In fruit flies normal wings are dominant to bent wings, and...

In fruit flies normal wings are dominant to bent wings, and at a different gene, normal bristles are dominant to shaven (very short looking) bristles. A cross was made between a homozygous normal wing shaven bristled fly with a homozygous bent wing, normal bristle fly. The F1 that resulted was crossed with a fly that was bent winged, shaven bristles. The following results were obtained: F2: 198 Normal wings, shaven bristles 4 Normal wings, normal bristles 2 Bent wings, shaven bristles 196 Bent wings, normal bristles Total: 400 1. If these genes were not linked, what were the expected values of each of those four phenotypes in the F2 (1 point)? 2. What is the evidence that these genes are linked? Explain using the chi-squared statistic (3 points). Chi squared table of cut-off values Degrees of freedom | Cut-off value at p=0.05 3.841

Transcript

00:01 So you have a fruit fly with two different genes involved.

00:06 One is we call b, the other is called s.

00:13 So the b stands for dominant allele stand for normal wing.

00:19 So in this case, we'll say normal, and lower b stand for bent wing.

00:31 Capital s stands for the normal.

00:37 And lower s stand for shaven bristle.

00:50 So let's take a look at this cross.

00:53 So for the parent generation, you'll have a homozygars, normal wing, and shaven bristle fly.

01:04 Capital b, homozygars, lower s.

01:08 Normal wing and shaven, this is the phenotype.

01:18 And there's other, the other fly has the, homozygars, bent wing, and normal bristle.

01:25 Lower b, homozygars, and capital s, bent, but normal bristle.

01:36 Now, since each parent will only have two same allele, so for the normal shaven parents, there's only one type of gametes, cat to s lower s, capital b lower s.

01:50 For the other parents, lower b, capital s.

01:53 So easily we can find out all f1 is going to be a heterozegers, capital b, lower b, capital s, lower s.

02:00 It's a normal phenotype.

02:02 Now f1 will cross with a bent and bent wind and normal burst of, bent wind and shaven burstle fly.

02:11 So that means both g will be homozeggars recessive because both a little bit recessive and show recessive phenotype.

02:19 So this is going to be bent wing and shaven bristle.

02:28 Now, if the two genes are not linked, in this case, f1 will produce different alleles, and you can tell they will follow mandalions ' inheritance of law of segregation and independent assortment of the two genes.

02:46 So b gene will separate into two alleles, capital b, lower b, and the same with s, capital s and lower s.

02:54 Each has 50 % of a chance to be passed down to the gametes, and they also will sort independently.

03:00 So say you have capital b, capital s, capital b, lower s, lower b, capital s, and last case, lower b, lower s.

03:12 They each should have one to one to one to one ratio, because this is the segregation and independent soarment.

03:19 Now for the bent wing and shaven bristle fly, there's only one type of gametes, lower b and lower s, because both alleles are the same.

03:28 Now let's take a look at f2.

03:31 So if again, assuming that the two g is not linked, you'll have the egg or i would say one gamete from one parent and the other gamete and the other gamete from the other parents.

03:44 So you pet them up together.

03:45 So you can see you can have capital b, lower b, capital s, lower s.

03:50 So one allele from each of the parents.

03:53 Or you can have the second case, capital b, lower s, lower s.

03:59 Third case, a mausagoras, lower b, capital s, lower s, and the last one, both recessive.

04:08 So the first one will have a normal phenotype, normal wing, and normal burstle.

04:14 So this is the first phenotype.

04:17 Second, you have dominant b, normal wing, but homozygars recessive, shaven, bristle.

04:28 The third one, homozygars lower b, so bent wing, but dominant s, normal bristle.

04:44 I guess i will just say.

04:49 And the last one, obviously, both recessive, bent wing, and shaven.

04:57 Now, they should be 1 to 1 to 1 to 1 ratio.

05:06 So each has 1 4th of the chance.

05:08 So in this case, if we have a total of, i'll guess, i'll write them on the side, total 400, the normal 1 4th of the total, 400 times 1 4th equals 100.

05:33 The second is also 100.

05:36 Same thing, 100 and 100.

05:40 So if these two gos are not linked, you'll have four different phenotype.

05:45 They should be all the same, 100.

05:48 However, what we observed obviously is not the case...

Question

Please give Ace some feedback