A dihybrid cross is performed in drosophila one parent is...

A dihybrid cross is performed in Drosophila - one parent is heterozygous for both genes (B/b ; F/f) and is test-crossed with a homozygous recessive type (b/b ; f/f). (where B = black body; b = brown body; F = forked bristles; f = unforked bristles.) The results are: black, forked 230 black, unforked 210 brown, forked 240 brown, unforked 250 How many degrees of freedom would you use in this case? What would be the expected number for each phenotype? Calculate the chi-square value for this cross. Assuming p=0.05, what critical value would you use to compare the calculated chi-square value against? Do you reject or not reject the expected ratio in this case

Transcript

00:01 So the question talks about a dihybrid cross between a heterozygous drosophila with a recessive drosophila for both the genes.

00:19 And the genes are for black body and four bristles, black body, four bristles.

00:43 So this is b and this is.

00:47 Now we are given that black forked 230 black unfurked 210 brown ford 240 as well as brown unfurked 250 so the very first question the very first part of the question asked for the degrees of freedom so since we have four phenotypes therefore the degree of freedom is 4 minus 1 is equal to 3 this is the degree of freedom now the next question next part of the question states what would be the expected number of each phenotype now in order to find out the expected number of each phenotype we'll have to quickly perform a dihabit cross between capital b small b capital f small f crossed with small b small b small b small f so the gametes here will be capital b capital f capital b small f small b capital f small b capital f just one gamete capital capital small b small f now let's quickly make a pounette square this is our square now we'll quickly write down the different gametes and then we'll perform the cross capital b small b small f small f small b capital f small b capital f small f small f small b small small b small f so this gives us a phenotype of one is to one as to one is to one is to one therefore, here this is the observed and the expected phenotype is 1 is to 1 is to 1.

03:23 Now in order to calculate the chi squared value, we'll again quickly make a table for it.

03:32 So this is the table.

03:34 This is black forked, black unfurged, brown forked and brown unforged.

03:40 Now here we'll write the observed.

03:43 Value and here we'll write the expected so the observed values are 230 210 240 250 250 230 210 240 250 let's make another column here wherein we'll write the total values so this is a total of 5 so this is a total of 930 now here the affirm here the is one is to one is to one so what we'll have to do is one by four into 930 one by four into 930 1 by 4 into 930 once again 1 by 4 into 930 which is equal to 232 .5 32 .5 and 32 .5 and 32 .5...

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