Question

You are studying two new traits that control tomato color (red v. albino) and shape (round v. long). You cross a red and round tomato plant with one displaying albino and long. The resulting F1 are all red and round. The F2 have the following phenotypes (8 points)

834 round, red

250 round, albino

272 long, red

80    long, albino

1. Which of these characteristics are dominant?
2. What are the most likely genotypes of the F2?
3. What is the mode of inheritance? (Use words like, autosomal, sex-linked, dominant, epistatic, incomplete dominant, etc.)

d. Test your hypothesis using the Chi-square statistic.

Answer 1

Ans-a)Red colour and round shape are dominant trait

Ansb)self cross btw RrSs

Where R for red, r for albino and S for round shape and s for long shape

Phenotype ratio of f2 From cross is

round, red: round, albino: long, red: long, albino

9   : 3     : 3   : 1

Ansc) autosomal dominant type of inheritance as phenptypic ratio of f2 is 9:3:3:1.

Ansd)

The trait for red colour(R) is dominant over alinios (r) and Round shape (S) is dominant to long (s)  

A dihybrid cross between two heterozygous pea plants is performed (RrSs xRrSs)

The following phenotypic frequencies are observed:

834 round, red

250 round, albino

272 long, red

80    long, albino

The expected frequenciesof F2 in dihybrid cross is in9:3:3:1 ratio

Step 1:  Identify hypotheses

A chi-squared test Used to distinguish between two distinct possibilities and hence requires two contrasting hypotheses:

Null hypothesis (H₀): There is no significant difference between observed and expected frequencies (i.e. genes are unlinked)

Alternative hypothesis (H₁): There is a significant difference between observed and expected frequencies (i.e. genes are linked)

Step 2: Construct a table of frequencies and calculate chi square value

A table must be constructed that compares observed and expected frequencies for each possible phenotype

The formula used to calculate a statistical value for the chi-squared test is as follows:

Where: ∑ = Sum ; O = Observed frequency ; E = Expected frequency.

Table:

• Phenotype	observed	Expected	O-E	(O-E)2	(O-E)2/E
  Red round	834	(9/16)x1436=807.75	26. 25	689	689/807.75=0.85
  Round albino	250	(3/16)x1436=269.25	-19.5	380.25	380.25/269.25=1.41
  Long red	272	(3/16)x1436=269.25	2.75	7.56	7.56/269.25=0.03
  long albino	80	(1/16)x1436=89.75	-9.75	95.1	95.1/89.75=1.06
  				(O-E)2/E=0.85+1.41+0.03+1.06=3.35

Based on these results the statistical value calculated by the chi-squared test is as follows:

X² =   3.35

Step 3: Determine the degree of freedom (df)

In order to determine if the chi-squared value is statistically significant a degree of freedom must first be identified

The degree of freedom is calculated from the table of frequencies according to the following formula:

df = (m – 1) (n – 1)

Where: m = number of rows ; n = number of columns

For all dihybrid crosses, the degree of freedom should be: (number of phenotypes – 1)=4-1=3In this Example, the degree of freedom is 3

​​​​​​​Step 4  Identify the p value

The final step is to apply the value generated to a chi-squared distribution table to determine if results are statistically significant

• A value is considered significant if there is less than a 5% probability (p < 0.05) the results are due to chance

• When df = 3, a value of greater than 7.815 is required for results to be considered statistically significant (p < 0.05)

• A value of 3.35 is less than table value at α=0.05 ie7.815 meaning there is a 10 – 25% probability results are caused by chance
• Hence, the difference between observed and expected frequencies are not statistically significant

• Intereference:As results are not statistically significant, the alternative hypothesis is rejected and the null hypothesis accepted:

• Null hypothesis (H₀): There is no significant difference between observed and expected frequencies (genes are unlinked) and independent assortment of gene occur.