12. A female critter with genotype AaBb is crossed with a double-recessive male (aabb). The progeny...
12. A female critter with genotype AaBb is crossed with a double-recessive male (aabb). The progeny included 54 individuals that were AaBb, 58 aabb, 462 Aabb, and 476 aaBb. Explain the layout of alleles on the female parent. Can you calculate a map distance (if you can what is it)?
Homework Answers
genotype AaBb crosses with the double recessive aabb (tester)
gametes formed by female critter AaBb = AB , Ab , ab, aB ( layout of the female alleles)
among all the gametes AB and ab are pareental type and others are recombinant type
double recessive aabb gives only one type of gamete= ab
AaBb * aabb
AB Ab aB ab * ab
| AB | Ab | aB | ab | |
| ab | AaBb | Aabb | aaBb | aabb |
AbBb & aabb are parental type progeny we get here
aaBb & Aabb are the recombinant progeny
from the result we get helps to calculate the physical distance by calculatinf the recombinant frequency. This method can be applycable for calculating the distance between the two genes
recombination frequency = recombinant frequency/total progeny*100
476+462/1050*100
=938/1050*100
=80.9% or 80.9 cm ( here the genes are very far apart to give large recombinant progeny )
father the genes more the recombination
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