A presumed dihybrid in Drosophila B/b ; F/f is testcrossed...

A presumed dihybrid in Drosophila B/b ; F/f is testcrossed to b/b ; f/f. (B = black body; b = brown body; F = forked bristles; f = unforked bristles.) The results were: Black, forked 230 Black, unforked 210 Brown, forked 240 Brown, unforked 250 Use the chi-square test to determine if these results fit the results expected from testcrossing the hypothesized dihybrid. Probability level (alpha) df 0.5 0.10 0.05 0.02 0.01 0.001 1 0.455 2.706 3.841 5.412 6.635 10.827 2 1.386 4.605 5.991 7.824 9.210 13.815 3 2.366 6.251 7.815 9.837 11.345 16.268 4 3.357 7.779 9.488 11.668 13.277 18.465 5 4.351 9.236 11.070 13.388 15.086 20.517

Transcript

00:01 In this question, we have the concept about the probability level of the alpha.

00:08 So based on this one, kai square test is very simple, or in terms, it's a very simple that gives an idea of whether the observed values are occurring because of chance or not.

00:22 So let us consider the cross between your heterozygose b and f times the homozygose recessive b and f.

00:33 So based on this one, we will have the f1 that would be black forked.

00:42 So we have black forked, which is one fourth.

00:49 Probability we have also black and fourth that is also one fourth we have brown fourth that is also one fourth and brown and fourth which is one fourth as as we can seen from the this cross all phenotypes have an equal probability of 25 % or 0 .25 which is the expected frequency.

01:23 The null hypothesis here take note that all phenotypes have equal probability, while the alternate hypothesis would be phenotypes do not have equal probabilities.

01:38 If your p value is less than or equal 0 .05, then the null hypothesis can be rejected as the result.

01:51 Will be statistically significant.

01:55 So let's see what is the p value that we are getting.

01:59 So the total numbers, that is our n, is equal to 230 plus 210 plus 240, and we have also 250, which is equal to 930.

02:12 So the phenotype that we have, we have black forked, and the number of samples in each category is 230 and the frequency that of or better call it as observed frequency so we have 230 divided by the divided by by the number of samples that is 930 so 930 so we have 0 .247 we have also black and forked and the black and fork the samples in each category is 210.

02:56 So 210 divided by 930, that would be 0 .226.

03:02 We have also brown forked.

03:06 Brown forked, we have 240, 240 divided by 930, that would be 0 .258.

03:15 And we have also brown and forked, which is equals to 250, and 250 divide 930 is equal to 0 .269.

03:30 Now let's find out the degree of freedom.

03:33 Degree of freedom, or that would be d .f is equals to the total number of sample categories, which is equal to 4 minus 1, that is 3.

03:54 So for the black fork, so we have black fork, the observed frequency is 0 .247...

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