Hypothesis the brown and red alleles of the eye color gene...

Hypothesis: the brown and red alleles of the eye color gene in f1 flies segregate into different gametes and fertilization is random. 1. State the phenotypic ratio predicted in the F2 generation of the eye color cross if the above hypothesis is true. 2. Phenotype Observed Expected red eyes 12 12 brown eyes 4 4 total 16 16 x^2 = 0 df = 1 p=1 3 Was the mismatch, if any, between your observation and expected values the result of chance? 4. was the hypothesis of segregation and random fertilization supported or not supported by your data?

Transcript

00:01 This question is all about chai square, and so we are going to walk through how to calculate the chai square, which is x squared here.

00:09 So in order to do this, we need to know the observed phenotypes for this genetic question and the expected phenotypes or genotypes.

00:22 So here we are looking at four different phenotypes.

00:27 There's both dominant traits.

00:30 There's just the first dominant trait, the other one's recessive, or vice.

00:34 Vice versa.

00:37 And then there's totally recessive.

00:46 So we are already told the observed number for each of these.

00:53 So observed with both dominant traits, there's 257, having just the first one, there's 88, having only the second one, 79, and then individuals having no dominant traits, just recessive, 31.

01:11 So if we're following this, equation here we have the observed so we need to subtract the expected number for each phenotype.

01:26 Now in order to find expected i'm going to assume that we're looking at a very classic heterozygous cross here because we're not told with this question or at least i don't have that information.

01:48 So i'm expecting that we have individuals like this that are crossed with each other and we end up getting this because that's what normally happens this would be your f1 generation and then what we have over here is the f2 generation so when you have a heterozygous cross with two genes you would expect this ratio nine out of 16 to have both dominant traits and then you would expect three out of 16 to have just one dominant trait, three out of 16 to have just the other dominant trait, and one out of 16 to have no dominant traits.

02:41 So out of this total population here where there are 455 individuals, i'm going to multiply that by each fraction to figure out what i would expect how many out of 455 to have each phenotype.

03:07 So, for example, this first one, you know, i expect nine sixteenths of the population to have both dominant traits.

03:16 So in my calculator, i do nine divided by 16 and then multiply your new number by 455.

03:23 When you do that, you should get 256.

03:27 So this is my expected number for that first phenotype.

03:35 Next up, where i have just one dominant trait, not the other, three divided by 16, then multiply your new number by 455, and i get 85 .3.

03:50 So that is my expected number here.

03:54 And luckily, since the math here and here is the same, i can write it in for this next one as well.

04:03 And then lastly, we have the phenotype of having no dominant traits, all recessive, one divided by 60, then multiply that by 455 and you end up with 28 .4.

04:17 So that is going to be my expected for this last one.

04:23 Now what happens in this particular equation is you're going to, in your calculator, do observed minus expected, and then whatever number you get, you're going to square it.

04:36 And then take your new number and divide it by expected.

04:40 So you get that expected number twice.

04:45 So when i do that with this first phenotype, the two dominant traits, 257 minus 256.

04:54 You get a really small number, square it.

04:57 You get just one, and so you square that, and it's still one.

05:01 And then divide one by 256.

05:04 You end up with a very small number.

05:07 0 .0039...