A. The Chi-square test (How well do my data fit my hypothesis?)B. Arithmetic tools in genetics2. Sum rule (additive probability of mutually exclusive events. Logical operator: EITHER / OR)4. Trick: If neither (1) or (2) seem to fit, try to look at it from another angle using ‘NOT’.5. Binomial expansion (multiple events in no particular order)P = (n over s) * a ^s * b^tn over s = n! / (s! * t!)s+t=n n = number of all eventss = number of events of type It = number of events of type IIa = probability of type Ib = probability of type IILECTURE 5CHI-SQUARE TESTPROBABILITIES IN GENETICSReading: Chapter 3.8, 3.7. Homework: Part I: Ch. 3 Q 18, 19, 23, 27, 28, 31, 32, 34. Study tips:- Put effort into solving the homework problems- Participate actively in the discussion class- Pre-read lecture material before lecture - Use the lecture notes as prompts to practice explaining the material- Active learning – Form a study group- Lecture notes contain ‘study guide’ questions. A. The Chi-square test (How well do my data fit my hypothesis?)Example: Mendel’s F2 generation of the dihybrid cross. a) Formulate a ‘Null hypothesis’ H0: “The genes for seed color and shape are inherited independently.”b) Formulate an expectation: The phenotypic ratios will be (ideally) 9:3:3:1. c) Do the experiment. Tabulate the observed values (O)d) Calculate the expected values (E) (ideal outcome)Observed (O) by Mendel Expected (E) (O-E) 2 /E315 round yellow 556 x 9/16 = 313 0.012108 round green 556 x 3/16 = 104 0.154101 wrinkled yellow 556 x 3/16 = 104 0.087 32 wrinkled green 556 x 1/16 = 35 0.257556 total X2 = 0.510; df = 4-1 = 3e) Calculate the X2-value as: X2 = Sum [ (O-E)2 / E ] f) Determine the ‘degrees of freedom’: df = number of data entries minus 1. g) Consult the X2 table (Table 3-1 in Griffiths) for your df and your choice of P-value (typically 0.05): 7.82. It states: “If this experiment was conducted 100 times, the X2 value would be below 7.82 approximately 95 times and above that only about 5 times.”Excerpt from X2-table: Probability...... 0.90 0.50 0.20 0.05 0.01 0.001df=3: X2..... 0.58 2.37 4.64 7.82 11.35 16.27f) Because the calculated X2 value (0.51) is much smaller than 7.82, Mendel’s data are clearly consistent with the null-hypothesis of independent segregation. Note: The P=0.05 (5%) cutoff is arbitrary. Q: Mathematically speaking, can the Chi-square test prove that a hypothesis is correct?A: No. For example, we have not specifically ruled out that the data might also match a different hypothesis. Statistical tests can only disprove false hypotheses (falsification). Q: Quality control: What will happen if you are routinely satisfied with P=0.05?B. Arithmetic tools in genetics1. Product rule (combined/serial probability of independent events. Logical operator: AND)Example: If a family has six children, what is the probability of first three girls and then three boys?2. Sum rule (additive probability of mutually exclusive events. Logical operator: EITHER / OR)Example: If a family has three children, what is the probability of only boys or only girls?3. The 'Not' rule (1 minus the probability that the event will occur). Example: If a family has three children, what is the probability of a mix of boys and girls?4. Trick: If neither (1) or (2) seem to fit, try to look at it from another angle using ‘NOT’. 5. Binomial expansion (multiple events in no particular order)Example: If a family has six children, which is the probability of three boys and three girls?P = (n over s) * a ^s * b^tn over s = n! / (s! * t!)s+t=nn = number of all eventss = number of events of type I t = number of events of type IIa = probability of type Ib = probability of type