Abaabababaa abaabbb bb bbaabaab 276 0 40 72 0 4 8 a the data...

aba+abababa+a+ aba+a++b+++b+b +++b+++++++b++ +++b+ba+aba+ab 276 0 40 72 0 4 8 a. The data support the conclusion that the a and b genes are linked. State the evidence in support of this conclusion. ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ b. Calculate the gene–centromere distance for each gene. ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ ______________________________________________________________________________ c. Calculate the map distance between the two genes.

Transcript

00:01 Data, two genes are linked, the other is on a different chromosome.

00:05 So calculate the map distance between them.

00:08 So as you can see that we have some of the genotype data are a lot more than the other.

00:20 So for example, abd 455, all dominant, lower abd, 470.

00:33 Then you have capital a, lower b, capital d, 460, and lower a, capital b, and lower d, which are 475.

00:50 So these are more common offspring, so they are considered as parental.

01:03 Now you have some of them are a lot less.

01:05 Capital a, lower b, lower d, 35.

01:12 And lower a, capital b, capital d, 33.

01:20 And capital a, capital b, and lower d, which is 37, and lastly, lower a, lower b, and capital d, which is 35.

01:36 So again, you can see these four are less common.

01:42 So when you have two gene linked, the less common offspring came from recombinant.

01:49 So their recombination, or we call it crossover between the gamete, between the two genes to produce recombinant gametes.

02:01 And the recombinant gametes produce recombinant offspring.

02:05 So they are the result of crossing over.

02:17 Now, let's decide which two genes are linked, which gene is on the different chromosome.

02:23 So if you look at the pattern of the parental, capital a and d, they're always together.

02:32 And then same with lower a and lower d, they're always together.

02:36 But b can switch in the middle.

02:39 So this tells us that a and d are linked.

02:46 And they are in a cis arrangement.

02:48 So capital a linked with capital d, and lower a with lower d.

02:54 B is on a different chromosome.

03:08 Again, we got this information by looking at all the parental genotype.

03:12 So we know that the capital a and d are together and lower a and d together.

03:17 B is assort independently with a and d.

03:21 So we can see that each has one force.

03:23 This is because b assort independently with a and d.

03:29 So this is a very important information.

03:31 So next step, we want to calculate the map distance between a and d.

03:36 So the map distance basically is the frequency of recombinant offspring...

Question

Please give Ace some feedback