《遗传学》第一次测验题解答

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1、遗传学第一次测验题解答使用班级： 生工1031 ，生工10321. 10% Suppose that the in breedi ng coefficie nt of I in the follow ing pedigree is0.25 .What is the in breedi ng coefficie nt of I s com mon an cestor , C3(1 + Fc )2.9 1.3.57.42.0 6.0E coR I解；Fi二工二、(1 + Fc )0.25 = ( 1/2 ) Fc = 12. 10% A lin ear DNA molecule is subjec

2、ted to sin gle and double digesti ons with restrictio n endonu cleases ,and the followi ng results are obta in ed:En zymesFragme nt Sizes ( in kb )EcoRIHind m2.9 , 4.5 , 7.4,8.03.9 , 6.0 , 12.9EcoRI and Hind 川 1.0 , 2.0 , 2.9,3.5,6.0,7.4Draw the restriction map defined by these data .解：Hindm3.10% Th

3、e order of markers transferred by four Hfr strainsto F- cells is as follows:Hfr StrainF eleme nt inOrder of Markers Don atedDraw a map showing the sites and orientation of insertion of the these four stra ins .解:暗暗4. 10%How isthe解：Ftran sferof因子通过接合管转移到34F similar tophage infection ? Howis it differ

4、entF阴性菌细胞内；噬菌体通过其尾部吸附细菌细胞壁及注射式注入自己的全部遗传物质；F因子 与噬菌体 都可以整合到细菌的主染色体上，也都可以游离出主染色体之外而存在于细胞质中，在游离过程中发生不精确分离而携带有部 分细菌的遗传物质；形成 F 因子或转导粒子。F因子无细胞外形式，只能在细胞内存在。噬菌体有细胞外形式。5. 13% An an alysis of tetrads in a cross ofChlamydo mon ainvolvi ng the locipf(paralyzed flagella ) and nic ( requireme nt for the vitam in n

5、i cot in amide ) gavethe following results: PD = 70 , NPD = 2 , and T = 28 . Calculate the linkage relati on ship betwee npf and nic .解：R( pf-nic ) = ( NPD+ 1/2 T ) /( PD+ NPD+ T ) *100%=2 + 14 / 100* 100%=16 %6.13% In Drosophila the sex-linkedgenes cut ( ct ) ,lozenge eye ( lz ) , and forkedbristle

6、s ( f ) are the following map distances apart:cut to lozenge , 7.7 units,and lozenge to forked , 29.0 units .Assuming that there is no interferenee, whatare the expected nu mbers of geno types out of 1000 female flies recovered from thecrossct lz f/ + + +早早 X SS ct lz f/ Y ?解:7.71 29.0cutlqze ngeeye

7、forkedbristles实际双交换/理论双交换=1实际双交换=0.077 * 0.29=0.02233ctlzf327.665+327.665ctlz+133.835+f133.835+lzf27.335ct+27.335ct+f11.165+lz+11.1657. 15% A population of mice is “ founded ” by two mice , one D/D ( normal coat color )and the other d/d ( dilute coat color ). Their offspri ng and succeedi ng offspri

8、 ng are allowed to mate ran domly for a nu mber of gen erati ons un til a large populati on of mice is established. Complete selectio n aga instdilute mice is nowimposed. What will be the change in gene frequencies after ten generations under the select ion program ?解： P 0 = q 0 = 0.5 n = 10=0.083P

9、10 = 11 / 12 = 0.9178. 7% Assu ming complete pen etra nee ,could the trait in dicated in the pedigree below be caused by ( in dicate yes or no ):a. an autosomal gene recessive in males and females?YESb. An autosomal gene dominant in males and females ? NOc. An autosomal gene dominant in males and re

10、cessive in females?NOd. An autosomal gene dominant in in females and recessive in males?YESe. A Y-li nked gene? NOf. An X-li nked recessive gene?_NOg. An X-li nked domi nant gene? _NO9. 5% Stem len gth in a pla nt species can vary from 6 to 46 cm ,anda polyge nic system with five pairs of genes(symb

11、olized A,a,B,b,D,d,E,e,F,f ) is responsible for variation in stem length.OH-rOOST)CROSS I plant 1CROSS n pla nt 2CROSS 川 pla nt 1x plant 1all F x plant 2all F *x pla nt 2all F*1 with 30 cm stems1 with 30 cm stems1 with 30 cm stemsCon sider the follow ing crosses:F 1 testcross ( F 1X a pla nt with 6

12、cm stems )Offspri ng vary ing from 10 to 26 cm stemsa. Give the complete gen otypes for pla nt 1 and pla nt 2 .AABBDDeeff;AAbbddEEFFb. In the F 1 testcross of cross川,what proport ion of the offspri ng is expectedto have 10 cm stems?1/16in stead of testcrossed46 cm14 cm28/256AABBDDEEFFAAbbddeeffc. If

13、 the F 1 plants of cross 川 had been self-fertilized give the followi ng expected results:a) stem len gth of pla nts with the Ion gest stemsb) stem len gth of pla nts with the shortest stemsc) proport ion of offspri ng with 22 cm stemsd) geno types of the offspri ng with the Ion gest stems10. 7%e) ge

14、no types of the offspri ng with the shortest stemsgarde n peas,tall pla nts ( T )are dominant over short ( t ),ro und seeds(R )are dominant over wrinkled ( r ),yellow seeds ( Y )are dominant over green ( y ),and purple flowers ( A )are dominant over white ( a ).Consider the following pea plants ,and

15、an swer the questi ons below: pla nt 1 =RryyAaTt , pla nt 2 = RrYYAattIn a cross of pla nt 1X pla nt 2 , what proporti on of the proge ny will:1/321/329/321/21/16a. have the geno typerrYyaatt ? b. have the phe no type wrin kled ,yellow ,white ,short ?c. be dominant for all four characteristics ?d. b

16、e pure breed ing ( homozygous )for seed shape ?e. be pure breedi ng for round and purple ?f.g.nonebe pure breed ing for all four characteristics ?What proportion of the round yellow white tall progeny will have the genotypeRrYyaaTt ? 2/31.( 10%)Describe how ,if at all ,each of the follow ing might p

17、romote the isolati on of subpopulati ons and the formatio n of subspecies.(a) Assortative mating(b) In breed ing(c) Hybrid dysgenesis解：(a)分类，同类型的交配；由于表型相近的个体往往具有相同或相近的基因型，它们的 交配会进一步促进基因型的纯合化，导致亚种的形成。(b) 近交个体间具有共同祖先传递的基因，基因纯合的可能性远远大于随机交配所导 致的纯合率，从而促进亚种的形成。(c) 杂合子不育，退化或衰退，活力不足，杂种不育。同样可导致群体分化，基因型 多态现象难

18、以维持。.(10% )Sex-limited (本题有误 )genes produce traits that are expressed in only one of the sexes, presumably because some horm onal attribute of female ness or male ness must be prese nt if the genes are tran scribed and / or the gene products expressed .The sex-limited human trait known as pattern ba

19、ldness is expressed as follows: menBheterozygous or homozygous for the autosomal gene H will lose the hair on top of their headsBbas they age ; women lose their hair only if they are H / H homozygotes .A man exhibits patter n bald ness ,but his pare nts do not .His wife is not baldi ng and bald ness

20、 is not found in her lineage. Give the genotypes of the husband HBHb, wife HbHb, and parents 男方母亲 HBHb 父 亲 HbHb,女方母亲父亲 HbHb, HbHb, . What proportio n, if any ,of the ma n so ns would you expect to become bald ? 1/2 His daughters? 0 Would the phe no types of the offspri ng differ ifBthe H gene were a

21、 sex-linked dominant gene but not sex limited ? yes Would they differ if it were a sex-l in ked and sex-limited dominant gene ? no3. ( 10 % )The figure below shows a section of polytene from Drosophila melanogaster .a). What tissue was this chromosome isolated from? Salivary gla nd b). Label a spot

22、on the chromosome where gene expressi on is very high . b4. (60% )(1). Suppose that the com mon an cestor Kin breedi ng coefficie nt in the followi ng pedigree is0.125 and L s is zerftA/hat is the inbreeding coefficient of I ?解：Fi = 3 ( 1/2 ) n( 1 + F a )7788=(1/2 )( 1 + 0.125 ) + ( 1/2 )( 1 + 0.125

23、 ) + ( 1/2 )+ ( 1/2 )三 0.025 = 13 / 512(2 ) . The graph below shows the results of an interrupted mating experiment for a close relative of E.coli .Using the data propose a mapof these genesndicate distances between all loci with the appropriate units .s_HfrG11 azi x F mal arg _ xyl _ pyr _ met_ pur

24、_Plating is on complete medium except for the following : No arginine is provided.Maltose is the primary carb on source. Azide is at selective concentration .Replica plates were then made to assess each colony s genotype at the xyl _ pyr _ met and pur _ loci.In conjugation mapping, distances between

25、 markers are determined by the differences in time of appeara nee of those markers .The time of appeara nee for a marker can be determ ined from the graph give n. Where the line represe nti ng the perce ntage ofrmal + arg + and azi emerges from the x axis for the marker in question of appeara nee fo

26、r that marker .For example, colonies with the ability to live on xylose as a carb on source are first see n at 9 minu tes after the begi nning of the mating. Abilityto synthesize pyrimidines is seen shortly thereafter, atHence, the loci xyl and pyr are 2 minutes apart. How high the percentage marker

27、 type rises is also a function of the dista nee from that marker to the origi nally selected markers ( mal + and arg + )in the vici nity being mapped. The xyl line, for example, reaches a plateau at 90% suggesti ng it is almost always recomb ined with the mal locus as long as it has successfully bee

28、 n tran sferred in to the cell ( which is what hasn t happened at 9 minutes ).The pur gene higher tha n 35% , reflect ing that there are many times whe n a crossover even separates the two loci because of their dista nee apart. Remember that all the bacteria being tested in this experiment are known

29、 to be mal _. ( pur + mal recomb inants would not be detected in this experime nt.以上说明不必答出，只需作图即可。is the time11 minu tes.of a givennever risesThe selected markersmal +arg+xyl911pur +1121272 pyr +10unordered tetras were formed.stock .The following(H f r g 11(3) .Individuals from a stock of Chlamydomo

30、nas containing genes a 另,and c weremated to those from a (1 )+wild-type(3 )bb)+cbac2681220180(5 )(6 )(7)(8 )(9 )a+c+a+bca+ca+c+ab+bcab+b+bca+c+bcab+abcabc+18121421820(a ) Are the genes lin ked ? If so ,determ ine the order and lin kage dista nee.(b ) Is it more correct to calculate the in termediate

31、 dista nee betwee n a groupof genes and then add them or calculate the distanee between the two outermost genes in one step ? Why ?解：厚垣红曲霉是非顺序四分子。子囊总数=6 9 0亲本组合孢子远远多于重组合孢子；故三基因有连锁关系。亲本一基因型a b c亲本二基因型 + + +R ( a-b ) = 12 + 18 + 0.5 * ( 180 + 142 + 18 + 20 ) / 6 9 0* 100 % = 30.43 %R ( a-c ) = 20 + 18

32、 + 12 + 0.5 * ( 12 + 180 + 18 + 20 ) / 6 9 0* 100 % = 23.91 %R ( b-c ) = 20 + 12 + 0.5 * ( 12 + 142 + 20 ) / 6 9 0* 100 % = 17.25 %ac b23.9117.25Yes , it is more correct to calculate the in termediate dista nee betwee n a group of genes and the n add them.因为两点测验无法察觉双交换发生；当两基因之间相距越远，区间发生双交换或偶数次交换的机会越

33、大；漏算的重组子越多，重组值与距离的偏差越大。(4) . In 1923,Bridges and Morgan reported the following data from a three-factorlin kage testcross in volv ing chromosome 3 inDrosophila mela no gaster.The threeloci are ca ( claret ),e ( ebony ),and ro ( rough ) .All three muta nt phe no typesare due to recessive alleles.(a )

34、 Are all three alleles in coupli ng con formati onin the heterozygous pare nt ?lfno t ,which allele is in repulsi on to the other two ?(b ) Con struct a gen etic map for these three loci show ing the dista nces betwee nloci in cM.(c ) Calculate the in terfere nee.Wild type49ebony1claret395rough119eb

35、ony , rough 370ebony , claret89rough , claret1ebony ,rough ,claret66Total1090解：重新排列：+ ,+, claret395ebony , rough+370+ ,rough ,+119ebony , +,claret89ebony , rough , claret66+ ,+ +49+ ,rough , claret1ebony,+ +1(a) 在双交换组中将rough换回去后所得的组合与亲本一样，三基因排列为：e_r_cclaret基因排斥另外两个基因，在三杂合子亲本中，三个隐性基因不排在一起。(b) R( r-e

36、) = 119 + 89 + 1 + 1 / 1090* 100 % = 19.27 %R(r-c )=66 + 49 +1+19.27 cM10.73 cM1/1090*100%=10.73%ercO.OOcM19.27 cM 30.00 cM(c) 符合 = 实双 / 理双 =2 / 1090 / 0.1972 * 0.1073 = 0.0887干扰 =1- 0.0887 = 0.9113即 91.13 %(5) . The Bedik people are n ative subsiste nee farmers in easter n Sen egal ,wheremalaria is

37、 com mon. They are the most ancient populati on in the area ,and there islittle in termarriage outside of the populatio n n 1975,Maura n-Sen drailand Boulouxreported the results of a study of 875 Bedik adults that represe nted 60% of theen tire Bedik population .They fou nd that 626 were homozygousH

38、BB*A/*A,249 wereheterozygous HBB*A/*S ,and none were homozygousHBB*S/*S.determine the frequency of the HBB*Sand HBB*Aallelesin this sample ,and determinewhether or not the populati on is in Hardy-We in berg equilibrium .解：Pa =626* 2 + 249 /875 * 2 = 0.858qs =249/ 875 * 2 =:0.142平衡群体为：(0.858*0.858,2

39、* 0.858 * 0.142,0.142 * 0.142)(0.7362,0.2437, 0.0201 )按875人数分布为(644,213,18)实际人数分布为:(626,249,0)显然是一个不平衡群体，杂合子偏多，纯合子偏少。根据X平方表的基本情况即可做出判断，以下可省略：2 2 2 2X = ( 626- 644 )/ 644 + ( 249- 213 )/ 213 + ( 0- 18 )/ 18=324/644 + 1296/213 + 324/18 = 0.503 + 6.084 + 18 = 24.587数值太大；已超过表中所能显示的数值。自由度等于1, P : 0.01差异极

40、显著存在。(6) n question (5),Determinethe values of s and t and the relative fitnessvaluesfor the three geno types un der the assumptio n that the populati on is in a bala need polymorphism equilibrium .解：假定所报道的群体是一个平衡群体，则626 / 644 = 0.9720249 / 213 = 1.1690 / 18 = 0选择系把适合度最高的1.169 定为100%即杂合子的相对适合度为1.000

41、0数为0.00000.9720 / 1.169 = 0.8315 =显性纯合子的相对适合度显性纯合子的选择系数s = 1- 0.8315 = 0.1685隐性纯合子的选择系数t= 1隐性纯合子的相对适合度=05. ( 10% )Choose the sin gle best an swer from the choices to the right .Choices may be used more tha n once .b_ will lead to the elimination of one allele from a populati on ,but one cannot predi

42、ct which one will be elim in ated . a. select ion aga inst homozygouscwill elimi nate an allele in b. select ion against heterozygotesa sin gle gen eratio n .c. select ion aga inst dominantsa will produce a stable equilibrium d. select ion aga inst recessives with both alleles still prese nt . e. mi

43、grati ondwill reduce the freque ncy of one f. mutati onallele in a population ,but very slowly . g. genetic driftgcauses cha nge in gene freque ncy h. n atura selecti onpredictable in amoun t ,but not direct ion . i. None of theseein troduces new variati on from extrapopulati onal sources.f introduces new variation from intrapopulational sources.iis effective only in large populati ons.eretards diverge nee among differe nt populati ons of the same species .widelygthe probable reason that small ,isolatedAmerican Indian tribes differ in blood group freque ncies .