遗传学课后习题及答案-刘祖洞

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刘祖洞遗传学习题答案13汇总讲解

第七章细菌和噬菌体的重组和连锁1.为什么说细菌和病毒是遗传学研究的好材料？2.大肠杆菌的遗传物质的传递方式与具有典型减数分裂过程的生物有什么不同？3.解释下列名词：（1）F-菌株，F+菌株，Hfr菌株；（2）F因子，F，因子，质粒，附加体；（3）溶源性细菌，非溶源性细菌；（4）烈性噬菌体，温和噬菌体，原噬菌体；（5）部分合子（部分二倍体）；4.部分合子在细菌的遗传分析中有什么用处？5.什么叫转导、普遍性转导、特异性转导（局限性转导）？6.转导和性转导有何不同？7.一个基因型为a+b+c+d+e+并对链霉素敏感的E.coliHfr菌株与基因型为a-b-c-d-e-并对链霉素耐性的F-菌株接合，30分钟后，用链霉素处理，然后从成活的受体中选出e+型的原养型，发现它们的其它野生型（+）基因频率如下：a+70％，b+-，c+85％，d+10％。

问a，b，c，d 四个基因与供体染色体起点（最先进入F-受体之点）相对位置如何？解：根据中断杂交原理，就一对接合个体而言，某基因自供体进入受体的时间，决定于该基因同原点的距离。

因此，就整个接合群体而论，在特定时间内，重组个体的频率反映着相应基因与原点的距离。

报据题目给定的数据，a、b、c、d与供体染色体的距离应该是：8.为了能在接合后检出重组子，必须要有一个可供选择用的供体标记基因，这样可以认出重组子。

另一方面，在选择重组子的时候，为了不选择供体细胞本身，必须防止供体菌株的继续存在，换句话说，供体菌株也应带有一个特殊的标记，能使它自己不被选择。

例如供体菌株是链霉素敏感的，这样当结合体（conjugants）在含有链霉素的培养基上生长时，供体菌株就被杀死了。

现在要问：如果一个Hfr菌株是链霉素敏感的，你认为这个基因应位于染色体的那一端为好，是在起始端还是在末端？解：在起始端9.有一个环境条件能使T偶数噬菌体（T-even phages）吸附到寄主细胞上，这个环境条件就是色氨酸的存在。

遗传学(第3版) 刘祖洞、乔守怡、吴燕华、赵寿元高等教育出版社 (2013-01)课后习题答案6

Chapter 6 Circulating Methylated DNA as Biomarkers for Cancer DetectionHongchuan Jin, Yanning Ma, Qi Shen andXian WangAdditional information is available at the end of the chapter/10.5772/514191. IntroductionIn addition to genetic alterations including deletion or point mutations, epigenetic changes such as DNA methylation play an important role in silencing tumor suppressor genes dur‐ing cancer development. By adding a methyl group from S-adenosyl-L-methionine to the cy‐tosine pyrimidine or adenine purine ring, DNA methylation is important to maintain genome structure and regulate gene expression. In mammalian adult tissues, DNA methyla‐tion occurs in CpG dinucleotides that often cluster in the genome as CpG islands in the 5’regulatory regions of the genes. Through recruiting transcriptional co-repressors including methyl-CpG-binding domain proteins (MBDs) and chromatin remodeling proteins like his‐tone deacetylases (HDACs) or impeding the binding of transcriptional activators, DNA methylation could suppress the transcription of many tumor suppressor genes critical to cancer initiation and progression [1-3].More and more results confirmed that cancer is a multi-stage process fuelled by many epige‐netic changes in addition to genetic changes in DNA sequence [4]. Chemical molecules like Trichostatin A (TSA) and 5-aza-2'-deoxycytidine (5-Aza-CdR) targeting epigenetic regula‐tors such as histone modifications and DNMTs (DNA methyltransferases) have been found to inhibit tumor growth both in vitro and in vivo. By reversing the epigenetic silencing of important tumor suppressor genes, an increasing number of epigenetic drugs such as 5-Aza-CdR, 5-Aza-CR and Vorinostat (SAHA) are currently investigated in the clinical trials for cancer treatment as a single drug or in combination with other epigenetic drugs or other ap‐proaches such as chemotherapy and showed very promising activities by offering signifi‐cant clinical benefits to cancer patients [5-13].© 2013 Jin et al.; licensee InTech. This is an open access article distributed under the terms of the CreativeCommons Attribution License (/licenses/by/3.0), which permits unrestricted use,distribution, and reproduction in any medium, provided the original work is properly cited.As one of the major epigenetic changes to inactivate tumor suppressor genes critical to hu‐man cancer development, DNA methylation was recognized as the biomarker for cancer de‐tection or outcome prediction in addition to the identification of novel tumor suppressor genes. DNA mutations will occur randomly in any nucleotides of one particular gene and the comprehensive determination of DNA mutations is thus very difficult and time-consum‐ing. In contrast, aberrant DNA hypermethylation usually takes place in defined CpG Islands within the regulatory region of the genes and it is much more convenient to detect DNA methylation in a quantitatively manner. In addition, DNA methylation can be amplified and is thus easily detectable using PCR-based approaches even when the DNA concentration af‐ter sample extraction is relatively low. Due to such advantages over DNA mutation- or pro‐tein-based biomarkers, DNA methylation-based biomarkers have been intensively investigated in the recent years. A large body of research reports has proved the value of DNA methylations in the prognosis prediction and detection of various cancers. DNAs used for such methylation analyses are usually extracted from tumor tissues harvested after sur‐gical operation or biopsy, thus limiting its wide application as the biomarkers for the early detection or screening of human cancers. Recently, it has been reported that there are certain amount of circulating DNAs in the peripheral blood of cancer patients, providing an ideal source to identify novel biomarkers for non-invasive detection of cancers. Both genetic and epigenetic changes found in the genomic DNAs extracted from primary tumor cells could be detected in the circulating DNAs, indicating that the detection of methylated DNAs in the circulation represents a new direction to develop novel biomarkers for cancer detection or screening in a non-invasive manner.2. Cell free DNA in the circulationAccording to the origin of circulating tumor-related DNA, it could be grouped into circulat‐ing cell free DNA or DNA from cells in the blood such as circulating tumor cells (CTC) in cancer patients (Figure 1).In 1869, the Australian physician Thomas Ashworth observed CTCs in the blood of a cancer patient. Therefore, it was postulated that CTCs were responsible for the tumor metastases in distal sites and should have important prognostic and therapeutic implications [14-16].However, the number of CTCs is very small compared with blood cells. Usually around 1-10CTCs together with several million blood cells could be found in 1 ml of whole blood, mak‐ing the specific and sensitive detection of CTCs very difficult [17-18]. Until recently, technol‐ogies with the requisite sensitivity and reproducibility for CTC detection have been developed to precisely analyze its biological and clinical relevance. The US Food and Drug Administration (FDA) approved the test for determining CTC levels in patients with meta‐static breast cancer in 2004. Currently, it has been expanded to other cancer types such as advanced colorectal cancer and prostate cancer. Although CTCs-counting based test have proven its value in predicting prognosis and monitoring therapeutic effects, the number of CTCs per ml of blood limited its sensitivity greatly [19]. With the development of high-sen‐sitive PCR-based methods, the detection of gene mutations or epigenetic changes such asMethylation - From DNA, RNA and Histones to Diseases and Treatment138DNA methylation within small amount of CTCs could be the next generation of CTC-based test for cancer detection. However, the cost of such tests will be greatly exacerbated, thuslimiting its wide application in the clinic [20-22].Figure 1. Circulating tumor cells and cell free DNA. Circulating Tumor cells (CTC) escape from primary sites and spread into the vessel to form metastases in the distal organs with. Cell free DNAs (cf-DNAs) are released into the circulation from dead cancer cells or proliferating tumor cells. RBC: red blood cell; WBC: white blood cell.Although its origin and biological relevance remains unknown, circulating cell free DNA (cf-DNA) is supposed to be valuable source to identify cancer markers with ideal sensitivity and specificity for non-invasive detection of cancer [23-24]. Early in 1948, two French scientists Mandel and Metais firstly reported the presence of cf-DNAs in human plasma [25]. Such an important discovery has been unnoticed for a long time until cell-free circulating nucleic acid was found to promote the spread and metastasis of crown gall tumor in plants [26]. Subse‐quently, increased level of cf-DNAs was found in patients with various diseases such as lupus erythematosus and rheumatoid arthritis cancer [27-28]. In 1977, Leon et al. reported that higher level of circulating DNA in the plasma of cancer patients when compared to healthy con‐trols. Moreover, greater amounts of cf-DNA were found in the peripheral blood of cancer patients with tumor metastases and cf-DNA levels decreased dramatically after radiothera‐py while persistently high or increasing DNA concentrations were associated with a lack of response to treatment [29], clearly revealing the potential value of cf-DNA as biomarker for cancer detection. Following studies confirmed that cf-DNAs in the plasma contains genetic and epigenetic changes specific to DNAs within the tumor cells from primary tissues, indicat‐ing that tumor specific cf-DNAs are originated from tumor cells rather than lymphocytes reacting towards the disease [30-31]. For example, K-Ras mutation was found in cf-DNA from 17 out of 21 patients with pancreatic adenocarcinoma and mutations were similar in corre‐sponding plasma and tissues samples. Importantly, such DNA alterations were found inCirculating Methylated DNA as Biomarkers for Cancer Detection/10.5772/51419139patients with pancreatitis who were diagnosed as pancreatic cancer 5-14 months later, indi‐cating that release of tumor-specific DNA into the circulation is an early event in cancer development and cf-DNA could be used as the biomarkers for early cancer detection [32].Treatment resulted in disappearance of K-Ras mutations in plasma DNA in six of nine pa‐tients. Three patients with a persistently positive K-Ras gene mutation in plasma samples from patients before and after treatment showed early recurrence or progression and pancreatic carcinoma patients with the mutant-type K-ras gene in plasma DNA exhibited a shorter survival time than patients with the wild-type gene, indicating the cf-DNA could be of value in monitoring disease progression or evaluating treatment response [31, 33].Through quantitatively analyzing plasma DNAs from patients with organ transplantation,Lo et al found that the majority of plasma DNAs was released from the hematopoietic sys‐tem. However, donor DNA could be detected in the plasma of recipients suffering from the graft rejection because of the large amount of cell death which promotes the release of donor DNAs into the peripheral blood of the recipients [34]. Therefore, it was postulated that cell-free tumor related DNA could originate from the apoptotic tumor cells since high-rate of apoptosis indeed occurs in primary and metastatic tumor tissues. However, cf-DNA quanti‐ties are significantly reduced in cancer patients after radiotherapy when a great number of tumor cells were believed to undergo apoptotic cell death and cf-DNAs in supernatants of cultured cancer cells increases with cell proliferation rather than apoptosis or necrosis, indi‐cating that proliferating tumor cells could actively release cf-DNA into the tumor microen‐vironment and circulation.In contrast to labile RNAs that were included into the actively secreted exosomes, the nature of cf-DNAs remains to be clarified. As negatively charged molecules, cf-DNA was bound by plasma proteins to escape from endonuclease-mediated degradation. Unfortunately, plasma proteins bound to cf-DNAs was not well characterized yet. Meanwhile, secreted exosomes could remodel microenviroments and promote tumor metastasis since RNAs within exo‐somes especially microRNA with high stability may influence gene expression in neighbor cells. The biological relevance of cf-DNAs remains unknown. DNA was believed to be more structural rather than functional. However, it was supposed that cf-DNA could play a role as vaccine in tumor microenvironment.3. Methods for the detection of methylated DNAIt is unclear so far whether serum or plasma is better for cf-DNA extraction. Although the DNA amount is significantly higher in the serum, the majority of the increase was due to the release of nuclear acids from destroyed blood cells during blood clotting [35]. In addition,the time gap between blooding drawing and DNA extraction as well as the methodologies used for DNA isolation contribute greatly to the amount of cf-DNA harvested. On an aver‐age, around 30 ng cf-DNA could be extracted from one ml of blood sample [36]. Therefore,in order to determine the quantity of potential cf-DNA-based biomarkers precisely and pro‐mote its wide application for cancer detection, it is very important to unify the source asMethylation - From DNA, RNA and Histones to Diseases and Treatment140well as the methodologies for cf-DNA extraction and use various internal controls to adjustpossible inter-laboratory variations.Figure 2. Schematic introductions of various methods for methylation analyses. MSP, BGS and COBRA are based on bisulfite-mediated conversion of unmethylated cytosines into uracils. CpG methylation could block DNA digestion by some restriction enzymes, making it possible to determine methylation status independent of bisulfite treatment by analyzing digestion products. Alternatively, DNA fragments containing methylated CpG sites could be enriched by an‐ti-methylcytosine antibody or methylation binding proteins. Advances in next generation genome sequencing tech‐nology led to the development of noel techniques such as SMRT which can specially analyze 5-methylcytosines with genome wide coverage.In general, the detection of DNA methylation could be bisulfite-dependent or -independent (Figure 2).The chemical reaction of sodium bisulfite with DNA could convert unmethylated cytosine of CpG into uracil or UpG but leave methylated cytosine of CpG unchanged. The following analyses such as methylation-and unmethylation specific polymerase chain reaction (M- and U-SP), bisulfite genome sequencing (BGS) or combined bisulfite restriction analysis (CO‐BRA) could determine the conversion of CpG sites of interest, thus reflecting their methyla‐tion status as methylated or unmethylated [37]. With varied resolution levels, different bisulfite-dependent DNA methylation analysis methods detect the conversion after bisulfite treatment of genomic DNA, which could have certain artificial effects such as incomplete conversion of unmethylated CpG into UpG, leading to high rate of false negative conclusion of DNA methylation status.Recently, some new modifications of cytosine in CpG dinucleotides have been discovered such as 5-hydoxymethylcytosine which was called the sixth base since 5-methylcytosine was named as the fifth base [38]. Generated from the oxidation of 5-methylcytosine by the Tet family of enzymes, 5-hydoxymethylcytosine was first found in bacteriophages and recentlyCirculating Methylated DNA as Biomarkers for Cancer Detection/10.5772/51419141shown to be abundant in human and mouse brains as well as in embryonic stem cells [39-40]. Although the exact relevance of 5-hydoxymethylcytosine in the genome is still not fully clarified, it has been found to regulate gene expression or promote DNA demethyla‐tion. The in vitro synthesized artificial oligonucleotides containing 5-hydoxymethylcyto‐sines can be converted into unmodified cytosines when introduced into mammalian cells,indicating that 5-hydoxymethylcytosine might be one of intermediate products during ac‐tive DNA demethylation [41]. Therefore, the increase of 5-hydoxymethylcytosine might re‐flect the demethylation of CpG dinucleotides. Unfortunately, 5-hydoxymethylcytosines,similar to 5-methylcytosines, appear to be resistant to bisulfite-mediated conversion and PCR could amplify DNA fragments containing 5-hydoxymethylcytosines or 5-methylcyto‐sines with similar efficiency [42-43]. Therefore, bisulfite-dependent methylation analyses could produce false positive results by counting 5-hydoxymethylcytosines into 5-methylcy‐tosines. In addition to 5-hydroxymethylcytosines, some forms of DNA modifications such as the seventh base, 5-formylcytosine and the eighth base, 5-carboxylcytosine, have been found in mammalian cells recently [44-47]. As the products of 5-hydoxymethylcytosine oxidation through TET hydroxylases, both 5-formylcytosine and 5-carboxylcytosine will be read as the uracil after bisulfite conversion, thus making it impossible for bisulfite-dependent analyses to distinguish unmodified cytosines from 5-formylcytosines and 5-carboxylcytosines.Bisulfite independent analyses such as MedIP (methylated DNA immunoprecipitation)could more or less detect DNA methylation specifically. In bisulfite independent analyses, 5-methylcytosines are differentiated from unmethylated cytosine by either enzyme digestion or affinity enrichment. DNA methylation analysis using restriction enzyme digestion is based on the property of some methylation-sensitive and -resistant restriction enzymes such as HpaII and MspI that target CCGG for digestion. HpaII fails to cut it once the second cyto‐sine was methylated while MspI-mediated digestion is not affected by DNA methylation,thus making it possible to determine the methylation status of CpG in the context of CCGG tetranucleotides by analyzing the products of DNAs digested by HpaII and MspI respective‐ly. As a primary method to analyze DNA methylation, it can only determine the methyla‐tion of CpG in the context of CCGG tetranucleotides and will overlook the majority of CpG dinucleotides in the genome.The development of monoclonal antibody specific to 5-methylcytosines revolutionized the analyses of DNA methylation [48-49]. Immunoprecipitated DNA by this antibody could be subject to DNA microarray or even deep sequencing to reveal novel sequences or sites con‐taining 5-methylcytosines [50]. This antibody specifically recognizes 5-methylcytosines but not 5-hydoxymethylcytosines. However, 5-methylcytosines could present not only in CpG dinucleotides but also in CHH or CHG trinucleotides, especially in plants, human embryon‐ic stem cells and probably cancer cells as well. CHH methylation indicates a 5-methylcyto‐sine followed by two nucleotides that may not be guanine and CHG methylation refers to a 5-methylcytosine preceding an adenine, thymine or cytosine base followed by guanine. Such non-CpG DNA methylations were enriched at transposons and repetitive regions, although the exact biological relevance remains unknown. However, antibody against 5-methylcyto‐Methylation - From DNA, RNA and Histones to Diseases and Treatment142sine may precipitate methylated CHH and CHG trinucleotide containing DNA fragments in addition to DNA sequences with methylated CpG sites.DNA methylation functions as the signal for DNA-interacting proteins to maintain genome structure or regulate gene expression. The proteins such as MBD1 (methyl-CpG binding do‐main protein 1), MeCP2 (methyl CpG binding protein 2) and MBD4 (methyl-CpG binding domain protein 4) bind methylated CpG specifically to regulate gene expression [51-52].Therefore, methyl-CpG binding domain could specifically enrich differentially methylated regions (DMRs) of physiological relevance [53]. Similar to MeDIP, MBD capture specifically enrich methylated CpG sites rather than hydroxymethlated CpG sites. The detailed analysis to compare MeDIP and MBD capture revealed that both enrichment techniques are sensitive enough to identify DMRs in human cancer cells. However, MeDIP enriched more methylat‐ed regions with low CpG densities while MBD capture favors regions of high CpG densities and identifies the greater proportion of CpG islands [49].Recently, the advance of next generation sequencing led to the development of several novel techniques, making it possible to quantitatively analyze DNA methylation at single nucleo‐tide resolution with genome wide coverage. Both the single molecule real time sequencing technology (SMRT) and the single-molecule nanopore DNA sequencing platform could dis‐criminate 5-methylcytosines from other DNA bases including 5-hydroxymethylcytosines even methyladenine independent of bisulfite conversion [54-55]. With many advantages such as less bias during template preparation, lower cost and better accuracy, such new techniques could offer more methods to detect DNA methylation with high specificity and sensitivity in addition to more potential DNA methylation based biomarkers for cancer de‐tection and screening.4. Potential DNA methylation biomarkers for cancer detectionIt has been questioned whether the methylated DNA in the circulation is sensitive to detect cancers early enough for curative resection. However, the development of sensitive detection methods confirmed the potential value of DNA methylation in cancer detection (Table 1).Most of DNA methylation biomarkers are well-known tumor suppressor genes silenced in primary tumor tissues. However, the biomarks do not have to be functional relevant. For ex‐ample, currently well-used biomarkers such as AFP (Alpha-Fetal Protein), PSA (Prostate-specific antigen) and CEA (Carcinoembryonic antigen) are not tumor suppressor genes with important biological functions. Profiling of methylated DNA in the circulation instead of primary tumor tissues with MeDIP or MBD capture or other methylation specific analyses methods would identify more potential biomarks rather than functional important tumor suppressor genes.Circulating Methylated DNA as Biomarkers for Cancer Detection/10.5772/51419143Cancer Markers Sensitivity Specificity Methods Ref.Bladder cancer CDKN2A (ARF) CDKN2A(INK4A)CDKN2A (INK4A)13/27 (48%)2/27 (7%)19/86 (22%)N/AN/A31/31 (100%)MSPMSPMSP[58][59]Breast cancer CDKN2A (INK4A)CDKN2A (INK4A)5/35 (14%)6/43 (14%)N/AN/AMS-AP-PCRMS-AP-PCR[56][57]Colorectal cancerMLH1CDKN2A (INK4A) CDKN2A(INK4A) CDKN2A (INK4A)ALX4CDH4NGFRRUNX3SEPT9TMEFF23/18 (17%)14/52 (27%)13/94 (11%)21/58 (36%)25/30 (83%)32/46 (70%)68/133 (51%)11/17 (65%)92/133 (69%)87/133 (65%)N/A44/44 (100%)N/AN/A36/52 (70%)17/17 (100%)150/179 (84%)10/10 (100%)154/179 (86%)123/179 (69%)MSPMSPMSPMSPMSPMSPMSPMSPMSPMSP[60][61][62][63][64][65][66][67][66]Esophageal cancer APCAPCCDKN2A (INK4A)13/52 (25%)2/32 (6%)7/38 (18%)54/54 (100%)54/54 (100%)N/AMSPMSPMSP[68][69]Gastric cancer CDH1CDKN2A (INK4A)CDKN2B (INK4B)DAPK1GSTP1Panel of five 31/54 (57%)28/54 (52%)30/54 (56%)26/54 (48%)18/54 (15%)45/54 (83%)30/30 (100%)30/30 (100%)30/30 (100%)30/30 (100%)30/30 (100%)30/30 (100%)MSPMSPMSPMSPMSPMSP[70]Head and neck cancer CDKN2A (INK4A)DAPK1MGMTPanel of threeDAPK18/95 (8%)3/95 (3%)14/95 (15%)21/95 (22%)N/AN/AN/AN/AN/AN/AMSPMSPMSPMSPMSP[71][72]Liver cancer CDKN2A (INK4A) CDKN2B(INK4B)13/22 (45%)4/25 (16%)48/48 (100%)35/35 (100%)MSPMSP[73][74]Lung cancer CDKN2A (INK4A)DAPK1GSTP1MGMTPanel of fourCDKN2A (INK4A)APC 3/22 (14%)4/22 (18%)1/22 (5%)4/22 (18%)11/22 (50%)N/A42/89 (47%)N/AN/AN/AN/AN/AN/A50/50 (100%)MSPMSPMSPMSPMSPMSPMSP[75][76][77]Methylation - From DNA, RNA and Histones to Diseases and Treatment 144Cancer Markers Sensitivity Specificity Methods Ref.CDKN2A (INK4A)CDKN2A (INK4A)77/105 (73%)12/35 (34%)N/A15/15 (100%)MSP MSP [78][79]Prostate cancer GSTP1GSTP123/33 (70%)25/69 (36%)22/22 (100%)31/31 (100%)MSP MSP[80][81]Table 1. Methylated DNA biomarkers in the literature.Most of the methods used for methylation biomarkers analyses are still bisulfite dependent.Few reports used MS-AP-PCR (methylation-sensitive arbitrarily primed PCR) which takes the advantage of methylation sensitive restriction endonucleases to distinguish methylated CpG from unmethylated form, although the sensitivity seems to be lower than MSP [56-57].Interestingly, combination of more than one methylated DNA as a methylation panel could great increase the sensitivity for cancer detection without significant reduction of specificity.Unfortunately, most of studies were performed in a retrospective manner. More prospective studies with large sample sizes will be warranted to compare different approaches especial‐ly bisulfite-independent methods in addition to confirm the value of DNA methylation for cancer detection.5. Conclusion and PerspectivesWith the development of the next generation genome sequencing as well as single molecular PCR, it became possible to analyze trace amount of DNAs including circulating cell-free DNA. Circulating tumor cells have been proven its value in prognosis predication even ear‐ly detection of various cancers. The analyses of methylated DNAs in the circulating will be the next promising epigenetic biomarkers for cancer detection. As one of the intermediate products of DNA demethylation, 5-hydroxymethlcytosines are resistant to bisulfite conver‐sion. Therefore, it should be carefully to interpret the data of methylation analyses based on bisulfite treatment due to potentially high rate of false positive results. Although some me‐thylated DNAs were found to valuable as a single biomarker for cancer detection, more po‐tential DNA methylations will be found after the wide application of SMRT and other sequencing platforms with high speed, depth and accuracy. DNA methylation signatures in‐cluding a panel of methylated DNAs will show the potential in the early diagnosis or screening and prognosis or therapy response prediction of many cancers. In addition, such DNA methylation biomarkers could be more sensitive and specific for cancer detection when combined with well-used biochemical biomarkers. However, unified methods with gold standards will be warranted to promote the development and clinical application of DNA methylation biomarkers.Circulating Methylated DNA as Biomarkers for Cancer Detection/10.5772/51419145AcknowledgementsThis work was supported by the National Natural Science Foundation of China (81071963;81071652), Program for Innovative Research Team in Science and technology of Zhejiang Province (2010R50046) and Program for Qianjiang Scholarship in Zhejiang Province (2011R10061; 2011R10073).Author detailsHongchuan Jin, Yanning Ma, Qi Shen and Xian Wang **Address all correspondence to: wangx118@Department of Medical Oncology, Laboratory of Cancer Epigenetics, Biomedical Research Center, Sir Runrun Shaw Hospital, Zhejiang University, ChinaReferences[1]Jones, P. A., & Baylin, S. B. (2007). The epigenomics of cancer. Cell , 128, 683-692.[2]Jones, P. A., & Baylin, S. B. (2002). The fundamental role of epigenetic events in can‐cer. Nat Rev Genet , 3, 415-428.[3]Baylin, S. B., Esteller, M., Rountree, M. R., Bachman, K. E., Schuebel, K., & Herman, J.G. (2001). Aberrant patterns of DNA methylation, chromatin formation and gene ex‐pression in cancer. Hum Mol Genet , 10, 687-692.[4]Baylin, S. B., & Herman, J. G. (2000). DNA hypermethylation in tumorigenesis: epige‐netics joins genetics. Trends Genet , 16, 168-174.[5]Oki, Y., & Issa, J. P. (2006). Review: recent clinical trials in epigenetic therapy. Rev Re‐cent Clin Trials , 1, 169-182.[6]Kelly, T. K., De Carvalho, D. D., & Jones, P. A. (2010). Epigenetic modifications astherapeutic targets. Nat Biotechnol , 28, 1069-1078.[7]Ramalingam, S. S., Maitland, M. L., Frankel, P., Argiris, A. E., Koczywas, M., Gitlitz,B., Thomas, S., Espinoza-Delgado, I., Vokes, E. E, Gandara, D. R., & Belani,C. P.(2010). Carboplatin and Paclitaxel in combination with either vorinostat or placebo for first-line therapy of advanced non-small-cell lung cancer. J Clin Oncol , 28, 56-62.[8]Braiteh, F., Soriano, A. O., Garcia-Manero, G., Hong,D., Johnson, MM, Silva Lde, P.,Yang, H., Alexander, S., Wolff, J., & Kurzrock, R. (2008). Phase I study of epigenetic modulation with 5-azacytidine and valproic acid in patients with advanced cancers.Clin Cancer Res , 14, 6296-6301.Methylation - From DNA, RNA and Histones to Diseases and Treatment146/10.5772/51419 [9]Font, P. (2011). Azacitidine for the treatment of patients with acute myeloid leukemiawith 20%-30% blasts and multilineage dysplasia. Adv Ther, 3(28), 1-9.[10]Fu, S., Hu, W., Iyer, R., Kavanagh, J. J., Coleman, R. L., Levenback, C. F., Sood, A. K.,Wolf, J. K., Gershenson, D. M., Markman, M., Hennessy, B. T., Kurzrock, R., & Bast, R. C., Jr. (2011). Phase 1b-2a study to reverse platinum resistance through use of a hypomethylating agent, azacitidine, in patients with platinum-resistant or platinum-refractory epithelial ovarian cancer. Cancer, 117, 1661-1669.[11]Silverman, L. R., Fenaux, P., Mufti, G. J., Santini, V., Hellstrom-Lindberg, E., Gatter‐mann, N., Sanz, G., List, A. F., Gore, S. D., & Seymour, J. F. (2011). Continued azaciti‐dine therapy beyond time of first response improves quality of response in patients with higher-risk myelodysplastic syndromes. Cancer.[12]Sonpavde, G., Aparicio, A. M., Zhan, F., North, B., Delaune, R., Garbo, L. E., Rousey,S. R., Weinstein, R. E., Xiao, L., Boehm, K. A., Asmar, L., Fleming, M. T., Galsky, M.D., Berry, W. R., & Von Hoff, D. D. (2011). Azacitidine favorably modulates PSA ki‐netics correlating with plasma DNA LINE-1 hypomethylation in men with chemo‐naive castration-resistant prostate cancer. Urol Oncol, 29, 682-689.[13]Keating, G. M. (2012). Azacitidine: a review of its use in the management of myelo‐dysplastic syndromes/acute myeloid leukaemia. Drugs, 72, 1111-1136.[14]Alix-Panabieres, C., Schwarzenbach, H., & Pantel, K. (2012). Circulating tumor cellsand circulating tumor DNA. Annu Rev Med, 63, 199-215.[15]Zhe, X., Cher, M. L., & Bonfil, R. D. (2011). Circulating tumor cells: finding the needlein the haystack. Am J Cancer Res, 1, 740-751.[16]Fidler, I. J. (2003). The pathogenesis of cancer metastasis: the ‘seed and soil’ hypothe‐sis revisited. Nat Rev Cancer, 3, 453-458.[17]Ghossein, RA, Bhattacharya, S, & Rosai, J. (1999). Molecular detection of micrometa‐stases and circulating tumor cells in solid tumors. Clin Cancer Res, 5, 1950-1960. [18]Pelkey, TJ, Frierson, H. F., Jr, & Bruns, D. E. (1996). Molecular and immunological de‐tection of circulating tumor cells and micrometastases from solid tumors. Clin Chem, 42, 1369-1381.[19]Mocellin, S., Keilholz, U., Rossi, C. R., & Nitti, D. (2006). Circulating tumor cells: the‘leukemic phase’ of solid cancers. Trends Mol Med, 12, 130-139.[20]Chimonidou, M., Strati, A., Tzitzira, A., Sotiropoulou, G., Malamos, N., Georgoulias,V., & Lianidou, E. S. (2011). DNA methylation of tumor suppressor and metastasis suppressor genes in circulating tumor cells. Clin Chem, 57, 1169-1177.[21]Garcia-Olmo, D. C., Gutierrez-Gonzalez, L., Ruiz-Piqueras, R., Picazo, M. G., & Gar‐cia-Olmo, D. (2005). Detection of circulating tumor cells and of tumor DNA in plas‐ma during tumor progression in rats. Cancer Lett, 217, 115-123.。

刘祖洞遗传学课后题答案之欧阳家百创编

第二章孟德尔定律1、欧阳家百（2021.03.07）2、为什么分离现象比显、隐性现象有更重要的意义？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。

9、真实遗传的紫茎、缺刻叶植株（AACC ）与真实遗传的绿紫茎缺刻叶紫茎马铃薯叶绿茎缺刻叶绿茎马铃薯叶 247 90 83 34（1）在总共454株F2中，计算4种表型的预期数。

（2）进行2测验。

（3）问这两对基因是否是自由组合的？紫茎缺刻叶紫茎马铃薯叶绿茎缺刻叶绿茎马铃薯叶观测值（O ）247 90 83 34 预测值（e ）(四舍五入) 255 85 85 29临界值81.7205.0.3=χ比较。

可见该杂交结果符合F 2的预期分离比，因此结论，这两对基因是自由组合的。

11、如果一个植株有4对显性基因是纯合的。

另一植株有相应的4对隐性基因是纯合的，把这两个植株相互杂交，问F2中：（1）基因型，（2）表型全然象亲代父母本的各有多少？解：(1) 上述杂交结果，F 1为4对基因的杂合体。

于是，F2的类型和比例可以图示如下：也就是说，基因型象显性亲本和隐性亲本的各是1/28。

(2) 因为，当一对基因的杂合子自交时，表型同于显性亲本的占3／4，象隐性亲本的占1／4。

所以，当4对基因杂合的F 1自交时，象显性亲本的为(3/4)4，象隐性亲本的为(1/4)4 = 1/28。

第三章遗传的染色体学说2、水稻的正常的孢子体组织，染色体数目是12对，问下列各组织的染色体数目是多少？（1）胚乳；（2）花粉管的管核；（3）胚囊；（4）叶；（5）根端；（6）种子的胚；（7）颖片；答；（1）36；（2）12；（3）12*8；（4）24；（5）24；（6）24；（7）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，比例如何？答：即下一代胚乳有八种基因型，且比例相等。

刘祖洞遗传学第三版答案-第9章-数量性状遗传

刘祖洞遗传学第三版答案-第9章-数量性状遗传出简单的比例，因此只能用统计方法分析。

（3）有可能出现超亲遗传。

数量性状遗传和质量性状遗传的主要区别：（1）数量性状是表现连续变异的性状，而质量性状是表现不连续变异的性状；（2）数量性状的遗传方式要比质量性状的遗传方式复杂的多，它是由许多基因控制的，而且它们的表现容易受环境条件变化的影响。

2.什么叫遗传率？广义遗传率？狭义遗传率？平均显性程度？解析：根据定义回答就可以了。

参考答案：遗传率指亲代传递其遗传特性的能力，是用来测量一个群体内某一性状由遗传因素引起的变异在表现型变异中所占的百分率，即：遗传方差/总方差的比值。

广义遗传率是指表型方差（Vp）中遗传方差（Ve）所占的比率。

狭义遗传率是指表型方差（Vp）中加性方差（V A）V V〔在数量/D A性状的遗传分析中，对于单位点模型，可以用显性效应和加性效应的比值d/a来表示显性程度。

但是推广到多基因系统时，∑d/∑a并不能说明任一位点上基因的显性性质。

因为∑d 和∑a 都可能因为有正有负而相消，除非两个亲本分别集中了所有显性和隐性等位基因。

但是∑d 2和∑a 2是显性效应和加性效应的积累，不会产生正负相消，因此在多对基因效应相等的假设下，221221k i i D kA i i d V kd d V ka a a =====∑∑， /D A V V 度的。

/D A V V /D A V V 时为无显性；/D A V V 时为部分显性或不完全显性；/D A V V 1/D A V V 为超显性。

〕 3.自然界中杂交繁殖的生物强制进行自交或其它方式近交时生活力降低，为什么自然界中自交的生物继续自交没有不良影响呢？解析：天然自交的生物其隐性的有害基因已经被淘汰。

参考答案：自然界中自交的生物通过自交使隐性基因暴露，在长期的进化过程中淘汰了隐性纯合的有害基因。

因此继续自交就没有不良影响了。

4.Johannsen 用菜豆做实验，得出纯系学说。

刘祖洞遗传学习题答案6

第六章染色体和连锁群1、在番茄中，圆形（O ）对长形（o ）是显性，单一花序（S ）对复状花序（s ）是显性。

这两对基因是连锁的，现有一杂交得到下面4种植株：圆形、单一花序（OS ）23 长形、单一花序（oS ）83 圆形、复状花序（Os ）85 长形、复状花序（os ）19 问O —s 间的交换值是多少？解：在这一杂交中，圆形、单一花序（OS ）和长形、复状花序（os ）为重组型，故O —s 间的交换值为：%20%100198583231923=⨯++++=r2、根据上一题求得的O —S 间的交换值，你预期杂交结果，下一代4种表型的比例如何？解：OS 0.1 Os 0.4 oS 0.4 os 0.1 OS 0.1 OOSS 0.01 OOSs 0.04 OoSS 0.04 OoSs 0.01 Os 0.4 OOSs 0.04 OOss 0.16 OoSs 0.16 Ooss 0.04 oS 0.4OoSS 0.04 OoSs 0.16 ooSS 0.16 ooSs 0.04 os 0.1 OoSs 0.01Ooss 0.04ooSs 0.04ooss 0.01O_S_ ：O_ss ：ooS_ ：ooss = 51% ：24% ：24% ：1%，即4种表型的比例为：圆形、单一花序(51％)，圆形、复状花序(24％)，长形、单一花序(24％)，长形、复状花序(1％)。

3、在家鸡中，白色由于隐性基因c 与o 的两者或任何一个处于纯合态有色要有两个显性基因C 与O 的同时存在，今有下列的交配：♀CCoo 白色 × ♂ccOO 白色↓子一代有色子一代用双隐性个体ccoo测交。

做了很多这样的交配，得到的后代中，有色68只，白色204只。

问o—c之间有连锁吗？如有连锁，交换值是多少？解：根据题意，上述交配：♀ CCoo白色⨯ ccOO白色♂↓有色CcOo ⨯ ccoo白色↓有色C_O_ 白色（O_cc，ooC_，ccoo）416820468=+4368204204=+此为自由组合时双杂合个体之测交分离比。

07__连锁交换与连锁分析-《遗传学》（第3版）（答案）刘祖洞乔守怡吴燕华赵寿元

第七章1. 在番茄中，圆形（O）对长形（o）是显性，单一花序（S）对复状花序（s）是显性。

这两对基因是连锁的，现有一杂交Os/oS×os/os得到下面4种植株：圆形、单一花序（OS）23，长形、单一花序（oS）83，圆形、复状花序（Os）85，长形、复状花序（os）19。

问：O—s间的重组率是多少？答案：O-S之间交换值为20%（42/210=0.2）。

2. 根据上一题求得的O—s间的重组率，你预期Os/oS×Os/oS杂交结果，下一代4种表型的比例如何？答案：雌雄配子均有四种类型，且比例为：Os∶oS∶os∶OS=4∶4∶1∶1，Os（0.4）oS（0.4）os（0.1）OS（0.1）Os（0.4）OOSS（0.16）OoSs（0.16）Ooss（0.04）OOSs（0.04）oS（0.4）OoSs（0.16）ooSS（0.16）ooSs（0.04）OoSS（0.04）os（0.1）Ooss（0.04）ooSs（0.04）ooss（0.01）OoSs（0.01）OS（0.1）OOSs（0.04）OoSS（0.04）OoSs（0.01）OOSS（0.01）因此：O_S_O_ss ooS_ooss园单园复长单长复0.51 0.24 0.24 0.013. 在家鸡中，白色由于隐性基因c与o的两者或任何一个处于纯合态，有色要有两个显性基因C与O的同时存在，今有下列的交配：♀白色CCoo×♂白色ccOO↓子一代有色子一代用双隐性个体ccoo测交。

做了很多这样的支配，得到的后代中，有色68只，白色204只。

问：o—c之间有连锁吗？如有连锁，重组率是多少？答案：F1代为CcOo，测交结果如下：CO Co cO co co CcOo（有色）Ccoo（白色）ccOo（白色）ccoo（白色）68 204由于有色∶白色=1∶3，因此反推全部配子中，CO型占1/4，恰巧符合自由组合定律，没有连锁；如果认为有连锁，计算重组率= 2/4 = 50%，亦可否认连锁。

刘祖洞遗传学习题答案5

第五章【2 】性别决议与伴性遗传1.哺乳动物中,雌雄比例大致接近1∶1,如何解释？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染色体的两类数量相等的配子,而女性只产生一种含X染色体的配子.精卵配子联合后产生含XY和X X两类比例雷同的合子,分别发育成男性和女性.是以,男女性比近于1 ：1.2.你如何差别某一性状是常染色体遗传,照样伴性遗传的？用例来解释.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.果蝇的性决议是XY型,雌蝇是XX,雄蝇是XY,问下列交配所产生的子代,基因型和表型若何？（l）WwVgvg×wvgvg （2）wwVgvg×WVgvg解：上述交配图示如下：(1) WwVgvg ⨯ wvgvg：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♂.(2) wwVgvg ⨯ WVgvg：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.子一代个别互订交配,问子二代的芦花性状与性别的关系若何？解：家鸡性决议为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,雄鸡全体为芦花羽,雌鸡1／2芦花羽,1／2非芦花.5.在鸡中,羽毛的显色须要显性基因C的消失,基因型cc的鸡老是白色.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掌握的,并且雌鸡是异配性别.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,假如这些子代个别互订交配,它们的子裔的表型分别比是如何的？注：基因型 C—Z b Z b和 C—Z b W鸡的羽毛长短芦花斑纹.解：依据题意,芦花公鸡的基因型应为CCZ B Z B,这一交配可图示如下：是以,假如子代个别互订交配,它们的子裔的表型分别比为芦花：非芦花＝9/16 ：7/16.若按性别统计,则在雄性个别中芦花：非芦花＝6/16 ：2/16;在雌性个别中芦花：非芦花＝3/16 ：5/16;6.在火鸡的一个优秀品系中,消失一种遗传性的白化症,养禽工作者把5只有关的雄禽进行测验,发明个中3只带有白化基因.当这3只雄禽与无亲缘关系的正常母禽交配时,得到229只幼禽,个中45只是白化的,并且满是雌的.育种场中可以进行一雄多雌交配,但在表型正常的184只幼禽中,育种工作者除了为清除白化基因外,想尽量多保存其他个别.你看火鸡的这种白化症的遗传方法如何？哪些个别应当镌汰,哪些个别可以宁神地保存？你如何做？解：229只幼禽是3只雄禽的子代个别的统计数字.因而,依据题意,这3只雄禽基因型雷同,所以,可视为统一亲本.因为雌禽为异配性别,又表现正常,于是揣摸,其基因型为ZW.雄禽为同配性别,又在子代中消失白化个别,并且满是雌的,所以这3只雄禽确定是白化基因杂合子,即ZZ a . 于是,上述交配可图示如下：基于上述剖析,可以以为,在火鸡中,这种白化症的遗传方法为性连锁隐性遗传.对于上述假定作χ2磨练：36.32]1[=χp ＞0.05,差异不明显.是以可以以为上述结论是准确的.如许,不难看出,184只表型正常的幼禽中,全体雌禽(ZW)可以宁神地保留,对于雄禽应进一步与表型正常的雌禽作一次交配,凡子代消失白化火鸡者应镌汰.7.有一视觉正常的女子,她的父亲是色盲.这个女人与正常视觉的汉子娶亲,但这个汉子的父亲也是色盲,问这对配头所生的后代视觉若何？解：依据题意,该女子的基因型为XX c,正常须眉的基因型为XY,这一婚配可图示如下：即这对配头所生的女儿都色觉正常,儿子有一半正常,一半色盲.8.一个没有血友病的汉子与表型正常的女人娶亲后,有了一个患血友病和Klinefelter分解症的儿子.解释他们两人的染色体构成和基因型.提醒：在形成卵子的第二次减数决裂时,X染色体可产生不离开现象.解：已知血友病为X连锁隐性遗传.因儿子的遗传构成中的Y染色体来自父方.而X染色体来自母方,所以,血友病患儿的表型正常的母亲,必定是血友病基因携带者,即XX h.又因为,Klinefelter患者染色体构成是XXY,故该患儿是h基因纯合体,X h X h Y.可见,来自杂合体(表型正常)母亲的成对基因为减数决裂第二次决裂不分别而成.于是这一婚配及患儿的形成可图示如下：9.植物Lychnisalba是雌雄异株.把阔叶雌株与窄叶雄株杂交,得到的F1代雌雄植株都是阔叶的,但F2雄性植株有两种类型——阔叶和窄叶,你如何解释？哪一共性别是异配性别（XY）,哪一共性别是同配性别？解：因为F1都是阔叶,母本的阔叶对父本的窄叶是显性.假如雌株是异配性别（XY）,而雄株是同配性别（XX）的话,就可以以为F1雌株半合子XY的X染色体来自同配性别XX的父亲,所所以窄叶.而F1雌株是阔叶,它们必定又从母亲那儿得到一条X染色体,这就是说,F1雌株精确是XX（同配性别）,但对阔叶和窄叶基因来说他们是杂合体,而雄株是XY.F2的成果支撑这一不雅点,即F2雄株有两种类型,他们是杂合的雌体的产物.F2雌株都是阔叶,因为他们的两条X染色体中有一条来自F1-XY父亲,F1父亲在X染色体上只携带一个阔叶等位基因.10.下面是患有肌养分不良个别的一个家系,是一个女人和两个不同的汉子在两次分别的婚姻中产生的.你以为那种遗传方法最有可能.请写出家系中各成员的基因型.解：这精确是因为X性连锁隐性基因引起的,因为这个家系的女儿中没有一个有病,但女性I-2正好一半的儿子有病,这女人很可能是杂合体.11.（1）双亲都是色盲,他们能生出一个色觉正常的儿子吗？（2）双亲都是色盲,他们能生出一个色觉正常的女儿吗？（3）双亲色觉正常,他们能生出一个色盲的儿子吗？（4）双亲色觉正常,他们能生出一个色盲的女儿吗？解：（1）不能（2）不能（3）能（4）不能12.在黑腹果蝇中,有一截刚毛（bobbedbristles）基因消失于X和Y的同源区域,所以X和Y上都有这基因.这基因记作bb,它对野生型基因（+）为隐性.隐性纯合体的刚毛短而细.若有一截刚毛雌蝇（X bb X bb）与纯合体正常刚毛雄蝇（X+Y+）交配,问F1和F2的基因型和表型的比例若何？解：13.火鸡卵有时能孤雌生殖.这有三个可能的机制：①卵没有经由减数决裂,仍为二倍体;②卵核被极体授精;③卵核染色体加倍.你预期每一假设机制所产生的子代的性比若何？（假定雏鸡要能活下去,一个Z染色体是必须消失的.）解：依据题意,将该雌鸡的卵母细胞三种可能的发育进程图示如下：(1a)(1b)O O ZW ZW O O ZW ZW极体卵卵极体（不成活）(2a)(2b)WM1M22（不成活）ZZ ZWZW ZWZW卵卵WW可见,当为机制⑴时,孤雌发育子代全体为雌性;机制⑵中,雌：雄为4/5 ：1/5;机制⑶发育之予代全体为雄性.(注：这里所示1a,1b,2a,2b 和3a,3b,都是随机产生.) 14.在小家鼠中,有一突变基因使尾巴曲折.如今有一系列杂交实验,成果如下：问：这突变基因是显性照样隐性？是常染色体遗传,照样伴性遗传？ZWZW(3a) (3b)Z W ZWM1M2Z W W ZZ WW ZZWW（不成活）卵卵加倍极体极体表中6个杂交中,亲代和子代的基因型各若何？解：该突变基因是X连锁显性遗传.用T表示该突变基因,t表示正常基因,则6个杂交的亲代和子代的基因型分别为：（1）：（2）：（3）：（4）：（5）：（6）：。

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第二章孟德尔定律之袁州冬雪创作1、为什么分离现象比显、隐性现象有更重要的意义？答：因为1、分离规律是生物界普遍存在的一种遗传现象，而显性现象的表示是相对的、有条件的；2、只有遗传因子的分离和重组，才干表示出性状的显隐性.可以说无分离现象的存在，也就无显性现象的发生.2、在番茄中，红果色（R）对黄果色（r）是显性，问下列杂交可以发生哪些基因型，哪些表示型，它们的比例如何（1）RR×rr（2）Rr×rr（3）Rr×Rr（4）Rr×RR（5）rr×rr3配子？杂种后代的基因型和表型怎样？（1）Rr×RR（2）rr×Rr（3）Rr×Rr粉红红色白色粉红粉红粉红4D）对球状（d）是显性，这两对基因是自由组合的.问下列杂交可以发生哪些基因型，哪些表型，它们的比例如何？（1）WWDD×wwdd （2）XwDd×wwdd（3）Wwdd×wwDd（4）Wwdd×WwDd2 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，1/8wwdd3/8白色、盘状，3/8白色、球状，1/8黄色、盘状，1/8黄色、球状5.在豌豆中，蔓茎（T）对矮茎（t）是显性，绿豆荚（G）对黄豆荚（g）是显性，圆种子（R）对皱种子（r）是显性.现在有下列两种杂交组合，问它们后代的表型如何？（1）TTGgRr×ttGgrr （2）TtGgrr×ttGgrr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.在番茄中，缺刻叶和土豆叶是一对相对性状，显性基因C节制缺刻叶，基因型cc是土豆叶.紫茎和绿茎是另外一对相对性状，显性基因A 节制紫茎，基因型aa的植株是绿茎.把紫茎、土豆叶的纯合植株与绿茎、缺刻叶的纯合植株杂交，在F2中得到9∶3∶3∶1的分离比.如果把F1：（1）与紫茎、土豆叶亲本回交；（2）与绿茎、缺刻叶亲本回交；以及（3）用双隐性植株测交时，下代表型比例各如何？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：(2) 绿茎缺刻叶对F1的回交：（3）双隐性植株对F l测交：AaCc × aaccAaCc Aacc aaCc aacc1紫缺：1紫马：1绿缺：1绿马(即两对性状自由组合形成的4种类型呈1：1：1：1.)7.在下列表中，是番茄的五组分歧交配的成果，写出每交配中亲本植株的最可以的基因型.（这些数据不是实验资料，是为了说明方便而假设的.）序号亲本基因型子代基因型子代表示型1 AaCc × aaCc紫茎缺刻叶 ×绿茎缺刻叶1/8AaCC，2/8AaCc，1/8Aacc1/8aaCC，2/8aaCc，1/8aacc3/8紫缺，1/8紫马3/8绿缺，1/8绿马2 AaCc × Aacc紫茎缺刻叶 ×紫茎土豆叶1/8AACc，1/8AAcc，2/8AaCc2/8Aacc，1/8aaCc，1/8aacc3/8紫缺，3/8紫马1/8绿缺，1/8绿马3 AACc × aaCc紫茎缺刻叶 ×绿茎缺刻叶1/4AaCC，2/4AaCc，1/4Aacc 3/4紫缺，1/4紫马4 AaCC × aacc紫茎缺刻叶 ×绿茎土豆叶1/2AaCc，1/2aaCc 1/2紫缺，1/2绿缺5 Aacc × aaCc紫茎土豆叶 ×绿茎缺刻叶1/4AaCc，1/4Aacc1/4aaCc，1/4aacc1/4紫缺，1/4紫马1/4绿缺，1/4绿马8、纯质的紫茎番茄植株(AA)与绿茎的番茄植株(aa)杂交，F1植株是紫茎.F1植株与绿茎植株回交时，后代有482株是紫茎的，526株是绿茎的.问上述成果是否符合1：1的回交比例.用2检验.解：根据题意，该回交子代个体的分离比数是：紫茎绿茎观测值（O）482 526预测值（e）504 504代入公式求2：这里，自由度df = 1.查表得概率值(P)：＜P＜.根据概率水准，认为差别不显著.因此，可以结论：上述回交子代分离比符合实际分离比1：1.9、真实遗传的紫茎、缺刻叶植株（AACC）与真实遗传的绿茎、土豆叶植株（aacc）杂交，F2成果如下：紫茎缺刻叶紫茎土豆叶绿茎缺刻叶绿茎土豆叶247 90 83 34（1）在总共454株F2中，计算4种表型的预期数（2）停止2检验（3）问这两对基因是否是自由组合的？紫茎缺刻叶紫茎土豆叶绿茎缺刻叶绿茎土豆叶观测值（O ） 247 90 83 34 预测值（e ） (四舍五入)255858529当df = 3时，查表求得：0.50＜P ＜81.7205.0.3=χ比较.可见该杂交成果符合F 2的预期分离比，因此结论，这两对基因是自由组合的.10、一个合子有两对同源染色体A 和A'及B 和B'，在它的生长期间（1）你预料在体细胞中是下面的哪类组合，AA'BB ？AABB'？AA'BB'？AABB ？A'A'B'B'？还是还有其他组合.（2）如果这个体成熟了，你预期在配子中会得到下列哪些染色体组合：（a ）AA'，AA ，A'A'，BB'，BB ，B'B'？（b ）AA'，BB'，（c ）A ，A'，B ，B'，（d ）AB ，AB'，A'B ，A'B'？（e ）AA'，AB'，A'B ，BB'？解：（1）在体细胞中是AA'BB'；（2）在配子中会得到（d ）AB ，AB'，A'B ，A'B' 11、如果一个植株有4对显性基因是纯合的.另外一植株有相应的4对隐性基因是纯合的，把这两个植株相互杂交，问F2中：（1）基因型，（2）表型全然象亲代父母本的各有多少？解：(1) 上述杂交成果，F 1为4对基因的杂合体.于是，F2的类型和比例可以图示如下：也就是说，基因型象显性亲本和隐性亲本的各是1/28.(2) 因为，当一对基因的杂合子自交时，表型同于显性亲本的占3／4，象隐性亲本的占1／4.所以，当4对基因杂合的F1自交时，象显性亲本的为(3/4)4，象隐性亲本的为(1/4)4 = 1/28.12、如果两对基因A和a，B和b，是独立分配的，而且A对a是显性，B对b是显性.（1）从AaBb个体中得到AB配子的概率是多少？（2）AaBb与AaBb杂交，得到AABB合子的概率是多少？（3）AaBb与AaBb杂交，得到AB表型的概率是多少？解：因形成配子时等位基因分离，所以，任何一个基因在个别配子中出现的概率是：(1) 因这两对基因是独立分配的，也就是说，自由组合之二非等位基因同时出现在同一配子中之频率是二者概率之积，即：(2) 在受精的过程中，两性之各类型配子的连系是随机的，因此某类型合子的概率是构成该合子的两性配子的概率的积.于是，AABB合子的概率是：(3) 在AaBb AaBb交配中，就各对基因而言，子代中有如下关系：但是，实际上，在形成配子时，非等位基因之间是自由组合进入配子的；而配子的连系又是随机的.因此同时思索这两对基因时，子代之基因型及其频率是：于是求得表型为AB的合子之概率为9/16.13、遗传性共济失调（hereditary ataxia）的临床表型是四肢运动失调，呐呆，眼球震颤.本病有以显性方式遗传的，也有以隐性方式遗传的.下面是本病患者的一个家系.你看哪类遗传方式更可以？请注明家系中各成员的基因型.如这病是由显性基因引起，用符号A；如由隐性基因引起，用符号a.解：在这个家系中，遗传性共济失调更可以是隐性遗传的.14、下面的家系的个别成员患有极为罕见的病，已知这病是以隐性方式遗传的，所以患病个体的基因型是aa.（1）注明Ⅰ-1，Ⅰ-2，Ⅱ-4，Ⅲ-2，Ⅳ-1和Ⅴ-1的基因型.这儿Ⅰ-1暗示第一代第一人，余类推.（2）Ⅴ-1个体的弟弟是杂合体的概率是多少？（3）Ⅴ-1个体两个mm 全是杂合体的概率是多少？（4）如果Ⅴ-1与Ⅴ-5成婚，那末他们第一个孩子有病的概率是多少？（5）如果他们第一个孩子已经出生，而且已知有病，那末第二个孩子有病的概率是多少？解：(1) 因为，已知该病为隐性遗传.从家系分析可知，II-4的双亲定为杂合子.因此，可写出各个体的基因型如下：(2) 由于V-1的双亲为杂合子，因此V-1，2，3，4任一个体为杂合子的概率皆为1/2，那末V-1的弟弟为杂合体的概率也就是1/2.(3)V-1个体的两个mm(V-2和V-3)为杂合体的概率各为1/2，由于二者独AA aa AaAaAaAa aaaaAa aa aaaaAaAaAaaaAaAaaaAaaaAa立，于是，她们全是杂合体的概率为：1/2 1/2 =1/4. (4) 从家系分析可知，由于V-5个体的父亲为患病者，可以必定V-5个体定为杂合子(Aa).因此，当V-1与V-5成婚，他们第一个孩子患病的概率是1/2. (5) 当V-1与V-5的第一个孩子确为患者时，因第二个孩子的出现与前者独立，所以，其为患病者的概率仍为1/2.15、假设地球上每对夫妇在第一胎生了儿子后，就停止生孩子，性比将会有什么变更？16、孟德尔的豌豆杂交试验，所以可以取得成果的原因是什么？第三章遗传的染色体学说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？那末，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：有丝分裂减数分裂发生在所有正在生长着的组织中从合子阶段开端，继续到个体的整个生活周期无联会，无交叉和互换使姊妹染色体分离的均等分裂每个周期发生两个子细胞，产品的遗传成分相同子细胞的染色体数与母细胞相同只发生在有性繁殖组织中高等生物限于成熟个体；许多藻类和真菌发生在合子阶段有联会，可以有交叉和互换后期I是同源染色体分离的减数分裂；后期II是姊妹染色单体分离的均等分裂发生四个细胞产品（配子或孢子）产品的遗传成分分歧，是父本和母本染色体的分歧组合为母细胞的一半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平均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，各具有对折的染色体（n）雌雄性细胞受精连系为合子，受精卵（合子），又恢复为全数的染色体2n.包管了亲代与子代间染色体数目标恒定性，为后代的正常发育和性状遗传提供了物质基础，包管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分向南北极是随机的，即一对染色体的分离与任何另外一对染体的分离不发生关联，各个非同源染色体之间都可以自由组合在一个子细胞里，n对染色体，便可以有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以组合数为212=4096.各个子细胞之间在染色体组成上将可以出现多种多样的组合.此外，同源染色体的非mm染色单体之间还可以出现各种方式的交换，这就更增加了这种差别的复杂性.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，问下列各组织的染色体数目是多少？（1）胚乳；（2）花粉管的管核；（3）胚囊；（4）叶；（5）根端；（6）种子的胚；（7）颖片；答；（1）36；（2）12；（3）12*8；（4）24；（5）24；（6）24；（7）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，比例如何？4、某生物有两对同源染色体，一对染色体是中间着丝粒，另外一对是端部着丝粒，以形式图方式画出：（1）第一次减数分裂的中期图.（2）第二次减数分裂的中期图5、蚕豆的体细胞是12个染色体，也就是6对同源染色体（6个来自父本，6个来自母本）.一个学生说，在减数分裂时，只有1/4的配子，它们的6个染色体完全来自父本或母本，你认为他的回答对吗？答：分歧错误.因为在减数分裂时，来自父本或母本的某一条染色体进入某个配子的概率是1/2，则6个完全来自父本或母本的染色体同时进入一个配子的概率应为2*1/2)6 = 1/32.6、在玉米中：（1）5个小孢子母细胞能发生多少配子？（2）5个大孢子母细胞能发生多少配子？（3）5个花粉细胞能发生多少配子？（4）5个胚囊能发生多少配子？答：（1）5个小孢子母细胞能发生20个配子；（2）5个大孢子母细胞能发生5个配子；（3）5个花粉细胞能发生5个配子；（4）5个胚囊能发生5个配子；7、马的二倍体染色体数是64，驴的二倍体染色体数是62.（1）马和驴的杂种染色体数是多少？（2）如果马和驴之间在减数分裂时很少或没有配对，你是否能说明马-驴杂种是可育还是不育？答：（1）马和驴的杂种染色体数是63.（2）如果马和驴之间在减数分裂时很少或没有配对，则马-驴杂种是不育的.8、在玉米中，与糊粉层着色有关的基因很多，其中三对是A—a，I—i，和Pr—pr.要糊粉层着色，除其他有关基因必须存在外，还必须有A基因存在，而且不克不及有Ⅰ基因存在.如有Pr存在，糊粉层紫色.如果基因型是prpr，糊粉层是红色.假使在一个隔离的玉米试验区中，基因型 AaprprII的种子种在偶数行，基因型 aaPrprii的种子种在奇数行.植株长起来时，允许天然授粉，问在偶数行生长的植株上的果穗的糊粉层颜色怎样？在奇数行上又怎样？（糊粉层是胚乳一部分，所以是3n）.9、兔子的卵没有受精，颠末刺激，发育成兔子.在这种孤雌生殖的兔子中，其中某些兔子对有些基因是杂合的.你怎样诠释？（提示：极体受精.）答：动物孤雌生殖的类型有一种是：雌性二倍体通过减数分裂发生单倍体卵和极核，卵和极核融合形成二倍体卵，再发育成二倍体个体.例如，AaBb通过减数分裂可发生AB、Ab、aB、ab四种卵和极核，AB卵和AB极核来自同一个次级卵母细胞，二者融合形成AABB卵，这是纯合的.如果是AB卵和aB极核融合，则个体对A位点是杂合的.如果是AB卵和Ab极核融合，则个体对B位点是杂合的.如果是Ab卵和aB极核融合，则个体对A位点和B位点都是杂合的.可以是第二极体与卵细胞连系，有些基因才有可以是杂合的.第四章基因的作用及其与环境的关系1、从基因与性状之间的关系，怎样正确懂得遗传学上内因与外因的关系？2、在血型遗传中，现把双亲的基因型写出来，问他们子女的基因型应该如何？（1）（2）（3）解： ABO血型为孟德尔式遗传的复等位基因系列，以上述各种婚配方式之子女的血型是：（1）（2）（3）3、如果父亲的血型是B 型，母亲是O 型，有一个孩子是O 型，问第二个孩子是O 型的机会是多少？是B 型的机会是多少？是A 型或AB 型的机会是多少？的基因型为：第二个孩子是O 型的机会是0.5，是B 型的机会也是0.5，是A 型或AB 型的机会是0.4、分析图4－15的家系，请根据分析成果注明家系中各成员的有关基因型.解：5、当母亲的表型是ORh -MN ，子女的表型是ORh +MN 时，问在下列组合中，哪个或哪几个组合不成能是子女的父亲的表型，可以被解除？ABRh +M ， ARh +MN ， BRh -MN ， ORh -N.解：ABO 、MN 和Rh 为平行的血型系统，皆遵循孟德尔遗传法则；ABO 血型是复等位基因系列，MN 血型是并显性，Rh 血型显性完全. 现对上述四类血型汉子停止分析如下：父亲，应予解除.6、某个女人和某个汉子成婚，生了四个孩子，有下列的基因型：iiRRL M L N ，I A iRrL N L N ，iiRRL N L N ，I B irrL M L M ，他们父母亲的基因型是什么解：他们父母亲的基因型是：I A iRrL M L N ，I B iRrL M L N7.兔子有一种病，叫做Pelger 异常（白血细胞核异常）.有这种病的兔子，并没有什么严重的症伏，就是某些白细胞的核不分叶.如果把患有典型Pelger 异常的兔子与纯质正常的兔子杂交，下代有217只显示Pelger 异常，237只是正常的.你看Pelger 异常的遗传基础怎样？解：从271：237数据分析，近似1：1.作2检验：当df = 1时，查表：0.10＜p ＜0.50.根据0.05的概率水准，认为差别不显著.可见，符合实际的1：1.现在，某类型与纯质合子杂交得1：1的子代分离比，断定该未知类型为一对基因差别的杂合子.8、当有Pelger 异常的兔子相互交配时，得到的下一代中，223只正常，439只显示 Pelger 异常，39只极度病变.极度病变的个体除了有不正常的白细胞外，还显示骨骼系统畸形，几乎生后不久就全部死亡.这些极度病变的个体的基因型应该怎样？为什么只有39只，你怎样诠释？解：根据上题分析，pelger 异常为杂合子.这里，正常：异常＝223：439 1：2.依此，极度病变类型(39)应属于病变纯合子：Pp ⨯ Pp↓PP 41正常Pp 21异常pp 41极度病变又，因39只极度病变类型生后不久死亡，可以推断，病变基因为隐性致死基因，但有显性效应.如果这样，不但39只的生后死亡不必费解，而且，病变纯合子比数这样低也是可以懂得的.原因是部分死于胚胎发育过程中.9、在小鼠中，有一复等位基因系列，其中三个基因列在下面：A Y= 黄色，纯质致死；A = 鼠色，野生型；a = 非鼠色（黑色）.这一复等位基因系各位于常染色体上，列在前面的基因对列在后面的基因是显性.A Y A Y个体在胚胎期死亡.现在有下列5个杂交组合，问它们子代的表型如何？a 、A Y a （黄）×A Y a （黄）b 、A Y a （黄）×A YA （黄）c 、A Y a （黄）×aa （黑）d 、A Ya （黄）×AA （鼠色）e 、A Ya （黄）×Aa （鼠色）10、假定停止很多A Ya×Aa 的杂交，平均每窝生8只小鼠.问在同样条件下，停止很多 A Y a×A Ya 杂交，你预期每窝平均生几只小鼠？解：根据题意，这两种杂交组合的子代类型及比例是：Aa a A Y ⨯↓A A Y 41a A Y 41Aa 41aa 412黄： 1灰： 1黑a A a A Y Y ⨯↓Y Y A A 41a A Y 21aa 41（死亡） 2黄： 1黑4黄2黑.11、一只黄色雄鼠（A Y_）跟几只非鼠色雌鼠（aa ）杂交，你能不克不及在子代中同时得到鼠色和非鼠色小鼠？为什么？12、鸡冠的种类很多，我们在图4-13中先容过4种.假定你最初用的是纯种豌豆冠和纯种玫瑰冠，问从什么样的交配中可以获得单冠？解：知鸡冠形状是基因互作的遗传形式.各基因型及其相应表型是：基因型表示型 R_P_ 胡桃冠 R_pp 玫瑰冠 rr P_豌豆冠 rrpp单片冠因此，RRpp rrPP ⨯↓RrPp↓⊗__169P R ，pp R _163，_163rrP ，rrpp 161胡桃冠：玫瑰冠：豌豆冠：单片冠13、Nilsson －Ehle 用两种燕麦杂交，一种是白颖，一种是黑颖，二者杂交，F1是黑颖.F2（F1×F1）共得560株，其中黑颖418，灰颖106，白颖36.1）说明颖壳颜色的遗传方式.（2）写出F2中白颖和灰颖植株的基因型.（3）停止2检验.实得成果符合你的实际假定吗？解：（1）从题目给定的数据来看，F 2分离为3种类型，其比例为：黑颖：灰颖：白颖＝418：106：3612：3：1.即9：3：3：1的变形.可见，颜色是两对基因节制的，在表型关系上，呈显性上位.（2）假定B 为黑颖基因，G 为灰颖基因，则上述杂交成果是：P bbgg BBGG ⨯ 黑颖白颖 ↓F 1BbGg 黑颖 ↓⊗F 2__169G B gg B _163_163bbG bbgg 16112黑颖：3灰颖：1白颖(3) 根据上述假定停止2检验：当df ＝2时，查表：＜p ＜.认为差别不显著，即符合实际比率.因此，上述假定是正确的.14、在家蚕中，一个结白茧的个体与另外一结白茧的个体杂交，子代中结白茧的个体与结黄茧的个体的比率是3：1，问两个亲体的基因型怎样？解：在家蚕中，黄茧与白茧由一对等位基因节制，Y —黄色，y —白色，Y 对y 显性.但是，有一与其不等位的抑制基因I ，当I 存在时，基因型Y_表示白茧.根据题目所示，白：黄＝ 3：1，标明在子代中，呈3：1分离.于是推论，就I —i 而言，二亲本皆为杂合子Ii ；就Y —y 而言，则皆表示黄色的遗传基础只是3／4被抑制.所以，双亲的基因型(交配类型)应该是：IiYY IiYY IiYY IiYy IiYY Iiyy IiYy iiyy15、在小鼠中，我们已知道黄鼠基因A Y对正常的野生型基因 A 是显性，别的还有一短尾基因T ，对正常野生型基因t 也是显性.这两对基因在纯合态时都是胚胎期致死，它们相互之间是独登时分配的.（1）问两个黄色短尾个体相互交配，下代的表型比率怎样？（2）假定在正常情况下，平均每窝有8只小鼠.问这样一个交配中，你预期平均每窝有几只小鼠？解：根据题意，此黄色短尾鼠为杂合子A yATt ，其子代情形可图示如下：（1）ATt A ATt A Y Y ⨯↓致死⎪⎪⎪⎪⎪⎪⎭⎪⎪⎪⎪⎪⎪⎬⎫AATT ATT A tt A A TtA A TT A A Y Y Y Y Y Y Y 161161161162161ATt A Y 164Att A Y 162AATt 162AAtt 161黄短黄常灰短灰常灰色短尾：1灰色常态尾.(2) 在上述交配中，成活率只占受孕率的9/16.所以，假定正常交配每窝生8只小鼠时，这样交配平均每窝生4—5只.16、两个绿色种子的植物品系，定为X ，Y.各自与一纯合的黄色种子的植物杂交，在每个杂交组合中，F1都是黄色，再自花授粉发生F2代，每个组合的F2代分离如下：X ：发生的F2代 27黄：37绿Y ：发生的F2代，27黄：21绿请写出每交配中二个绿色亲本和黄色植株的基因型.解：F1的表型说明，决议黄色的等位基因对决议绿色的等位基因呈显性.F2的成果符合有若干对自由组合的基因的假设，当这些基因中有任何一对是纯合隐性时，发生绿色表型.黄色和绿色的频率计算如下：1品系X ：aabbcc ，黄色品系AABBCC ，F1为AaBbCc假如在一个杂交中唯一一对基因分离（如Aa Aa ），；另外一些影响黄色的基因对都是纯合的（AaBBCC AaBBCC ）.这一杂交发生黄色子代的比率是绿色比率是如果这个杂交有两对基因分离（如AaBbCC AaBbCC ），那末黄色子代的比率是：绿色比率是三对基因分离（AaBbCc AaBbCc ）时，黄色子代的比率是：绿色比率是A 位点B 位点C 位点F2基因型F2表示型 AA 41BB 41CC 41 AABBCC6411黄 Cc 42 AABBCc6422黄cc 41 AABBcc6411绿 Bb 42CC 41 AABbCC 6422黄Cc 42 AABbCc6444黄cc 41 AABbcc6422绿 bb 41CC 41 AAbbCC 6411绿Cc 42 AAbbCc6422绿cc 41 AAbbcc6411绿 A 位点 B 位点 C 位点F2基因型F2表示型 Aa 42BB 41CC 41 AaBBCC 6422黄Cc 42 AaBBCc6444黄cc 41 AaBBcc6422绿 Bb 42CC 41 AaBbCC 6444黄Cc 42AaBbCc6488黄cc 41 AaBbcc6444绿 bb 41CC 41 AabbCC 6422绿Cc 42 AabbCc6444绿cc 41 Aabbcc6422绿 A 位点B 位点C 位点F2基因型F2表示型 aa 41BB 41CC 41 aaBBCC 6411绿 Cc 42 aaBBCc 6422绿cc 41 aaBBcc 6411绿 Bb 42CC 41 aaBbCC 6422绿 Cc 42 aaBbCc 6444绿cc 41 aaBbcc 6422绿 bb 41CC 41 aabbCC 6411绿 Cc 42 aabbCc 6422绿cc 41aabbcc 6411绿汇总黄：绿 =27：37(2) 品系Y ：aabbCC ，黄色品系AABBCC ，F1为AaBbCC ，这个杂交有两对基因分离（如AaBbCC AaBbCC ），的比率是：16/9)4/3(2=绿色比率是16/7)4/3(12=-9：绿7. 第五章性别决议与伴性遗传1、哺乳动物中，雌雄比例大致接近1∶1，怎样诠释？解：以人类为例.人类男性性染色体XY ，女性性染色体为XX.男性可发生含X 和Y 染色体的两类数目相等的配子，而女性只发生一种含X 染色体的配子.精卵配子连系后发生含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？用例来讲明.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.果蝇的性决议是XY 型，雌蝇是XX ，雄蝇是XY ，问下列交配所发生的子代，基因型和表型如何？（l）WwVgvg×wvgvg （2）wwVgvg×WVgvg 解：上述交配图示如下： (1) WwVgvg wvgvg：Ww ⨯ wY↓Vgvg ⨯ vgvg↓1/4 Ww 1/2 Vgvg = 1/8 WwVgvg 红长♀1/2 vgvg = 1/8 Wwvgvg 红残♀1/4 ww 1/2 Vgvg = 1/8 wwVgvg 白长♀1/2 vgvg = 1/8 wwvgvg 白残♀1/4 WY 1/2 Vgvg = 1/8 WYVgvg 红长♂1/2 vgvg = 1/8 WYvgvg 红残♂1/4 wY 1/2 Vgvg = 1/8 wYVgvg 白长♂1/2 vgvg = 1/8 wYvgvg 白残♂即基因型：等比例的WwVgvg，WwVgvg， wwVgvg，wwvgvg，WYVgvg，WYvgvg， wYVgvg， wYvgvg.表示型：等比例的红长♀，红残♀，白长♀，白残♀，红长♂，红残♂，白长♂，白残♂.(2) wwVgvg WVgvg：ww ⨯ WY↓Vgvg ⨯ Vgvg↓1/2 Ww 1/4 VgVg = 1/8 WwVgVg 红长♀1/2 Vgvg = 1/4 WwVgvg 红长♀1/4 vgvg = 1/8 Wwvgvg 红残♀1/2 wY 1/4 VgVg = 1/8 wYVgVg 白长♂1/2 Vgvg = 1/4 wYVgvg 白长♂1/4 vgvg = 1/8 wYvgvg 白残♂即，基因型：1WwVgVg ：2WwVgvg ：1Wwvgvg ：1wYVgVg ：2wYVgvg ：1wYvgvg.表示型： 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此外关系如何？解：家鸡性决议为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：P ♀ Z b W ⨯ Z B Z B♂↓F1♀ Z B W ⨯ Z B Z b♂↓F2 1Z B Z B：1Z B Z b：1Z B W ：1Z b W芦花♂芦花♀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，羽毛的芦花花纹是由伴性（或Z连锁）显性基因B 节制的，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，子一代都是芦花花纹，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鸡的羽毛是非芦花花纹.解：根据题意，芦花公鸡的基因型应为CCZ B Z B，这一交配可图示如下：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只有关的雄禽停止检验，发现其中3只带有白化基因.当这3只雄禽与无亲缘关系的正常母禽交配时，得到 229只幼禽，其中45只是白化的，而且全是雌的.育种场中可以停止一雄多雌交配，但在表型正常的184只幼禽中，育种工作者除了为消除白化基因外，想尽可以多保管其他个体.你看火鸡的这种白化症的遗传方式怎样？哪些个体应该淘汰，哪些个体可以放心地保管？你怎样做？解：229只幼禽是3只雄禽的子代个体的统计数字.因而，根据题意，这3只雄禽基因型相同，所以，可视为同一亲本.由于雌禽为异配性别，又表示正常，于是推断，其基因型为ZW.雄禽为同配性别，又在子代中出现白化个体，而且全是雌的，所以这3只雄禽必定是白化基因杂合子，即ZZ a.于是，上述交配可图示如下：。