This is an already answered question, but I don´t understand where they get the

Question

This is an already answered question, but I don´t understand where they get the new numbers from in Answer 2 and 3 .

I understand Answer 1:

https://www.chegg.com/homework-help/questions-and-answers/series-two-point-map-crosses-involving-five-genes-located-chromosome-ii-drosophila-followi-q6691105

I don´t understand Answer 2 & 3:

https://www.chegg.com/homework-help/questions-and-answers/two-point-crosses-carried-analysis-five-genes-chromosome-ii-drosophila-following-results-g-q10629119

https://www.chegg.com/homework-help/questions-and-answers/two-point-crosses-carried-analysis-five-genes-chromosome-ii-drosophila-following-results-g-q10658201

Copy of the question:

Two-point crosses were carried out for analysis of five genes on chromosome II in Drosophila, with the following results:

a) If adp is located near the end of chromosome II, construct a map of these five genes.

b) In another set of crosses, gene d was mapped against b and pr. The results were 17% recombination for d vs. b and 23% for d vs. pr. Predict the distance between d and c.

Gene pair used recombinant frequency (%) pr, adp 29 pr, vg 13 pr, c 21 pr, b 6 adp, b 35 adp, c 8 adp,vg 16 vg,b 19 vg, c 8 c,b 27

Explanation / Answer

Hi
I could do not have access to the links you have mentioned. But I have solved the question and this is one of the right way to arrive at the answer.
In the table, distance in terms of frequeny is given. Lets us start from the largest distance of 35. The gene pair is adp and b. So they must be situated at farthest poins. As I exlain, please right down on a sheet of paper. The next large distance is 29, between pr and adp. plot pr close to b. Look at the distance between pr and b, it is 6. 35-29 = 6, so it is correct. Now go to next largest number 27, the c and b genes. lot the c close to adp. Now check adp and c, it is 8. look at adp and vg, the vg is located at 16 from adp. Now the mas faily looks fine. FIll the rest of distance using remaining pairs. The map would look like this:
adp--8---c---8---vg-------13--------pr--6--b.

Now the another gene d, is mapped with b. The frequency is 17%, i. e distance is 17 mU,.the d and pr = 23. 23-17 = 6 is equal to the distance between b and pr. So now the map would look like this:
adp--8---c---8---vg-------13--------pr--6--b-----17-----d.

Distnce between d and c = count all numbers between them = 44 mU.

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