# Does one need to perform two Chi-square tests?

A question from my University's Question paper 2016:

A newly identified fruitfly mutant, cyclope eye –large and single in the middle of the head is hypothesised to be autosomal dominant. The experiment started with homozygous wild type females and heterozygous cyclopes males. The data from the F2 generation was 44 wild type males males, 60 wild type females, 110 cyclops males and 150 cyclops females. Does this support or reject the hypothesis$^1$? Use chi square to prove.

I think the following should be a proper approach at solving it:

H0= The proportion of cyclope eyed male and female is not that same,i.e. the gene responsible is not autosomal. $or$ There is an association between the sex of fruitfly and the phenotype.$^2$

H1= The proportion of cyclope eyed male and female is the same, the gene is present on autosome $or$ There is no association between the sex and the phenotype.$^2$

Chi square contingency test with Male and Female (in the columns) and Wildtype and Mutant phenotype (in rows).

Once the H0 is proved wrong (if it is proved wrong), we consider a next Chi-square test, goodness-of fit, where :

H0= The proportion of mutant phenotype is the same as the wildtype i.e. 1:1 ratio is followed

H1=The proportion and the mutant phenotype is greater i.e. the allele responsible is dominant.

When the final H1 is accepted the $^1$hypothesis is correct or else wrong.

$^2$ I'm not sure which one is better / proper.

Is it alright?

• The wording of the cited question is strange. Hypothesis testing does not "support" anything, and definitely does not "prove" anything. All you can do is reject the null hypothesis. This means the alternative hypothesis is consistent with the data, but does not prove it. Hypothesis testing is asymmetrical in this sense, and that is very important to grasp. Apr 3 '17 at 20:59

You should just have 1 chi-square test to test the hypothesis.

H0 in a chi-square test is your prediction based on a model; H1 is that the real outcome differs from what is expected, so you have that a little backwards. H0 should be the expected counts knowing the initial proportions and expecting autosomal dominant inheritance.

F1 generation will be half heterozygous males/females, half WT males/females, so F2 generation has 3 types of mating events:

Cc x Cc occurs 25% of the time, and 75% of offspring are affected

Cc x cc occurs 50% of the time, and 50% of offspring are affected

cc x cc occurs 25% of the time, and 0% of offspring are affected

Multiply these probabilities and you should have 43.75% expressing the phenotype; your prediction for the chi-square test is that you will see this same percentage regardless of sex.

From there, doing a chi-square test is simple and there are many many resources online for this (here is just one).

• Is it a two-tailed test? Because the observed data can be significantly more or less than the expected 43.75% i.e 7/16? Apr 12 '17 at 20:54
• @SanjuktaGhosh Chi-square is always one-tailed. It isn't a two-sided distribution like the normal, values that are off to the left are basically just "unexpectedly great fits" like if you flipped a coin 1000 times and got exactly 500. If you look at the math in the calculations they are based on squared differences, so 7/16 and 9/16 are treated exactly the same: if you predicted 8/16 in one case you have 1^2 in the other case you have (-1)^2. Apr 12 '17 at 21:03