Question

In Drosophila, these genes occur on chromosome III

+ wild h hairy (extra hairs on scutellars and head)

+ wild fz frizzled (thoracic hairs turn inward)

+ wild eg eagle (wings spread and raised)

The cross between + + + / h fz eg and h fz eg / h fz eg yielded this F1:

+ + + = 393

h fz eg = 409

+ fz eg = 58

h + + = 80

+ + eg = 28

h fz + = 30

+ fz + = 1

h + eg = 1

a. Show the gene map with correct gene order and the distances between them.

b. Calculate the amount of recombination observed.

c. Calculate interference and coefficient of coincidence.

Answer 1

a) Parental (Non-recombinant) type will be highest in number & double cross-over type will be least in number. Now, from the table, we find that +++ & h fz eg will be non-recombinant as they are highest in number; + fz + & h + eg will be double cross-over type as they are least in number. Now, parental & double cross-over type will differ in middle gene only. By comparing +++ with + fz + (or, h fz eg with h + eg), we find that fz locus will be middle. Thus, correct gene order will be h-fz-eg (Or, eg-fz-h).

Now, position of cross-over (shown by "/") is shown in below table.

Genotype	Number	Cross-over type
+ + +	393	Non-recombinant
h fz eg	409	Non-recombinant
+ / fz eg	58	Single cross-over
h / + +	80	Single cross-over
+ + / eg	28	Single cross-over
h fz / +	30	Single cross-over
+ / fz / +	1	Double cross-over
h / + / eg	1	Double cross-over
Total	1000

Now, recombination frequency = (Number of recombinant progeny / Total number of progeny)

So, h-fz recombination frequency = (58+80+1+1) / 1000 = 0.14 = 14%

So, fz-eg recombination frequency = (28+30+1+1) / 1000 = 0.06 = 6%

Now, 1% recombination frequency = 1 map unit (m.u.)

So, h-fz map distance = 14 m.u

So, fz-eg map distance = 6 m.u

- 14 m.u. 20 1.

b) So, recombination frequency = (58+80+28+30+1+1) / 1000 = 0.198 = 19.8%

c) Now, number of expected double cross-over = Frequency of double cross-over x Total number of offspring = Frequency of single cross-over between 'h' & 'fz' x Frequency of single cross-over between 'fz' & 'eg' x Total number of offspring = 0.14 x 0.06 x 1000 = 8.4

Given, observed double cross-over = 1+1 = 2

So, coefficient of coincidence = Number of observed double cross-over / Number of expected double cross-over = 2 / 8.4 = 0.238 (Up to 3 decimals)

Now, Interference = 1 - coefficient of coincidence = 1 - 0.238 = 0.762