2. Three gene loci A, B and C are all located on chromosome 2 in corn. The map b
ID: 146089 • Letter: 2
Question
2. Three gene loci A, B and C are all located on chromosome 2 in corn. The map below indicates the map distances of these loci. The interference between these genes is 0.4. 20 m.u 12 m.u. The recessive mutant alleles of these loci are designated a, b, and c. A pure-breeding triple recessive corn plant (a, b, c) was crossed with a corn plant homozygous for the wild-type traits. The resulting F1 plant was testcrossed with a triple homozygous recessive plant and 1000 progeny were counted. List the expected classes of progeny, the recombinant class they represent and the expected # of progeny in that class. Progeny class Recombinant class Expected #Of progenyExplanation / Answer
In the above problem, 3 gene loci are stated, A, B and C.
Now, according to the problem please follow the cross.
A pure breeding triple recessive corn plant (aabbcc) was crossed with a cron plant homozygous for the wild type traits (AABBCC). The resulting F1 plant (AaBbCc) was testcrossed with a triple homozygous recessive plant (aabbcc).
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
abc
AaBbCc
AaBbcc
AabbCc
Aabbcc
aaBbCc
aaBbcc
aabbCc
aabbcc
Progeny Class:
AaBbCc
AaBbcc
AabbCc
Aabbcc
aaBbCc
aaBbcc
aabbCc
aabbcc
Recombinant class:
AaBbcc
AabbCc
Aabbcc
aaBbCc
aaBbcc
aabbCc
Expected no of progeny:
Now, Consider our example of a, b, and c linkage. We can calculate the probability of a double crossover by using the Law of the Product rule. When a crossover in one region can not affect the probability of a crossover in a different region, the probability of a double crossover will be simply the product of their separate probabilities. In our case, the probability of single crossover between a and b = 0.2 (corresponds to 20 map units) Probability of single crossover between b and c = 0.12 (corresponds to 12 map units) Probability of both is 0.2 x 0.12 = 0.024 or 2.4%. This is the expected frequency of double cross over and the double recombinants.As, 1000 progeny were counted, expected number of double recombinants was 24. The single recombinant between a and b will be 20% (corresponds to 20 map units) that is 200 and single recombinant between b and c will 12.0% (corresponds to 12 map unit) that is 120.
ABC
ABc
AbC
Abc
aBC
aBc
abC
abc
abc
AaBbCc
AaBbcc
AabbCc
Aabbcc
aaBbCc
aaBbcc
aabbCc
aabbcc
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