Female Drosophila heterarygous for three recessive mutations b (black bodyl. da
ID: 3166429 • Letter: F
Question
Female Drosophila heterarygous for three recessive mutations b (black bodyl. da (dumpy wings), and be (haoked bristles) were testcrossed and the follawing progeny were obtained: 1. Phenotyne wild-type black black, hooked dumpy, haoked hooked hooked, dumpy, black dumpy, black dumpy Number 169 19 301 21 172 304 What indicates the genes are linked? What is the order of the genes? What was the genatype of the original heterazygaus female? What is the map distance between dg and b? What is the map distance between b and What is the overall map distance acrass the 3 genes? What is the coefficient of coincidence?Explanation / Answer
+++ 169
b dp hk 172
b + + 19
+ dp hk 21
dp b + 6
+ + hk 8
b + hk 301
+ dp + 304
Total population = 1000
In any double cross, the maximum number of progeny is the progenies which have the same genotype as that of the parents and the progeny which have the least number is the progeny which are the result of a double crossover. So looking at the progenies we can easily conclude that b + hk and + dp + are the original parent type and dp + hk and + + hk are the double recombinants.
This also gives us the idea about the gene which is present in the middle as only the middle gene is transferred to the other chromosome during the double crossover. So the middle gene is b.
The distance between b and hk
If we want to find the distance between the b and hk then we need to find out the crossover frequency and divide it by the total number of progeny
b + + 19
+ dp hk 21
dp b + 6
+ + hk 8
Linkage distance = total number of recombinant * 100 / total number of progeny
54 * 100/ 1000
= 5.4
So, map distance between b and hk is 5.4 map units or 5.4 cM
The distance between b and dp
If we want to find the distance between the ct and v then we need to find out the crossover frequency and divide it by the total number of progeny
+++ 169
b dp hk 172
dp b + 6
+ + hk 8
Linkage distance = total number of recombinant * 100 / total number of progeny
355*100/ 1000
= 35.5
So, map distance between b and dp is 35.5 map units or 35.5cM
Since the flies have undergone the double recombination so we can expect the recombination interference
In order to find out the recombination interference we need to find out the coefficient of coincidence (c.o.c.)
The coefficient of coincidence is given by
the observed frequency of double recombinant/ expected frequency of double recombinant
expected frequency of double recombinant is given by
Map distance between b and dp /100 * map distance between b and hk /100 = 5.4/100*35.5/100 = 0.01917
The observed frequency of double recombinant = 6+8 /1000
0.014
Coefficient of coincidence = 0.014/0.01917= 0.73
A- If the genes would not have linked then we would have got all the progress in the almost equal number but in the given cross the all the progress number differs from each other. So genes are linked
B- Gene order is + + dp and b hk +
C- Original heterozygote was + + dp and b hk +
D- the distance between dp and b is 35.5 map unit
E- the distance between b and hk is 5.4 map unit
F- overall map distance is 35.5 +5.4 = 40.9
G- coefficient of coincidence (c.o.c.) is 0.73
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