1、Biostatistics 余思扬 3100100227 2012 年 6 月 - 1 - 浙 江 大 学 2010 2011 学年 秋冬 学期 生 物 统 计( 学) 与 实 验 设 计 课 程 期 末 考 试 试 卷 A 卷 Problem 1 (25 points): One experiment is conducted for studying the influence of plant condition on the content of Nicotine in leaf of tobacco. Two tobacco varieties and two different p
2、lant cultivations (fertilization, no fertilization) were arranged, 2 pieces of leaves were sampled for each variety. The obtained experiment data are shown in the following table. 叶片 1 叶片 2 叶片 1 叶片 2 施肥 施肥 不施肥 不施肥 品种 1 9 5 19 20 4 7 14 16 品种 2 15 17 15 17 12 11 12 11 (1) What is this experiment desi
3、gn? This is a typical Crossed Nested Design. (2) Write out the ANOVA model of this experiment for analysis; define each factor in the model. Error : Leaf : Variety : ion Fertilizat : Mean : 2 , 1 2 , 1 2 , 1 2 , 1 n k j i Y ijkn j ik ij j k j i ijkn (3) For the above ANOVA model, write out the formu
4、la of degree freedom and corresponding sum of square for factors in the model. Source Degree of Freedom (df) Sum of Squares (SS) A a-1 = 1 a i i Y Y bcr 1 2B b-1 = 1 b j j Y Y acr 1 2C(B) b(c-1) = 2 b j c k jk Y Y ar 1 1 2AB (a-1)(b-1) = 1 a i b j j i ij Y Y Y Y cr 1 1 2AC(B) (a-1)b(c-1) = 2 a i b j
5、 c k jk ij ijk Y Y Y Y r 1 1 1 2Error abc(r-1) = 8 a i b j c k r n ijk ijkn Y Y 1 1 1 1 2Note: A, B, C and Error are referred to factor , , and Error . And a, b, c and r are the number tested in this design of factor , , and Error . 浙江大学 2010 2011 学年秋 冬学期 生物统计( 学) 与实验 设计课程期末考试试卷 A 卷 - 2 - (4) Write
6、out the SAS program for analysis of this data. (5) According to the following output of SAS analysis, draw appropriate statistical conclusion. From the SAS result, we could find that the PrF of model is 0.18610.05, the model is not reach significant level, need to be adjusted. And from Type III SS w
7、e can find that only the factor that significant is fertilization. So the conclusion is in the model, fertilization is significant but the model itself is not reach the significant level, we need to change model. And also we can find in this model, all the interaction factors are missing, so the cha
8、nge of model we can add all the interactions as I write in Question (4) above, and we may get a better result. DATA Nicotine; INPUT Fertilization $ Variety $ Leaf $ Content ; DATALINES; A1 B1 C1 9 A1 B1 C2 5 A2 B1 C1 19 A2 B1 C2 20 A1 B1 C1 4 A1 B1 C2 7 A2 B1 C1 14 A2 B1 C2 16 A1 B2 C1 15 A1 B2 C2 1
9、7 A2 B2 C1 15 A2 B2 C2 17 A1 B2 C1 12 A1 B2 C2 11 A2 B2 C1 12 A2 B2 C2 11 ; PROC GLM; CLASS Fertilization Variety Leaf; MODEL Content = Fertilization Variety Leaf(Variety) Fertilization*Variety Fertilization*Leaf(Variety); RANDOM Leaf(Variety) /TEST; CONTRAST Leaf in B1 Leaf(Variety) 1 -1 0 0; CONTR
10、AST Leaf in B2 Leaf(Variety) 0 0 1 -1; MEANS Fertilization /TUKEY; MEANS Variety /TUKEY; LSMEANS Leaf(Variety) /STDERR PDIFF; LSMEANS Fertilization*Variety /STDERR PDIFF; LSMEANS Fertilization*Leaf(Variety) /STDERR PDIFF; RUN; 浙江大学 2010 2011 学年秋 冬学期 生物统计( 学) 与实验 设计课程期末考试试卷 A 卷 - 3 - Problem 2 (30 po
11、ints): One experiment of rice variety is conducted for studying the relationship of yield and planting density. The experiment has 2 varieties (A1, A2) and 3 planting densities 10 (B1), 20 (B2), 30 (B3). The observation data are shown in the following table. B1 B2 B3 B1 B2 B3 B1 B2 B3 A1 8 8 8 7 7 6
12、 6 5 6 A2 9 9 8 7 9 6 8 7 6 (1) Is it a factorial design or nested design? It is a typical Two-way Factorial Design. (2) Write out the ANOVA model for this experiment. Error : Density : Variety : Mean : 3 , 2 , 1 3 , 2 , 1 2 , 1 k j i Y ijk ij j i ijk (3) Can the interaction of variety and planting
13、density be analyzed ？Why? Yes. Because in this experiment we had three replicates, so we can conduct the analysis of interaction. (4) Assume the variety is fixed, planting density is random; write out the SAS program to analysis this data. (5) How to use the SAS to test the difference between B3 and
14、 the average of B1 and B2? We should adapt Linear Contrast command in this test: Add this term into PROC: (6) For this set of experiment data, how to construct a regression model of rice yield on planting density, so that it can be used to predict rice yield for other planting density. CONTRAST Line
15、ar Density 1 1 -2; DATA Rice; INPUT Variety $ Density $ Yield ; DATALINES; A1 B1 8 A1 B2 8 A1 B3 8 A1 B1 7 A1 B2 7 A1 B3 6 A1 B1 6 A1 B2 5 A1 B3 6 A2 B1 9 A2 B2 9 A2 B3 8 A2 B1 7 A2 B2 9 A2 B3 6 A2 B1 8 A2 B2 7 A2 B3 6 ; PROC GLM; CLASS Variety Density; MODEL Yield = Variety | Density; MEANS Variety
16、 /TUKEY; MEANS Density /TUKEY; LSMEANS Variety*Density /STDERR PDIFF; RUN; 浙江大学 2010 2011 学年秋 冬学期 生物统计( 学) 与实验 设计课程期末考试试卷 A 卷 - 4 - Problem 3 (15 points): The following data is about the height (X 1 ), weight (X 2 ) and body surface area (Y) of ten infants. Male Female X 1X 2Y X 1X 2Y 54 3 2446.2 54
17、 3 2117.3 50.5 2.25 1928.4 53 2.25 2200.2 51 2.5 2094.5 51.5 2.5 1906.2 56.5 3.5 2506.7 51 3 1850.3 52 3 2121 51 3 1632.5 (1) How to test the difference in body surface area between male and female? In this problem, we should employ the Analysis of Covariance (ANOCOVA). (2) Write out the SAS program
18、 for the above statistical testing. (3) For the following results of analysis, draw appropriate statistical conclusion. From the SAS result, we could find that the PrF of model is 0.00140.01, the model is very significant, and we can do further analysis. And from Type III SS we can find that the fac
19、tor that significant are Height and Sex. DATA Infants; INPUT X1 X2 Sex Y ; DATALINES; 54 3 S1 2446.2 54 3 S2 2117.3 50.5 2.25 S1 1928.4 53 2.25 S2 2200.2 51 2.5 S1 2094.5 51.5 2.5 S2 1906.2 56.5 3.5 S1 2506.7 51 3 S2 1850.3 52 3 S1 2121 51 3 S2 1632.5 ; PROC GLM; CLASS Sex; MODEL Y = Sex X1 X2; LSME
20、ANS Sex /STDERR PDIFF ADJUST=TUKEY ETYPE=3; RUN; 浙江大学 2010 2011 学年秋 冬学期 生物统计( 学) 与实验 设计课程期末考试试卷 A 卷 - 5 - So the conclusion is in the model, Height and Sex are significant. And also we can find in this model, the factor Weight is not reach significant level. And from all above the final conclusion i
21、s the body surface is related with height and sex. Problem 4 (30 points): The simulation data of QTL mapping was analyzed by a mixed model approach using software of QTLNetwork. The QTL parameters and their estimates (SE) are presented in Table 4-1 for QTL positions and effects, and in Table 4-2 for
22、 QTL-QTL interaction. Table 4-1. Parameters and their estimates of QTL positions and genetic effects (additive and additive by environment interaction) QTL Linkage Marker Interval Distance (cM) Estimate (cM) a SE a 1 ae SE ae 11 1 3 3.0 2.0 4.70 4.83(0.25) -4.47 -4.49(0.35) 2 1 7 8.0 6.0 0.00 0.00 0
23、.00 0.00 3 2 4 5.0 4.0 -4.10 -4.13(0.25) 0.00 0.00(0.02) 4 3 6 2.0 3.0 0.00 0.00 0.00 0.00 5 3 3 1.0 2.0 3.50 3.84(0.25) -3.16 -3.39(0.35) Table 4-2. Parameters and their estimates of epistasis effects for pair-wise QTLs i QTL j QTL aa SE aa 1 aae SE aae 11 2 3.2 3.30(0.26) 0 -0.54(0.34) 1 3 0 0.13(
24、0.26) -4.2 -4.37(0.37) 2 4 -3 -3.02(0.26) 3.16 3.04(0.36) (1) Based on Tables 4-1 and 4-2, how many QTLs were detected? There are totally 5 QTLs detected in this test, they are: QTL 1(hasa ,ae effects andaa effect with QTL 2 ,aae effect with QTL 3 ); QTL 2(hasaa effect with QTL 1and QTL 4 ,aae effec
25、t with QTL 4 ); QTL 3(hasa effects andaae effect with QTL 1 ); QTL 4(hasaa andaae effect with QTL 2 ); QTL 5(hasa andae effects). (2) If one QTL has onlya effects and another has botha andae effects, describe the difference between these two QTLs. The QTL has onlya effects means the QTL doesnt respo
26、nse to environment factors and its a stable trait in any environment, which is very useful in breeding; The QTL has botha andae effects means the QTL not only has its own effect but also an interaction with environment, which means this QTL may express differently in strength in different environmen
27、t, this kind of gene sometimes give extremely good yield and sometimes does the opposite. (3) As compared the estimates with the true values of QTL parameters, how do you think about the estimation results? The results are mostly accurate, and in all cases the estimates are very near the true value
28、with ranges including the true value. And the distance estimates are also accurate. 浙江大学 2010 2011 学年秋 冬学期 生物统计( 学) 与实验 设计课程期末考试试卷 A 卷 - 6 - The same data was also analyzed for marker effects by regression methods. The results are listed in Table 4-3 for simple regression analysis and in Table 4-4 f
29、or stepwise regression analysis. Table 4-3. Estimated coefficients of regression and correlation for singular marker to one environment phenotypes Marker Environment 1 Environment 2 b r p-value b r p-value M1_3 0.553 0.040 0.572 14.76 0.51 F Estimate Error Pr F Intercept 88.32 1.39 .0001 89.53 1.51
30、.0001 M1_3 16.70 1.43 .0001 16.17 1.44 .0001 M2_5 -8.12 1.43 .0001 -8.61 1.44 .0001 M3_4 -4.18 2.10 0.0481 M3_6 13.86 1.44 .0001 16.87 2.08 .0001 (4) Based on results from Table 4-3, draw conclusion on the number and location of QTLs, and also the effects of markers. The number of QTLs detected is 4
31、. They are located near M1_3 Chrome1 Marker3 (Environment 2), M2_3 Chrome2 Marker3 (Environment 1 2), M2_5 Chrome2 Marker5 (Environment 1 2), M3_6 Chrome3 Marker6 (Environment 2). Marker effects can be described asb (coefficient) andr (correlation coefficient) and p-value. (5) Do you think the stepw
32、ise regression is better than simple regression? Why do you think in this way? Yes. The stepwise regression is including all the variables together in one model while the simple regression is only including one variable one time. (6) Draw general conclusion on advantage and disadvantage of these three methods for statistical analysis: simple regression, multiple regression, and mixed model approach. See Textbook.
