The linkage relationship between loci “A” and “B” is being studied in an F1 indi
ID: 58421 • Letter: T
Question
The linkage relationship between loci “A” and “B” is being studied in an F1 individual produced from a cross between two true breeding parents (AABB and aabb). Given that the two loci are 10 cM apart, what is the expected frequency of Ab gamete produce from this heterozygous F1 individual (AaBb)? The linkage relationship between loci “A” and “B” is being studied in an F1 individual produced from a cross between two true breeding parents (AABB and aabb). Given that the two loci are 10 cM apart, what is the expected frequency of Ab gamete produce from this heterozygous F1 individual (AaBb)?Explanation / Answer
According to the Mendel law of independent assortment, the gametes formed from the heterozygous parent (AaBb) will be as follows:
AB- Parental- 1/4
ab- Parental- 1/4
Ab- Recombinant- 1/4
aB- Recombinant- 1/4
Thus, the percentage of recombinant to parental combination is 50:50.
According to the given information, the loci “A” and “B” are linked and the distance between them is 10 cM.
The genetic map shows the order of genes on a chromosome. The distance between the genes is directly proportional to the frequency of recombination.
In the given case, the distance between the two loci is 10 cM. Thus, the frequency of recombination is 10 percent.
The frequency of total recombination is 10 percent, which includes 5 percent aB and 5 percent Ab recombinants.
Hence, the expected frequency of Ab gamete from a heterozygous F1 (AaBb) individual is 5 percent.
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