Please be very detailed and show step by step. Thank you. You have a donor strai
ID: 301158 • Letter: P
Question
Please be very detailed and show step by step. Thank you.
You have a donor strain that is a+, b+, c+ and infect it with a phage. You take the phage and infect a strain of bacteria that are a-, b-, c-. In a transduction mapping experiment, you get the following results:
experiment selected marker unselected marker
1 a+ 25% b+
2 a+ 3% c+
Draw the relative positions of a, b and c.
Explanation / Answer
Parents genotype is AABB and aabb
F1 Progeny genotype is AaBb
Test cross parents genotype is aabb,test cross parent is recessive.
AABB X aabb parents genotype
AaBb F1
AaBb X aabb test cross
AaBb Aabb aaBb aabb F2 progeny
Distance between A and B 15. 7 map unit,It means recombination frequency is 15.7 percent, so probability of recombinant progeny i.e Aabb and aaBb is 15. 7 %
probability of Aabb =15.5/2=7.85% and out of 3000, expected progeny of Aabb is 7.85/100*3000=235.5
probability of aaBb= 15.5/2=7.85 ,expected progeny of aaBb is 7.85/100*3000=235.5
probability of parental genotype =100-15.7=84.3% so out of 3000 progeny ,genotype of parental progeny=84.3/100*3000=2529
Expected progeny of genotype AABB is 2529/2=1264.5
Expected progeny of genotype aabb is 2529/2=1264.5
The cotransduction frequency of a+b+ is 25%
cotransduction frequency of a+c+ is 3%
It means gene a anb are closer
so the position of a, b and c is abc i.e a-------b--------------c