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2. Two particular genes are linked and 14 map units apart. A GBIGB individual is mated with a gblgb individual. a) Show the g
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The genes G and B are located 14 map units apart. That means the probability of cross over between them is 14%.
a. Genotype of F1 offspring.
parents : GB/GB x gb/gb
Gamete: GB x gb
F1: GB/gb


b. Gametes by F1.
Remember F1 cannot produce all combinations in equal frequency. The genes G and B are linked, so only 14% will show recombination between G and B. ALways there are two parental and two recombinant types possible in 2 gene cross, so parenatl frequency will be 100-14 = 86; 86/2 = 43. The frequency of each recombinant will be 14/2 = 7%
The gametes will be:
GB - 43%
gb - 43%
Gb - 7%
gB - 7%

c. If ther F1 mate among themself, then the there is 7% chance of having a gamete with Gb genotype. In a cross two parents are invovled, so 7% x 7% - 0.49% chance of obtaining GGbb F2 offspring.

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