The paper “Linkage Studies of the Tomato” (Transactions of the Canadian Institute, 1931) reported the following data on phenotypes resulting from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes. We wish to investigate whether the following frequencies are consistent with genetic laws, which state that the phenotypes should occur in the ratio 9:3:3:1.

The data were produced in such a way that the Random and Independent conditions are met. Carry out a chi-square goodness-of-fit test using these data. What do you conclude?

Short Answer

Expert verified

There is sufficient evidence to conclude that phenotypes are occurring in ratio of9:3:3:1

Step by step solution

01

Given information

Given the nature of the query, The following data on phenotypes arising from crossing tall cut-leaf tomatoes with dwarf potato-leaf tomatoes was published in the publication "Linkage Studies of the Tomato" (Transactions of the Canadian Institute, 1931). We want to see if the following frequencies are in accordance with genetic principles, which specify that phenotypes should occur in a 9:3:3:1ratio.

Using these data, we must do a chi-square goodness-of-fit test.

02

Explanation

PhenotypeFrequency
Tall cut926
Tall potato288
Dwarf cut293
Dwarf potato104

Ratio is 9:3:3:1.

The test statistic will be calculated using the formula

χ2=(O-E)2E

The proportions are:

localid="1650620358889" p1=99+3+3+1=0.5625

p2=316

p3=316

p4=516

The null and alternative hypotheses is given below:

H0:p1

H0:p2

H0:p3

H0:p4

H0:p5

Ha:At least one of piis different

03

Calculation for test statistic 

The test statistic is calculated as follows:


Observed value

Expected value

(O-E)

(O-E)2
(O-E)2E
926
906.1875
19.8125
392.5352
0.4332
288
302.0625
14.0625
197.7539
0.6547
293
302.0625
-9.0625
82.1289
0.2719
104
100.6875
3.3125
10.9727
0.109




(O-E)2E2=1.4687

The test statistic is:

localid="1650620512430" χ2=(O-E)2E=1.4687

The degree of freedom is calculated as:

Degreeoffreedom=Numberofcategories-1=2-1=1

The p-value using chi-square table at 1 degree of freedom is 0.6895

The p-value is above significance level. The null hypothesis does not get rejected. Thus, there is sufficient evidence to conclude that phenotypes are occurring in ratio of 9:3:3:1

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