Question

"These are the results of a three point test cross in Drosophila for the three loci cv f v: cv + + = 212, + v + = 48, + + + = 606, cv v + = 771, cv f v = 299, cv + f = 52, + f v = 229, + + f = 850. What is the map distance to 2 decimal places between the cv and f loci? Hint: Establish gene order first."

Answer 1

Let's formulate the information in table form

cv f v cv + +	+ cv +	212 single cross over	2​​
+ v +	+ + v	48 double cross over	4
+ + +	+ + +	606 single cross over	3
cv v +	+ cv v	771 Parental	1
cv f v	f cv v	299 single cross over	3
cv + f	f cv +	52 Double cross over	4
+ f v	f + v	229 single cross over	2
+ + f	f + +	850 Parental	1
Total		3067

In the above table, the last column number indicates reciprocal

First identify the recombination I.e Parental, single cross over and double cross over.

The parental progeny are those have the largest number because no crossing over occurs, the double cross over recombinant progeny are those having least number and between that number lies are single cross over progeny

Now let’s find the middle allele, which will add one when we compare parental with double cross over.

++f Parental

cv+f Double cross over now, here +f is common in Both genotype so the first gene or allele must have to the middle place and it is cv.

New gene order is f cv v

But in question, it will little bit confuse you because of some time v placed at last or sometimes f. It doesn't matter we will choose either one i.e mentioned above i.e f cv v

Find the first recombination frequency between f & cv

FORMULA

SCO+DCO/TOTAL PROGENY *100

606+299+52+48/3067*100

=32.76 cM

RECOMBINANT PERCENTAGE BETWEEN cv & v

=229+212+52+48/3067*100

=17.64 cM

Let's construct a gene map

GENE MAP

31.76 cM 17.64 cM

f............................................cv............................v

Distance between f and v is 49.40 cM (total)

Hope you understand it thank you