1、第一章 遺傳的細(xì)胞學(xué)基礎(chǔ)一、 遺傳學(xué)名詞解釋染色體chromosome 姊妹染色單體sister chromatid 同源染色體homologous chromosome 染色體組genome 二價(jià)體bivalent 聯(lián)會(huì)synapsis 受精fertilization 雙受精double fertilization 胚乳直感xenia 果實(shí)直感metaxenia 無融合生殖apomixis 細(xì)胞周期cell cycle 無性生殖asexual reproduction 染色體組型karyotype 減數(shù)分裂meiosis 有絲分裂mitosis 3. If an organism has a

2、 diploid number of 16, how many chromatids are visible at the end of mitotic prophase? 32How many chromosomes are moving to each pole during anaphase of mitosis?166. Contrast the end results of meiosis with those of mitosis.7. Define and discuss these terms: (a) synapsis, (b) bivalents, (c) chiasmat

3、a, (d) crossing over, (e) chromomeres, (f) sister chromatids, (g) tetrads, (h) dyads, (i) monads.8. An organism has a diploid number of 16 in a primary oocyte.(a) How many tetrads are present in the first meiotic prophase?8(b) How many dyads are present in the second meiotic prophase?8(c) How many m

4、onads migrate to each pole during the second meiotic anaphase?810. Explain why meiosis leads to significant genetic variation while mitosis does not.11. Describe the role of meiosis in the life cycle of a vascular plant.12. How are giant polytene chromosomes formed?第二章 遺傳物質(zhì)的分子基礎(chǔ)一、 遺傳學(xué)名詞解釋遺傳密碼genetic

5、 code 簡(jiǎn)并degeneracy 中心法則centraldogma 轉(zhuǎn)錄transcription 翻譯translation二、 簡(jiǎn)答題5. List three main differences between DNA and RNA.9. During translation, what molecule bears the codon? The anticodon?1. Contrast the positive and negative control systems.2. Contrast the role of repressor in an inducible system

6、 and in a repressible system.第三章 孟德爾遺傳一、 遺傳學(xué)名詞解釋性狀character 單位性狀unit character 顯性性狀dominant character 隱性性狀recessive character 基因座locus 等位基因allele 純合體homozygote 雜合體heterozygote 測(cè)交test cross 完全顯性complete dominance 不完全顯性imcomplete dominance 共顯性co-dominance 鑲嵌顯性mosaic dominance 基因型genotype 表現(xiàn)型phenotype

7、互補(bǔ)作用complementary effect 返祖現(xiàn)象atavism 積加作用additive effect 重疊作用duplicate effect 上位作用epistatic effect 顯性上位epistatic dominance 隱性上位epistatic recessiveness 抑制作用inhibiting effect 多因一效multigenic effect 一因多效pleiotropism 二、 簡(jiǎn)答題1、試述分離規(guī)律.獨(dú)立分配規(guī)律和連鎖交換規(guī)律的實(shí)質(zhì)？2、試分析孟德爾成功的原因，對(duì)你有何啟迪？5. Why was the garden pea a good ch

8、oice as an experimental organism in Mendels work?50. With regard to the ABO blood types in humans, determine the genotypes of the male parent and female parent: Male parent：Blood type B whose mother was type O Female parent: Blood type A whose father was type B Predict the blood types of the offspri

9、ng that this couple may have and the expected proportion of each.1/4、1/4、1/4、1/452. A husband and wife have normal vision, although both of their fathers are red-green color-blind, which is inherited as an X-linked recessive condition. What is the probability that their first child will be (a) a nor

10、mal son?1/4 (b) a normal daughter? 1/2(c) a color-blind son? 1/4(d) a color-blind daughter?053. In cats, yellow coat color is determined by the b allele, and black coat color is determined by the B allele. The heterozygous condition results in a coat pattern known as tortoise shell. These genes are

11、X-linked. What kinds of offspring would be expected from a cross of a black male and a tortoise-shell female? What are the chances of getting a tortoise-shell male?055. In cattle, coats may be solid white, solid black, or black and white spotted. When tree-breeding solid whites are mated with true-b

12、reeding solid blacks, the F1 generation consists of all solid white individuals. Following many F1 F1 matings, the following ratio was observed in the F2 generation: 12/16 solid white 3/16 black and white spotted 1/16 solid black Explain the mode of inheritance governing coat color and pattern by de

13、termining how many gene pairs are involved and which genotypes yield which phenotypes. Is it possible to isolate a true-breeding strain of black and white spotted cattle? If so, what genotype would they have? If not, explain why not.56. Three autosomal recessive mutations in Drosophila, all with tan

14、 eye color(r1, r2, r3) are independently isolated and subjected to complementation analysis. The results are shown here. r1 r2F1: all wild-type eyes, r1 r3F1: all tan eyes. Then r1 and r3 are alleles, the F1 phenotype of r2r3 is wild-type eyes .21. Consider three independently assorting gene pairs,

15、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 an offspring that is AABbCc from parents that are AaBbCC and AABbCc ?1/8第四章 連鎖遺傳一、 遺傳學(xué)名詞解釋連鎖linkage 干擾interference 符合系數(shù)coefficient of coincidence 連鎖群linkage

16、 group 性染色體sex-chromosome 常染色體autosome 性連鎖sex linkage sex-linked inheritance 交叉遺傳criss-cross inheritance 限性遺傳sex-limited inheritance 從性遺傳sex-influenced inheritance 二、 簡(jiǎn)答題2. Why does more crossing over occur between two distantly linked genes than between two genes located very close together on the

17、same chromosome? 39. Using two pairs of genes P/p and Z/z), a test-cross parent (ppzz) is crossed to an organism of unknown genotype. Analysis of the data indicates that the gametes are produced in the following proportions: PZ, 42,4% Pz, 6.9% pZ, 7.1% pz, 43.6% Draw all possible conclusions about t

18、he location of these genes.40. In Drosophila, females expressing the three X-linked recessive traits, scute (sc) bristles, sable body (s), and vermilion eyes (v) are crossed with wild-type males. In the F1 generation, all females are wild type, while all males express all three mutant traits. The cr

19、oss is carried to the F2 generation, and 1000 off spring are counted, with the results shown here. No determination of sex has been made in the F2 data. Phenotype Offspringsc s v 314+ + + 280+ s v 150sc + + 156+ s + 30sc + v 46sc s + 10+ + v 14(a) Determine the genotypes of the P1 and Fi parents, us

20、ing proper nomenclature.(b) Determine the sequence of the three genes and the map distance between them.(c) Are there more or fewer double crossovers than expected? Calculate the coefficient of coincidence. Does this represent positive or negative interference?41. In a theoretical diploid organism,

21、a female of genotype a b c + + +produces 100 meiotic tetrads. Of these, 68 show no crossover events. Of the remaining 32, 20 show a crossover between a and b, 10 show a crossover between b and c, and 2 show a double crossover between a and b and between b and c. Of the 400 meiotic products, how many

22、 of each of the 8 different genotypes will be produced?(42% 5% 2.5% 0.5 ) ( 168 20 10 2)Assuming the order a-b-c and the allele arrangement shown above, what is the map distance between these loci?11 6 . Determine the genotypes and proportion of gametes from double crossover, the single crossovers b

23、etween a and b, single crossovers between b and c, and noncrossover.43. Three genes are in the order d-b-c. The heterozygote db+/+c forms 8 categories of gametes. Identify the noncrossover-gametes , the single crossover gametes, and the double crossover gametes. Which category is likely to have the

24、fewest? Which will have the most?42. In a cross inNeurospora, involving two alleles B and b, the following tetrad patterns are observed. Calculate the distance between the locus and the centromere.10 Tetrad Pattern number BBbb 36 bbBB 44 BbBb 4 bBbB 6 BbbB 3 bBBb 744. An organism of genotype AaBbCc

25、is test-crossed. The genotypes of the progeny are as follows:20 Aa Bb Cc 20 Aa Bb cc 20 aa bb Cc 20 aa bb cc 5 Aa bb Cc 5 Aa bb cc 5 aa Bb Cc 5 aa Bb cc(a) If these three genes are all assorting independently of each other, how many genotypic classes would you expect in the progeny of the testcross?

26、8(b) If these three genes are so tightly linked that crossover never occurs, how many genotypic classes would you expect in the progeny of the test cross?2(c) What can you conclude from the actual data?ab linkage 20%45. A, B, and D are linked, and the sequence is A-B-D, assuming ra-b=16%, rb-d=8%, I

27、=0.5, what are the genotypes and ratio of the gametes of the heterozygote AbD/aBd？第五章 染色體變異一、 遺傳學(xué)名詞解釋缺失deletion或deficiency 重復(fù)duplication 位置效應(yīng)position effect 劑量效應(yīng)dosage effect 倒位inversion 易位translocation 假顯性pseudo-dominant 染色體組genome 整倍體Euploid 一倍體monoploid 單倍體haploid 二倍體diploid 同源多倍體autopolyploid 異源

28、多倍體雙二倍體 Allopolyploid 非整倍體aneuploid 超倍體hyperploid 亞倍體hypoploid 缺體nullisomic 單體monosomic “Turner氏綜合癥” 三體trisomic 先天愚型 Down 氏綜合征Down syndrome1. For a species with a diploid number of 18, indicate how many chromosomes will be present in the somatic nuclei of individuals that are haploid, triploid, tetr

29、aploid, trisomic, and monosomic.13. Define the following pairs of terms, and distinguish between them: aneuploidy/euploidy monosomy/trisomy 13 Patau syndrome/Edwards syndrome 18 autopolyploidy/allopolyploidy autotetraploid/amphidiploid 內(nèi) paracentric inversion/pericentric inversion 間14. Contrast the

30、relative survival times of individuals with Down, Patau, and Edwards syndromes. Speculate as to why such differences exist.15. What evidence suggests that Down syndrome is more often the result of nondisjunction during oogenesis rather than during spermatogenesis?16. Contrast the fertility of an all

31、otetraploid with an autotriploid and an autotetraploid.2. A human female with Turner syndrome also expresses the X-linked trait hemophilia, as did her father. Which of her parents underwent nondisjunction during meiosis, giving rise to the gamete responsible for the syndrome?mother produce non-X oot

32、id3. The primrose, Primula kewensis, has 36 chromosomes that are similar in appearance to the chromosomes in two related species, Primula fioribunda (2n = 18) and Primula verticillata (2n = 18). How could P kewensis arise from these species? How would you describe P kewensis in genetic terms?allotet

33、raploid or amphidiploid4. Indicate the expected number of Barr bodied in interphase cells of the following individuals: klinefelter syndrome; turner syndrome; karyotypes 47, XYY; 47, XXX; and 48, XXXX. 第六章 細(xì)菌和病毒的遺傳一、 遺傳學(xué)名詞解釋影印培養(yǎng)法 原噬菌體prophage 溶原性細(xì)菌lysogenic bacteria 轉(zhuǎn)化transformation 感受態(tài)competence 接合

34、conjugation 性導(dǎo)sexduction F因子fertility factor F因子F prime factor 轉(zhuǎn)導(dǎo)transduction 普遍性轉(zhuǎn)導(dǎo)generalized transduction 轉(zhuǎn)導(dǎo)體transducing particle 共轉(zhuǎn)導(dǎo)cotransduction 流產(chǎn)轉(zhuǎn)導(dǎo) 局限轉(zhuǎn)導(dǎo)restricted transduction F菌株 F 菌株 Hfr菌株 F菌株 質(zhì)粒plasmid 附加體episome二、 簡(jiǎn)答題3. Distinguish between the three modes of recombination in bacteria.4. D

35、escribe the observations that led Zinder and Lederberg to conclude that the prototrophs recovered in their transduction experiments were not the result of F-mediated conjugation.5. Define plaque, lysogeny, and prophage.噬菌斑 溶源性 原噬菌體6. List the major differences between (a) the F+ F- and the Hfr F- bacterial crosses, and (b) F+, F-, Hfr, and F bacteria.4. If a single bacteriophage infects one E. colt cell present on a lawn of bacteria and upon lysis yields 200 viable viruses, how many phages will exist in a single plaque if only three more lyric cycles occur?（200）4第七章 基因突變一、 遺傳學(xué)名詞解釋復(fù)等位基因