Calculating the Chi-Square Statistic The chi-square statistic, 2 , is determined

Question

Calculating the Chi-Square Statistic

The chi-square statistic, 2, is determined using the following equation:

2 = [(O – E)2/E]

where O is the observed number of individuals in a defined class, and E is the expected number of individuals in a defined class based on the hypothesis. The symbol indicates that you should sum the results for each defined class.

For example, suppose you hypothesize that the gene for seed color follows Mendelian inheritance. After you perform a cross between two monohybrids, you expect to get 75 yellow seeds and 25 green seeds in a population of 100, following the Mendelian 3:1 ratio (F2 in Figure 4). However, you observe 73 yellow seeds and 27 green seeds. To determine the 2, youwould use the following steps.

First defined class: yellow seeds

[(O – E)2/E] = [(73 – 75)2/75] = (–2)2/75 = 4/75

Second defined class: green seeds

[(O – E)2/E] = [(27 – 25)2/75] = (2)2/75 = 4/25

2 = (4/75 + 4/25) = 16/75 = 0.21

Does the Data Support the Hypothesis?

To determine the probability that the calculated 2 would have occurred by chance given that the hypothesis is true, you first need to determine the degrees of freedom by subtracting one from the total number of defined classes as follows:

df = c – 1

where df is the degrees of freedom and c is the number of defined classes.

For this example, two defined classes, the phenotypes, are being tested, so the degrees of freedom is equal to 1.

Consult the table of critical values of 2 (Figure 6) below to determine the probability of the calculated 2 occurring by chance if the hypothesis is true. Locate the 2 closest to your calculated 2 in the row of the degrees of freedom for the test, which is the first row for the example. Most likely, the calculated value will lie between two values in the row, in this case, 0.15 and 0.46. Move up the columns of these two values and read the probability numbers at the top of the table. This is the p value, which in this case, lies between 0.70 and 0.50.

This p value is greater than the generally accepted 0.05 cutoff meaning the data supports the hypothesis that the gene follows Mendelian inheritance. This does not prove the hypothesis; it simply means that this particular data set is likely to occur if the hypothesis is true. If the p value had been below the 0.05 cutoff, the hypothesis would have been rejected. Again, this would not have disproved the hypothesis; it simply means the data are unlikely to occur as such if the hypothesis is true.

QUESTION:

In a test population of 1000 people, 210 display hemophilia. You hypothesize that this trait follows Mendelian genetics with a 3:1 ratio of non- hemophilia to hemophilia. Use the 2 test to determine whether the data support the hypothesis or not. Show your steps and explain your reasoning.

Probabilit Degrees of Freedonm 1 2 0.90 0.80 0.70 0.30 0.20 0.05 .01 0.001 0.50 0.46 1.39 2.41 3.22 4.605.99 9.21 13.82 2.37 0.95 0.10 0.02 0.21 0.58 1.06 1.61 0.004 0.06 0.15 1.07 1.64 2.71 3.84 6.64 10.83 0.10 0.35 0.71 1.14 1.63 2.20 3.073.83 5.35 7.23 8.56 10.64 12.59 16.81 22.46 2.17 2.73 3.494.59 5.53 7.349.52 11.03 13.36 15.51 20.09 26.12 3.32 3.94 1.42 3.66 6.25 4.64 1.65 2.20 3.36 4.88 5.99 7.789.49 13.28 18.47 7.29 1.01 7.8211.34 16.27 4 2.34 3.00 4.35 6.06 9.24 11.0715.09 20.52 6 7 8 9 10 2.83 3.82 4.67 6.35 8.38 9.80 12.02 14.07 18.48 24.32 4.17 4.86 5.38 6.18 6.39 7.27 8.34 10.66 12.24 14.6816.92 21.67 27.88 9.34 11.78 13.44 15.99 18.31 23.21 29.59 Non-significant Significant

Explanation / Answer

In case of Mendel's monohybrid cross the F2 generation has 3:1 ratio where 3 part is the population has at least one dominant allele and 1 part of the population have two recessive alleles.

Hemophilia is an X-linked disease. The male has one X- chromosome and female have two X- chromosome. So in case of a male if X- chromosome is affected then he will show the phenotype but in case of female if both the chromosomes are affected then only she will show the phenotype. Since no information is given about the affected male and female population so we will take that take that all the population belongs to a female.

So Chi-square value is 8.5

Degree of freedom = number of alleles- 1

= 2-1 = 1

When we look at the table we will find the value 8.5 is between the p-value 0.01 and 0.001.

Since the cutoff p-value is 0.05 which is higher than the 0.01 and 0.001 so we need to reject the hypothesis as our data doesn't fit the Mendelian ratio of 3:1

O E O-E (O-E)2 (O-E)2/E Normal 790 750 40 1600 2.133333 Hemophilia 210 250 -40 1600 6.4 Total 1000 0 8.533333

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