Please provide the expected f2 phenotypic proportions ie 34...

Please provide the expected F2 phenotypic proportions (i.e. 3/4). Show your work. Hint: It could be easier to perform the fork-line method when you have a tri-hybrid cross or try drafting individual-gene Punnett squares and use the probability rules. F1 females: w+ st+//st bw+//bw x F1 males: W st+//st bw+//bw W Phenotypic proportions (reduce the fractions to lowest terms): - Wild type: - Brown: - Scarlet: - White: Only brown females Brown or scarlet flies Scarlet flies or wild type males Wild type females or scarlet males Wild type or white flies Female brown or female scarlet flies

Transcript

00:03 So we're starting with an f1 generation where we know is completely heterogeneous, so a dominant leal recessive at each gene, and we're crossing them with an individual who is homozygous recessive at both genes.

00:14 So what we should do to figure out, you know, what would we expect for the f2 generation is to do a punnet square, a punnet square where we look at the possible gamutic combinations of each individual in the cross.

00:26 So for the first individual, the f1 individual, we would do possible community a combination of first v second first v and first c first v second c second v first c second v first c and second of both and we see that with this individual there are four possible comedic combinations and they're all different when we do the same thing for our test cross you know first v first c first v second c second v first c and second of both we end up with four possible convenient combinations that are all the same all of them are recessive and recessive c.

01:07 So while normally we would do a four by four punnet square because there's four possible combative combinations for each one, right? we know that if we just do this part of our punnet square, everything below that is going to be the same because all of the possible community combinations from this parent here are the same.

01:26 So we're actually going to just shorten our punit square to just be a one by four to save us a little bit of work and we will complete it as normal looking up and to the left to fill each probability box.

01:38 And once we've done this, we can see that there are four possible combinations for our f2 generation.

01:45 And all of them happen in equal measure.

01:47 There is a one in four or a 25 % chance of having both dominant traits.

01:51 There is a one in four or a 25 % chance of having the first dominant trait, not the second.

01:57 25 % chance of having the second, not the first, and a 25 % of having no dominant traits, just recessive.

02:04 So all of these numbers are our expected numbers.

02:09 So we can compare our expected to our observed numbers.

02:12 Now our observed numbers are just the numbers we are told happen in the f2 generation.

02:17 We are told that 287 have both dominant traits.

02:21 216 have the first, 204 have the second, and 293 have neither dominant traits, giving us a total of 1 ,000 offspring.

02:30 So those are our observed numbers, and then we can find the expected numbers.

02:35 By taking this 25 % and multiplying that by the total number of offspring 1 ,000 to see that i would expect 250 offspring to have each of these different phenotypes.

02:54 Because according to the law of independent assortment, they should happen in equal measure, each being 25%.

03:00 So now that we have all of those numbers, we can figure out how close or how far away we are from, the expected numbers to are observed by doing a chide square test.

03:11 Now a chide square test, we do a set of math for each of these four different phenotypes and then find the sum of those to see how statistically different are observed and our expected numbers are.

03:27 And the way that we calculate this is by taking the observed number minus the expected square that and then whatever number you get, you divide that by expected...

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