genes A and B are 20 map units apart. If you cross Ab/aB xab/ab, how many each progeny type would you expect if total progeny was 1000.
a. AaBb(AB/ab)
b. Aabb ( Ab/ab)
c. aaBb(aB/ab)
d. aabb(ab/ab)
is b and c the parental ? would that mean that b and c are 20 ? Thanks
Answer:
Distance between genes (m.u.) = Recombination frequency (%)
Ab/aB genotype produces 4 types of gametes as below:
Recombinant gametes= 20%
AB = 10%
ab = 10%
Non-recombinant gametes = 80%
Ab = 40%
aB = 40%
ab/ab genotype produces only one type of gametes, ab.
Ab/aB xab/ab –Parents
Ab |
|
Ab(40%) |
Ab/ab(40%) |
aB(40%) |
aB/ab (40%) |
AB (10%) |
AB/ab (10%) |
ab(10%) |
ab/ab (10%) |
a. AaBb(AB/ab) = 10/100 * 1000 = 100
b. Aabb ( Ab/ab) = 40/100 * 1000 = 400
c. aaBb(aB/ab) = 40/100 * 1000 = 400
d. aabb(ab/ab) = 10/100 * 1000 = 100
The b & c are parental.
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