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Chapter 3 Mendel’s Genetic Analysis and Extension 第三章 孟德尔式遗传分析及其扩展

Chapter 3 Mendel’s Genetic Analysis and Extension 第三章 孟德尔式遗传分析及其扩展 3.1 Mendel’s laws 孟德尔遗传规律 3.1.1 Law of segregation 分离律 3.1.2 Law of independent assortment 自由组合律 3.2 多对相对 性状杂交的遗传分析 3.3 二项式分布公式 3.4 Tests for goodness of fit: chi-square tests 适合度检验 - 卡 方检验.

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Chapter 3 Mendel’s Genetic Analysis and Extension 第三章 孟德尔式遗传分析及其扩展

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  1. Chapter 3 Mendel’s Genetic Analysis and Extension 第三章 孟德尔式遗传分析及其扩展 3.1 Mendel’s laws 孟德尔遗传规律 3.1.1 Law of segregation 分离律 3.1.2 Law of independent assortment 自由组合律 3.2多对相对性状杂交的遗传分析 3.3二项式分布公式 3.4 Tests for goodness of fit: chi-square tests 适合度检验-卡 方检验

  2. 3.5 Interaction among genes 基因互作 3.5.1 Interaction between alleles 等位基因间互作 3.5.2 Interaction between non-alleles 非等位基因 间的相互作用 3.6 Genotype, phenotype and environmental 基因型、表型与环境

  3. Terms:C1,C2 Inherited trait Phenotype Genotype Dominant Recessive Pure-breeding Cross Test cross Back cross Self-cross Reverse cross Hybrid Homozygous Heterozygous Progeny

  4. Alleles Co-dominance Lethal Viability Multiple alleles Monohybrid cross Dihybrid cross Epistasis Gametogenesis Goodness-of-fit χ2 statistic hypothesize

  5. 3.1 Mendel’s laws Reasons of choosing pea for material: stable true-breeding easy distinguishable characteristics self and close pollination easy counting Law of segregation (分离规律) Law of independent assortment (自由组合规律 ）

  6. 3.1.1 Law of segregation Mendel studied inheritance of seven phenotypes in pea, we concentrate on only one of these - Petal color 红花×白花 F1红花 F2 929株 (705红/224白，接近3∶1) 白花×红花 与正交结果完全一致

  7. why 705 224 F1 和F2 性状表现不受亲本组合方式影响

  8. Presumption for the phenomenon produced As red is dominant to white then F1 plants had red flowers. When F1 plants self-fertilizing three different classes of genotype, CC Cc and cc are possible. There arise in the ratio 1:2 :1. This is how the 3:1 phenotypic ratio is established.

  9. Validation and Detection (假设的验证和杂合体的检测) One simple extension of the 3:1 phenotype ratio produced when an F1 individual is crossed to the homozygous recessive parent. F1 can produce gametes with dominant and recessive alleles, and so the progeny of the cross have the dominant and recessive phenotype in equal numbers, a 1:1 phenotype ratio. This type of cross is termed a testcross, and is useful in any situation where it is necessary to determine if an individual is heterozygous. C c c 85 81

  10. Essence of law of segregation: 决定某一性状的等位基因在配子形成过程中彼此分离互不干扰，使得配子中仅含成对基因中的一个，产生数目相等的两种类型配子。

  11. Single autosomal dominant gene disorders: 常染色体单基因显性遗传病的传递： 家族性多发性结肠息肉(polypus)、先天性白内障(cataract)、短肢症、Huntington’s disease 等1460多种性状。 特点：双亲之一杂合态患者 子女1/2得病，机率相等 代代相传

  12. Single autosomal recessive gene disorders: 常染色体单基因隐性遗传病的传递： 白化病(albinism)、侏儒症、唇裂、腭裂、Sickle cell anemia等1230多种性状。 特点：双亲表型正常均携带致病基因 1/4子女患病，机率均等 隔代遗传

  13. Male male sufferer couple twins from separate zygotes Female female sufferer close relative couple twins from same zygotes I II no progeny 1 2 first sufferer decedent carrier unknown sex sex-linked recessive carrier Symbol for drawing human pedigree

  14. 3.1.2 Law of independent assortment 3:1 is the basic Mendelian ratio, everything that follows depends on it Dihybrid cross (双因子杂交):A cross between the two pure-breeding parental lines yielded an F1 generation which consisted only of round yellow. Of 556 seeds, found 315 round yellow, 108 round green, 101 wrinkled yellow, 32 wrinkled green. This is close to a ratio of 9:3:3:1, which is referred to as thedihybrid ratio(双因子杂种比例)

  15. The predicted ratios of phenotypes and genotypes can be determined by using Punnett square (庞纳特方格). The 9:3:3:1 ratio is simply two 3:1 ratios combined, and shows that the alleles of the two genes behave independently of each other. Take the seeds shape gene first: in the F2 ratio of 3 round to 1wrinkled. Then the color: 3 yellow to 1 green. If the ratios for the two phenotypes are multiplied across the 9:3:3:1 ratio is obtained. So any number of genes can be calculated by the expression (3:1)n, n is the number of alleles involved. (108+32):(101+315)=1:3

  16. 3 Y_ : 1 yy 3 R_ : 1 rr 9Y_R_ : 3Y_rr : 3yyR_ : 1yyrr (9黄圆 : 3黄皱 : 3绿圆 : 1绿皱) F2 ratio of phenotypes (3:1) (3:1)…= (3:1)n ratio of genotypes (1:2:1) (1:2:1)…= (1:2:1)n n= no. of allels

  17. Four kinds of gamete with equal amount produced after testcross, and therefore four types of progeny with equal ratio of phenotypes and genotypes produced. 测交产生四种等比例配子，产生四种等比例表型及基因型后代。 p77 Table 4-4

  18. Essence of law of independent assortment: 位于非同源染色体上的等位基因，各自分离互不干扰，并且自由组合到不同的配子中去。

  19. 事实上杂交后代遗传比例并非与理论值完全吻合,为什么呢?事实上杂交后代遗传比例并非与理论值完全吻合,为什么呢? Observed and estimated value are not matched completely?

  20. 3.2 多对相对性状杂交的遗传分析

  21. Probabilities (概率):When we toss a coin we cannot say which way up it will land because each landing is a random or stochastic event. In science, our hypothesis predicts the probabilities of particular results, and allows us to test that hypothesis.

  22. The sums rule (加法法则):The sums rule (addition) is applied to combine probabilities of events which are mutually exclusive. A genetic example would be a cross where we expected to get 1/4 red :2/4 pink :1/4 white flowers.The probability of any particular flower being red or pink is 1/4+2/4=3/4 用于互斥事件总的概率的计算。 一个遗传学上的例子是：我们预测某次杂交后代的花色之比是1/4 红:2/4粉红:1/4白，则某花为红或粉红的概率是1/4+2/4=3/4)

  23. The products rule (乘法法则): The products rule is applied to independent events and events happening in a specific order. 用于独立的且以一定顺序发生的事件。 A genetic example is a marriage between two people heterozygous for a recessive genetic disease, say phenylketonuria (PKU) . If they have three children, what is the probability (chance) of all three being affected? The probability of a child being homozygous for the recessive allele is ¼, therefore: P = ¼ × ¼ × ¼ = (1/4)3 = 1/64

  24. 用分支法计算多对性状杂交后代基因型和表现型用分支法计算多对性状杂交后代基因型和表现型 出现的概率 (Mendelian Problems – problems 6 Mid-term examination paper etc.)

  25. 3.3 二项式分布公式：n为子代数 r为某一基因型的子代 p 为一基因型的概率 q为另一基因型的概率 n-r为另一基因型的子代数目 n ! Pr qn-r r ! ( n-r) ! 用以计算杂交后代某一特定基因型或表型出现的概率（适合小样本的一对质量性状）

  26. 3.4 Tests for goodness of fit: chi-square tests ( 适合度检验： X2卡方检验) Goodness of fit test calculate the probability of observed data being obtained by chance if that data is sampled from a hypothesized population or distribution.The usual test is to calculate chi-square. 适合度检验可计算从一假设的分布群体或区域中随机观察到某数据的概率。 常用检测方法 - 卡平方 X2检验法：以指数值表示实得数与理论数的差异

  27. The chi-square test (卡方检验): Chi-square=X2 =sum[(observed-expected)2 / expected] =Σ(Oi - Ei)2 / Ei i=1 p78 n

  28. e.g.在具有一个相对性状的红、黄番茄之间进行杂交，其F2出现了红番茄305株和黄番茄95株，试问：它们是否符合3:1 的理论预期数？ 解: 被考察类型的实得数(o)为：红番茄305 黄番茄 95 被考察类型的预期数(e)为： 红番茄 (305+95)×3/4=300 黄番茄 (305+95)×1/4=100 自由度(N)=2-1=1 （被考察的项目有两项，N=1） 将上述个数代入公式得： x2=∑( 305-300)2 / 300 + ( 95-100)2/100 = 0.27

  29. X2表

  30. Using chi-square probability tables: The probability of getting a result by chance is looked up in tables of X2. Looking along the row for the appropriate degrees of freedom, find the values greater and smaller than that calculated, and read the probability at the top. 从X2表中可查到随机得到某一结果的概率。 X2表中不同的横行代表不同的自由度，找到正确的自由度，再找到大于和小于计算值的数值，最后从表顶行读出对应的概率

  31. Level of significance (显著水平) The level of significance is the probability below which we assume the data observed does not fit expectations. This is conventionally 0.05. When there is a probability of less than 0.05, then to reject the null hypothesis as probably being false and say the data does not fit. 显著水平就是一个概率，当低于此概率时，我们便认为观察值不符合预期值。显著性水平在习惯上取0.05。如果某数据的概率低于0.05则我们认为数据与预期值不符而放弃（即差异显著）。

  32. P>0.05, 试验值与理论值相符, 差异不显著; 0.01<=P<=0.05, 试验值与理论值不符, 差异显著; P<=0.01, 试验值与理论值极不相符, 差异极显著. 查表得概率(P) 介于0.5~0.10 之间,说明这种偏差50%以上由机会造成.试验值与理论值相符, 差异不显著.

  33. 3.5 Interaction among genes 基因互作 3.5.1Interaction between alleles等位基因间互作： 1. Incomplete dominance (不完全显性)：the phenotype of F1 is between two parents. 杂交后代表型介于双亲中间的类型。 e.g. 红花紫茉莉 (CC) × 白花紫茉莉(cc) C对c不完全显性 Cc 粉红紫茉莉 自交 1 CC : 2Cc : 1cc 基因型比=表现型比=1：2：1

  34. 2. Co-dominance (共显性): The heterozygote displays both alleles. 一对等位基因在杂合体中都表现出性状的现象 e.g. In humans the MN blood group is controlled by a single gene. only two alleles exist, LM and LN. 人类中MN血型由单基因控制，只有LM and LN两个等位基因存在。 LMLM- - M型 LMLN- - MN型 LNLN- - N型 LM、LN 基因分别控制红细胞表面M抗原、N抗原的产生，在杂合体中红细胞表面既有M抗原又有N抗原，属共显性现象。

  35. 3. Mosaic dominance (镶嵌显性)：双亲性状在后代个体的某一部位上都得到了表现，形成镶嵌图式，这种显性遗传现象为镶嵌显性 瓢虫鞘翅色泽 AA x EE 前缘黑 后缘黑 AE 前后缘均黑 1AA 2AE 1EE 前缘黑 前后缘均黑 后缘黑

  36. 4. Lethal alleles (致死等位基因):Some alleles affect the viability of individuals that carry them. There are lethal dominant and recessive genes. In most cases, the homozygous recessive does not survive but the heterozygotes may have normal life-span for the recessive gene carrier. Individuals with lethal dominant gene die before grown-up without any progeny.(p93) 黄鼠 X 黄鼠 Aa Aa 1黑鼠 : 2黄鼠 : 1胚胎死亡 AA Aa aa

  37. 5.Multiple alleles (复等位基因):在群体中，同源染色体同一基因座上存在两个以上的决定同一性状的基因。 A well known example of multiple alleles is the human ABO blood group system. IA , IB , i复等位基因决定红细胞表面抗原的特异性 O----ii Phenotype A---- IA IA IAi B---- IB IB IBi 可用于亲子间的鉴定 AB--- IA IB

  38. 3.5.2 Interaction between non-alleles 非等位基因间的相互作用: • Complementary effect (互补型)： 1-3 叠加效应；4-6 上位效应 e.g. 用白花品种的豌豆(CCpp)与另一白花品种的豌豆(ccPP) 杂交 白花(CCpp) × 白花(ccPP) 紫花(PpCc) 自交 紫花(C_P_) 白花(C_pp) 白花(ccP_) 白花(ccpp) 9/16 3/16 + 3/16 + 1/16 C 花青素前体 P 花青素 紫色

  39. ·两显性基因同时存在，共同决定某一新性状 · 缺少其中任何一种显性基因，不表现新性状 ·F2表型比 9:（3：3：1）= 9：7

  40. 2. Additive effect (累加型): e.g. 用两个不同品种的圆形南瓜进行杂交 圆形(AAbb) × 圆形(aaBB) 扁形(AaBb) 自交 扁形(A_B_) 圆形(A_bb+aaB_) 长形(aabb) 9/16 3/16 +3/16 1/16

  41. ·两显性基因单独存在，表现相似性状 ·两显性基因共同存在，作用累加表现另一性状 ·无显性基因时表现另一性状 · F2表型比9：（3：3）：1 = 9:6:1

  42. 3. Duplicate effect (重叠型): e.g. 三角形蒴果的荠菜 与卵形蒴果的荠菜杂交 三角形 (T1T1T2T2) × 卵形 (t1t1t2t2) 三角形 (T1t1T2t2) 自交 三角形 (T1_T2_) 三角形 (T1_ t2t2) 三角形(t1t1 T2_) 卵形(t1t1t2t2) 9/16 + 3/16 + 3/16 1/16

  43. ·两种或两种以上显性基因互作，产生相同表型，·两种或两种以上显性基因互作，产生相同表型， 其中任何一种显性基因存在，则性状表现一致 ·无显性基因则表现另一性状 · F2表型比（9：3：3）：1 = 15:1

  44. 4. Dominant epistasis (显性上位)

  45. ·显性白皮基因(W)对显性黄皮基因(Y)具上位性作用·显性白皮基因(W)对显性黄皮基因(Y)具上位性作用 · W上位基因存在，阻碍Y基因作用，表型白色 ·缺少W基因，Y基因表现黄色 · W、Y均都不存在，yy绿色表型显现 · F2表型（9：3）：3：1 = 12:3:1

  46. 5. Recessive epistasis(隐性上位 ): 黑

  47. ·隐性纯合cc阻止色素形成表现白化 · 无cc则R、 r基因控制的黑色、淡黄色性状分别表现 · F2 表型比 9：3：（3：1）= 9:3:4.

  48. Epistasis (上位效应): Epistasis refers to situations where the expected ratio of phenotype is not observed due to some form of physiological interaction between the genes involved. This is usually seen as a distortion of the 9:3:3:1 ratio with a reduction in the number of different phenotype observed. ·两对独立遗传基因共同对某性状发生作用 · 一对基因对另一对基因表达具遮盖作用

  49. 6. 抑制型（Restrain type): 白羽温德鸡iicc X 白羽莱杭鸡IICC 白羽IiCc X 白羽IiCc I_C_ I_cc iicciiC_ F2 白羽 白羽 白羽有色羽 9/16 + 3/16 + 1/16 : 3/16 I - 抑制基因C - 有色基因

  50. · 两对互作基因，一对不表现性状但对另一对显性基因 具抑制效应 · F2表型比（9：3：1）：3 = 13:3

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