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LSM 1102 (MOLCULAR GENETICS) (PLANT GENETICS) - PRACTICAL 6 – Mendelian Genetics

NOTES ON THE CHI-SQUARE TEST The purpose of the chi-square (χ2) test is to determine whether experimentally obtained data constitute a good fit to, or a satisfactory approximation of, a theoretical expected ratio. Thus the χ2 test enables one to determine whether it is reasonable to attribute deviations of a perfect fit to chance. Obviously, if deviations are small, they can be more reasonably attributed to chance than if they are large. The question we try to answer with the χ2 test is “How small the deviations be attributed to chance alone?” The fomula for the χ2 test is as follows:

χ2 = ∑ −E

EO 2)(

Where O = the observed number of individuals in a particular phenotype E = the expected number of individuals in that phenotype Σ = the summation of all possible values of (O-E)2/E for the various phenotypic categories EXAMPLE In a cross of tall tomato plants to dwarf ones, the F1 consisted entirely of tall plants. The F2 consisted of 102 tall and 44 dwarf plants. Do the F2 data fit a ratio of 3:1? To answer this question the χ2 was calculated; the calculations are summarized in Table 1 below. Set the 2 hypotheses: (Null hypotthesis) H0 = the observed data fit the 3:1 ratio (Alternate Hypothesis) H1 = the observed data do not fit the 3:1 ratio Table 1. Summary of the calculation of χ2 No. of plants Phenotype Exp. ratio O E (O - E ) (O - E)2 (O - E)2/E Tall (T_) ¾ 102 109.5 -7.5 56.25 0.5137 Dwarf (tt) ¼ 44 36.5 7.5 56.25 1.5411 Total 1 146 146.0 χ2 = 2.0548

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The calculated χ2 value is 2.0548, but what does this χ2 value mean? If the observed numbers (O) were exactly equal to the theoretcial expected numbers (E), the fit would be perfect and χ2 would be zero. Thus a small value of χ2 indicates that the observed and the expected ratios are in close agreement, whereas a large value indicates marked deviation from the expected ratio. Because chance deviations from the theoretical values are expected to occur, the question to be answered is “Are the observed deviations within the limits expected by chance?” Generally, statisticians have agreed on the arbitrary limits of odds of 1 chance in 20 (probability = 0.05) for drawing the line between acceptance and rejection of the hypothesis as a satisfactory explanation of the data tested. A χ2 value of 3.841 for a two-class ratio corresponds to a probability of 1 in 20, or 0.05. One would expect a χ2 value of 3.841 due to chance alone is only about 5% of similar trials if the hypothesis is true. When χ2 exceeds 3.841 for a two-class ratio, the probability that the deviations can be attibuted to chance alone is less than 1 in 20. The hypothesis of fit between the observed and expected ratios (Ho) is thus rejected. In the practice example, χ2 = 2.0548 which is considerably less than 3.841, therefore you can attribute the deviations to chance alone and accept the null hypothesis (Ho) that the data fit the 3:1 ratio. Where did the value of 3.831come from from? Statisticians have published extensive tables listing χ2 values (see Table 2). Notice that the first row are probability (P) values. Down the left handside (first column) are “degrees of freedom” values (df = 1,2,…,30). The number of degrees of freedom is generally one less than the number of phenotypic classes. In the example, with two classes in the ratio (3:1), you have one degree of freedom in interpreting χ2. Thus on the one-degree-of freedom row and under the 0.05 column you find the χ2 value of 3.841, which is the maximum value of χ2 that we are willing to accept and still attribute the deviations observed to chance alone. How to conclude? If the calculated χ2 value is smaller than the critical χ2 value, we accept the null hypothesis and conclude that the observed data give support to the null hypothesis. If the the calculated χ2 value is greater than the critical χ2 value, we reject the null hypothesis and conclude that the observed data do not give support to the null hypothesis. In the example, χ2 was observed to be 2.0548. Looking at the one-degree-of freedom row, you will see that this value of χ2 falls between the 0.05 (χ2 = 3.841) and the 0.20 (χ2 = 1.642) probability columns.This means that the probability that the deviations may be attributed to chance alone is between 5% to 20%; that is, if you were to do this same experiment 100 times, you would expect to observe deviations as large as you have obtained or larger in between 5 and 20 of the 100 trials due to chance alone. Thus the null hypothesis is accepted. The observed data of this experiment fit the 3:1 F2ratio. With this background in the meaning, calculation and interpretation of χ2 you can solve similar problems.

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Table 2. Table of χ2 values

Df

P=.99

.95

.80

.50

.20

.05

.01

1

.000157

.00393

.0642

.4550

1.642

3.841

6.639

2 .0201 .103 .446 1.386 3.219 5.991 9.210 3 .115 .352 1.005 2.366 4.642 7.815 11.345 4 .297 .711 1.649 3.357 5.989 9.488 13.277 5 .554 1.145 2.343 4.351 7.289 11.070 15.086 6 .872 1.635 3.070 5.348 8.558 12.592 16.812 7 1.239 2.167 3.822 6.346 9.803 14.067 18.475 8 1.646 2.733 4.594 7.344 11.030 15.507 20.090 9 2.088 3.325 5.380 8.343 12.242 16.919 21.666 10 2.558 3.940 6.179 9.342 13.442 18.307 23.209 15 5.229 7.261 10.307 14.339 19.311 24.996 30.578 20 8.260 10.851 14.578 19.337 25.038 31.410 37.566 25 11.524 14.611 18.940 24.337 30.675 37.652 44.314

Reference Gardner E. J., Mertens T. R. and Hammersmith R. L. (1985) Laboratory Investigations, Investigation 4 ; the Chi-square test. Pp 25-37. 8th Edition, MacMillan Publishing Company, New York.

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SORGHUM GENETICS (one-tray per group) Some of the traits obey Mendelian principles Monhybrid cross, R locus (3 red leaves : 1 green leaves F2ratio) Each group of students are provided with a petri dish containing seedlings of F2 generation of the cross between a plant with red leaves and green leaves. R locus determines oresence of red pigment (anthocyanin) in leaves R (dominant) = red leaves, r (recessive) = green leaves P Red (RR) × green (rr) F1 Red (Rr) F2 ¾ Red (R_) : ¼ green (rr) You will count the number of red and green seedlings and record your data in the table below: Table 1. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of plants Deviation Deviation2 Phenotype Exp. ratio O E (O - E ) (O - E)2 (O – E)2/E Red (R_) ¾ Green (rr) ¼ Total 1 χ2 = (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test (set probability at 0.05) (iii) Draw your conclusions regarding the inheritance of red pigment in sorghum. Do your observed data fit the 3 : 1 expected ratio? Which hypothesis do you accept?

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TOBACCO GENETICS 1. Monhybrid cross, A locus (3 green leaves : 1 albino leaves F2 ratio) Complete dominance

Two petri-dishes per group of tobacco seedlings Parents : Green leaf (AA) vs albino leaf (aa)

P Green (AA) × Yellow (aa) F1 Yellow-green (Gg) F2 ¾ green (A_) : ½ albino (aa) You will count the number of green and albino seedlings and record your data in the table below: Table 2. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of plants Deviation Deviation2 Phenotype Exp. ratio O E (O - E ) (O - E)2 (O – E)2/E Green (AA) ¾ Albino (aa) ¼ Total 1 χ2 = (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test (set probability at 0.05) (iii) Draw your conclusions regarding the inheritance of leaf color in tobacco. Do your observed data fit the 3 : 1 expected ratio? Which hypothesis do you accept? If the observed ratio did not fit the expected give some possible reasons for this deviation.

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TOBACCO GENETICS 2. Monhybrid cross, G locus (1 : 2 : 1 ratio F2 ratio) – Incomplete dominance Two petri-dishes per group Tobacco seedlings (F2 ratio ¼ green: ½ yellow-green: ¼ yellow ratio)

Parents : Green leaf (GG) vs yellow leaf (gg) GG = green leaves, Gg = yellow-green leaves, gg = yellow leaves

P Green (GG) × Yellow (gg) F1 Yellow-green (Gg) F2 ¼ green (GG) : ½ yellow green (Gg) : ¼ yellow (gg) You will count the number of green, yellow-green and yellow seedlings and record your data in the table below: Table 2. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of plants Deviation Deviation2 Phenotype Exp. ratio O E (O - E ) (O - E)2 (O – E)2/E Green (GG) ¼ Yellow-green (GG)

½

Green (rr) ¼ Total 1 χ2= (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test (set probability at 0.05) (iii) Draw your conclusions regarding the inheritance of leaf color in tobacco. Do your observed data fit the 1 : 2 : 1 expected ratio? Which hypothesis do you accept?

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TOMATO GENETICS 1. Tomato seedlings (F2 3 hairy : 1 hairless ratio)

Parents: Normal plants (hairy on stems and leaves, HH) vs Mutant (hairless, hh) Note: Seedlings should be 21 days or older. The character is best seen in good light against a black background. The hairy gene shows complete dominance.

P Hairy (HH) × hairless (hh) F1 Hairy (Hh) F2 ¾ Hairy (H-) : ¼ hairless (hh) You will count the number of hairy and hairless seedlings and record your data in the table below: Table 4. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of plants Phenotype Exp. ratio O E (O - E ) (O - E)2 (O – E)2/E Hairy (H-) ¾ Hairless (hh) ¼ Total 1 χ2 = (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test (set probability at 0.05) (iii) Draw your conclusions regarding the inheritance of hairy/hairless stem in tomato. Do your observed data fit the 3 : 1 expected ratio? Which hypothesis do you accept? If it does not fit, offer some explanations as to the possible reasons for the deviation.

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2. Tomato seedlings (F2 ¼ Green leaf : ½ Green-yellow leaf: ¼ yellow leaf ratio) Parents : Green leaf (GG) vs yellow leaf (gg) GG = green leaves, Gg = yellow-green leaves, gg = yellow leaves This locus shows incomplete dominance.

P Green (GG) × Yellow (gg) F1 Yellow-green (Gg) F2 ¼ green (GG) : ½ yellow green (Gg) : ¼ yellow (gg) You will count the number of green, yellow-green and yellow seedlings and record your data in the table below: Table 5. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of plants Phenotype Exp. ratio O E (O - E ) (O - E)2 (O – E)2/E Green (GG) ¼ Yellow-green (Gg)

½

Green (gg) ¼ Total 1 χ2 = (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test (set probability at 0.05) (iii) Draw your conclusions regarding the inheritance of leaf color in tomato. Do your observed data fit the 1 : 2 : 1 expected ratio? Which hypothesis do you accept?

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NOTES ON CORN GENETICS The substantial genetic variability in corn (maize) obeys Mendelian principles; both seedling and aleurone characters may be used. The patterns of mendelian inheritance can be demonstrated by studying segregation ratios in F2 generation from standard monohybrid and dihybrid crosses. Monohybrid crosses involving one gene factor (one pair of contrasting characters) give rise to the 3:1 F2 segregating ratio. Dyhybrid crosses involvce two factors (two pairs of segregating characters) give rise to 9:3:3:1 F2 segregating ratio (no linkage or epistasis). Epistasis gives rise to various modifications of the 9:3:3:1 ratio. I. SEGREGATING CORN SEEDLINGS Because maize has a relatively long life cycle, taking 3 or 4 months to complete, you will not be able to conduct actual experimental matings with corn. Your Instructor has provided you with trays of of corn seedlings germinated from F2 kernels purchased from a number of biological supply companies. You will count the number of seedlings in the different phenotypes and then to interpret the data in terms of Mendelian principles. Dihybrid cross, T and G loci (9:3:3:1 F2 ratio) T = Tall, t = dwarf G = Green, g = albino This is a two-factor cross between a homozygous Tall/Green parent (genotype TTGG) with a dwarf/albino parent (genotype ttgg). The cross results in an F1 generation which is heterozygous (TtGg). The F1 plants are selfed, resulting in a F2 generation. P Tall/Green (TTGG) × dwarf /albino (ttgg) F1 Tall/Green (TtGg) F2 9/16 Tall/Green : 3/16Tall/albino : 3/16 dwarf/Green : 3/16 dwarf/albino (T_G_) (T_gg) (ttG_) (ttgg) You will be given a tray of corn seedlings containing F2 tall/green, tall/albino, dwarf/green and dwarf/albino plants. You will count the number of the 4 types of seedlings and record your data in the table below: Table 4. Record of F2 data and summary of the calculation of χ2 from your group totals. No. of seedlings Phenotype Exp. ratio O (Obs) E (Exp) (O - E ) (O - E)2 (O - E)2/E Tall/Green 9/16 Tall/albino 3/16 Dwarf/Green 3/16 Dwarf/albino 1/16 Total 1 χ2 = (i) Set the Null (Ho) and Alternate Hypotheses (H1). (ii) Carry out the Chi-square test. (iii) Draw your conclusions regarding the inheritance of height and colour of corn plants. Do your observed data fit the 9:3:3:1 expected ratio? Which hypothesis do you accept?