D. 以上都不对; 11. 下列疾病除______________外都是多基因病. A. 原发性高血压; B. 精神分裂症; C. 强直性脊柱炎; D. 血友病. 12. 一对夫妇已生出两个苯酮尿症(常染色体隐性遗传病)患儿,这对夫妇再生育 时,生出不患病婴儿的概率是:_______ A. 0; B. 25%; C. 100%; D. 75%;
遗传学实验指导
遗传学实验指导 实验1 细胞有丝分裂与减数分裂 实验1.1 植物根尖细胞有丝分裂过程的制片与观察 目的要求 学习和掌握植物细胞有丝分裂制片技术；观察植物细胞有丝分裂过程中染色体的形态特征及染色体的动态行为变化。 实验原理 有丝分裂是植物体细胞进行的一种主要分裂方式。有丝分裂的目的是增加细胞的数量而使植物有机体不断生长。在有丝分裂过程中，细胞核内的遗传物质能准确地进行复制，然后能有规律地均匀地分配到两个子细胞中去。植物有丝分裂主要在根尖、节间、茎的生长点、芽及其它分生组织里进行。将生长旺盛的植物分生组织经取材、固定、解离、染色、压片等处理即可以观察到细胞内的有丝分裂图象。如若需要进行染色体计数，则需进行前处理，即取材之后采用物理的或化学的方法，阻止细胞分裂过程中纺锤体的形成，使细胞分裂停止在中期。这时，染色体不排到赤道板上，而是散在整个细胞质中。这十分便于对染色体的形态、数目进行观察。 试剂和器材 1材料 均可以大蒜(Allium sativum 染色体数目2n=16)、玉米(Zea mays 染色体数目2n=20)、洋葱(Allium cepa染色体数目2n=16)或蚕豆(V icla faba染色体数目2n=12) 等根尖为实验材料。 2试剂 95％乙醇、冰乙酸、石炭酸品红、l mol／L HCl。 3器材 恒温培养箱、显微镜、水浴锅、载玻片、盖玻片、单面刀片、镊子、培养皿、量筒、吸水纸。 操作方法 1生根 植物根尖是植物的分生组织，取材容易，操作方便。植物根尖细胞分裂旺盛，因此，它是细胞有丝分裂相制备与观察的理想选取部位。大蒜、洋葱易于在水培、沙培、土培条件下生根。采用水培时要注意在暗处培养，以满足根生长条件，使根系生长旺盛。玉米和蚕豆种子可先用温水浸泡1天之后，再转入铺有多层吸水纸或纱布的培养皿中，上面盖双层湿纱布置于24～26℃温箱中培养，每天换水二次。 2取材 待根长至l.5~2.0 cm时，将根取下。若实验只需观察细胞有丝分裂的过程和各时期的特征，可将根尖直接放入Carnoy固定液(95％乙醇：冰乙酸=3：1)中固定；如果要观察染色体形态和数目，则必须对根尖进行前处理后才能固定。取材和固定必须要在细胞分裂高峰期进行，即分裂细胞占细胞总数最大值时进行，这样分裂细胞比例大，便于选择和观察。 不同的植物在不同的环境条件，其细胞分裂高峰的时间是不同的。大蒜和洋葱的细胞分裂高峰期通常
南开大学《遗传学》练习题-第6章答案.doc
性 * 弟八早 1 重组极性杂合DNA 模型中的异常分离现象最早是在酵母不同交配型的杂交中发现的。合子减数 分裂产生的4个子囊抱子除了正常的2A:2a 分离外，出现了 3J:U 或者1/1:3〃的分离，试用某种DNA 重组模型加以说明，并附图解。 答：Holliday 模型，发生单链交换以后的局部异源双链在细胞内的错配修复系统识别下将其 中的一条链切除。前提：酶系统只选择性的切除两条异源双链中的一条，而不是随机切除一 条链。图参见课件。 2 子囊通纲的一种真菌Ascobulus 的一些突变产生浅色子囊泡子，成为a 突变体。不同a 突变体进行了以 下的杂交，观察具有黑色野生型子囊饱子的子囊，杂交中出现的黑色饱子的基因型如下： 答：三次转换，总计：％—1次；⑶一0 次；々3—2次。基于“同线分析”原理，与断点连锁越 1 2 J 紧密，越易发生基因转变。由于。3离断裂位点最近，气次之，灼最远，基因转变具有方向 J 1 L 性，因此推断三个位点的顺序为：（重组位点）勺一％一。2。 3 下表表示人一鼠杂种无性系以及它们所包含的人类染色体。检验了人的4种酶：TK （lhymidine-kinase, 胸腺激酶）,只在无性系C 中有活性;LDH （lactate dehydrogenase,乳酸脱氢酶）,在A 、C 3个无性 系中都有活性；PGK （phpsphoglycerate kinase?磷酸甘油酸激酶），在无性系A 和C 中有活性；AHH （aryl 无性系 人的染色体 X 2 11 17 A + — + — B — + + — C + — + + 答：TK 只有C 系有活性，即A 、B 系无活性，也就是说，A 、B 系是突变体，由于基因与 染色体具有平行关系，因此与17号染色体同线，即TK 基因位于17号染色体。 同理，LDH 位于11号染色体，PGK 位于X 染色体，AHH 位于2号染色体。 1 1 + a\ + 您 + + + 白1 + + + a 2 + + + + + 白2 + "3 + "3 试说明这些结果。有们、。2和。3这三个突变位点的可能次序如何？ "|X 〃2 〃]X ?2X ?3
刘祖洞遗传学习题答案5
第五章性别决定与伴性遗传 1、哺乳动物中，雌雄比例大致接近1∶1，怎样解释？ 解：以人类为例。人类男性性染色体XY，女性性染色体为XX。男性可产生含X和Y 染色体的两类数目相等的配子，而女性只产生一种含X染色体的配子。精卵配子结合后产生含XY和XX两类比例相同的合子，分别发育成男性和女性。因此，男女性比近于1 ：1。 2、你怎样区别某一性状是常染色体遗传，还是伴性遗传的？用例来说明。 3、在果蝇中，长翅（Vg）对残翅（vg）是显性，这基因在常染色体上；又红眼（W）对白眼（w）是显性，这基因在X染色体上。果蝇的性决定是XY型，雌蝇是XX，雄蝇是XY，问下列交配所产生的子代，基因型和表型如何？ （l）WwVgvg×wvgvg （2）wwVgvg×WVgvg 解：上述交配图示如下： 即基因型：等比例的WwVgvg，WwVgvg，wwVgvg，wwvgvg，WYVgvg，WYvgvg，wYVgvg，wYvgvg。 表现型：等比例的红长♀，红残♀，白长♀，白残♀，红长♂，红残♂，白长♂，白残♂。 即，基因型：1WwVgVg ：2WwVgvg ：1Wwvgvg ：1wYVgVg ：2wYVgvg ：1wYvgvg。 表现型：3红长♀：1红残♀：3白长♂：1白残♂。
4、纯种芦花雄鸡和非芦花母鸡交配，得到子一代。子一代个体互相交配，问子二代的芦花性状与性别的关系如何？ 解：家鸡性决定为ZW型，伴性基因位于Z染色体上。于是，上述交配及其子代可图示如下： 可见，雄鸡全部为芦花羽，雌鸡1／2芦花羽，1／2非芦花。 5、在鸡中，羽毛的显色需要显性基因C的存在，基因型cc的鸡总是白色。我们已知道，羽毛的芦花斑纹是由伴性（或Z连锁）显性基因B控制的，而且雌鸡是异配性别。一只基因型是ccZ b W的白羽母鸡跟一只芦花公鸡交配，子一代都是芦花斑纹，如果这些子代个体相互交配，它们的子裔的表型分离比是怎样的？ 注：基因型C—Z b Z b和C—Z b W鸡的羽毛是非芦花斑纹。 B B 因此，如果子代个体相互交配，它们的子裔的表型分离比为芦花：非芦花＝9/16 ：7/16。若按性别统计，则在雄性个体中芦花：非芦花＝6/16 ：2/16；在雌性个体中芦花：非芦花＝3/16 ：5/16； 6、在火鸡的一个优良品系中，出现一种遗传性的白化症，养禽工作者把5只有关的雄禽进
遗传学实验指导
《遗传学实验》简介 课程名称：遗传学实验 课程性质：随课实验 课程属性：基础课 学时学分：学时16学分1 面向专业： 一、课程的任务和基本要求 遗传学实验是生物学所有专业的一门专业基础课程，是一门独立的实验课程，同时也是为了配合遗传学的教学而开设的一门实验课程,本课程由基础操作性、综合性和设计性等多层次实验内容构成，目的在于巩固和加深学生对遗传学知识的理解、验证遗传学理论、并初步掌握遗传学研究所必需的基本实验技术。在此基础上，进行一定比例的综合性、设计性实验，将实验理论和实验技能融为一体，从而培养学生的基本实验思想、实验方法、实验技能和综合应用能力。本课程的任务是从个体、细胞水平揭示遗传学的基本现象与规律，培养学生牢固掌握经典遗传学研究方法与技术，并初步掌握现代遗传学实验操作技能，熟悉遗传学分析方法，同时要求学生初步具备进行遗传学创新性研究的基本能力与素质。 二、实验项目 植物染色体核型分析 果蝇的培养及生活史和形态观察 小鼠骨髓染色体的制片及观察 人群中基因频率的调查 人类巴氏小体观察 三、有关说明 1、与其它课程和教学环节的联系 先修课程：动物学、植物学、生物化学、微生物学 后续课程：分子生物学、基因工程、人类遗传学 平行开设课程：细胞生物学 2、教材和主要参考书目 （1）教材：刘祖洞江绍慧遗传学实验高等教育出版社1987年第二版 （2）主要参考书目 ①张文霞等编遗传学实验指导高等教育出版社2005 年版 ②张贵友等编普通遗传学实验指导清华大学出版社2003 年 实验一1、洋葱根尖有丝分裂染色体标本制备及观察（３学时） 一、教学基本要求： 1. 掌握植物染色体压片法。 2. 掌握有丝分裂各时期的特征。 3. 掌握用秋水仙素诱发多倍体的一般方法 二、教学内容： 1、洋葱根尖有丝分裂染色体标本制备及观察 〖目的和要求〗 本次实验通过对植物根尖的制片和观察, 要求学生初步学会对植物组织、细胞的固定、离析和压片方法，了解有丝分裂的全过程及其染色体的动态变化情况,掌握有丝分裂各时期的特征。〖实验原理〗 有丝分裂是细胞均等增殖的过程, 是体细胞分裂的主要方式.在有丝分裂过程中.细胞内每条染色体都能复制一份, 然后分配到子细胞中,因此两个子细胞与母细胞所含的染色体在数目,形态和性质上均是相同的，在各种生长旺盛的植物组织中均存在着有丝分裂。分生组织→固定→解离→染色→压片→观察（间期、早期、中期、后期、末期） 〖材料和方法〗 洋葱或大蒜或种子根尖 水浴锅、显微镜 卡诺氏固定液、石炭酸品红（或醋酸洋红）染色液，1N盐酸（或0.5%果胶酶+纤维素酶）,秋水仙素（0.1%）1、取材：大蒜、洋葱根尖，恒温箱
普通遗传学第五章-连锁遗传--自出试题与答案详解第一套
连锁遗传 一、名词解释 1、完全连锁与不完全连锁 2、相引性与相斥性 3、交换 4、连锁群 5、基因定位 6、干涉 7、并发系数 8、遗传学图 9、四分子分析 10、原养型或野生型 11、缺陷型或营养依赖型 12、连锁遗传 13、伴性遗传 14、限性遗传 15、从性遗传 16、交换 17、交换值 18、基因定位 19、单交换 20、双交换 二、填空题 1、有一杂交：CCDD × ccdd，假设两位点是连锁的，而且相距20个图距单位。F2中基因型(ccdd)所占比 率为。 2、在三点测验中，已知AbC和aBc为两种亲本型配子,在ABc和abC为两种双交换型配子,这三个基因在染 色体上的排列顺序是____________。 3、基因型为AaBbCc的个体，产生配子的种类和比例： （1）三对基因皆独立遗传_________种，比例为___________________________。 （2）其中两对基因连锁，交换值为0，一对独立遗传_________种，比例为________________。 （3）三对基因都连锁_______________种，比例___________________________。 4、A和B两基因座距离为8个遗传单位，基因型AB/ab个体产生AB和Ab配子分别占 %和 %。 5、当并发系数C=1时，表示。当C=0时，表示，即；当1＞C＞0时，表示。即第一次见换后引起邻近第二次交换机会的。C＞1时，表示，即第一次见换后引起邻近第二次交换机会的。常在中出现这种现象。 6、存在于同一染色体上的基因，组成一个。一种生物连锁群的数目应该等于，由性染色体决定性别的生物，其连锁群数目应于。 7、如果100个性母细胞在减数分裂时有60个发生了交换，那麽形成的重组合配子将有个，其交换率 为。 8、在脉孢菌中，减数分裂第一次分裂分离产生的子囊属型的，第二次分裂分离产生的子囊 属型的。 三、选择题 1、番茄基因O、P、S位于第二染色体上，当F1 OoPpSs与隐性纯合体测交，结果如下：+++ 73， ++S 348， +P+ 2， +PS 96， O++ 110， O+S 2， OP+ 306，OPS 63 ，这三个基因在染色体上的顺序是 （） A、o p s B、p o s C、o s p D、难以确定 2、如果干涉为33.3％，观察到的双交换值与预期的双交换值的比例应为（）
【免费下载】医学遗传学练习题
《医学遗传学》练习题 一、名词解释 1、遗传学 遗传学是研究生物遗传与变异的科学。其研究的主要内容：（1）遗传物质的本质；（2）遗传物质的复制及传递规律；（3）遗传信息的实现，包括基因的表达与调控等；（4）遗传物质的改变，包括基因突变、染色体畸变等。这些研究是在细胞、分子、个体、群体等不同水平上进行的。 2、嵌合体 一个个体同时存在两种或两种以上不同核型的细胞系，称为嵌合体。嵌合体是在受精卵早期卵裂时，某一细胞中染色体发生畸变形成的。嵌合体也包括数目畸变嵌合体和结构畸变嵌合体。常见的嵌合体大多为46条和47条染色体的嵌合。 3、分子病 是指由基因突变造成蛋白质结构或合成量异常所引起的疾病。 4、药物遗传学 是研究人体药物代谢反应的遗传基础和生化本质的，是药理学和遗传学相互结合的边缘学科、交叉学科，属于生化遗传学范畴。 5、遗传病 是指由于生殖细胞或受精卵的遗传物质在结构或功能上发生了改变所引起的疾病，并按一定方式在上下代间传递。 6、延迟显形 是指某些带有显形致病基因的杂合体，在生命的早期不表现相应症状，当发育到一定年龄时，致病基因的作用才表现出来。 7、限性遗传 是指位于染色体上的基因，由于性别限制，只在一种性别表现，而在另性别完全不能表现，这种现象称为限性遗传。 8、染色体 是生物遗传物质的载体，具有贮存遗传信息，表达遗传信息的功能。 9、染色体综合征 染色体病常表现为综合症，涉及生长迟缓，多发畸形、智力障碍和皮纹改变等。正因为如此，染色体病也常被称为染色体综合症。 10、携带者 是指表型正常但带有致病基因或异常染色体的个体。 11、克隆 是指单个细胞经过有丝分裂形成的细胞群。 12 、交叉遗传 X连锁遗传中男性的致病基因只能从母亲传来，将来只能传给女儿，不存引男性向男性的传递，称为交叉遗传。 13、非整倍体 是指体细胞中的染色体，在2n的基础上增加或减少了一条或几条。 14、常染色质和异染色质 常染色质螺旋化程度低，染色较浅而均匀，含有单一或重复顺序，具有转录活性，叫常染色质。异染色质在间期核中仍处于凝集状态，即螺旋化程度较高，着色较深，很少转录或无转录活性。 15、易患性 在多基因遗传病中，由遗传基础和环境因素共同作用，决定一个个体是否易患病，称为易患性。 16、亲缘系数 指近亲的两个个体在一定基因座位上具有共同祖先的同一等位基因的概率又称血缘系数。 17、遗传性酶病 由于基因突变导致酶蛋白缺失或酶活性异常所引起的遗传性代谢紊乱,称为遗传性酶病，又称先天性代谢缺陷。 18、医学遗传学 是将遗传学基本理论与医学实践相结合的一门新学科，是遗传学知识在医学领域中的应用。 19、单一序列 是单拷贝的序列，在一个基因组内只出现一次或很少几次，构成编码细胞中的蛋白质或酶的基因。 20、重复序列 指这种DNA在基因组中有多数拷贝，有些与染色体的结构有关，多数的生物学功能不详。 21、自杀基因与多药抗药性基因 指该基因编码的酶可以将某种无毒的底物转化成有毒产物的基因。 指该基因编码蛋白质可以使细胞具有抵抗化疗药物的基因。二者都用于基因治疗。 二、填空 1、DNA由脱氧核糖核苷酸组成，后者由（磷酸）、（脱氧核糖核）和含（氮碱基）（腺嘌呤A、鸟嘌呤 G、胸腺嘧啶T或胞嘧啶C）组成。 2、根据着丝粒的位置，人类染色体分为三种类型：（中着丝粒染色体）、（亚中着丝粒染色体）和（近端着丝粒染色体）。 3、有丝分裂期分为（四个时期）：前期、中期、后期和（末期）。
遗传学期末复习题
遗传学复习题 一、名词解释 1、前导链与后随链：DNA复制的两条新链中，有一条链是沿5′→ 3′方向连续合成的，合成的速度相对较快，故称为前导链；另一条则是沿5′→ 3′方向先合成一些比较短的片段，然后再由连接酶将它们连接起来，其合成是不连续的，合成的速度相对较慢，故称为后随链。 2、转录的模板链：DNA转录中作为转录模板的DNA一条链称为模板链，另外一条则称为非模板链。 3、密码子与反密码子：mRNA上的每3个相邻碱基组成一个密码子，也称为三联体密码，一个密码子决定一种氨基酸。翻译过程中负责转运氨基酸的tRNA 的分子结构中具有三个与密码子相配对的碱基组成的反密码子。 4、简并：一种氨基酸可由一个以上密码子决定的现象称为简并。 5、基因家族：真核生物的有些来源相同、DNA序列相似、所编码的蛋白质具有互相关联的功能的基因，这样的一组基因称为“基因家族”。 6、重叠基因：有的噬菌体存在不同基因共用一部分DNA序列的现象，具有这种共用序列的基因称为重叠基因。 7、单交换与双交换：两对基因之间距离较小，这个区段只能发生一个交换，即为单交换。当基因间距离比较大时，同一个性母细胞可能在这个区段发生两个交换，即称为发生双交换。 8、干扰与符合系数：一个单交换的发生影响了另一个单交换的发生，这种现象称为干扰。干扰程度的大小通常用符合系数或并发系数表示。 9、超亲遗传：是指在数量性状的遗传中，F2及以后的分离世代群体中，出现超越双亲性状的新表型值的现象。
10、狭义遗传率：是加性方差在表现型方差中的百分数。 11、亲缘系数：两个个体都带有同一祖先某一特定等位基因的概率。 12、纯系：纯系是指一个群体中只存在一种基因型，并且这种基因型是纯合的。自花授粉的一个植株的自交后代可得到纯系。 13、细胞质遗传：真核细胞中的线粒体、叶绿体中也存在DNA，它所组成的基因也能决定生物某些性状的表现和遗传。这类遗传现象，称为细胞质遗传。 14、细胞质基因组：分布于细胞质的全部DNA序列。 15、表观遗传变异：是指DNA序列不发生变化但基因表达却发生了可遗传的变化，最终导致表型的改变，即基因型未发生变化而表型发生了可遗传的变化。 16、质核互作雄性不育：由细胞质基因和核基因相互作用控制的雄性不育类型，简称质核型雄性不育，又称为胞质不育型。 17、孢子体不育：雄性不育的花粉育性受母体的基因型（孢子体基因型）控制，与花粉（配子体）本身的基因无关。花粉败育发生在孢子体阶段。 18、配子体不育：是指花粉育性直接由雄配子体(花粉)本身的基因决定，花粉败育发生在雄配子阶段。 19、基因频率：一个群体里，A基因在A、a基因总数中的比率，称为A的基因频率。一个群体里，a基因在A、a基因总数中的比率，称为a的基因频率。 20、基因型频率：就是指具有特定基因型的个体数，占群体全部基因型个体总数的比率，也是特定基因型在群体中出现的概率。 21、随机交配：是指在一个有性繁殖的生物群体中，任何一个雌性或雄性个体与任何一个相反性别的个体交配的概率都相同。 22、基因突变：也称点突变，是DNA分子结构上微小的改变，它是由于碱
遗传学实验指导
遗传学实验指导
实验须知 一、实验课的目的 1．培养锻炼科学的思维方法，实事求是的科学态度和严格的科学作风，提高分析问题解决问题的能力。 2．通过实验加深对理论知识的理解，学习掌握基本的实验技术和实验技能，为今后学习和研究打下一定的基础。。 3．培养学生爱好公物，爱护集体，团结互助的优良道德品质。 二、实验课要求 1．课前必须预习，了解基本原理和主要步骤及意义，写出预习报告。 2．上课必须穿白大褂，带实验讲义和实验报告纸。 3．遵守实验室制度，注意安全（水、电、暖、强酸、强碱、有毒物质等）。 4．实验中要正规操作，动手动脑主动进行实验；掌握关键，力求得出准确结果；仔细观察，认真思考，及时做好记录；综合分析得出正确的结果。 5．在实验过程中不诉大声喧哗，有问题及时请教老师或同学。 6．器材、药品等按规定使用，严禁乱用乱放。 7．爱护仪器，在实验过程中因个人未能安实验要求操作而导致的器材损坏，按规章制度进行赔偿。 8．实验结束后，相关器材要彻底清洗干净，放到指定原位。 9．废弃物按要求分类收集、处理。 10．值日生要打扫干净实验台、地面，并负责关好门窗、检查水、电等。 11．因故不能上实验者应有请假手续。
实验1 植物染色体压片技术 一、实验原理 植物根尖的分生细胞的有丝分裂，每天都有分裂高峰时间，此时把根尖固定，经过染色和压片，再置放在显微镜下观察，可以看到大量处于有丝分裂各时期的细胞和染色体。 二、实验目的 学习植物材料的固定方法和常规压片技术；观察染色体的动态变化。 三、实验材料 洋葱（Aillum cepa）根尖 四、实验器具和药品 1．用具：染色板，载玻片，盖玻片，指管，温度计，试剂瓶，滴瓶，镊子，解剖针，毛边纸。 2．药品：无水酒精，70%酒精，冰醋酸，对二氯苯或秋水仙素，醋酸钠，碱性品红，石炭酸，甲醛，山梨醇，二甲苯。 a卡诺固定液的配制：用3份无水酒精，加入1份冰醋酸（现配现用）。 b染色液的配制： 配方Ⅰ.石炭酸品红（Carbol.fuchsin），先配母液A和B。 母液A：称取3g碱性品红，溶解于100mL的70%酒精中（此液可长期保存）。 母液B：取母液A 10mL，加入90mL的5%石炭酸水溶液（2周内使用）。 石炭酸品红染色液：取母液B 45mL，加入6mL冰醋酸和6mL 37%的甲醛。此染色液含有较多的甲醛，在植物原生质体培养过程中，观察核分裂比较适宜，后来在此基础上，加以改良的配方Ⅱ，称改良石炭酸品红，可以普遍应用于植物染色体的压片技术。 配方Ⅱ：改良石炭酸品红。取配方Ⅰ石炭酸品红染色液2-10ml，加入 90-98ml45%的醋酸和1.8g山梨醇（sorbitol）。此染色液初配好时颜色较浅，放置二周后，染色能力显著增强，在室温下不产生沉淀而较稳定。 1N Hcl ：浓HCl82.5ml→1000ml 五、实验说明 1．有丝分裂主要发生在根尖、茎生长点及幼叶等部位的分生组织，根尖取材容易，操作和鉴定方便。根尖由于取材方便，是观察植物染色体最常用的材料，有些植物种子难以发芽，或仅有植株而无种子，也可以用茎尖作为材料； 2．植物细胞分裂周期的长短不尽相同，通常在十到几十小时之间，温度明显地影响分裂周期，对于一个不太熟悉的实验材料，最好在特定温度下长根，掌握有丝分裂高峰期，以便得到更多的有丝分裂的细胞。 3．预处理是降低细胞质的粘度，使染色体缩短分散，防止纺锤体形成，让更多的细胞处于分裂中期。预处理的方法有低温预处理和药物预处理。(1)低温预处理将材料浸在蒸馏水中，放在1-4℃冰箱内离体处理24h。此法效果很好，对染色体无破坏作用，染色体缩短均匀，简便易行，各种作物都适用。(2)药物预处理：①0.05-0.2%秋水仙素溶液与室温下处理2-4h，对抑制纺锤体活动效果明显，易获得较多的中期分裂相，且染色体收缩较直，有利于染色体结构的研究。②饱和对二氯苯溶液室温下处理3-5h，对阻止纺锤体活动和缩短染色体效果也较好，但对染色体小而多的植物来说不利于染色体的计数。③0.002-0.00mol/L8-羟基喹啉18℃条件下处理5-6h，可以引起细胞粘度的改变，导致纺锤体活动受阻，使中期染色体在赤道面上保持相应的排列位置。缢痕区也较为清晰。一般认为对中等或长染色体的植物较合适。④70ppm放线菌酮和250ppm8-羟基喹啉的混合液于25℃条件
《医学遗传学》期末重点复习题
2．与苯丙酮尿症不符的临床特征是（1）。 A 患者尿液有大量的苯丙氨酸 B 患者尿液有苯丙酮酸 C 患者尿液和汗液有特殊臭味 D 患者智力发育低下 E 患者的毛发和肤色较浅 3．细胞在含BrdU的培养液中经过一个复制周期，制片后经特殊染色的中期染色体（）两条姊妹染色单体均深染 4．DNA分子中脱氧核糖核苷酸之间连接的化学键是（）磷酸二酯键 5．HbH病患者的可能基因型是（5）。 A ――/―― B －a/－a C ――/aa D －a/aa E aacs/―― 6．下列不符合常染色体隐性遗传特征的是（4）。 A．致病基因的遗传与性别无关，男女发病机会均等 B．系谱中看不到连续遗传现象，常为散发 C．患者的双亲往往是携带者 D．近亲婚配与随机婚配的发病率均等 E．患者的同胞中，是患者的概率为1/4，正常个体的概率约为3/4 7．人类a珠蛋白基因簇定位于（5）。 A 11p13 B 11p15 C 11q15 D 16q15 E 16p13 8．四倍体的形成可能是由于（3）。
A 双雄受精 B 双雌受精 C 核内复制 D 不等交换 E 部分重复9．在蛋白质合成中，mRNA的功能是（3）。 A 串联核糖体 B 激活tRNA C 合成模板 D 识别氨基酸 E 延伸肽链10．在一个群体中，BB为64％，Bb为32％，bb为4％，B基因的频率为（4）。 A B C D E 11．一个个体中含有不同染色体数目的三种细胞系，这种情况称为（3）。 A 多倍体 B 非整倍体 C 嵌合体 D 三倍体 E 三体型 12．某基因表达的多肽中，发现一个氨基酸异常，该基因突变的方式是（5）。 A 移码突变 B 整码突变 C 无义突变 D 同义突变 E 错义突变13．一种多基因遗传病的群体易患性平均值与阈值相距越近（1）。 A 群体易患性平均值越高，群体发病率也越高 B 群体易患性平均值越低，群体发病率也越低 C 群体易患性平均值越高，群体发病率越低 D 群体易患性平均值越低，群体发病率迅速降低 E 群体易患性平均值越低，群体发病率越高 14．染色质和染色体是（4）。
遗传学复习题整理
第一章绪论 一、名词解释： 1、遗传病（genetic disease）：是指遗传物质改变(基因突变或染色体畸变)所引起的 疾病。 2、先天性疾病:是指个体出生后即表现出来的疾病。 3、家族性疾病：是指某些表现出家族性聚集现象的疾病，即在一个家族中有多人患同 一种疾病。 二、简答 （１）遗传病的主要特征： ①垂直传递：遗传病是在上、下代之间垂直传递。 ②基因突变或染色体畸变是发生遗传病的根本原因，也是遗传病不同于其他疾病的 主要特征。 ③生殖细胞或受精卵发生的遗传物质改变才能遗传，而体细胞中遗传物质的改变， 并不能向后代传递。 ④遗传病常有家族性聚集现象。遗传病患者家系中，亲缘关系越近，发病机率越高， 随着亲缘关系疏远，发病率降低。 （2）遗传病的分类： 分类依据：根据遗传物质改变的不同和遗传的特点不同。 ㈠单基因病1．常染色体显性遗传病（AD);2.常染色体隐性遗传病(AR);3．X连锁隐性遗传病; 4．X连锁显性遗传病;5.Y连锁遗传病6.线粒体遗传病㈡多基因病㈢染色体病㈣体细胞遗传病 第二章基因 第一节基因的结构与功能
一、名词解释： 1、基因（gene）：是合成一种有功能的多肽链或者RNA分子所必需的 一段完整的DNA序列。 2、断裂基因：真核生物结构基因的DNA顺序包括编码顺序和非编码 顺序两部分。编码顺序在DNA分子中是不连续的，被非编码顺序 分隔开，形成镶嵌排列的断裂形式，因此称为断裂基因。 3、外显子（exon）：真核生物结构基因的DNA编码顺序称为外显子。 4、内含子(intron)：真核生物结构基因的DNA非编码顺序称为内子。 5、多基因家族（multigene family）：是指由某一共同祖先基因经过重复和变异所产 生的一组基因。根据基因在染色体的分布，可分为基因簇和基因超家族两种类型。 6、假基因（pseudogene）：其基因序列与具有编码功能的基因序列类似，因为不能编 码蛋白质，所以称为假基因。2简答 二、问答 1、人类DNA的存在形式有哪几种？ （1）高度重复顺序（卫星ＤＮＡ，反向重复顺序） （2）中度重复顺序（短分散元件，长分散元件） （3）单一顺序 第三节基因突变 一、名词解释 1、基因突变（gene mutation）：是指DNA分子中的核苷酸顺序发生改变，使遗传密 码编码产生相应的改变，导致组成蛋白质的氨基酸发生变化，以致引起表型的改变。
园艺植物遗传学部分实验指导
植物基因组DNA的提取（CTAB法） 一、原理 植物组织中绝大部分是核DNA，它和组蛋白、非组蛋白结合在一起，以核蛋白（即染色质或染色体）的形式存在于细胞核内。CTAB是一种阳离子去污剂，可溶解细胞膜，它能与核酸形成复合物，在高盐溶液中（0.7mol/L NaCl）是可溶的，当降低溶液盐的浓度到一定程度（0.3mol/L NaCl）时从溶液中沉淀，通过离心就可将CTAB与核酸的复合物同蛋白、多糖类物质分开，然后将CTAB与核酸的复合物沉淀溶解于高盐溶液中，再加入乙醇使核酸沉淀，CTAB能溶解于乙醇中。 二、实验材料、仪器及试剂 1、仪器 水浴锅；混匀器；高速冷冻离心机；移液器；754紫外分光光度计；液氮罐；水平电泳槽；电泳仪；一次性手套；吸头；1.5 mL离心管。 2、材料与药品 植物材料，以幼苗、嫩叶含DNA高。 液氮；氢氧化钠；CTAB；Tris；氯仿；乙醇；RNA酶；硼酸；溴酚蓝；异戊醇；β-巯基乙醇；盐酸；异丙醇；EDTA；GoldView核酸染料；琼脂糖；甘油；DNA分子量标记（DNA marker）。 三、操作方法 （1）按1 g样品3 ml DNA提取液的比例，取相应的园艺植物叶片，加入DNA提取液后，在研钵中充分磨碎；取3ml左右研磨后的液样于5 mL离心管中，65℃水浴35分钟以上，其间颠倒混匀一次。 （2）加等体积的氯仿：异戊醇（24：1）溶液，上下颠倒摇匀后，10000 rpm下离心10分钟，取吸上清液于一新的离心管中。 （3）再加两倍体积的预冷异丙醇，上下颠倒混匀后，10000 rpm下离心10分钟，弃上清。 （4）用3 mL 70%乙醇洗涤沉淀，重复一次，在室温下倒置晾干。 （5）沉淀用100 uL TE溶液溶解，用TE（pH8.0）作空白对照，于紫外分光光度计上测定260nm、280nm和230nm吸收值，计算A260/A230和A260/A280的比值，并换算出核酸的含量。 （6）琼脂糖凝胶板的制备 （7）加样：将样品与溴酚蓝按4：1的比例混合，用移液器缓慢加入凝胶样品孔中，点样量为每孔5 μgDNA。 （8）电泳：点样端接负极，电泳开始时，可用较高的电压，如100伏特，这样可使样品很快进入胶内，而减少样品扩散，待样品进入胶内即可保持每厘米5伏特左右的电压，DNA 在电泳中向正极移动。 当溴酚蓝的区带移至距凝胶底部1-2cm，停止电泳，当胶板为10-15cm的长度时，电泳时间一般为2小时左右。 （9）观察与鉴定
水产生物遗传育种学：遗传学试题和答案
遗传学答案(朱军教材) 遗传学 习题与参考答案 第二章遗传的细胞学基础（练习） 一、解释下列名词:染色体染色单体着丝点细胞周期同源染色体异源染色体无丝分裂有丝分裂单倍体联会胚乳直感果实直感 二、植物的10个花粉母细胞可以形成：多少花粉粒?多少精核?多少管核?又10个卵母细胞可以形成：多少胚囊?多少卵细胞?多少极核?多少助细胞?多少反足细胞? 三、玉米体细胞里有10对染色体，写出下列各组织的细胞中染色体数目。 四、假定一个杂种细胞里含有3对染色体，其中A、B、C来自父本、A’、B’、C’来自母本。通过减数分裂能形成几种配子？写出各种配子的染色体组成。 五、有丝分裂和减数分裂在遗传学上各有什么意义？ 六、有丝分裂和减数分裂有什么不同?用图解表示并加以说明。 第二章遗传的细胞学基础（参考答案） 一、解释下列名词: 染色体：细胞分裂时出现的，易被碱性染料染色的丝状或棒状小体，由核酸和蛋白质组成，是生物遗传物质的主要载体，各种生物的染色体有一定数目、形态和大小。 染色单体：染色体通过复制形成，由同一着丝粒连接在一起的两条遗传内容完全一样的子染色体。 着丝点：即着丝粒。染色体的特定部位，细胞分裂时出现的纺锤丝所附着的位置，此部位不染色。 细胞周期：一次细胞分裂结束后到下一次细胞分裂结束所经历的过程称为细胞周期(cell cycle)。 同源染色体：体细胞中形态结构相同、遗传功能相似的一对染色体称为同源染色体(homologous chromosome)。两条同源染色体分别来自生物双亲，在减数分裂时，两两配对的染色体，形状、大小和结构都相同。 异源染色体：形态结构上有所不同的染色体间互称为非同源染色体，在减数分裂时，一般不能两两配对，形状、大小和结构都不相同。 无丝分裂：又称直接分裂，是一种无纺锤丝参与的细胞分裂方式。 有丝分裂：又称体细胞分裂。整个细胞分裂包含两个紧密相连的过程，先是细胞核分裂，后是细胞质分裂，核分裂过程分为四个时期；前期、中期、后期、末期。最后形成的两个子细胞在染色体数目和性质上与母细胞相同。 单倍体：指具有配子染色体数(n)的个体。 联会：减数分裂中同源染色体的配对。 联会复合体——减数分裂偶线期和粗线期在配对的两个同源染色体之间形成的结构，包括两个侧体和一个中体。 胚乳直感：又称花粉直感。在3n胚乳的性状上由于精核的影响而直接表现父本的某些性状。果实直感：种皮或果皮组织在发育过程中由于花粉影响而表现父本的某些性状 二、可以形成：40个花粉粒，80个精核，40个管核；10个卵母细胞可以形成：10个胚囊，10个卵细胞，20个极核，20个助细胞，30个反足细胞。 三、(1)叶(2)根(3)胚乳(4)胚囊母细胞(5)胚 (6)卵细胞(7)反足细胞(8)花药壁(9)花粉管核 (1)叶：20条；(2)根：20条；(3)胚乳：30条；(4)胚囊母细胞：20条；(5)胚：20条；
遗传学课后习题及答案
Chapteｒ 1 ＡｎIｎｔｒoｄuction tｏGeneｔｉcs （一）名词解释: 遗传学:研究生物遗传和变异的科学。 遗传：亲代与子代相似的现象。 变异：亲代与子代之间、子代个体之间存在的差异. （二)选择题：?1．１９00年(2))规律的重新发现标志着遗传学的诞生. ?（1）达尔文（２）孟德尔(３) 拉马克（4)克里克 ２．建立在细胞染色体的基因理论之上的遗传学, 称之（ 4 )。 (1）分子遗传学（2)个体遗传学(3）群体遗传学(4)经典遗传学?３．遗传学中研究基因化学本质及性状表达的内容称（1 ）。 (1)分子遗传学（2）个体遗传学（3）群体遗传学 (4）细胞遗传学 ４. 通常认为遗传学诞生于(３)年。?（１）1859 (2)1865 （3） 1900 （4）1910?5．公认遗传学的奠基人是（３)： (1）Ｊ·Lａmarck （２)Ｔ·H·Ｍorgａｎ(3)G·J·Menｄｅl （4)C·R·Dａrwin?6．公认细胞遗传学的奠基人是（2)：?(１)J·Lamaｒck （2)Ｔ·H·Moｒｇaｎ(3)Ｇ·J·Mendｅｌ（4）C·R·Darｗiｎ Ｃhaptｅｒ２Mitosiｓand Meiｏsis 1、有丝分裂和减数分裂的区别在哪里？从遗传学角度来看,这两种分裂各有什么意义？那么,无性生殖会发生分离吗?试加说明。 答:有丝分裂和减数分裂的区别列于下表：
有丝分裂的遗传意义： 首先:核内每个染色体,准确地复制分裂为二,为形成的两个子细胞在遗传组成上与母细胞完全一样提供了基础。其次，复制的各对染色体有规则而均匀地分配到两个子细胞的核中从而使两个子细胞与母细胞具有同样质量和数量的染色体。 减数分裂的遗传学意义 首先,减数分裂后形成的四个子细胞,发育为雌性细胞或雄性细胞,各具有半数的染色体(n)雌雄性细胞受精结合为合子，受精卵(合子）,又恢复为全数的染色体2ｎ。保证了亲代与子代间染色体数目的恒定性,为后代的正常发育和性状遗传提供了物质基础,保证了物种相对的稳定性。 其次,各对染色体中的两个成员在后期I分向两极是随机的,即一对染色体的分离与任何另一对染体的分离不发生关联，各个非同源染色体之间均可能自由组合在一个子细胞里,n对染色体,就可能有2n种自由组合方式。 例如,水稻ｎ=12，其非同源染色体分离时的可能组合数为2１2 ＝4０96。各个子细胞之间在染色体组成上将可能出现多种多样的组合。 此外,同源染色体的非妹妹染色单体之间还可能出现各种方式的交换，这就更增加了这种差异的复杂性。为生物的变异提供了重要的物质基础。 ２。水稻的正常的孢子体组织,染色体数目是1２对，问下列各组织染色体数是多少? 答:（１)胚乳：32；(2)花粉管的管核:１2;(3)胚囊:12；（4)叶：2４;(5)根端:24；（6)种子的胚：24;(7）颖片：２4。 ３。用基因型Ａabｂ的玉米花粉给基因型ＡaＢb的玉米雌花授粉，你预期下一代胚乳的基因型是什么类型,比例为何？ 答:胚乳是三倍体,是精子与两个极核结合的结果。预期下一代胚乳的基因型和比例为下列所示： 4. 某生物有两对同源染色体，一对是中间着丝粒，另一对是端部着丝粒,以模式图方式画出：(1)减数第一次分裂的中期图; (2）减数第二次分裂的中期图。
遗传学课后题答案章
遗传学课后题答案章
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第二章孟德尔定律 1、答：因为 （1）分离规律是生物界普遍存在的一种遗传现象，而显性现象的表现是相对的、有条件的；（2）只有遗传因子的分离和重组，才能表现出性状的显隐性。可以说无分离现象的存在，也就无显性现象的发生。 2、解： 序号杂交基因型表现型 1 RR×rr Rr 红果色 2 Rr×rr 1/2Rr，1/2rr 1/2红果色，1/2黄果色 3 Rr×Rr 1/4RR，2/4Rr，1/4rr 3/4红果色，1/4黄果色 4 Rr×RR 1/2RR，1/2Rr 红果色 5 rr×rr rr 黄果色 3、解： 序号杂交配子类型基因型表现型 1 Rr × RR R，r；R 1/2RR，1/2Rr 1/2红色，1/2粉红 2 rr × Rr r；R，r 1/2Rr，1/2rr 1/2粉红，1/2白色 3 Rr × Rr R，r 1/4RR，2/4Rr，1/4rr 1/4红色，2/4粉色，1/4白色 4、解： 序号杂交基因型表现型 1 WWDD×wwdd WwDd 白色、盘状果实 2 WwDd×wwdd 1/4WwDd，1/4Wwdd，1/4wwDd，1/4wwdd，1/4白色、盘状，1/4白色、球状，1/4黄色、盘状，1/4黄色、球状 2 wwDd×wwdd 1/2wwDd，1/2wwdd 1/2黄色、盘状，1/2黄色、球状 3 Wwdd×wwDd 1/4WwDd，1/4Wwdd，1/4wwDd，1/4wwdd，1/4白色、盘状，1/4白色、球状，1/4黄色、盘状，1/4黄色、球状 4 Wwdd×WwDd 1/8WWDd，1/8WWdd，2/8WwDd，2/8Wwdd，1/8wwDd，1/8wwdd 3/8白色、盘状，3/8白色、球状，1/8黄色、盘状，1/8黄色、球状 5.解：杂交组合TTGgRr × ttGgrr： 即蔓茎绿豆荚圆种子3/8，蔓茎绿豆荚皱种子3/8，蔓茎黄豆荚圆种子1/8，蔓茎黄豆荚皱种子1/8。 杂交组合TtGgrr ×ttGgrr： 即蔓茎绿豆荚皱种子3/8，蔓茎黄豆荚皱种子1/8，矮茎绿豆荚皱种子3/8，矮茎黄豆荚皱种子1/8。 6.解：题中F2分离比提示：番茄叶形和茎色为孟德尔式遗传。所以对三种交配可作如下分析： (1) 紫茎马铃暮叶对F1的回交： AaCc×AAc c→AACc、AAcc、AaCc、Aacc 表型二种，比例为1：1 (2) 绿茎缺刻叶对F1的回交： AaCc×aaC C→AaCC、AaCc、aaCC、aaCc 表型二种，比例为1：1 （3）双隐性植株对Fl测交： AaCc×aacc AaCc Aacc aaCc aacc 1紫缺：1紫马：1绿缺：1绿马
遗传学练习题
遗传练习题 （各章节重点还要复习，这里只是部分的模拟题，并不全面） A 型题（单项选择题） 1．染色质和染色体是_____ A．同一物质在细胞的不同时期的两种不同的存在形式 B．不同物质在细胞的不同时期的两种不同的存在形式 C．同一物质在细胞的同一时期的不同表现 D．不同物质在细胞的同一时期的不同表现 E．两者的组成和结构完全不同 2．常染色质是间期细胞核中_____ A．螺旋化程度高，有转录活性的染色质 B．螺旋化程度低，有转录活性的染色质 C．螺旋化程度高，无转录活性的染色质 D．螺旋化程度低，无转录活性的染色质 E．螺旋化程度低，很少有转录活性的染色质 3．按照ISCN 的标准系统，1 号染色体，短臂，3 区，1 带第3 亚带应表示为_____ A．1p31.3 B．1q31.3 C．1p3.13 D．1q3.13 E．1p313 4．人类生殖细胞的染色体数目是。 A、22条 B、23条 C、46条 D、92条 E、93条 5．着丝粒位于或靠近染色体中央，可称为_____ A、亚中着丝粒染色体 B、中央着丝粒染色体 C、端着丝粒染色体 D、近端着丝粒染色体 E、中间着丝粒染色体 6.含有两种或两种以上不同核型细胞系的个体称为_____ A、单倍体 B、多倍体 C、嵌合体 D、多体 E、二倍体 7. 根据丹佛体制，人类染色体依次从1编到22号可分组。 A、2组 B、5组 C、7组 D、9组 E、10组 8．X染色体归入组。 A、 A 组 B、B组 C、C 组 D、D组 E、F组 9. Y染色体归入组。 A、 A 组 B、B组 C、C 组 D、G组 E、F组 10．除S期以外的整个细胞周期的各个阶段均处于凝聚状态的染色质称为_____ A．常染色质B．结构异染色质C．兼性异染色质D．异染色质E．染色体11．由常染色质凝聚并丧失基因转录活性后转变而来的染色质称为_____ A．结构异染色质B．兼性异染色质C．异染色质D．常染色质E．以上都不是 12．染色质的基本结构单位是_____ A．核小体B．螺线管C．超螺线管D．染色单体E．放射环 13．_____时期的染色体是进行染色体研究分析的最佳时期 A．前期B．中期C．后期D．末期E．整个周期 14．有丝分裂的哪个时期染色体最为粗大、清晰_____ A．前期B．中期C．后期D．末期E．以上都不是 15．精细胞发生过程所经历的四个时期是_____ A．生长期、增殖期、成熟期、变形期B．成熟期、生长期、增殖期、变形期C．生长期、成熟期、增殖期、变形期D．增殖期、生长期、成熟期、变形期