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**In this article we will discuss about the chi-square method for testing the goodness of seed samples. **

Application of statistical methods in biological experiments is called Biometry. A very important use of biometry in Genetics is for testing whether an observed ratio may be taken as fit for an expected ratio.

For examples, the expected F_{2} phenotypic monohybrid ratio is 3: 1. But the actual observed ratio as obtained by Mendel himself for the Tall X Dwarf monohybrid cross in pea plant was 2.84: 1 and for the coloured and white seed coat cross was 3.15: 1.

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Some deviations between observed and expected ratios is always expected while experimenting with living organisms due to accidents and chance.

The problem is — how much deviation can be allowed due to these reasons, or, in other words, whether 2.84: 1 and 3.15: 1 can be taken as equivalent to 3: 1? This can be found out by a process called Testing Goodness of Fit arid there are two methods of finding it out: (a) The Standard Error Method and (b) The Chi-square (X^{2}) Method.

In order to understand the application of the Chi-square method for testing goodness of fit of a sample to a particular ratio, one must have a clear conception about the Law of Probability. The Law of Probability enables us to forecast the frequency of occurrence of a particular event when 2 or more alternative results are possible.

As for example, if a coin is tossed the result is either ‘head’ or ‘tail’, both the events having an equal chance of occurrence. Thus, if the coin is tossed 10,000 times, there will be near about 5,000 heads and 5,000 tails, i.e. the two events ‘head’ and ‘tail’ will occur in 1: 1 ratio.

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Accuracy in Biometry depends upon the size of the sample, which should be sufficiently large. Thus, if the coin is tossed 5, 10 or 50 times the result may not be exactly in 1: 1 ratio.

If one takes two coins and tosses them simultaneously for a large number of times, the result will be — two Heads: one Head and one Tail: Two Tails in the ratio 1: 2: 1. Probability in these cases is given by the coefficients of expansion of the binomial formula (a + b)^{n}, where ‘n’ represents the number of objects.

So, if four coins are tossed simultaneously, the probabilities can be calculated as — (a + b)^{n}, i.e. (a + b)^{4}, i.e., a^{4} + 4a^{3}b + 6a^{2}b^{2} + 4ab^{3} + b^{4}. Therefore, the proportion of 4 Heads : 3 Heads and 1 Tail: 2 Heads and 2 Tails : 1 Head and 3 Tails: 4 Tails will be — 1: 4: 6: 4: 1.

**The Chi-square (X ^{2}) Method**

**:**

This method, named after the Greek letter X (Chi), is more easily — and, therefore, more frequently — applicable to a number of biological problems.

**The formula used in this method is:**

X^{2} = ∑d^{2}/e or ∑ (O-e)^{ 2}/e which is a summation of the square of deviation in each class (d^{2}), i.e., (O – e)^{2} where O = the observed number and e = the expected number in each class and is divided by the expected number (e) for each class.

For example, let us consider Mendel’s Tall X Dwarf cross with pea plant. The observed numbers of plants in F_{2} were 787 tall and 277 dwarf. For an exact 3: 1 ratio, the expected number should have been 798 tall and 266 dwarf. Now, considering the two classes tall and dwarf

X^{2} = (787 -789)^{2}/798 + (277-266)^{2}/266

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= 121/798 + 121/266

= 0.61

To find out whether this X^{2} value fits a 3: 1 ratio, the attached Chi-square table is to be consulted (Table in Appendix).

Degree of freedom (d.f) means the number of alternative classes which will be one less than the number of actual class (i.e., n – 1). So, for a 3: 1 ratio the d.f. is 1. The right side column of the table shows the probability of getting a high value of x^{2} because of chance alone if the sample is from a population conforming to this particular ratio.

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Consulting this table it is found that the x^{2} value of 0.61 has P between 0.** **30 and 0.50, i.e., the probability of getting such a x^{2} value lies between 30% and 50%. So the result obtained is a very good fit for 3: 1 ratio. A ‘P’ value above 0.05 (about 5% level, i.e., 5 times in 100 trials) is considered a good fit. Higher the value of P, i.e. smaller the Chi-square value, better is the fit.

It should always be remembered that in statistical analysis the sample taken must be reasonably large. Otherwise the conclusion may become erroneous. The ratio may be found to be good fit for 2 or 3 different genetic factors. In class-work, if such a problem arises, all the genetic factors should be discussed.

If the P value lies between 5% and 70% or even 80% it should be considered a good fit. If the P value falls below 5% level then it is a bad fit. And if the P value lies above the 80% limit then it indicates artificiality or some other defect in the experiment.

For class-work, usually artificial seed samples composed of different pulses and beans are supplied. These samples are often so small that they appear to be good fits for more than one genetical ratio.

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**Some genetically ratios are discussed:**

**1. Seed Sample No. 1****:**

**The given sample (pea seeds) shows the following classes of phenotypes: **

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The observed number of seeds appear to be very close to 3: 1 ratio.

The observed no. of seeds = 87 + 33 = 120. This no. is divisible by 4, then the expected numbers may be 90 and 30 which will give the genetic ratio 3: 1.

The no. of classes in this experiment is 2. So, degree of freedom (d.f.) is (n – 1) = 1. For d.f. 1, a Chi-square value of 0.40 has P (probability) lying between 0.50 and 0.70.

Hence, the probability of getting such a Chi-square value lies between 50% and 70%. So the given seed sample is a very good fit for 3 : 1 ratio, which is the F_{2} phenotypic ratio for a monohybrid cross, i.e. a cross between two homozygous parents involving one pair of alleles.

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**2. Seed Sample No. 2****:**

**The given sample shows the following classes of phenotypes:**

The observed number of seeds appear to be very close to 9: 7 ratio. The observed number of seeds = 67 + 45 = 112. This number is divisible by 16 and the quotient is 7. For 9: 7 ratio, the expected numbers will be 7 × 9 = 63, 7 × 7 = 49.

The number of classes in this experiment is 2. So d.f. is n – 1. For, d.f. 1, a Chi-square volume of 0.58 has P (Probability) lying between 0.30 and 0.50. Hence, the probability of getting such a Chi-square value lies between 30% and 50%.

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So the given seed sample is a good fit for 9: 7 ratio, which is the F_{2} phenotypic ratio for complementary factor. When two pairs of alleles have no individual effect, but together they complement each other to produce a phenotypic effect, it is called a complementary factor.

**3. Seed Sample No. 3****:**

**The given sample shows the following classes of phenotypes:**

The observed number of seeds appear to be very close to 12: 3: 1 ratio. The observed number of seeds is 92 + 27 + 9 = 128. This number is divisible by 16 and the quotient is 8. For 12: 3: 1 ratio, the expected number will be 8 × 12 = 96, 8 x 3 = 24 and 8 × 1 = 8.

The number of classes in this experiment is 3. So, d.f. is n – 1 = 3 – 1 = 2. For d.f. 2, achi-square value of 0.666 has P (Probability) lying between 0.70 and 0.80.

Hence, the probability of getting such a chi-square value lies between 70% and 80%. So the given seed sample is a good fit for 12: 3: 1 ratio, which is F_{2} phenotypic ratio for Dominant Epistasis. When the dominant allele belonging to one allelic pair is also dominant over both the alleles of another allelic pair it is called a case of dominant epistasis.

**In this way artificial seed samples may be made and tested for the different genetic ratios, such as:**

1. 3: 1 — Monohybrid F_{2} phenotypic ratio.

2. 1: 1 — Monohybrid test cross F_{2} phenotypic ratio.

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3. 9: 7 — F_{2} phenotypic ratio for complementary factors.

4. 13: 3 — F_{2} phenotypic ratio for inhibitory factor.

5. 15: 1 — F_{2} phenotypic ratio for duplicate genes (multiplication of factors).

6. 63: 1 — F_{2} phenotypic ratio for triplicate genes (multiplication of factors).

7. 2: 1 — F_{2} phenotypic ratio for dominant lethal factor.

8. 1: 2: 1 — F_{2} phenotypic ratio for incomplete dominance or blending inheritance.

9. 27: 37 — F_{2} phenotypic ratio for complementary and supplementary factors.

10. 9: 3: 4 — F_{2} phenotypic ratio for recessive epistasis or supplementary factor.

11. 12: 3: 1 — F_{2} phenotypic ratio for dominant epistasis.

12. 9: 6: 1 — F_{2} phenotypic ratio for complimentary factors where the genes also have some expression singly.

13. 9: 3: 3: 1 — F_{2} phenotypic ratio for di-hybrid cross.

14. 1: 1: 1: 1 — F_{2} phenotypic ratio for di-hybrid test cross.

**N.B.:**

The most important point in solving these problems is to correctly guess the appropriate genetic ratio after counting the total number of seeds. One clue to do this is to divide the total number of seeds by 4 and 16.

If it is divisible by 4 only then it usually involves one pair of genes; otherwise, it is likely to be a case involving 2 pairs of alleles. With actual samples, fractional number can also be obtained — then this clue will not work. However, for class-work, for the sake of simplicity, such fractional numbers are ordinarily avoided.