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Chi-squared (χ2) test: How to decide if the data fits any of the Mendelian ratios?

1 Chi-squared (χ2) test can test out the ratios.

2 Let's test the following data to determine if it fits a 9:3:3:1 ratio:

3 Assume there is no difference between observed numbers & expected numbers. This assumption of no difference is called the null hypothesis.

Phenotype Predicted

ratio

Observed

(O)

Expected(E)

(O-E) 2 E

Round, yellow seeds

9 315 (9/16)(556) = 312.75

0.016

Round, green seeds

3 108 (3/16)(556) = 104.25

0.135

Wrinkled, yellow seeds

3 101 (3/16)(556) = 104.25

0.101

Wrinkled, green seeds

1 32 (1/16)(556) = 34.75

0.218

Total seeds 556 556 Σ 0.470

- 1 - TEA

= 0.470

4 Degrees of freedom (df) = n-1, where n is the no of classes.

5 No of classes (n) = 4, so df = 3 (df = n-1 = 4-1 = 3); 6 In biology, we use 0.05 (5%) probability level as our critical value.

7 Calculated X2 value (0.470) is found to be lower than value (7.82) given in x2 table (P0.05, df=3).

Therefore the deviation is not significant & the result fits 9:3:3:1 ratio.

(If χ2 value is less than 0.05 value, we accept the null hypothesis.)

8 Results would not due to chance & the results would be significant.Remember:

1 When Mendel self-crossed heterozygous tall plants, he obtained 787 tall and 277 dwarf plant. On the basis of his law of segregation, the expected ratio is 3:1. Test the result using chi square test.

Chi-square TableDegrees of

freedomProbability

0.95 0.90 0.5 0.1 0.051 0.01 0.02 0.46 2.71 3.842 0.10 0.21 1.39 4.61 5.993 0.35 0.58 2.37 6.25 7.824 0.71 1.06 3.36 7.78 9.495 1.15 1.61 4.35 9.24 11.07

A value less than the number in the table means the Null Hypothesis is accepted.

A value greater than that in the table means that the ‘results are significant at the X% level’ and the Null Hypothesis is rejected (i.e. some other explanation must be sought).

Phenotype Predicted ratio

Observed (O)

Expected (E) (O-E) 2 E

Total seeds Σ

X2 = Degree of freedom =

Conclusion:

2 In pea plants, the allele for smooth seed, W, is dominant over the allele for wrinkled seed, w, while the allele for yellow seed, G, is dominant over the allele for green seed, g. In one of Mendel’s dihybrid crosses, pea plants of genotypes wwGG were crossed with WWgg. The F1 obtained were selfed and yielded 315 F2 plants with phenotypes of smooth and yellow, 108 smooth and green, 101 wrinkled and yellow, and 32 wrinkled and green seeds. Using the X2 test at 5% level, determine whether

(STPM 2007)

(a) the result fits a 9:3:3:l ratio [7 m]

(b) the number of phenotypes of smooth seeds to wrinkled seeds fits a 3:1 ratio. [4 m]

(c) the number of phenotypes of yellow seeds to green seeds fits a 3:1 ratio.

[4 m]

Table of chi-square values at 5% level

Degree of freedom

5% level

1 3.8412 5.9913 7.8154 9.4885 11.0706 12.5927 14.0678 15.5079 16.91910 18.307

3 The inheritance of body colour in fruit flies was investigated. Two fruit flies with grey bodies were crossed.

Of the offspring, 152 had grey bodies and 48 had black bodies.

(a) Using suitable symbols, give the genotypes of the parents. Explain your answer.

[2]

Genotypes : ...............................................................................................................

................................

Explanation: ...............................................................................................................................................

(b) (i) Explain why a statistical test should be applied to the data obtained in this investigation. [2]

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(ii) The chi-squared (2) test was applied to the data obtained. The formula is given below.

Use the formula to determine the value of 2 for the results of this investigation. [3]

Degrees of freedom Probability value0.99 9.95 0.1 0.05 0.01 0.001

1 0.0002 0.0039 2.71 3.84 6.63 10.832 0.020 0.103 4.61 5.99 9.21 13.823 0.115 0.352 6.25 7.81 11.34 16.274 0.297 0.711 7.78 9.49 13.28 18.47

(iii)The null hypothesis in this investigation predicted that there would be no difference between the observed and expected values. Use the table to determine whether this hypothesis can be supported.

Explain how you arrived at your answer. [2]

4 When a heterozygous black rat is crossed with another heterozygous black rat, 43 black, 15 creams, 22 albino offspring are produced. Using Chi-square test the genetic hypothesis 9:3:4 is consistent with the data. - Null hypothesis - the genetic hypothesis 9:3:4 is consistent with the data. - Level of significance – 5% - Degrees of freedom = n-1 = 3-1 = 2 - Determining expected frequencies – ( E ) Black- 9 Total offspring- 80 Black = E = 80 X 9 / 16 = 45 Cream = 80 X 3 / 16 = 15 Albino = 80 X 4 / 16 = 20 Calculated value = 0.28 Table value = 5.96

Inference – The calculated value is less than the table value . Therefore the hypothsis is accepted . In other words the observed value is in consistent with the ratio 9:3:4.

5 When two heterozygous pea plants are crossed, 1600 plants are produced in the F 2 generation out of which 940 are yellow round, 260 are yellow wrinkled, 340 are green round and 60 are green wrinkled. By means of chi- square test whether these values are deviated from Mendel’s dihybrid ratio of 9:3:3:1 or by means of chi-square test prove whether it is a real independent assortment.

- Null hypothsis – There is real independent assortment . Levels of significance 5%. - Degrees of freedom – n-1 = 4-1=3 - Determining expected frequencies (E) Mendel’s dihybrid ratio = 9:3:3;1

Yellow round = 9 . Total = 1600 Expected Yellow round = (E) = 1600 X 9 / 16 = 900. Yellow wrinkled = 3 Expected Yellow wrinkled = 600 X 3 / 16 = 300 Green round = 3 Expected Green round = 1600 X 3 / 16 = 300 Green wrinkled = 1 Expected Green wrinkled = 16600 X 1 / 16 = 100 5 ) Calculated value = 27.43. Table value = 7.83 Inference : The calculated value ( 27.43 ) is greater than the table value. Therefore the hypothesis is rejected, there is no real independent assortment or the observed values are deviated from Mendel’s dihybrid ratio of 9:3:3:1.

6 Snapdragons are a type of garden plant. A pure breeding strain of a white flowering variety was obtained and crossed with a pure breeding red flowered strain. The two strains were crossed producing F1 plants all with pink flowers. The F1 plants were then interbred to produce F2 plants with the following flower colours: red 62 pink 131 white 67

The following hypothesis was proposed: Flower colour in snapdragons is controlled by a single gene with two codominant alleles.

(a) Complete the genetic diagram to explain this cross. Use the following symbols to represent the alleles:

R = red flowers W = white flowers

Parental phenotypes: Red flowers x White flowers

Parental genotypes [1m]

Gametes. [1m]

F1, genotypes [1m]

F1, phenotypes

[1m]

Gametes [1m]

F2, genotypes [1m]

F2, phenotypes [1m]

Expected F2 phenotype ratio[1m]

(b) A chi-squared test is carried out on the data to determine whether the hypothesis is supported or rejected. The statistic is calculated in the following way:

(i) Complete the following table. [2m]

(ii) Calculate the X2 value for the above data. Show your working. [1m]

(c) A suitable Null hypothesis would be that there is no significant difference between the observed and expected numbers. Biologists consider that if the probability is greater than 5% the deviation is statistically non significant.

(i) Using the figures from the table for 2 degrees of freedom, explain whether you would accept or reject the Null hypothesis.

[2m]

(ii) What does this suggest about the inheritance of flower colour in snapdragons? [1m]

GCE A-June 2010

Answers2 a) Refer to above example b)

Phenotype Expectedratio

Observed number

(o)

Expected number

(e)

Divergence(o-e)

Divergence2

(o-e)2(o-e) 2

e

Smooth 3 423 417.000 6.000 36.000 0.086Wrinkled 1 133 139.000 -6.000 36.000 0.259

Total 556 Σ = 0.345

The calculated X2 value (0.345) is found to be lower than the value (3.841) given in the X2 table (P0.05, degree freedom, df = l). Therefore the deviation is not significant and the result fits the 3:1 ratio.

c)

The calculated X2 value (0.009) is found to be lower than the value (3.84 1) given in the X2 table (P0.05, degree freedom, df =1). Therefore the deviation is not significant and the result fits the 3:1 ratio.

3 (a) Gg / suitable equivalent;Grey : black about 3: 1; 2[Note: Can be in table/ diagram]

(b) (i) To determine the probability; [Accept: Likelihood]Of the results being due to chance; [Accept: Coincidence] 2

(ii)

O E O-E (O-E)2 (O–E) 2 E

152 150] 2 4 0.027]

48 50] 2 4 0..08] method ignore calc'n errors);

[Note: Alternative showing of E and method]

50

])50–48(

150

)150–152[( 22

2 = 0. 107 / 0.11; 3

(iii) df = 1 and p = 0.05 / 95% level or critical value / described = 3.84;[Accept: Ringed/ indicated on table]

Accept hypothesis because 2 is less than (table / critical) value / thereis no significant difference / difference is due to chance; 2

[Note: Check carry forward of 2 value or critical value for interpretation or converse argument]

Phenotype Expectedratio

Observed number

(o)

Expected number

(e)

Divergence(o-e)

Divergence2

(o-e)2(o-e) 2

e

Yellow 3 416 417.000 -1.000 1.000 0.002Green 1 140 139.000 1.000 1.000 0.007Total 556 Σ = 0.009

6 (a) RR WW (allow: Cr Cw/key) 1 R W RW Pink R W RR RW RW WW

Red Pink White 1 2 1

(b) (i) E Column: 65 130 65

O-E2 column 9 1 4 (1 mark) O.21 (A. O.207 O.208. ) (not: 0.05)

(c) (i) Accept null hypothesis Less than critical value / 0.90 probability / 90% probability/ deviation from expected due to chance/< 95%/> 5% ref. chance or significance needed If calcn wrong e.g. 5.99 or above then reverse above i.e. ecf ∴ reject null hypothesis etc

(ii) Snapdragon flower colour is controlled by a single gene with 2 codominant alleles (not genes)